1961:
consider. How things get "mixed together," as you put it, is completely irrelevant. So while technically invalid as a solution, counting cases can be shown to work if the cases you count are the coins, but not if the cases you count are the boxes. If I had my druthers, the Box
Paradox would be separated from the puzzles it is similar to, but those would be linked back to the Box Paradox to help support the correct answer. But it will never happen if people like you keep trying to treat it as a puzzle, and refuse to acknowledge how other problems (Monty Hall, Three Prisoners, and Boy or Girl) are related. In particular, that means the "right" answer to the Boy or Girl "Paradox" is 1/2 unless is is made clear how P(problem statement's "one is a boy"|BG or GB)=1. Every source that has ever considered this question has agreed, but most sources fall into the trap Bertrand warned against and say the answer is 1/3 without addressing how the observation was made. And none of the controversies surrounding any of these wikipedia pages will ever be resolved without acknowledging what I am trying to say.
2302:
is quite literally the Box
Paradox, with the fourth box containing different coins). If you "know that at least one is a boy" because you asked "are any of the puppies male?" (note how I improved the question by not saying "one"), then there is a 1/3 chance both are boys because the trait has not been associated with a specific offspring. If you "know that at least one is a boy" because you meet father and son walking in the street, you have associated the information with a specific child, even though no means of "distinguishing" the children is evident. The answer is 1/(1+2X), where X is the probability a man will choose to walk with his son instead of his daughter. And if you simply "know" the information? Then Bertrand's solution applies, but with four cases. The answer must be the same as "what is the probability the two children in a family share the same gender?" That is, 1/2.
2222:
distinguishable to the host. The host shuffles the cups. You pick one. He shows you that one of the other two is empty and offers you the choice to switch to the third cup. There is no conditional probability at all involved in solving this problem: the third cup hides the coin if and only if your initial choice doesn't, which has chance 2/3. The natural probability theory description of the three cups problem has an outcome space of just two outcomes with probabilities 1/3 and 2/3. For many people the three cups and the three doors problem are obviously the same, but most probabilists would only call them similar: the three doors problem can be reduced to the three cups problem by invoking symmetry. But you can't "reduce" the three cups problem to the three doors problem. The relationship between the two problems is asymmetric.
381:
black. ThreeCards(1).Upface = True ThreeCards(1).Downface = True 'Card two, both sides are white. ThreeCards(2).Upface = False ThreeCards(2).Downface = False 'Card three have two positions, so it will be defined later. For ii = 0 To 1000000 Pullout = Int(Rnd * 3) + 1 If
Pullout = 3 Then 'Card three, upface could be either one, black or white. If CInt(Rnd) = 1 Then ThreeCards(3).Upface = True ThreeCards(3).Downface = False Else ThreeCards(3).Upface = False ThreeCards(3).Downface = True End If End If If ThreeCards(Pullout).Upface = True Then n = n + 1 If ThreeCards(Pullout).Downface = True Then BothBlack = BothBlack + 1 End If End If Next ii Print "Probability is " & Str((BothBlack / n) * 100) & "%" End Sub
2729:, people watching a video or possibly a silent video insist a phoneme or word is what they expect it to be; possibly lip-reading a P or B sound. The point is that the "naive" solution of 1/2 is hard to understand for someone who has already agreed to the answer of 1/3. For someone asking "what is the color/material of the second coin", they won't see any reason to think the probability is 1/2. That only happens if you ignore the previous answer of 1/3 and don't think about the unseen coin, and just ask yourself "which boxes have a coin like the one I just removed". I do think the current description could be "revisionist" and not reflect the original statement.
2149:
might see door 2 opened or you might see door 3 opened. The mechanisms which lead to these possibilities are quite different. In the three boxes problem it is pure chance. The sample space has six elements of equal probability (three boxes, two coins per box). In the Monty Hall problem the action of the host is (partly) forced. The sample space has four elements of unequal probability. (Car is behind door 1 host opens door 2; car is behind door 1 host opens door 3; car is behind door 2; car is behind door 3). By splitting outcomes in one description, or merging them in the other, one can finally make the two problems correspond in a one-to-one fashion.
4183:
1077:
drawing the black/white card, and a 1/3 chance of drawing the white/white card. After drawing a card that you don't see, that chance is still 33% either way, as there are 3 possibilities. After looking at the table with the card on it, and seeing the black card, one card gets eliminated as a possibility, so it can only be one of two cards. As point of fact, any information we can gain about this card will only serve to either make the odds 100% (if we were told that the sides are different) or 50% (both sides are the same). If we only know one side, then we eliminate one of three options, and since two options are left to us, it's still 50/50.
2756:
the gold one. The word "if" means we are speculating about a subset of hypothetical scenarios, and it's easy to think we should give the hypothetical scenario of selecting the mixed box and then the gold coin the same probability or weight as selecting the box with gold coins. (For example, maybe the gold coin is closer to the opening to the box.) This suggests the range of answers could vary with a practical experiment compared to asking about the hypothetical situation. Instead of repeatedly trying until they get a gold coin, the experimenter could just ask about whatever type of coin they happen to randomly pick out.
1977:
hand, to count coins as starting cases is incorrect, and this is what my examples were about. If you have 100 golden coins in GG box and 1 G coin in GS box, you do not choose equally from 101 G coins. So, your statement that 'Either approach is technically invalid' is quite meaningless. The same refers to your statement that 'How things get "mixed together," as you put it, is completely irrelevant'. On the contrary, it is exactly what matters. The fact that golden coins are physically separated by boxes prevents you from chosing from the whole pool of 101 G coins and invalidates the 'coin solution'.
2138:
most popular mathematization of the two problems, they are indeed mathematically identical. I don't see a natural translation of the three boxes into the three doors. The host opens door 2 or door 3 and reveals a goat. The player is interested in which of the three situations holds: door 2 and door 3 contain goat and goat, or goat and car or car and goat. When the host opens door 3, situation 2 (goat and car) is ruled out. What is left is goat and goat, or car and goat. In the former situation he is only 50% sure to open door 3, while in the latter situation he is 100% sure.
3825:
1021:
the white/white card, we're left with a 2/3 chance to draw one of the two black faces, and a 1/3 chance to draw the white face, and that's where the error lies. Because we do not draw the other side of the card randomly, in essence, the way the problem is worded would give a 50% chance of either because the white/white card is extraneous. The card can be one of 2 options. Try doing the experiment without the white/white card, since that's what the wording in this problem suggests, and the trend will tend towards 50/50 unless you have a fluke.
1864:. Why? Let X represent 100 silver coins in a drawer. If "each silver coin is replaced by 100 silver coins", then the problem simply changes from GG/GS/SS to GG/GX/XX and is otherwise unchanged. I imagine this is not what the author intended; if so can I suggest a more detailed explanation? This would be better placed in new section titled something like "alternative explanations if you still don't get it", which could also have many of the discussions in these talk pages written up and added to it. (Although, if you believe in
74:
53:
173:
4178:{\displaystyle {\frac {1}{2}}\times \left({\frac {1}{2}}\times \left({\frac {1}{3}}\times {\frac {1}{1}}+{\frac {1}{3}}\times {\frac {1}{2}}+{\frac {1}{3}}\times {\frac {0}{2}}\right)+{\frac {1}{2}}\times \left({\frac {1}{3}}\times {\frac {1}{1}}+{\frac {1}{3}}\times {\frac {1}{2}}+{\frac {1}{3}}\times {\frac {0}{2}}\right)\right)+{\frac {1}{2}}\times \left({\frac {1}{1}}\times \left({\frac {1}{3}}\times {\frac {2}{2}}+{\frac {1}{3}}\times {\frac {0}{1}}+{\frac {1}{3}}\times {\frac {0}{2}}\right)\right)}
163:
142:
2250:
numbered, unchosen goat door. That doesn't mean different doors in all cases.) The probability is always a function of how the host chooses a door/cup to reveal. But Morgan, et al, are wrong in claiming that, based on the knowledge gleaned from the problem statement, you can assume a bias for the host. I'm not saying you must claim no bias, but any bias you can assume is equally likely to favor either door/cup, and that means the probability the host chooses either one is the same.
928:
important in population probabilities, used by drug developers, marketing execs, and polititians. A card could be either one colour or two. But if one pulls it from a bag with 2/3 of the card having one colour, there is a 2/3 chance the card is monochromatic. As expressed here the problem helps one move to the same problem, say, where there are 5 cards in a bag. If we are told one is all white, one is all black, and the other 3 are W/B, we can solve this directly.
3591:
948:
of losing. Mathematically speaking, the underlying problem is exactly what you described in your original comment: what matters is the probability of getting a match, not a particular color. However, the scam (hypothetically) works because I have shown you a color, and I ask an intentionally misleading question about a color. I am hoping that you don't think about generalities, but you are confused by the situation at hand and assume the probability is 1/2.
2183:, and I is the information you have, each of these three problems can be solved by P(C1|I) = P(I|C1)*P(C1)/ = /=2/3 and the equivalent statements for C2 and C3. That makes them mathematically equivalent, regardless of alternate solutions (which will, btw, also have equivalents across all three). Most references that mention both state this equivalence. And they can't be mathematically equivalent unless there is some logical equivalence somewhere.
2199:
revealed; say, G. The probability it is the G-only box is 2/3 based on the formula I gave above. Similarly in the MHP, your door must have at least one, but can have both, of two properties: the door to its right (imagine they are laid out in a circle) has a goat, or the door to its left does. Parallel properties, and the door can be R-only, L-only, and Both. Monty Hall reveals one of these properties to you.
3350:
439:
look at one side, we could at look at the top of card A, the bottom of card A, the top of B or the bottom of B. Of these, 3 are green and one is purple. But we look at see that it is not purple. So we could be looking at one of 3 sides. Top of A, Bottom of A, or Top of B. So there is a 2/3 chance we are looking at card A, and thus 2/3 chance that the bottom is green.
22:
251:
1708:'s talk page about a similar Lewis Carroll problem. Also, apparently the three cards article was nominated for deletion in January 2006 on the grounds of it being original research—so adding Bertrand and Carroll to the article, perhaps in a "History of the problem" section, would give the article more credibility. I'll add this comment to the
2838:
changed from 1/2 to 1 - for example if, beforehand, by the balls were arranged so each box had 1 in front and 1 in back and the 1-of-each box had gold in front, and you chose the front ball of whichever box you picked - then the answer would be 1/2, since in that case, picking gold would not be any evidence as to which box you picked.
3586:{\displaystyle {\frac {1}{2}}\times \left({\frac {1}{2}}\times \left({\frac {1}{3}}+{\frac {1}{3}}+{\frac {1}{3}}\right)+{\frac {1}{2}}\times \left({\frac {1}{3}}+{\frac {1}{3}}+{\frac {1}{3}}\right)\right)+{\frac {1}{2}}\times \left({\frac {1}{1}}\times \left({\frac {1}{3}}+{\frac {1}{3}}+{\frac {1}{3}}\right)\right)}
2189:
symmetry between the two obvious kinds of results, car and goat, like there is between the gold and silver coins. So the equivalence isn't equating prizes to kinds of coins, or even what's behind doors to in boxes. And in the BPP, there is no knowledgeable entity that forces the information to go a certain way.
2098:
This is the change I meant, and GG is the one that doesn't change in combination with OG. Then, the conditional probability is found by normalizing these probabilities with the factor 1/2, the sum of those probabilities. So P(GG|OG)=2/3, P(GS|OG)=1/3, and P(SS|OG)=0. Let me think about a way to word it better.
2890:
silver coin? All this is extremely poorly worded, making the reading of the rest of the article tedious because we also have to be guessing what the rules were meant to be from the results, which is almost impossible to achieve in an article that is about things being counterintuitive. Can someone fix that?
