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38:
510:
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868:{\displaystyle \mathrm {P(GG\mid see\ gold)={\frac {P(see\ gold\mid GG)\times {\frac {1}{3}}}{P(see\ gold\mid GG)\times {\frac {1}{3}}+P(see\ gold\mid SS)\times {\frac {1}{3}}+P(see\ gold\mid GS)\times {\frac {1}{3}}}}} ={\frac {\frac {1}{3}}{\frac {1}{3}}}\times {\frac {1}{1+0+{\frac {1}{2}}}}={\frac {2}{3}}}
955:
Bertrand's purpose for constructing this example was to show that merely counting cases is not always proper. Instead, one should sum the probabilities that the cases would produce the observed result; and the two methods are equivalent only if this probability is either 1 or 0 in every case. This
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The Monty Hall and Three
Prisoners problems are identical mathematically to Bertrand's Box paradox. The construction of the Boy or Girl paradox is similar, essentially adding a fourth box with a gold coin and a silver coin. Its answer is controversial, based on how one assumes the "drawer" was
169:
Bertrand's box paradox: the three equally probable outcomes after the first gold coin draw. The probability of drawing another gold coin from the same box is 0 in (a), and 1 in (b) and (c). Thus, the overall probability of drawing a gold coin in the second draw is
494:
83:
A coin withdrawn at random from the three boxes happens to be a gold coin. If you now examine the *other* coin in that box, what is the probability it will also be a gold coin?
86:
A veridical paradox is a paradox whose correct solution seems to be counterintuitive. It may seem intuitive that the probability that the remaining coin is gold should be
238:
The problem can be reframed by describing the boxes as each having one drawer on each of two sides. Each drawer contains a coin. One box has a gold coin on each side (
153:
This simple but counterintuitive puzzle is used as a standard example in teaching probability theory. The solution illustrates some basic principles, including the
1140:
322:
The flaw is in the last step. While those two cases were originally equally likely, the fact that you are certain to find a gold coin if you had chosen the
250:). A box is chosen at random, a random drawer is opened, and a gold coin is found inside it. What is the chance of the coin on the other side being gold?
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In a survey of 53 Psychology freshmen taking an introductory probability course, 35 incorrectly responded
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The three remaining possibilities are equally likely, so the probability that the drawer is from box
1008:
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box, means they are no longer equally likely given that you have found a gold coin. Specifically:
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1122:, Lawrence Erlbaum. Ch. 5, "Some instructive problems: Three cards", pp. 157–160.
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The two remaining possibilities are equally likely. So the probability that the box is
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condition is applied correctly by the second solution method, but not by the first.
37:
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The paradox starts with three boxes, the contents of which are initially unknown
489:{\displaystyle \left(\mathrm {i.e.,P(GG)=P(SS)=P(GS)} ={\frac {1}{3}}\right)}
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246:), and the other a gold coin on one side and a silver coin on the other (
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box, but are only 50% sure of finding a gold coin if you had chosen the
49:
165:
36:
253:
The following faulty reasoning appears to give a probability of
1120:
Cognition and Chance: The psychology of probabilistic reasoning
1077:(1982). "Some teasers concerning conditional probabilities".
275:
Originally, all three boxes were equally likely to be chosen.
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Originally, all six coins were equally likely to be chosen.
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were correct, it would result in a contradiction, so
500:the conditional probability that the chosen box is
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a box containing one gold coin and one silver coin.
1004:Other veridical paradoxes of probability include:
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30:For other paradoxes by Joseph Bertrand, see
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980:; only 3 students correctly responded
901:The chosen coin cannot be from drawer
300:, and the other coin is also gold, is
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242:), one a silver coin on each side (
32:Bertrand's paradox (disambiguation)
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894:can also be obtained as follows:
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1170:Probability theory paradoxes
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1118:Nickerson, Raymond (2004).
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924:, or either drawer of box
118:. Bertrand showed that if
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916:So it must come from the
18:Three cards and a top hat
1053:"Bertrand's box paradox"
69:There are three boxes:
56:. It was first posed by
1029:Sleeping Beauty problem
1019:Three Prisoners problem
63:Calcul des Probabilités
878:The correct answer of
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46:Bertrand's box paradox
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1134:Paradoxes from A to Z
1024:Two envelopes problem
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350:The probability that
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1175:Probability problems
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27:Mathematical paradox
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388:are equally likely
150:cannot be correct.
1014:Monty Hall problem
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285:So it must be box
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54:probability theory
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16:(Redirected from
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1057:Oxford Reference
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498:Bayes' rule
1164:Categories
1075:Falk, Ruma
1039:References
376:Initially
1079:Cognition
819:×
779:×
767:∣
721:×
709:∣
663:×
651:∣
606:×
594:∣
528:∣
1107:44509163
1035:chosen.
161:Solution
1099:7198956
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1124:ISBN
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384:and
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935:is
289:or
1166::
1101:.
1093:.
1083:11
1081:.
1073:;
1055:.
996:.
933:GG
926:GG
922:GS
911:SS
907:GS
502:GG
386:GS
382:SS
380:,
378:GG
352:GS
345:SS
338:GG
328:GS
324:GG
298:GG
291:GS
287:GG
280:SS
269::
248:GS
244:SS
240:GG
218:=
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186:+
157:.
66:.
1143:.
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1089::
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991:3
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913:.
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858:2
853:=
845:2
842:1
837:+
834:0
831:+
828:1
824:1
813:3
810:1
805:3
802:1
796:=
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784:1
776:)
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764:d
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749:e
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567:(
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558:=
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534:e
531:s
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519:(
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474:1
469:=
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456:(
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438:(
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97:2
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34:.
20:)
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