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34:
Suppose a mathematician carries two matchboxes at all times: one in his left pocket and one in his right. Each time he needs a match, he is equally likely to take it from either pocket. Suppose he reaches into his pocket and discovers for the first time that the box picked is empty. If it is assumed
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be the number of matches removed from this one before the left one is found to be empty. When the left pocket is found to be empty, the man has chosen that pocket
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Feller, William, An
Introduction to Probability Theory And Its Applications, Third Edition, Wiley, 1968, Chapter VI, section 8
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Without loss of generality consider the case where the matchbox in his right pocket has an unlimited number of matches and let
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27:. Feller says that the problem was inspired by a humorous reference to Banach's smoking habit in a speech honouring him by
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Returning to the original problem, we see that the probability that the left pocket is found to be empty first is
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624:{\displaystyle P=P=2P={\binom {2N-k}{N-k}}\left({\frac {1}{2}}\right)^{2N-k}}
31:, but that it was not Banach who set the problem or provided an answer.
338:{\displaystyle P={\binom {N+m}{m}}\left({\frac {1}{2}}\right)^{N+1+m}}
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matches, the expected number of matches in the second box is
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because both are equally likely. We see that the number
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matches, what is the probability that there are exactly
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The expectation of the distribution is approximately
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35:that each of the matchboxes originally contained
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438:of matches remaining in the other pocket is
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736:List of things named after Stefan Banach
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236:matches remaining in the other pocket.
232:Distribution of probability of having
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669:{\displaystyle 2{\sqrt {N/\pi }}-1}
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187:failures in Bernoulli trials with
155:is the number of successes before
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680:.) So starting with boxes with
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223:negative binomial distribution
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75:matches in the other box?
678:Stirling's approximation
19:is a classic problem in
676:. (This is shown using
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17:Banach's match problem
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214:{\displaystyle p=1/2}
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180:{\displaystyle (N+1)}
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128:{\displaystyle (N+1)}
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797:Probability problems
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699:{\displaystyle N=40}
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792:Applied probability
411:{\displaystyle 1/2}
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763:Feller, page 238.
719:{\displaystyle 6}
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431:{\displaystyle K}
383:{\displaystyle P}
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148:{\displaystyle M}
96:{\displaystyle M}
68:{\displaystyle k}
48:{\displaystyle N}
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221:, which has the
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771:External links
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29:Hugo Steinhaus
23:attributed to
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390:which equals
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135:times. Then
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25:Stefan Banach
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18:
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777:Java applet
21:probability
786:Categories
742:References
661:−
656:π
614:−
573:−
562:−
535:−
485:−
225:and thus
730:See also
79:Solution
499:<
366:<
788::
726:.
694:40
714:6
691:=
688:N
664:1
652:/
648:N
643:2
631:.
617:k
611:N
608:2
603:)
598:2
595:1
590:(
582:)
576:k
570:N
565:k
559:N
556:2
550:(
544:=
541:]
538:k
532:N
529:=
526:M
523:[
520:P
517:2
514:=
511:]
508:1
505:+
502:N
496:M
492:|
488:k
482:N
479:=
476:M
473:[
470:P
467:=
464:]
461:k
458:=
455:K
452:[
449:P
426:K
406:2
402:/
398:1
378:]
375:1
372:+
369:N
363:M
360:[
357:P
345:.
331:m
328:+
325:1
322:+
319:N
314:)
309:2
306:1
301:(
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288:m
284:m
281:+
278:N
272:(
266:=
263:]
260:m
257:=
254:M
251:[
248:P
234:k
209:2
205:/
201:1
198:=
195:p
175:)
172:1
169:+
166:N
163:(
143:M
123:)
120:1
117:+
114:N
111:(
91:M
63:k
43:N
Text is available under the Creative Commons Attribution-ShareAlike License. Additional terms may apply.