2945:
if (rndInt != 3) //you still have a chance of getting gold { ball = randomIntFromInterval(1, rndInt); //draw a ball from the box //if 1 then gold, if 2 then silver if (ball == 1) { gold += 1; //increment the game counter if (rndInt != 2) //you are in the box with
2683:
The issue is, that the "paradox" which Joseph
Bertrand refers to is not the problem itself, which appears to be a veridical paradox to some. It is the way that Bertrand demonstrated that the naive solution, which concludes that the probability is 1/2, could not be right. The problem itself was stated
2483:
In the well formulated MHP we know at the beginning of the game that we have "chosen" the car with probability 1/3; and we know that the host will open another door with a goat. Here again another chance than 1/3 for the chosen door is not possible if we assume that there is no difference whether the
2301:
The point that makes it an "indistinguishable problem" is that you can't associate the information with a single "container," not that you can't distinguish the containers. Push Door #2 and Door #3 together, and you have an "indistinguishable problem." The same applies to the Two Child
Problem (which
2291:
Three Cups
Problem requires that each cup receive a label. The "indistinguishable problem" you want to describe is really the "combined doors" one. The two unchosen doors are pushed together, and a goat is led from behind them in such a way that you can't tell which door it came from. You can't do it
2198:
of gold coins, only. The second contains any number of silver, and the third contains equal numbers of each. Said another way, there are two parallel properties G and S. One box has property G only, one has property S only, and one has Both. A box is chosen at random, and somehow a single property is
2188:
No, I don’t know any reference that details the logical one. But it is there. They are logically equivalent because you can symbolically describe each in the same way, although it isn't the most intuitive representation for any of them that makes it so. Part of that is because in the MHP, there is no
1907:
The only correct way to solve the problem (Bertrand) is by treating the boxes as the individual cases, and by summing the probabilities that the cases would produce. This works for all these three versions literally in the same way as it does for the original one, and results in the same answer, 2/3,
1891:
Regarding original formulation with boxes and drawers (or simply coins), I would pay everybody's attention that standard arguments about chosing drawers (coins) rather than boxes are quite wrong, because coins are not physically mixed together, and in fact we chose a box at first and a coin at second
1812:
I agree that the wording strongly suggested the card has two black faces. I was about to make your clarifying changes, but then I realized the whole sentence was dubious. So I have gotten rid of the "face which is facing you" bit altogether. More complete explanations are in the section below it.
947:
Sorry, I forgot to respond earlier. The scam story I was talking about goes something like this: I draw one of the three cards from a hat. No matter what it shows, I invite you to bet money that the other side is the opposite color. Of course, regardless of the color, if you bet you have a 2/3 chance
895:
with your proposal because in all the references I've seen, the problem is given as it currently appears here. I think it would be better to expand "Symmetry" into a top-level section and mention Simon
Templar there. It would contain the alternate statement of the problem, as well as the story of the
438:
Here is another way to consider the problem using the colour and side rather than the card. We see a green card so we know the double purple is gone. So there are only 2 cards in the running. How many permutations and combinations are there so we can assess the probability? Of these 2 cards, if we
314:
it is the white/black card. If you see black (as the problem states) you should still change your bet. This new information changes the odds. It is now 2/3 as this article explains - since you see a black side, 2 out of the three black sides are on the B/B card, so this card is more likely. From
2837:
It is crucial that there _was_ a chance of getting a silver ball; even though that is not what actually happened. Specifically, given the choice of box, the chance of getting gold from from the 2-gold box was 2/2, whereas the chance of getting gold from the 1-of-each box was 1/2. If the latter was
2755:
illustrates the difficulty of defining a random probability. I think what is happening here is we are primed to think of 'fate': we didn't select the box with silver coins. It is not hard to go from there to saying that if the 'random' box was the one with mixed coins, it was 'fate' that we selected
2699:
The answer is trivial: 1/3. But what if you remove one coin from the chosen box, without looking at it? The naive solution says that if you were to take one coin out and look at it, the answer would change to 1/2 no matter what you saw. So you don't actually need to look at it - simply taking a coin
1929:
It depends only on ratio Ng/Ns of gold and silver coin numbers in GS box. The posterior odds GG vs GS box are (Ng + Ns) : Ng and have nothing to do with number X of gold coins in GG box. If Ng = Ns, these odds are 2 : 1, and it is mere coincidence that X : Ng gives the same 2 : 1 for X=2 and Ng
1236:
The problem is that once you have drawn a card, and can see the black side staring at you...by then, right away, you know that you are much more likely to have drawn the black card than the mixed card. For that reason, it is not equally likely to see the other side of the card white as it is black.,
1120:
I'm glad 68.227.203.149 got it. Here's the safe approach to this type of problem: List all outcomes of the fundamental experiment in a way where it's clear that there is symmetry, i.e. that they all have the same probability. E.g., flipping two identical dice simultaneously, you can only distinguish
1013:
Okay, if you're dealing with three cards, the probability of drawing a card with 2 of the same faces is 2/3. That much we know. But drawing from the three cards, you actually, without knowing any information about the card you've drawn, have a 1/3 or 33.33...% chance of drawing any single card. Now,
927:
The 2 coin problem is similar - perhaps identical mathmatically by some solutions. I think, though, the third card adds to the common sense wrong answer. And considering pulling one card out of a bag of 3 helps illustrate the value of considering the sample from which a items is drawn. This is very
909:
I'm afraid I don't know the story; fill me in? As far as how the problem appears in the references, Schuyler credits
Gardner as his source, but Schuyler specifies that the card face looked at is black, which Gardner does not. I can't check the Nickerson or Bar-Hiller and Falk references right now,
832:
As I said before, the "Formal approach" section still needs work; historically, it's an outgrowth of an incorrect solution, and it should probably be scrapped entirely. As for the simple explanation, by all means, add it! The expert tag is not meant to discourage non-experts. Heck, I'm not an expert
818:
Why exactally do you feel an expert is needed? This is a very very simple probability question that prays on people ignoring the first part of the information so that they pick the 'obvious' answer of 1/2. I don't seen how an expert can provide more through an answer than the long winded one already
325:
Read the problem description carefully! "The side facing up is black. What are the odds that the other side is also black?" There is only one card that is black from other side too. From two cards (you don't have to count white/white as I explain earlier) probability to get B/B is 1/2. What I meant
2268:
We need a more subtle (in fact: relative) notion of indistinguishability. Of course when there are three cups on a table you can distinguish them by their locations. The point of the three cups story is that the player can't recognise which cup is which after they have been shuffled, while the host
2077:
The last sentence of the above section reads "It can be resolved only by recognizing how observing a gold coin changes the probability that the box is GS or GG, but not GG." Now I'm not an expert here, but it seems to me this is a typo and should in fact read "It can be resolved only by recognizing
1223:
We draw a card, and find a black side showing. This eliminates W/W as a possibility. Only now the question is asked, "What is the chance that the other face is Black?". With only B/B and B/W as possibilities, is the probability not 1/2? It seems false to imply the "probability of seeing a black
2925:
odds = Math.round((winner / gold) * 100); //alert("the odds of drawing any gold is " + gold.toString() + " in " + NumberofRounds.toString() + ". The odds of drawing two is : " + winner.toString() + " out of " + NumberofRounds.toString() + " rounds in total, of which "+ gold.toString() + "
2799:
In the original situation: if you picked box A, you got a gold ball. If you picked box B, you got a gold ball. There was no chance of getting a silver ball. You get a gold ball no matter what. So now there are 2 possibilities: either you're in box A and there's another gold, or you're in box B and
2724:
I think this is related to an effect I can't find a reference to. In one example of it, subjects in an experiment were slower to correctly identify an object that gradually became less blurred if they initially saw it in a blurred state and misidentified it. In another example, possibly related to
2372:
Surely in my simple view of the problem (which is equivalent to yours) "case" corresponds to a coin. There are three gold coins. For two of them, say G1 and G2 (which are in the same box), the other coin in the same box is a gold coin: For G1 this is G2, and for G2, this is G1. Only for one of the
2243:
It is not whether the cups, when viewed in isolation, are distinguishable that matters. It is whether the cups, in the situation described, are distinguished. Since one is revealed, and one is not, they are always distinguished to both you AND the host, so the point you are trying to make is moot.
1976:
I do not count cases as you try to implicate , I calculate probabilities of various outcomes. I treat boxes as equiprobable starting cases (each separates farther into subcases) which is absolutely valid (randomly chosen box, as formulated) and I do not at all count boxes as outcomes. On the other
1954:
think the "right" solution is. What matters for a wikipedia page is what is in literature. Second, Bertrand didn't write his paradox as a puzzle where the reader is supposed to figure out the right answer (and it really shouldn't share a page with the three card puzzle for this reason). He pointed
1933:
Argument that the chosen box has coins of the same type 2⁄3 of the time remains in force for any numbers of coins. But the next statement "So regardless of what kind of coin is in the chosen drawer, the box has two coins of that type 2⁄3 of the time" works only if Ng = Ns, and is not that trivial.
1386:
Now, we are told we are watching a black side, which means we can discard possibilities 3, 4 and 6 in my list, leaving us with just the three possibilities listed by 68.227.203.149 above. There's no information making any of these possibilities more or less probable than the other two, so we still
1183:
Now, we are told we are watching a black side, which means we can discard possibilities 3, 4 and 6 in my list, leaving us with just the three possibilities listed by 68.227.203.149 above. There's no information making any of these possibilities more or less probable than the other two, so we still
1020:
Looking at it as a matter of faces is an error, because that would mean you could draw each face randomly, and then you would divide the problem into 6ths, with a 2/5 chance of drawing another black face, and a 3/5 chance of drawing a white face. However, if we remove 2 white faces, IE eliminating
876:
I suggest that we might describe the problem in what I believe is closer to its original form: instead of specifying that we see a card face that's black, and asking the chance that the other side is black, we simply ask what the chances are that the other side is the same color as the side we're
380:
Private Type Card Upface As
Boolean Downface As Boolean End Type Private Sub Command1_Click() Dim ThreeCards(1 To 3) As Card Dim Pullout As Long Dim n As Long Dim ii As Long Dim BothBlack As Long Randomize Timer 'First we define cards... 'Lets say that True means black. 'Card one, both sides are
294:
Think. What are the odds that the other side is also black? So, the card what is just picked up, can't be white/white, which means you don't need white/white card at all to perform this "problem". Then probability that you just picked up white/black or black/black, is of course 50/50! I agree only
2479:
There is another nice correspondence between BBP and MHP: In BBP we know at the beginning that the chance to get the mixed box is 1/3. And (not new here) we know with certainty that we'll get a gold or a silver coin. If we now had a chance to have the mixed box other than 1/3 (maybe 1/2) we would
2148:
But to say that the two problems are logically equivalent should entail, in my opinion, finding a one-to-one correspondence between all components of the two problems. I think it can't be done. In the three boxes problem you might see either a silver or a gold coin. In the three doors problem you
2137:
Monty Hall and the Three
Prisoners certainly are more than similar, though whether or not they are mathematically equivalent depends on how one chooses to mathematize the problems. Not everyone makes the same translation from a verbal puzzle to a formal problem in probability theory. But with the
2097:
each event in {GG, GS, SS} is different than the probability of just each event itself. For example, if "OG" is the event "observe a gold coin," the probability is 1/3 for each of the events {GG, GS, SS}. But the probability is 1/3, 1/6, and 0, respectively, for {GG&OG, GS&OG, SS&OG}.
867:
Obviously, this is the same puzzle, in all functional respects, as the three cards problem as currently described in the article: simply substitute "cards" for coins", "heads" for "black" and "tails" for "white". Which highlights the puzzling part about the current presentation of the problem:
2889:
I don't understand if the first coin mentioned is withdrawn from the box – itself randomly chosen – or from any of the three boxes. I also do not understand the sequence of events, as it is not clear. Should we pick a coin, look at it and put it back in its original box would it turn out to be a
2779:
For some reason, this seemed a lot less paradoxical than the Monty Hall problem. The answer is obviously 2/3 and the error in the reasoning giving 1/2 is immediately apparent. As people's intuitive understanding of probability becomes more Bayesian, I suspect paradoxes of this kind will dissolve
1960:
But in order to satisfy the masses who want to treat it as a puzzle, it is important to point out what conditions allow "counting cases" to get the right numeric answer while being technically invalid. That happens when the probability of getting a gold coin is either 0 or 1 in every case you do
1609:
You put all of the cards in a hat, pull one out at random, and blindly pass it to an assistant, with the instruction that he should place it on the table with a black side facing up, if possible. The assistant takes the card, looks at both sides, and places it on the table. The side facing up is
863:
Carroll's problem was as follows: You have a bag, and in this bag are one regular coin with a heads and a tails, and one double-headed coin. You shake the bag, reach in and pull out a coin and look at only one side. The side you look at is heads; what is the chance that the other side is also
2357:
The problem with this formulation is that many will assume that a "case" corresponds to a box, not a coin. They'll say "There's one case where the other one is silver, and one case where the other one is gold." So the key additional step is "After choosing a random box, you next choose a random
2322:
I added the Bayes' rule version of the solution by Bayes theorem, and filled in the formal details of a solution by symmetry. I also corrected the wording and the notation in the first solution (refering to the original silver and gold coins version of the problem) . The probabilities which are
2249:
Like Morgan, et al, claim, the correct solution is always conditional. (Or alternately, based on a unconditional solution that includes what they call the condition in the "initial" setup. By that I mean the host could have randomly chosen, ahead of time, to reveal the higher numbered, or lower
2221:
Nice construction (the doors in a circle). Here is another equivalence question. Vos Savant explained Monty Hall by translating it to what I like to call the three cups problem: the host hides a coin underneath one of three upside-down cups on a table. The cups are indistinguishable to you, but
2021:
The article currently explains the "box" problem, and then restarts from scratch with the "card" problem, explaining much of the same probability stuff all over again, with different wording and some extra bits thrown in. Everything that is generic should be explained once only, under whichever
1076:
Yes, but if there is a card in front of you that is black, there are only 2 cards it could be, thus there's a 50% chance that it is the black/black card, and a 50% chance it's the black/white card. Before the draw, there was a 1/3 chance that you would draw the black/black card, a 1/3 chance of
2857:
You may reason that similarly, the fact that we subsequently choose a ball does not change the probabilities for the box. Indeed, that is correct; the fact that we subsequently choose a ball does not change the probabilities for the box. However, the coin's _type_ very much _does_ change the
1638:
Now if your asking what is the probability that the other side is black that is 2/3rds. To those down below that think it is 1/2: the "trick" to this is labeling the faces 1 through 6. With 1,2 being the black card, and 3 being the black face of the black/white card. You have been told one of
1524:
is the exact same problem, so I think that these two articles should be consolidated. In addition, some Knowledge users discussed deleting the Three cards article in January 2006 because they thought it was original research. Therefore, including Lewis Carroll (as discussed above) and Joseph
1499:
Actually, there's a nice explanation of the correct answer using the third card! After somebody answers you 50/50, just ask them what's the probability, if you take a random card and place it on the table, that the hidden color is the same as the shown color... With three cards, the answer is
2759:
I think this is more accurate than saying that people have been trained to think abstractly about similar objects, and that this causes them to 'merge' the two gold coins in the same box, and the scenarios of selecting each of those coins, in their mind. I mention this because just like most
2056:
Well, it already said as much; and the change you made is not quite logically consistent since it states the conclusion - that the two remaining possibilities are not equally likely - as a reason for the conclusion. But I'm going to try to clean it up a little to address your intended point.
877:
looking at, whichever that is. The answer remains the same: the chances are 2/3 that we're looking at one of the faces of a card with the same color on both sides. (This form of the puzzle also has a literary reference that can be mentioned -- it was used in a Leslie Charteris story about
2843:
For the original situation, you were probably focusing on it averaging over the gold coins. The reason it did this is, the coins were equally likely, since there were an equal number of coins in each box. In particular, if one of the boxes contains infinite golden balls _and_ the other(s)
1980:
Next, looking at history of revisions, I have realized that it is your addition that counting cases can give the correct answer 'when the probability of getting a gold coin is either 0 or 1 in every case you do consider'. This sound like a puzzle to me. What probability do you keep in mind?
3229:
2921:
if (rndInt != 3) //you drew a gold ball, congrats { gold += 1; //increment the game counter if (rndInt != 2) //you are in the box with 2 gold balls { winner += 1; //winner winner chicken dinner } } i++; }
2703:
What I intend, is to change the problem statement to what Bertrand actually said, and highlight the differences between how subsequent authors have re-interpreted it. It may take a bit of time to do so, so I'll save drafts until I'm done. And I'm open to comments and suggestions.
2926:
contain valid gold. The thought experiment is your odds of drawing gold again. For that, your odds are " + winner.toString() + "/" + gold.toString() + ". The answer, therefore, is : " + odds.toString() + " %."); // show popup with the ratio return odds; } </script: -->
1265:
This is also an interesting point. If someone else pulls the card, and says "one of the sides is black", we need to know if said person has looked on both sides, (or at least know the probability that he did,) before we can accurately estimate the probability that both sides are
819:
shown. There are three cards, BB WW BW. If one card is showing B then its either the BB card one way up, the BB card the other way up or the BW card with B up. Thus 2/3 of those possible situations have B on the under side. Very very very simple, User_talk:Dacium|(talk)]] --
2917:
function Ball() { var gold = 0; var winner = 0; var i = 0; var rndInt = 0; var NumberofRounds = 100000; while (i < NumberofRounds) { rndInt = randomIntFromInterval(1, 3); //if 1, then box contains 2 gold balls, if 2, then 1, if 3, then 0.
1892:(however, in case of cards this way of argueing is OK, since cards are indeed mixed together in one hat). I cite from the allegedly correct solution: "So it must from the G drawer of box GS, or either drawer of box GG. The three remaining possibilities are equally likely (
3330:
1253:
The problem with your reasoning, Adl116, is that we have more information than "one of the sides is black", when we take the card out of the bag and place it on the table, we're labelling the sides on that card, we are indeed drawing sides, and we know that "this side is
2511:
Suppose a cook has two pancakes. He says that one pancake is brown on both sides, and the other is brown on one side and golden on the other. He places one pancake on your plate, and it is brown on the side up. What is the probability that the other side is also
452:
The article cites a poll of 53 students, of which 35 guess 1/2 and only 3 guessed 2/3. That leaves 15 unaccounted for. I find it hard to believe that so many guessed neither the right answer nor the naive wrong answer. Does anybody have more detail on that?
1899:
Consider, however, following three versions of the problem with the same protocol (randomly chosen box, then randomly chosen coin). It is supposed that there is a mechanism to dispense coins, so that you can not feel by fingers there are many of them in a box.
2675:
I've been looking for a translation of the original text to base a major re-write upon (as opposed to the poorly-worded section I provided in 2014), and hadn't found one. But I did just (re-)locate the source that cites it and is what I was recalling then.
1841:, what is the probability that the opposite side is heads." Answer to the latter problem is 1/3. The idea being that the coin "game" is a carnival game where the odds/probability are stacked in favor of the house at 2/3. Apologies if I goofed something.
1393:
Assuming the person drew a card at random and did not fix the draw in either favor, does the above formula change simply because we weren't the ones to draw the card? No. The person who draws the card should be irrelevant. The formula should hold true.
2403:
applies to." That way, they can assign equal probabilities to each "case." And the controversy comes because both "boxes" and "coins" fulfill this definition, so both sides feel they have applied it properly even if it was done only intuitively. But the
1955:
out that it is technically incorrect (as you discuss) to merely count the cases, no matter whether you use the three boxes or the six coins as your cases. Either approach is technically invalid. That is what the wikipedia page on this subject should say.
859:
There is something puzzling for me in the way this problem is presented; it differs from how I've seen the same problem presented before, in a way that does not change the answer but does actually duplicate a second, related problem from Lewis Carroll.
1984:
At last, I do not find construction 'if people like you keep trying to treat ...' an appropriate style, the more so that I have not even mentioned word puzzle in my original comment. Would you, please, be less personal and more specific in essence?.
493:
Improvement well done! Fine to leave and explain how 1/2 can seem like the quick answer, but even qual approach leads to 2/3 if probability is considered right. As said on talk page we must find a reference or this page will not survive as original
1224:
face" has anything to do with the problem, as the black face has already been seen, akin to saying that a coin flip has a 50% chance of being tails after the coin has already been flipped and tails is showing. Any help here will be appreciated.
734:
It has also been called the Three Card Swindle. This name is applied when the swindler tries to get the mark to bet on the outcome on the theory that the relevant probability is 1/2 rather than 1/3. But this name doesn't find many pages on Google.
1663:
The odds aren't "2" but "2 to 1". Since such odds aren't stated as simple numbers, there is no danger of confusion between the two formats, which are logically equivalent. And the difference between them is not as interesting as the paradox under
1058:
I'm afraid you're wrong. Just look at the symmetry argument. If the probability that you draw a card with the same color on both sides is 2/3, then that probability cannot magically become 1/2 if you specify the color you observe on the top. —
2844:
contain(s) infinite silver balls then, conditioned on choosing a gold coin, it would be 100% sure that the box is the all-gold box. On the other hand, if there were 2 boxes, one with 6 balls and the other with 12 balls, then this would be like
1086:
Even if we look at the sides. How many sides are white? 3. How many sides are black? 3. So what are the odds of drawing a card with the white side up? 50%. With the black side up? 50%. With the white side down? 50%. Black side down? Yep, still
1796:
I realize this may seem redundant, as the only way to know the first part of the statement is by knowing the second part. But anyway prefer this to the current sentence, which strongly suggests the card has two black faces. Comments?
463:
As a former professor, I can answer this; it's easy. 35 students thought the question was a no-brainer. 3 students had heard this problem before, and 15 students didn't want to look like idiots, so they didn't raise their hands.
2126:(Steve Selvin, Marilyn vos Savant). I changed the words "logically equivalent" to "similar" because I do not believe the original statement is correct. Certainly there are strong similarities but I do not see that the problems are
1313:
In our case, there are three cards of equal probability, and each card has two sides of equal probability, so even though the two sides of the B/B card are indistinguishable (and likewise the two sides of the W/W card), we list
1124:
In our case, there are three cards of equal probability, and each card has two sides of equal probability, so even though the two sides of the B/B card are indistinguishable (and likewise the two sides of the W/W card), we list
414:
When assessing the cards on the table, you must assess where they came from, the cards in the hat. Just like guessing whether a person on a team is a man or a women, it is important to know whether this is a rugby team or not.
1101:
Okay, I think I get it now. Basically, what's happening, is that when there is a black side up, there are 3 possible scenarios 1) Side 1 of the B/B card is up 2) Side 2 of the B/B card is up 3) Side 3 (Black side of B/W) is up
2022:
variant is covered first (which presumably should be the box, since it is called the "box paradox"). The "card" explanation then just needs to demonstrate how it is exactly the same problem (which is pretty obvious really).
530:
by John Schuyler. The pdf you found is a copy, probably illegal, of "Science Puzzlers", first published by Scholastic in 1960, and republished by Dover in 1981 as "Entertaining Science Experiments with Everyday Objects".
2909:
It would appear even experts in this particular field don't know anything about this field. This doesn't bode well for the rest of us either. I will let my work stand for itself. Feel free to run it and look for errors.
2828:
I don't know which part of the article you're referring to (or even if you're referring to a specific section), so I'm not ruling out that the article's _explanation_ and/or _reasoning_ might be bad and/or wrong, but
1903:
Version 1, GG box contains 100 gold coins, others like before Version 2, GS box contains 50 gold and 50 silver coins, others like before Version 3 (my favorite) GG box contains only 1 gold coin, others like before
3122:
2450:
The simple solution which I proposed also occured before on this discussion page and in the article itself. But as it has been somewhat hidden within other considerations I have, maybe superfluously, opened this
1415:
True, if the assistant looks at ONLY ONE side of the card he chooses and then places a black side up if possible, then yes, it changes nothing and the formula still holds true precisely as stated in the original
309:
at either side, then there is a 50/50 chance it can be either. But there is more to this puzzle. Now I let you look one side of it. This is more information and changes your "bet". If you see white, there is
1292:
If, on the other hand, someone else, which we couldn't see, pulled out the card, looked at both sides, and said "(at least) one of the sides is black", your reasoning would be correct, and the chances would be
1259:
If, on the other hand, someone else, which we couldn't see, pulled out the card, looked at both sides, and said "(at least) one of the sides is black", your reasoning would be correct, and the chances would be
1024:
Now, I do agree that the probability of drawing a card with the same color on either side is 2/3, or 66.666%, but if the problem was presented as shown in this article, then the kids who said 50% were correct.
910:
but I do find it rather hard to believe that the majority of commentators discussing "the three cards problem" are actually using a form of it where the existence of the third card is absolutely irrelevant. --
872:
cards problem", one of those three cards is absolutely irrelevant. Because we specify that the card face we're looking at is black, the card that's white on both sides can never be the one we're looking at.
442:
Anyone who still has doubts - this is a very easy experiment to try. Try it 20 times with, say, 3 business cards with P or G written on each side. Record how many times the colour under matches the top.
3069:
2887:
The 'paradox' is in the probability, after choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, of the next coin drawn from the same box also being a gold coin.
1014:
if you look at one side of the card, you are left with 2 possible conclusions 1) The other side of your card has the same color as the side you see 2) The other side of your card has the another color
2796:
What if in the beginning, one of the boxes contains infinite golden balls. Does that make it 100% sure that one chooses the box with all gold? If I understood the 2/3 reasoning correctly, it would.
3238:
124:
2292:
with empty cups, as you describe; one needs to have a coin, and the other two need peas. Two are combined, and a pea is revealed. And the point is that the answer to the indistinguishable problem
1205:
Ok, now I need some further help from you smart people. It seems that the above argument assumes that we are drawing sides, not cards. If we are drawing cards, there are three possibilities:
1121:
three outcomes (no heads, one head, or two heads), but to have symmetry you must list four outcomes (HH, HT, TH, and TT, where H=head and T=tail, and pretending you can distinguish HT from TH).
315:
a common sense point of view, since you change your bet when you get to look and see white, you should also change your bet if you get to look and see black. More information changes the odds.
3113:
1569:
If we are picking a card that happens to be black on one side, then there are 2 possible cards that have at least one black side. Therefore, choosing the all-black card has 1/2 probability.
372:
Ok, I was wrong, I've done error, sorry to all. Result is 66,666...% = 2/3. And for future idiots, here is Vb code to simulate problem (just make form and command button to the right edge):
229:
1606:
Since we are observing only the side facing up, which turns out to be black, we have picked a black side as you put it. Picking a black card requires a much more elaborate setup, such as:
268:
2465:) the first step in BBP is to eliminate one of the two boxes the player did not choose to open a drawer: If he got a gold coin the SS box is eliminated, the GG box otherwise. So he will
4233:
600:. Sure, it's ambiguous, but I can't see anybody typing in "The Three Cards" and being disappointed by the result. Of course, that's because I can't see anybody typing it in at all...
3595:
Finally the second coin is chosen based on the box chosen and the result from the previous draw. The probability distribution of pulling the gold coin from each box looks like this:
2041:
I made a minor change in the informal description to reflect more specifically the probability distinction for the casual reader who may still assume even probability distribution.
1782:"However, this reasoning fails to exploit all of your information; you know not only that the card on the table has a black face, but also that one of its black faces is facing you."
3745:
3670:
3817:
2296:(for the two remaining containers) 1/3 and 2/3. The answer to the "distinguishable problem" is X/(1+X) and 1/(1+X); but unless you are told what X is, you can only assume X=1/2.
2399:
What probably causes the most confusion in this sort of probability problem, is what "case" should mean. Without realizing it, most people want it to mean "something that the
1635:
Ok first off: "What are the odds that the other side is also black?" I recommend learning the difference between odds and probability. Refer to wikipedia. The odds are 2
1566:
If we are picking a black side, then there are 3 possible black sides, 2 of which belong to the all-black card, rendering the probability of choosing the all-black card 2/3.
1344:
1155:
2179:, because as you noted probability problems can usually be solved by different approaches) be solved by the same mathematical formulation. If C1, C2, and C3 are the three
326:
by "False article?" is, are this article trying to describe real paradox, or is it just attempt to terrorize wikipedia? We don't have to argue about this, just try it!←
1833:
I was introduced to this problem with the three coins in a bag version. However, the difference is that here in this article the problem is stated as, paraphrased, "
2565:
2617:
2613:
2599:
1911:
If we consider a general case, GG box with X gold coins, GS box with Ng gold and Ns silver coins, and SS box with Y silver coins, then absolute probabilities are
578:
might be desirable since that seems to be Gardner's title. I'm not sure about the definite article or the capitalization, though, and maybe the name is ambiguous.
1525:
Bertrand or anyone else in this article under something like a "History of the problem" section could be worthwhile for the article's credibility and interest. —
418:
There are 2 single colour cards in the hat, and only one dual colour card. The chance of a single colour cards will be 2/3, regardless of what colour you see.
2722:
I believe you mean he said "if you take one coin out, look at it, and see that it's gold", which would appear to lead to the 1/2 conclusion and the paradox.
1923:(gold coin is extracted from GS box with probability Ng / (Ng + Ns), and conditional probability that randomly chosen golden coin originates from GG box is
408:'Thus, there are only two possible cards, double green-faced or purple/green-faced, and each has an equally likely probability of being the one you chose.'
4264:
1105:
In cases 1 and 2, the other side will also be black. I would delete everything I've said, but the discussion may help someone else comprehend this better.
114:
2694:
and the third contains a coin of each kind, one gold and one silver. A box is chosen at random. What is the probability that it contains the unlike coins?
4279:
2194:
In fact, the BPP contains more information than is needed, and that seems to get in the way of seeing the logical equivalence. Let the first box contain
219:
1381:
These are all equally probable outcomes of the fundamental experiment, and as there are no other possibilities, each has an initial probability of 1/6.
1180:
These are all equally probable outcomes of the fundamental experiment, and as there are no other possibilities, each has an initial probability of 1/6.
968:
In a survey of 53 Psychology freshmen taking an introductory probability course, 35 incorrectly responded 1/2; only 3 students correctly responded 2/3.
1694:, since they are exactly the same problem, only with cards substituted for boxes, sides and their colors substituted for drawers and their coins. --
4269:
790:
white) tells us anything about psychology, but I'm going to do a find-and-replace to follow Gardner. Yes, I know I have too much time on my hands!
1598:
It is tempting to think that the confusion is due to an ambiguous statement of the problem. But this is not the case. The current article text is
1028:
993:"Two of the 3 black faces belong to the same card. Given that you have choosen a black face, the chance of choosing one of those 2 faces is 2/3."
2484:
host opens door 2 or door 3. This last something strange consideration also is not necessary with BBP: We don't need at all the help of a host.--
2145:
in that the prior odds are 1:1, then comes some information which has a likelihood ratio of 1:2, hence the posterior odds are 1:2 (Bayes rule).
2803:
It seems like all the reasoning for 2/3 doesn't understand the question correctly. It would be correct only in a somewhat different situation.
344:
it's incorrect. Please check the earliest revisions of the article for that argument. As for the anon, all the arguments given in the article
3224:{\displaystyle {\frac {1}{2}}\times \left({\frac {1}{2}}+{\frac {1}{2}}\right)+{\frac {1}{2}}\times \left({\frac {1}{2}}+{\frac {1}{2}}\right)}
2885:
Hello. I am not familiar with this paradox and it seems to me that a sentence in the introduction is poorly written to people in my situation.
564:
What is the puzzle called in the sources you're using? It'd be nice to have a name for the article which gets at least *some* Google hits :) -
195:
4274:
3345:
Now it's time for the 2nd draw, including all 3 boxes. The draw will be different based on the result of the previous draw, so we have this:
1639:{1,2,3} is showing with equal probability. What is the probability that the face underneath is one of {1,2,3}. That probability is 2/3rds.
1401:
90:
4259:
1032:
2093:
Then I guess I worded it poorly, since you misunderstood. No probability actually changes; but the combination of a specific observation
1495:
What does the white/white card have to do with anything? It's not clear to me why it's included in the problem at all. Historic reasons?
2810:
2761:
2730:
2359:
2023:
1572:
This problem is based on the ambiguity of whether we know that we are picking a black side or not, not on its counterintuitive nature.
411:
I think this is not true. The history of where the cards came from is key. How do we know there is an equal chance of those 2 cards?
2914:
function randomIntFromInterval(min, max) { // min and max included return Math.floor(Math.random() * (max - min + 1) + min) }
186:
147:
2469:(now that means having a gold coin in the other drawer of his box) in two of three cases, namely if he didn't choose the mixed box.
3021:
2998:
2566:
https://web.archive.org/web/20060420063730/http://harrisschool.uchicago.edu:80/About/publications/working-papers/pdf/wp_05_14.pdf
81:
58:
3325:{\displaystyle {\frac {1}{2}}\times \left({\frac {1}{2}}+{\frac {1}{2}}\right)+{\frac {1}{2}}\times \left({\frac {1}{1}}\right)}
2689:
Three boxes are identical in external appearance. The first box contains two gold coins, the second contains two silver coins,
2752:
1587:
808:. I think it's mainly "Formal approach" that needs logical cleanup, and we also need to add a correct, informal explanation.
2937:
if (rndInt != 3) //You chose a box with at least 1 gold ball { gold += 1; //, but have not yet drawn a ball from it.
1520:
I don't know anything about this problem other than what I've read on Knowledge and the discussion pages. But obviously the
2858:
probabilities for the box. For example, if the coin is silver, then the new probability for the all-gold box will be 0%.
2660:
1443:
3) Assistant looks at both sides of W/W card and rejects the card (Repeats process until seeing at least one BLACK side)
33:
3073:
However since 1 box doesn't contain any gold coins, that possibility is discarded, so it's 1 out of 2, which would be:
2726:
2569:
804:
I think this article still needs a tag at the top indicating that it's not "done", whatever that means, so I'm adding
3117:
After that, in both cases, 1 out of 2 coins is chosen. The result depends on the box chosen, so it looks like this:
1981:
Aposteriori probability is 0 or 1 for any possible case, while apriori one is never (generally) 1, though may by 0.
2408:
given about the selections only applies the version where the cases are coins, since all we learn about is a coin.
1786:
However, this reasoning fails to exploit all of your information; you know not only that the card on the table has
760:
By listing green first I mean that green is the color shown after the draw. The external links are divided between
616:
3233:
But again, one of those possibilities (silver coin being chosen) has to be discarded, so it's actually like this:
3078:
1601:"You put all of the cards in a hat, pull one out at random, and place it on a table. The side facing up is black."
1473:
So now it IS 50/50 that the other side is WHITE, but of course these are not the conditions stated in the article.
670:
2400:
257:
2700:
out changes the answer to Bertrand's actual question. That absurd conclusion is the paradox Bertrand refers to.
2616:
to delete these "External links modified" talk page sections if they want to de-clutter talk pages, but see the
648:, with colors gold and silver.) Not helpful, I'm afraid, but I thought I'd report progress (or lack thereof).
2046:
1730:
1709:
1405:
4241:
1653:
1036:
4187:
2814:
2549:
2363:
2269:(clearly) can. No doubt the philosophers have some technical terms connected to the theory of naming things.
2244:
Both problems are the same, whether or not you can make a distinction in isolation. You can in the situation.
2027:
1758:
1747:
1684:
1521:
1419:
However, if the assistant looks at BOTH sides and places a black side up if possible, it in fact DOES matter.
712:
666:
2765:
2734:
2651:
2557:
2119:
693:
2323:
defined there are various conditional probabilities and the solution is actually the Bayes' rule solution.
2225:
I am not aware of any source which discusses these "sort of equivalences" in a careful and systematic way.
1908:
thus clearly indicating that usual speculations about absolute number of gold coins are quite irrelevant.
3002:
2895:
2785:
2118:
The lead of the article stated that Bertand's box paradox is logically equivalent is with the more famous
1500:
obviously 2/3. Now you can explain that the fact the shown color is black doesn't change the probability.
1064:
653:
484:
353:
296:
4237:
1649:
1106:
392:
364:
327:
2963:
2870:
2635:
If you have discovered URLs which were erroneously considered dead by the bot, you can report them with
2623:
1865:
1695:
1583:
987:"Two of the 3 black faces belong to the same card. The chance of choosing one of those 2 faces is 2/3."
911:
885:
882:
720:
454:
39:
2556:. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit
2476:
disappears completely: that is to find the statement of task which has really the claimed 2/3 solution.
457:
447:
422:
172:
3676:
3601:
2959:
2866:
73:
52:
3751:
2979:
2806:
2590:
2426:
Of course in searching the best solution we have to take account of both sides of Einstein's advice:
2328:
2274:
2230:
2157:
1990:
1986:
1939:
1935:
1575:
1397:
597:
575:
2946:
2 gold balls { winner += 1; //winner winner chicken dinner } } }
2375:
after choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin
1721:
I came to the discussion page to make exactly this same suggestion. I tagged the article for merge.
250:
21:
2083:
2042:
1754:
1743:
1724:
1705:
1691:
1579:
1238:
1188:
689:
688:
I'd still prefer Gardner's title but either of those will do too. We could also use something like
283:
1770:
1698:
194:
on Knowledge. If you would like to participate, please visit the project page, where you can join
89:
on Knowledge. If you would like to participate, please visit the project page, where you can join
2852:
in that, the fact that we subsequently roll a die does not change the probabilities for the coin.
2527:
2485:
2431:
2382:
2343:
2123:
1766:
1672:
1620:
1540:
1478:
1242:
1227:
1192:
1110:
1090:
1049:
305:
See if this helps. Let us assume that you have only the 2 cards as you say. If you draw one and
178:
2620:
before doing mass systematic removals. This message is updated dynamically through the template
2377:
I got G1, G2 or G3. It seems for me that the key for the whole understanding is to see the fact
2342:
There are two cases that the other coin is a gold coin, and one case that it is a silver coin.--
1317:
1128:
550:
I've taken out the sections on the "qualitative approach" because they basically make no sense.
162:
141:
2636:
1842:
1017:
There are no thirds to consider, there's a 50% chance that the color will come up either way.
984:
Hi - It might be just me, but I feel that in the "Solutions/Intuition" section, where it says:
2891:
2781:
2709:
2413:
2307:
2255:
2208:
2103:
2078:
how observing a gold coin changes the probability that the box is GS or GG, but not SS". ????
2062:
2006:
1966:
1060:
649:
480:
479:
I think I fixed the article, leaving the naive 1/2 in, but showing the correct number is 2/3.
349:
2531:
2489:
2435:
2386:
2347:
1877:
1818:
1505:
752:
This is a trivial issue, but it's easiest to fix trivial issues. The article currently uses
2643:
508:
2324:
2270:
2226:
2153:
2152:
Maybe I'm wrong but in any case the statement needs a literature reference to support it.
1802:
1761:. I think this is consistent with the comments in the above pre-merge talk page sections.
1282:
1271:
805:
2602:, "External links modified" talk page sections are no longer generated or monitored by
2079:
1846:
1713:
1526:
1000:
736:
673:. I'll change to one of these in a day or 2 and add a few links to here as mentioned.
2642:
If you found an error with any archives or the URLs themselves, you can fix them with
1437:
1) Assistant looks at both sides of B/B card and places one of the BLACK sides face up
4253:
2992:
1762:
1668:
1616:
1536:
1474:
949:
897:
878:
843:
834:
809:
791:
697:
601:
584:
565:
551:
532:
514:
507:
I found that Martin Gardner has written about it under the name of "The three cards"
465:
429:
4245:
3006:
2982:
2967:
2899:
2874:
2818:
2769:
2738:
2713:
2665:
2535:
2493:
2439:
2417:
2390:
2367:
2351:
2332:
2311:
2278:
2259:
2234:
2212:
2161:
2107:
2087:
2066:
2050:
2031:
2010:
1994:
1970:
1943:
1881:
1850:
1822:
1806:
1735:
1716:
1676:
1657:
1624:
1591:
1544:
1529:
1509:
1482:
1409:
1274:
1246:
1230:
1199:
1113:
1093:
1067:
1052:
1040:
1003:
974:
952:
932:
914:
900:
846:
837:
823:
812:
794:
739:
723:
700:
677:
656:
627:
604:
587:
568:
554:
535:
517:
498:
487:
468:
432:
395:
367:
356:
330:
319:
299:
2800:
the other ball is silver. 50/50, it's strictly about the probability of the boxes.
2705:
2677:
2570:
http://harrisschool.uchicago.edu/About/publications/working-papers/pdf/wp_05_14.pdf
2409:
2303:
2251:
2204:
2099:
2058:
2002:
1962:
2463:
The contestant determines two doors of which the host has to open one with a goat
1109:
07:06, 14 February 2007 (UTC) (Sorry about the IP, forgot to re-log in. Me again
2609:
1873:
1814:
1501:
820:
644:(sophomore math). (And he uses red/white cards, with an odd reference to three
620:
191:
970:
What did the other 15 psychology students suggest? Anyone knows? Just curious.
348:
correct, even if we don't explain why the argument proposed above is wrong. —
2608:. No special action is required regarding these talk page notices, other than
1798:
1440:
2) Assistant looks at both sides of B/W card and places the BLACK side face up
929:
674:
624:
495:
444:
419:
316:
168:
86:
996:
Without that, I at least was still thinking in terms of the overall cards...
619:. No caps per Wiki norm - is not a proper name! And put a link to it on the
1869:
971:
715:, which should really be merged here. Bertrand's problem, though, involved
261:
2848:
Flip a coin. If heads, roll a 4-sided die. If tails, roll a 8-sided die.
2520:
Tell the smallest extension of the problem which has really a 2/3 solution!
2373:
three gold coins, G3, the other coin in the same box is a silver coin. And
1452:
We throw out number 3 and we are left with 1 and 2 still equal probability:
1428:
In the latter case only three possible scenarios each with 1/3 probability:
1305:
1196:
511:
1868:, then I suppose that section I have just described probably belongs in
363:
How do you explain that the real world result are 50/50? PLEASE TRY IT!
637:
295:
one thing in this article; try it!!! I did try, probability was 50/50.
3016:
In the first draw, you pick 1 out of 3 boxes. This can be written as:
2001:
Discussion moved to a more appropriate place, Michaeldsp's talk page.
1299:
Why would this change anything? Just because we didn't pick the card?
1837:, what is the probability the opposite side is heads" compared to, "
1667:
I like the labeling trick; fortunately it's already in the article.
788:
I don't know if the apparent consensus on the ordering (black : -->
2978:
Due to Vaidman, Aranohov et al. should it be mentioned? cheers,
1237:
because there is no "equal opportunity" for lack of better words.
574:
Duh, actually reading the above shows that a move to a title like
3334:
As a result, the boxes now look like either of those 3 variants:
2502:
1009:
I think there is a logical error involved here somewhere (Or not)
527:
523:
264:
245:
15:
2522:. But completing the joke we could also add the cook saying:
1926:
p(GG box | random gold coin) = 1/ = ( Ng + Ns)/( 2*Ng + Ns)]
1648:
I would edit the thread but its just to darn much fun as is.
640:. He couldn't find a name or a reference, but he used it in
3064:{\displaystyle {\frac {1}{3}}+{\frac {1}{3}}+{\frac {1}{3}}}
2575:
When you have finished reviewing my changes, please set the
719:
coins contained in three boxes, rather than three coins. --
1896:) so the probability the drawer is from box is GG is 2⁄3."
2560:
for additional information. I made the following changes:
2792:
A honest question from a mathematically illiterate person
2461:
Just as in the well formulated Monty Hall Problem (MHP) (
2130:(I understand what that means); I don't really know what
1048:
Added: I'm signing this because I just made an account.--
2480:
have known this at the beginning, which is not possible.
2553:
1861:
4190:
3828:
3754:
3679:
3604:
3353:
3241:
3125:
3081:
3024:
2472:
The great advantage of BBP as opposed to MHP is that
1917:(gold coin is extracted from GG box with certainty),
1615:
This is clearly not the problem under consideration.
1320:
1131:
2524:
You'll win if you guess the color of the other side.
777:
Schuyler and Gardner, listed on this talk page, use
190:, a collaborative effort to improve the coverage of
85:, a collaborative effort to improve the coverage of
3822:Which means the final calculation looks like this:
2671:
The current article is based on revisionist history
2612:using the archive tool instructions below. Editors
2428:
Make things as simple as possible, but not simpler.
1555:The question boils down to: Are we picking a black
4227:
4177:
3811:
3739:
3664:
3585:
3324:
3223:
3107:
3063:
2993:https://github.com/isghe/Paradosso_delle_tre_carte
2775:Bridge players and Bayesians get this straight off
1690:This article should probably be consolidated with
1387:have symmetry, with a probability of 1/3 for each.
1338:
1184:have symmetry, with a probability of 1/3 for each.
1149:
855:This problem and Lewis Carroll's two coins problem
711:The "odd reference" sounds like he is referencing
2958:was declared along with the other declarations.
2526:And we could place the story on a game show ...--
2454:But as we have this section now, here some other
1704:I agree. Also, there was some discussion on the
891:I agree with you on mathematical grounds, but I
340:argument which the anons give above and explain
990:it would be clearer if it said something like:
833:in probability myself, and it doesn't stop me!
2760:psychology majors, my initial answer was 1/2.
2598:This message was posted before February 2018.
2503:Answer to the "The Three Cards" Teaser Problem
1187:With this approach, try your hand at e.g. the
3108:{\displaystyle {\frac {1}{2}}+{\frac {1}{2}}}
2287:I think you missed my point. The solution to
896:scammer, if you know what I'm talking about.
405:The key sentence in this analysis I think is
8:
2933:Your error is in your main conditional part.
2505:above mentioned John R. Schuyler presents a
1950:I must warn you that it doesn't matter what
391:You can use this code for free as you like.
2073:Typo in "The paradox as stated by Bertrand"
1559:that goes with a card, or are we picking a
3012:The solution in the article makes no sense
2804:
2548:I have just modified one external link on
522:Good find! Indeed it does give leads; see
336:I told people we should have retained the
136:
47:
4194:
4189:
4155:
4142:
4129:
4116:
4103:
4090:
4072:
4054:
4031:
4018:
4005:
3992:
3979:
3966:
3948:
3930:
3917:
3904:
3891:
3878:
3865:
3847:
3829:
3827:
3798:
3778:
3761:
3753:
3726:
3706:
3686:
3678:
3651:
3631:
3611:
3603:
3563:
3550:
3537:
3519:
3501:
3478:
3465:
3452:
3434:
3416:
3403:
3390:
3372:
3354:
3352:
3308:
3291:
3273:
3260:
3242:
3240:
3206:
3193:
3175:
3157:
3144:
3126:
3124:
3095:
3082:
3080:
3051:
3038:
3025:
3023:
2379:For G1 this is G2, and for G2, this is G1
1319:
1130:
636:I contacted my statistics professor from
4228:{\displaystyle ={\frac {5}{12}}=0,41(6)}
2678:Eugene Northrop, Riddles in Mathematics
1920:p(GS box and random gold coin)=(1/3)*
1829:Minor Concern How the Problem Is Stated
1792:it has a black face which is facing you
138:
49:
19:
1642:Probability: 1/3*1+1/3*1 +1/3*0 = 2/3
1563:that happens to be black on one side?
2587:to let others know (documentation at
7:
1914:p(GG box and random gold coin)=1/3
1780:I would like to change the sentence
475:From Heraclesprogeny to Arthur Rubin
184:This article is within the scope of
79:This article is within the scope of
1856:Removing the "100 silver coins" bit
669:to make it like ]. I still prefer
583:Stupid Gardnerian puzzlecruft ;) -
510:. This should give more leads. --
38:It is of interest to the following
4265:Low-importance Statistics articles
2881:The introduction is poorly written
1464:2) B/W card with the BLACK side up
14:
4280:Low-priority mathematics articles
3740:{\displaystyle (1/1);(1/2);(0/2)}
3665:{\displaystyle (1/1);(1/2);(0/2)}
2747:Consolation for psychology majors
2552:. Please take a moment to review
1535:Agreed, I'll merge the articles.
854:
204:Knowledge:WikiProject Mathematics
3812:{\displaystyle (2/2;(0/1);(0/2)}
1930:=1 in the original formulation.
1645:The odds are (2/3) / (1/3) = 2.
1461:1) B/B card with a BLACK side up
400:
249:
207:Template:WikiProject Mathematics
171:
161:
140:
99:Knowledge:WikiProject Statistics
72:
51:
20:
4270:WikiProject Statistics articles
2988:It looks we need just TWO cards
2358:coin." Or something like that.
1710:Three cards problem's talk page
1157:outcomes of equal probability:
256:This article was nominated for
224:This article has been rated as
119:This article has been rated as
102:Template:WikiProject Statistics
4222:
4216:
3806:
3792:
3786:
3772:
3755:
3734:
3720:
3714:
3700:
3694:
3680:
3659:
3645:
3639:
3625:
3619:
3605:
2753:Bertrand paradox (probability)
2318:Bayes rule, symmetry arguments
1887:Correct answer, wrong solution
1882:10:10, 16 September 2009 (UTC)
1823:20:41, 16 September 2009 (UTC)
1346:outcomes of equal probability:
1231:20:11, 14 September 2007 (UTC)
980:Minor clarification suggestion
1:
3007:21:16, 13 November 2022 (UTC)
2167:Two probability problems are
1592:02:54, 14 November 2007 (UTC)
1510:22:04, 28 November 2007 (UTC)
1410:00:37, 25 February 2009 (UTC)
1200:11:33, 14 February 2007 (UTC)
1114:15:29, 14 February 2007 (UTC)
1094:02:56, 14 February 2007 (UTC)
1068:01:45, 14 February 2007 (UTC)
1053:01:18, 14 February 2007 (UTC)
1041:01:14, 14 February 2007 (UTC)
868:even though it's called the "
847:09:28, 13 February 2006 (UTC)
838:09:07, 13 February 2006 (UTC)
824:08:41, 13 February 2006 (UTC)
617:The three cards (probability)
469:15:23, 27 February 2007 (UTC)
396:19:49, 14 December 2006 (UTC)
368:18:33, 14 December 2006 (UTC)
357:17:06, 14 December 2006 (UTC)
331:16:29, 14 December 2006 (UTC)
320:13:12, 14 December 2006 (UTC)
300:06:28, 14 December 2006 (UTC)
198:and see a list of open tasks.
93:and see a list of open tasks.
4275:B-Class mathematics articles
2968:05:08, 22 October 2021 (UTC)
2875:06:05, 22 October 2021 (UTC)
2770:03:22, 15 October 2017 (UTC)
2739:03:22, 15 October 2017 (UTC)
2666:20:09, 31 October 2016 (UTC)
2032:18:40, 4 December 2011 (UTC)
1839:No matter what side is drawn
1699:05:22, 10 January 2007 (UTC)
1275:21:02, 11 October 2007 (UTC)
1004:12:43, 2 February 2007 (UTC)
975:02:40, 10 October 2006 (UTC)
813:18:18, 14 January 2006 (UTC)
795:16:25, 12 January 2006 (UTC)
740:03:44, 5 November 2007 (UTC)
724:18:50, 2 February 2007 (UTC)
701:14:29, 14 January 2006 (UTC)
678:14:13, 14 January 2006 (UTC)
671:The three cards (probabilty)
657:01:35, 13 January 2006 (UTC)
628:20:06, 12 January 2006 (UTC)
605:16:39, 12 January 2006 (UTC)
588:16:23, 12 January 2006 (UTC)
569:12:24, 12 January 2006 (UTC)
555:07:50, 12 January 2006 (UTC)
536:09:08, 12 January 2006 (UTC)
518:08:49, 12 January 2006 (UTC)
499:00:13, 12 January 2006 (UTC)
488:00:04, 12 January 2006 (UTC)
448:18:20, 11 January 2006 (UTC)
433:12:28, 11 January 2006 (UTC)
423:12:26, 11 January 2006 (UTC)
4260:B-Class Statistics articles
2727:Phonemic restoration effect
1851:00:31, 28 August 2009 (UTC)
1741:
1682:
1483:18:13, 28 August 2010 (UTC)
1339:{\displaystyle 3\times 2=6}
1150:{\displaystyle 3\times 2=6}
458:04:54, 30 August 2006 (UTC)
4296:
4246:19:26, 18 March 2023 (UTC)
2900:09:52, 9 August 2020 (UTC)
2629:(last update: 5 June 2024)
2545:Hello fellow Wikipedians,
2333:12:10, 19 April 2013 (UTC)
2213:21:44, 21 April 2013 (UTC)
2162:07:42, 19 April 2013 (UTC)
2141:So we have a mathematical
2108:21:16, 10 April 2012 (UTC)
2088:16:38, 10 April 2012 (UTC)
2011:20:45, 3 August 2011 (UTC)
1995:19:57, 2 August 2011 (UTC)
1971:01:49, 1 August 2011 (UTC)
1790:black face, but also that
1771:01:17, 30 March 2008 (UTC)
1736:00:43, 24 April 2007 (UTC)
1717:03:57, 30 March 2007 (UTC)
1677:01:24, 30 March 2008 (UTC)
1658:21:59, 29 March 2008 (UTC)
1625:00:45, 30 March 2008 (UTC)
1545:00:47, 30 March 2008 (UTC)
1530:04:22, 30 March 2007 (UTC)
1191:(easy), or maybe even the
2974:Quantum three box paradox
2819:13:25, 24 July 2020 (UTC)
2714:12:09, 26 July 2017 (UTC)
2509:claiming a 2/3 solution:
2401:Principle of Indifference
2128:mathematically equivalent
2122:(Martin Gardner) and the
2067:21:29, 5 March 2012 (UTC)
2051:17:13, 4 March 2012 (UTC)
1944:04:21, 26 July 2011 (UTC)
1807:23:02, 19 June 2009 (UTC)
953:19:28, 4 March 2006 (UTC)
933:11:50, 4 March 2006 (UTC)
915:00:19, 2 March 2006 (UTC)
901:21:11, 1 March 2006 (UTC)
886:15:47, 1 March 2006 (UTC)
223:
156:
118:
67:
46:
2983:19:12, 3 July 2021 (UTC)
2680:, starting on page 165.
2536:10:59, 4 June 2013 (UTC)
2494:12:36, 30 May 2013 (UTC)
2440:10:57, 30 May 2013 (UTC)
2418:13:16, 27 May 2013 (UTC)
2391:11:51, 26 May 2013 (UTC)
2368:15:31, 23 May 2013 (UTC)
2352:10:13, 17 May 2013 (UTC)
2312:12:14, 10 May 2013 (UTC)
2279:11:25, 10 May 2013 (UTC)
1551:Ambiguity in the Problem
1247:07:25, 9 July 2009 (UTC)
230:project's priority scale
2541:External links modified
2518:is the most important:
2260:22:55, 9 May 2013 (UTC)
2235:07:59, 9 May 2013 (UTC)
2120:Three Prisoners problem
2114:Similar, not equivalent
667:The Three Cards problem
665:One more possible name
401:What's the answer (2/3)
187:WikiProject Mathematics
4229:
4179:
3813:
3741:
3666:
3587:
3326:
3225:
3109:
3065:
2550:Bertrand's box paradox
1759:Bertrand's box paradox
1748:Bertrand's box paradox
1685:Bertrand's box paradox
1522:Bertrand's Box Paradox
1516:Add a history section?
1340:
1151:
713:Bertrand's Box Paradox
694:The three cards puzzle
82:WikiProject Statistics
28:This article is rated
4230:
4180:
3814:
3742:
3667:
3588:
3327:
3226:
3110:
3066:
2725:but not mentioned on
1860:I am about to remove
1341:
1152:
1031:comment was added by
842:Eh, I did it myself.
4188:
3826:
3752:
3677:
3602:
3351:
3239:
3123:
3079:
3022:
2751:The similarly-named
2610:regular verification
2132:logically equivalent
1318:
1129:
210:mathematics articles
2600:After February 2018
2579:parameter below to
2171:equivalent if they
1755:Three cards problem
1744:Three cards problem
1706:Three cards problem
1692:Three cards problem
1189:Boy or Girl paradox
690:Three cards problem
284:Three cards problem
105:Statistics articles
4225:
4175:
3809:
3737:
3662:
3583:
3322:
3221:
3105:
3061:
2654:InternetArchiveBot
2605:InternetArchiveBot
2124:Monty Hall Problem
1366:#other side of W/W
1356:#Other side of B/B
1336:
1193:Monty Hall problem
1147:
179:Mathematics portal
34:content assessment
4202:
4163:
4150:
4137:
4124:
4111:
4098:
4080:
4062:
4039:
4026:
4013:
4000:
3987:
3974:
3956:
3938:
3925:
3912:
3899:
3886:
3873:
3855:
3837:
3571:
3558:
3545:
3527:
3509:
3486:
3473:
3460:
3442:
3424:
3411:
3398:
3380:
3362:
3316:
3299:
3281:
3268:
3250:
3214:
3201:
3183:
3165:
3152:
3134:
3103:
3090:
3059:
3046:
3033:
2905:The answer is 1/2
2821:
2809:comment added by
2630:
2338:A simple solution
2017:Poor organisation
1835:If heads is drawn
1753:I've just merged
1594:
1578:comment added by
1400:comment added by
1169:other side of W/W
1163:Other side of B/B
1044:
279:
278:
244:
243:
240:
239:
236:
235:
135:
134:
131:
130:
4287:
4234:
4232:
4231:
4226:
4203:
4195:
4184:
4182:
4181:
4176:
4174:
4170:
4169:
4165:
4164:
4156:
4151:
4143:
4138:
4130:
4125:
4117:
4112:
4104:
4099:
4091:
4081:
4073:
4063:
4055:
4050:
4046:
4045:
4041:
4040:
4032:
4027:
4019:
4014:
4006:
4001:
3993:
3988:
3980:
3975:
3967:
3957:
3949:
3944:
3940:
3939:
3931:
3926:
3918:
3913:
3905:
3900:
3892:
3887:
3879:
3874:
3866:
3856:
3848:
3838:
3830:
3818:
3816:
3815:
3810:
3802:
3782:
3765:
3746:
3744:
3743:
3738:
3730:
3710:
3690:
3671:
3669:
3668:
3663:
3655:
3635:
3615:
3592:
3590:
3589:
3584:
3582:
3578:
3577:
3573:
3572:
3564:
3559:
3551:
3546:
3538:
3528:
3520:
3510:
3502:
3497:
3493:
3492:
3488:
3487:
3479:
3474:
3466:
3461:
3453:
3443:
3435:
3430:
3426:
3425:
3417:
3412:
3404:
3399:
3391:
3381:
3373:
3363:
3355:
3331:
3329:
3328:
3323:
3321:
3317:
3309:
3300:
3292:
3287:
3283:
3282:
3274:
3269:
3261:
3251:
3243:
3230:
3228:
3227:
3222:
3220:
3216:
3215:
3207:
3202:
3194:
3184:
3176:
3171:
3167:
3166:
3158:
3153:
3145:
3135:
3127:
3114:
3112:
3111:
3106:
3104:
3096:
3091:
3083:
3070:
3068:
3067:
3062:
3060:
3052:
3047:
3039:
3034:
3026:
2997:Isidoro Ghezzi.
2980:Michael C. Price
2913:<script : -->
2888:
2664:
2655:
2628:
2627:
2606:
2594:
1734:
1727:
1696:Antaeus Feldspar
1683:Talk archive of
1573:
1491:Why three cards?
1412:
1361:#One side of W/W
1351:#One side of B/B
1345:
1343:
1342:
1337:
1156:
1154:
1153:
1148:
1026:
912:Antaeus Feldspar
883:Antaeus Feldspar
721:Antaeus Feldspar
388:
385:
383:
376:
282:Talk archive of
267:. The result of
253:
246:
212:
211:
208:
205:
202:
181:
176:
175:
165:
158:
157:
152:
144:
137:
125:importance scale
107:
106:
103:
100:
97:
76:
69:
68:
63:
55:
48:
31:
25:
24:
16:
4295:
4294:
4290:
4289:
4288:
4286:
4285:
4284:
4250:
4249:
4186:
4185:
4089:
4085:
4071:
4067:
3965:
3961:
3864:
3860:
3846:
3842:
3824:
3823:
3750:
3749:
3675:
3674:
3600:
3599:
3536:
3532:
3518:
3514:
3451:
3447:
3389:
3385:
3371:
3367:
3349:
3348:
3304:
3259:
3255:
3237:
3236:
3192:
3188:
3143:
3139:
3121:
3120:
3077:
3076:
3020:
3019:
3014:
2990:
2976:
2955:
2947:
2938:
2927:
2923:
2919:
2915:
2907:
2886:
2883:
2849:
2794:
2777:
2749:
2673:
2658:
2653:
2621:
2614:have permission
2604:
2588:
2558:this simple FaQ
2543:
2507:similar problem
2340:
2320:
2116:
2075:
2039:
2019:
1927:
1921:
1915:
1905:
1889:
1858:
1831:
1778:
1751:
1725:
1722:
1688:
1633:
1553:
1518:
1493:
1402:142.177.153.177
1395:
1316:
1315:
1166:One side of W/W
1160:One side of B/B
1127:
1126:
1027:—The preceding
1011:
982:
965:
857:
806:Template:Expert
802:
750:
598:The Three Cards
576:The Three Cards
562:
548:
477:
403:
387:
384:
382:
378:
375:
297:195.163.176.146
292:
287:
209:
206:
203:
200:
199:
177:
170:
150:
104:
101:
98:
95:
94:
61:
32:on Knowledge's
29:
12:
11:
5:
4293:
4291:
4283:
4282:
4277:
4272:
4267:
4262:
4252:
4251:
4238:Archimedes5000
4224:
4221:
4218:
4215:
4212:
4209:
4206:
4201:
4198:
4193:
4173:
4168:
4162:
4159:
4154:
4149:
4146:
4141:
4136:
4133:
4128:
4123:
4120:
4115:
4110:
4107:
4102:
4097:
4094:
4088:
4084:
4079:
4076:
4070:
4066:
4061:
4058:
4053:
4049:
4044:
4038:
4035:
4030:
4025:
4022:
4017:
4012:
4009:
4004:
3999:
3996:
3991:
3986:
3983:
3978:
3973:
3970:
3964:
3960:
3955:
3952:
3947:
3943:
3937:
3934:
3929:
3924:
3921:
3916:
3911:
3908:
3903:
3898:
3895:
3890:
3885:
3882:
3877:
3872:
3869:
3863:
3859:
3854:
3851:
3845:
3841:
3836:
3833:
3820:
3819:
3808:
3805:
3801:
3797:
3794:
3791:
3788:
3785:
3781:
3777:
3774:
3771:
3768:
3764:
3760:
3757:
3747:
3736:
3733:
3729:
3725:
3722:
3719:
3716:
3713:
3709:
3705:
3702:
3699:
3696:
3693:
3689:
3685:
3682:
3672:
3661:
3658:
3654:
3650:
3647:
3644:
3641:
3638:
3634:
3630:
3627:
3624:
3621:
3618:
3614:
3610:
3607:
3581:
3576:
3570:
3567:
3562:
3557:
3554:
3549:
3544:
3541:
3535:
3531:
3526:
3523:
3517:
3513:
3508:
3505:
3500:
3496:
3491:
3485:
3482:
3477:
3472:
3469:
3464:
3459:
3456:
3450:
3446:
3441:
3438:
3433:
3429:
3423:
3420:
3415:
3410:
3407:
3402:
3397:
3394:
3388:
3384:
3379:
3376:
3370:
3366:
3361:
3358:
3343:
3342:
3340:
3338:
3320:
3315:
3312:
3307:
3303:
3298:
3295:
3290:
3286:
3280:
3277:
3272:
3267:
3264:
3258:
3254:
3249:
3246:
3219:
3213:
3210:
3205:
3200:
3197:
3191:
3187:
3182:
3179:
3174:
3170:
3164:
3161:
3156:
3151:
3148:
3142:
3138:
3133:
3130:
3102:
3099:
3094:
3089:
3086:
3058:
3055:
3050:
3045:
3042:
3037:
3032:
3029:
3013:
3010:
2989:
2986:
2975:
2972:
2971:
2970:
2954:var ball = 0;
2953:
2952:
2951:
2944:
2943:
2942:
2936:
2935:
2934:
2929:
2924:
2920:
2916:
2912:
2906:
2903:
2882:
2879:
2878:
2877:
2862:
2860:
2859:
2854:
2853:
2847:
2846:
2845:
2840:
2839:
2833:
2831:
2830:
2824:
2793:
2790:
2776:
2773:
2748:
2745:
2744:
2743:
2742:
2741:
2692:
2691:
2672:
2669:
2648:
2647:
2640:
2573:
2572:
2564:Added archive
2542:
2539:
2499:
2498:
2497:
2496:
2481:
2477:
2470:
2459:
2452:
2445:
2444:
2443:
2442:
2421:
2420:
2396:
2395:
2394:
2393:
2339:
2336:
2319:
2316:
2315:
2314:
2298:
2297:
2284:
2283:
2282:
2281:
2263:
2262:
2246:
2245:
2240:
2239:
2238:
2237:
2223:
2216:
2215:
2201:
2200:
2191:
2190:
2185:
2184:
2181:a priori cases
2169:mathematically
2115:
2112:
2111:
2110:
2074:
2071:
2070:
2069:
2043:Flying Hamster
2038:
2035:
2018:
2015:
2014:
2013:
1974:
1973:
1957:
1956:
1925:
1919:
1913:
1902:
1888:
1885:
1866:WP:NOTTEXTBOOK
1857:
1854:
1830:
1827:
1826:
1825:
1777:
1774:
1750:
1740:
1739:
1738:
1726:brighterorange
1719:
1687:
1681:
1680:
1679:
1665:
1650:Jeremiahrounds
1632:
1629:
1628:
1627:
1613:
1612:
1611:
1604:
1603:
1602:
1552:
1549:
1548:
1547:
1517:
1514:
1513:
1512:
1492:
1489:
1488:
1487:
1486:
1485:
1468:
1467:
1466:
1465:
1462:
1456:
1455:
1454:
1453:
1447:
1446:
1445:
1444:
1441:
1438:
1432:
1431:
1430:
1429:
1423:
1422:
1421:
1420:
1417:
1390:
1389:
1383:
1378:
1376:#W side of B/W
1373:
1371:#B side of B/W
1368:
1363:
1358:
1353:
1348:
1335:
1332:
1329:
1326:
1323:
1310:
1301:
1300:
1296:
1295:
1288:
1287:
1278:
1277:
1268:
1267:
1262:
1261:
1256:
1255:
1250:
1249:
1221:
1220:
1219:
1218:
1215:
1212:
1203:
1202:
1185:
1181:
1178:
1177:
1176:
1173:
1170:
1167:
1164:
1161:
1146:
1143:
1140:
1137:
1134:
1122:
1107:68.227.203.149
1099:
1098:
1097:
1096:
1088:
1081:
1080:
1079:
1078:
1071:
1070:
1047:
1033:68.227.203.149
1010:
1007:
981:
978:
964:
961:
960:
959:
958:
957:
956:
955:
940:
939:
938:
937:
936:
935:
920:
919:
918:
917:
904:
903:
856:
853:
852:
851:
850:
849:
840:
827:
826:
801:
798:
786:
785:
782:
775:
774:
771:
768:
765:
758:
757:
749:
746:
745:
744:
743:
742:
729:
728:
727:
726:
706:
705:
704:
703:
683:
682:
681:
680:
660:
659:
633:
632:
631:
630:
610:
609:
608:
607:
591:
590:
580:
579:
561:
558:
547:
544:
543:
542:
541:
540:
539:
538:
502:
501:
476:
473:
472:
471:
436:
435:
402:
399:
393:212.16.102.117
379:
365:212.16.102.117
362:
360:
359:
328:212.16.102.117
323:
322:
291:
290:False article?
288:
286:
280:
277:
276:
269:the discussion
254:
242:
241:
238:
237:
234:
233:
222:
216:
215:
213:
196:the discussion
183:
182:
166:
154:
153:
145:
133:
132:
129:
128:
121:Low-importance
117:
111:
110:
108:
91:the discussion
77:
65:
64:
62:Low‑importance
56:
44:
43:
37:
26:
13:
10:
9:
6:
4:
3:
2:
4292:
4281:
4278:
4276:
4273:
4271:
4268:
4266:
4263:
4261:
4258:
4257:
4255:
4248:
4247:
4243:
4239:
4235:
4219:
4213:
4210:
4207:
4204:
4199:
4196:
4191:
4171:
4166:
4160:
4157:
4152:
4147:
4144:
4139:
4134:
4131:
4126:
4121:
4118:
4113:
4108:
4105:
4100:
4095:
4092:
4086:
4082:
4077:
4074:
4068:
4064:
4059:
4056:
4051:
4047:
4042:
4036:
4033:
4028:
4023:
4020:
4015:
4010:
4007:
4002:
3997:
3994:
3989:
3984:
3981:
3976:
3971:
3968:
3962:
3958:
3953:
3950:
3945:
3941:
3935:
3932:
3927:
3922:
3919:
3914:
3909:
3906:
3901:
3896:
3893:
3888:
3883:
3880:
3875:
3870:
3867:
3861:
3857:
3852:
3849:
3843:
3839:
3834:
3831:
3803:
3799:
3795:
3789:
3783:
3779:
3775:
3769:
3766:
3762:
3758:
3748:
3731:
3727:
3723:
3717:
3711:
3707:
3703:
3697:
3691:
3687:
3683:
3673:
3656:
3652:
3648:
3642:
3636:
3632:
3628:
3622:
3616:
3612:
3608:
3598:
3597:
3596:
3593:
3579:
3574:
3568:
3565:
3560:
3555:
3552:
3547:
3542:
3539:
3533:
3529:
3524:
3521:
3515:
3511:
3506:
3503:
3498:
3494:
3489:
3483:
3480:
3475:
3470:
3467:
3462:
3457:
3454:
3448:
3444:
3439:
3436:
3431:
3427:
3421:
3418:
3413:
3408:
3405:
3400:
3395:
3392:
3386:
3382:
3377:
3374:
3368:
3364:
3359:
3356:
3346:
3341:
3339:
3337:
3336:
3335:
3332:
3318:
3313:
3310:
3305:
3301:
3296:
3293:
3288:
3284:
3278:
3275:
3270:
3265:
3262:
3256:
3252:
3247:
3244:
3234:
3231:
3217:
3211:
3208:
3203:
3198:
3195:
3189:
3185:
3180:
3177:
3172:
3168:
3162:
3159:
3154:
3149:
3146:
3140:
3136:
3131:
3128:
3118:
3115:
3100:
3097:
3092:
3087:
3084:
3074:
3071:
3056:
3053:
3048:
3043:
3040:
3035:
3030:
3027:
3017:
3011:
3009:
3008:
3004:
3000:
2995:
2994:
2987:
2985:
2984:
2981:
2973:
2969:
2965:
2961:
2957:
2956:
2949:
2948:
2940:
2939:
2932:
2931:
2930:
2911:
2904:
2902:
2901:
2897:
2893:
2880:
2876:
2872:
2868:
2865:
2864:
2863:
2856:
2855:
2851:
2850:
2842:
2841:
2836:
2835:
2834:
2827:
2826:
2825:
2822:
2820:
2816:
2812:
2811:82.128.133.96
2808:
2801:
2797:
2791:
2789:
2787:
2783:
2774:
2772:
2771:
2767:
2763:
2762:23.121.191.18
2757:
2754:
2746:
2740:
2736:
2732:
2731:23.121.191.18
2728:
2723:
2720:
2719:
2718:
2717:
2716:
2715:
2711:
2707:
2701:
2697:
2696:
2690:
2687:
2686:
2685:
2681:
2679:
2670:
2668:
2667:
2662:
2657:
2656:
2645:
2641:
2638:
2634:
2633:
2632:
2625:
2619:
2615:
2611:
2607:
2601:
2596:
2592:
2586:
2582:
2578:
2571:
2567:
2563:
2562:
2561:
2559:
2555:
2551:
2546:
2540:
2538:
2537:
2533:
2529:
2525:
2521:
2517:
2514:I think here
2513:
2508:
2504:
2495:
2491:
2487:
2482:
2478:
2475:
2471:
2468:
2464:
2460:
2457:
2453:
2449:
2448:
2447:
2446:
2441:
2437:
2433:
2429:
2425:
2424:
2423:
2422:
2419:
2415:
2411:
2407:
2402:
2398:
2397:
2392:
2388:
2384:
2380:
2376:
2371:
2370:
2369:
2365:
2361:
2360:23.30.218.182
2356:
2355:
2354:
2353:
2349:
2345:
2337:
2335:
2334:
2330:
2326:
2317:
2313:
2309:
2305:
2300:
2299:
2295:
2290:
2286:
2285:
2280:
2276:
2272:
2267:
2266:
2265:
2264:
2261:
2257:
2253:
2248:
2247:
2242:
2241:
2236:
2232:
2228:
2224:
2220:
2219:
2218:
2217:
2214:
2210:
2206:
2203:
2202:
2197:
2193:
2192:
2187:
2186:
2182:
2178:
2174:
2170:
2166:
2165:
2164:
2163:
2159:
2155:
2150:
2146:
2144:
2139:
2135:
2133:
2129:
2125:
2121:
2113:
2109:
2105:
2101:
2096:
2092:
2091:
2090:
2089:
2085:
2081:
2072:
2068:
2064:
2060:
2055:
2054:
2053:
2052:
2048:
2044:
2036:
2034:
2033:
2029:
2025:
2024:109.151.57.10
2016:
2012:
2008:
2004:
2000:
1999:
1998:
1996:
1992:
1988:
1982:
1978:
1972:
1968:
1964:
1959:
1958:
1953:
1949:
1948:
1947:
1945:
1941:
1937:
1931:
1924:
1918:
1912:
1909:
1901:
1897:
1895:
1886:
1884:
1883:
1879:
1875:
1871:
1867:
1863:
1862:this addition
1855:
1853:
1852:
1848:
1844:
1840:
1836:
1828:
1824:
1820:
1816:
1811:
1810:
1809:
1808:
1804:
1800:
1795:
1793:
1789:
1783:
1775:
1773:
1772:
1768:
1764:
1760:
1756:
1749:
1745:
1737:
1732:
1728:
1720:
1718:
1715:
1711:
1707:
1703:
1702:
1701:
1700:
1697:
1693:
1686:
1678:
1674:
1670:
1666:
1662:
1661:
1660:
1659:
1655:
1651:
1646:
1643:
1640:
1636:
1630:
1626:
1622:
1618:
1614:
1608:
1607:
1605:
1600:
1599:
1597:
1596:
1595:
1593:
1589:
1585:
1581:
1577:
1570:
1567:
1564:
1562:
1558:
1550:
1546:
1542:
1538:
1534:
1533:
1532:
1531:
1528:
1523:
1515:
1511:
1507:
1503:
1498:
1497:
1496:
1490:
1484:
1480:
1476:
1472:
1471:
1470:
1469:
1463:
1460:
1459:
1458:
1457:
1451:
1450:
1449:
1448:
1442:
1439:
1436:
1435:
1434:
1433:
1427:
1426:
1425:
1424:
1418:
1414:
1413:
1411:
1407:
1403:
1399:
1392:
1391:
1388:
1384:
1382:
1379:
1377:
1374:
1372:
1369:
1367:
1364:
1362:
1359:
1357:
1354:
1352:
1349:
1347:
1333:
1330:
1327:
1324:
1321:
1311:
1309:
1307:
1306:Niels Ø (noe)
1303:
1302:
1298:
1297:
1294:
1290:
1289:
1286:
1284:
1280:
1279:
1276:
1273:
1270:
1269:
1264:
1263:
1258:
1257:
1252:
1251:
1248:
1244:
1240:
1235:
1234:
1233:
1232:
1229:
1225:
1216:
1213:
1210:
1209:
1208:
1207:
1206:
1201:
1198:
1197:Niels Ø (noe)
1194:
1190:
1186:
1182:
1179:
1175:W side of B/W
1174:
1172:B side of B/W
1171:
1168:
1165:
1162:
1159:
1158:
1144:
1141:
1138:
1135:
1132:
1123:
1119:
1118:
1117:
1115:
1112:
1111:Maveric Gamer
1108:
1103:
1095:
1092:
1091:Maveric Gamer
1089:
1085:
1084:
1083:
1082:
1075:
1074:
1073:
1072:
1069:
1066:
1062:
1057:
1056:
1055:
1054:
1051:
1050:Maveric Gamer
1045:
1042:
1038:
1034:
1030:
1022:
1018:
1015:
1008:
1006:
1005:
1002:
997:
994:
991:
988:
985:
979:
977:
976:
973:
969:
963:And the rest?
962:
954:
951:
946:
945:
944:
943:
942:
941:
934:
931:
926:
925:
924:
923:
922:
921:
916:
913:
908:
907:
906:
905:
902:
899:
894:
890:
889:
888:
887:
884:
880:
879:Simon Templar
874:
871:
865:
861:
848:
845:
841:
839:
836:
831:
830:
829:
828:
825:
822:
817:
816:
815:
814:
811:
807:
799:
797:
796:
793:
783:
780:
779:
778:
772:
769:
766:
763:
762:
761:
756:green/purple.
755:
754:
753:
747:
741:
738:
733:
732:
731:
730:
725:
722:
718:
714:
710:
709:
708:
707:
702:
699:
695:
691:
687:
686:
685:
684:
679:
676:
672:
668:
664:
663:
662:
661:
658:
655:
651:
647:
643:
639:
635:
634:
629:
626:
622:
618:
614:
613:
612:
611:
606:
603:
599:
595:
594:
593:
592:
589:
586:
582:
581:
577:
573:
572:
571:
570:
567:
560:Article name?
559:
557:
556:
553:
545:
537:
534:
529:
525:
521:
520:
519:
516:
513:
509:
506:
505:
504:
503:
500:
497:
492:
491:
490:
489:
486:
482:
474:
470:
467:
462:
461:
460:
459:
456:
455:171.64.71.123
450:
449:
446:
440:
434:
431:
427:
426:
425:
424:
421:
416:
412:
409:
406:
398:
397:
394:
389:
377:
373:
370:
369:
366:
358:
355:
351:
347:
343:
339:
335:
334:
333:
332:
329:
321:
318:
313:
308:
304:
303:
302:
301:
298:
289:
285:
281:
274:
270:
266:
263:
259:
255:
252:
248:
247:
231:
227:
221:
218:
217:
214:
197:
193:
189:
188:
180:
174:
169:
167:
164:
160:
159:
155:
149:
146:
143:
139:
126:
122:
116:
113:
112:
109:
92:
88:
84:
83:
78:
75:
71:
70:
66:
60:
57:
54:
50:
45:
41:
35:
27:
23:
18:
17:
4236:
3821:
3594:
3347:
3344:
3333:
3235:
3232:
3119:
3116:
3075:
3072:
3018:
3015:
2996:
2991:
2977:
2941:It should be
2928:
2908:
2892:Thierry Caro
2884:
2861:
2832:
2823:
2805:— Preceding
2802:
2798:
2795:
2778:
2758:
2750:
2721:
2702:
2698:
2695:
2693:
2688:
2682:
2674:
2652:
2649:
2624:source check
2603:
2597:
2584:
2580:
2576:
2574:
2547:
2544:
2523:
2519:
2515:
2510:
2506:
2500:
2473:
2467:win the game
2466:
2462:
2455:
2427:
2405:
2378:
2374:
2341:
2325:Richard Gill
2321:
2293:
2288:
2271:Richard Gill
2227:Richard Gill
2195:
2180:
2176:
2172:
2168:
2154:Richard Gill
2151:
2147:
2142:
2140:
2136:
2131:
2127:
2117:
2094:
2076:
2040:
2020:
1983:
1979:
1975:
1951:
1932:
1928:
1922:
1916:
1910:
1906:
1898:
1893:
1890:
1859:
1838:
1834:
1832:
1791:
1788:at least one
1787:
1785:
1781:
1779:
1752:
1689:
1647:
1644:
1641:
1637:
1634:
1571:
1568:
1565:
1560:
1556:
1554:
1519:
1494:
1385:
1380:
1375:
1370:
1365:
1360:
1355:
1350:
1312:
1304:
1291:
1281:
1226:
1222:
1204:
1104:
1100:
1061:Arthur Rubin
1046:
1023:
1019:
1016:
1012:
999:Comments?
998:
995:
992:
989:
986:
983:
967:
966:
892:
875:
869:
866:
862:
858:
803:
787:
784:black/white.
781:black/white,
776:
770:black/white,
759:
751:
716:
650:Arthur Rubin
645:
641:
563:
549:
481:Arthur Rubin
478:
451:
441:
437:
417:
413:
410:
407:
404:
390:
386:
374:
371:
361:
350:Arthur Rubin
345:
341:
337:
324:
311:
306:
293:
272:
226:Low-priority
225:
185:
151:Low‑priority
120:
80:
40:WikiProjects
2960:JumpDiscont
2867:JumpDiscont
2684:like this:
2591:Sourcecheck
2406:information
2294:can only be
1664:discussion.
1574:—Preceding
1396:—Preceding
1195:(harder)!--
621:Probability
307:do not look
201:Mathematics
192:mathematics
148:Mathematics
4254:Categories
2999:2.39.50.17
2661:Report bug
2458:arguments:
2196:any number
2143:similarity
1987:Michaeldsp
1936:Michaeldsp
1283:Bogfjellmo
1272:Bogfjellmo
800:Expert tag
773:red/white.
767:red/white,
764:black/red,
615:How about
546:Approaches
428:Agreed. -
96:Statistics
87:statistics
59:Statistics
2644:this tool
2637:this tool
2080:Starfiend
1870:Wikibooks
1714:Rafi Neal
1527:Rafi Neal
1001:Gwynevans
789:red : -->
737:Jim 14159
494:research.
338:incorrect
312:certainty
262:January 8
2807:unsigned
2650:Cheers.—
2451:section.
2134:means.
1763:Melchoir
1669:Melchoir
1617:Melchoir
1588:contribs
1580:Aznthird
1576:unsigned
1537:Melchoir
1475:Racerx11
1416:problem.
1398:unsigned
1239:Rock8591
1029:unsigned
950:Melchoir
898:Melchoir
893:disagree
844:Melchoir
835:Melchoir
810:Melchoir
792:Melchoir
602:Melchoir
552:Melchoir
533:Melchoir
524:question
466:Rklawton
258:deletion
2950:, where
2829:anyway:
2706:JeffJor
2577:checked
2554:my edit
2501:In his
2410:JeffJor
2304:JeffJor
2252:JeffJor
2205:JeffJor
2100:JeffJor
2059:JeffJor
2037:Wording
2003:JeffJor
1963:JeffJor
1776:Clarity
1742:Merged
1610:black."
1254:black".
864:heads?
638:CalTech
596:I like
228:on the
123:on the
30:B-class
2585:failed
2528:Albtal
2516:step 0
2512:brown?
2486:Albtal
2474:step 0
2456:simple
2432:Albtal
2383:Albtal
2344:Albtal
1874:Open4D
1815:Open4D
1502:Ratfox
1308:Wrote:
1293:50-50.
1285:Wrote:
1266:black.
1260:50-50.
1228:Adl116
1065:(talk)
881:.) --
821:Dacium
748:Colors
698:Haukur
654:(talk)
585:Haukur
566:Haukur
528:answer
515:(Talk)
485:(talk)
430:Haukur
354:(talk)
36:scale.
2175:(not
1799:RomaC
1757:into
1746:into
930:Obina
870:three
675:Obina
646:coins
625:Obina
623:page.
496:Obina
445:Obina
420:Obina
317:Obina
4242:talk
3003:talk
2964:talk
2896:talk
2871:talk
2815:talk
2786:talk
2766:talk
2735:talk
2710:talk
2581:true
2532:talk
2490:talk
2436:talk
2414:talk
2387:talk
2364:talk
2348:talk
2329:talk
2308:talk
2289:your
2275:talk
2256:talk
2231:talk
2209:talk
2177:must
2158:talk
2104:talk
2095:with
2084:talk
2063:talk
2047:talk
2028:talk
2007:talk
1991:talk
1967:talk
1940:talk
1894:why?
1878:talk
1847:talk
1843:Ij3n
1819:talk
1803:talk
1784:to "
1767:talk
1731:talk
1673:talk
1654:talk
1621:talk
1584:talk
1561:card
1557:side
1541:talk
1506:talk
1479:talk
1406:talk
1243:talk
1087:50%.
1037:talk
972:INic
696:. -
642:Ma 2
526:and
273:keep
271:was
265:2006
2618:RfC
2595:).
2583:or
2568:to
2381:.--
2173:can
1952:you
1872:.)
1712:. —
1631:Hmm
1217:W/W
1214:B/W
1211:B/B
1063:|
717:six
692:or
512:C S
352:|
346:are
342:why
260:on
220:Low
115:Low
4256::
4244:)
4214:41
4200:12
4153:×
4127:×
4101:×
4083:×
4065:×
4029:×
4003:×
3977:×
3959:×
3928:×
3902:×
3876:×
3858:×
3840:×
3530:×
3512:×
3445:×
3383:×
3365:×
3302:×
3253:×
3186:×
3137:×
3005:)
2966:)
2898:)
2873:)
2817:)
2788:)
2782:JQ
2768:)
2737:)
2712:)
2631:.
2626:}}
2622:{{
2593:}}
2589:{{
2534:)
2492:)
2438:)
2430:--
2416:)
2389:)
2366:)
2350:)
2331:)
2310:)
2277:)
2258:)
2233:)
2211:)
2160:)
2106:)
2086:)
2065:)
2049:)
2030:)
2009:)
1997:)
1993:)
1969:)
1946:)
1942:)
1880:)
1849:)
1821:)
1805:)
1794:."
1769:)
1723:—
1675:)
1656:)
1623:)
1590:)
1586:•
1543:)
1508:)
1481:)
1408:)
1325:×
1245:)
1136:×
1116:)
1039:)
652:|
483:|
4240:(
4223:)
4220:6
4217:(
4211:,
4208:0
4205:=
4197:5
4192:=
4172:)
4167:)
4161:2
4158:0
4148:3
4145:1
4140:+
4135:1
4132:0
4122:3
4119:1
4114:+
4109:2
4106:2
4096:3
4093:1
4087:(
4078:1
4075:1
4069:(
4060:2
4057:1
4052:+
4048:)
4043:)
4037:2
4034:0
4024:3
4021:1
4016:+
4011:2
4008:1
3998:3
3995:1
3990:+
3985:1
3982:1
3972:3
3969:1
3963:(
3954:2
3951:1
3946:+
3942:)
3936:2
3933:0
3923:3
3920:1
3915:+
3910:2
3907:1
3897:3
3894:1
3889:+
3884:1
3881:1
3871:3
3868:1
3862:(
3853:2
3850:1
3844:(
3835:2
3832:1
3807:)
3804:2
3800:/
3796:0
3793:(
3790:;
3787:)
3784:1
3780:/
3776:0
3773:(
3770:;
3767:2
3763:/
3759:2
3756:(
3735:)
3732:2
3728:/
3724:0
3721:(
3718:;
3715:)
3712:2
3708:/
3704:1
3701:(
3698:;
3695:)
3692:1
3688:/
3684:1
3681:(
3660:)
3657:2
3653:/
3649:0
3646:(
3643:;
3640:)
3637:2
3633:/
3629:1
3626:(
3623:;
3620:)
3617:1
3613:/
3609:1
3606:(
3580:)
3575:)
3569:3
3566:1
3561:+
3556:3
3553:1
3548:+
3543:3
3540:1
3534:(
3525:1
3522:1
3516:(
3507:2
3504:1
3499:+
3495:)
3490:)
3484:3
3481:1
3476:+
3471:3
3468:1
3463:+
3458:3
3455:1
3449:(
3440:2
3437:1
3432:+
3428:)
3422:3
3419:1
3414:+
3409:3
3406:1
3401:+
3396:3
3393:1
3387:(
3378:2
3375:1
3369:(
3360:2
3357:1
3319:)
3314:1
3311:1
3306:(
3297:2
3294:1
3289:+
3285:)
3279:2
3276:1
3271:+
3266:2
3263:1
3257:(
3248:2
3245:1
3218:)
3212:2
3209:1
3204:+
3199:2
3196:1
3190:(
3181:2
3178:1
3173:+
3169:)
3163:2
3160:1
3155:+
3150:2
3147:1
3141:(
3132:2
3129:1
3101:2
3098:1
3093:+
3088:2
3085:1
3057:3
3054:1
3049:+
3044:3
3041:1
3036:+
3031:3
3028:1
3001:(
2962:(
2894:(
2869:(
2813:(
2784:(
2764:(
2733:(
2708:(
2663:)
2659:(
2646:.
2639:.
2530:(
2488:(
2434:(
2412:(
2385:(
2362:(
2346:(
2327:(
2306:(
2273:(
2254:(
2229:(
2207:(
2156:(
2102:(
2082:(
2061:(
2045:(
2026:(
2005:(
1989:(
1985:(
1965:(
1938:(
1934:(
1876:(
1845:(
1817:(
1801:(
1765:(
1733:)
1729:(
1671:(
1652:(
1619:(
1582:(
1539:(
1504:(
1477:(
1404:(
1334:6
1331:=
1328:2
1322:3
1241:(
1145:6
1142:=
1139:2
1133:3
1043:.
1035:(
275:.
232:.
127:.
42::
Text is available under the Creative Commons Attribution-ShareAlike License. Additional terms may apply.
