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it is a well known game played in there any theory for winning this is possible where it falls How it works:in a shuffled deck the dealer offers a cut. You then bet on that card will be dealt on the inside or the outside. He then deals the cards one a time alternating between two spots; one spot called inside, the other spot called outside. He keeps dealing until the card that was originally cut appears. If it appears on the outside those who bet that spot win and the others lose; same is true if it appear on the inside (inside bettors win / outside bettors lose).

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If I understand the description correctly, the probability of winning is $\frac12$ for either bet: it’s equivalent to betting on the toss of a fair coin. –  Brian M. Scott Nov 17 '12 at 8:20
do you explain elobarately.if i loss then where should i stand on same side or simply walkout –  mohanasundaram Nov 17 '12 at 8:25
It makes no difference: each play is independent of the previous plays, just as each coin toss is independent of the previous tosses. Your probability of winning is always $\frac12$ (unless the dealer is cheating somehow). –  Brian M. Scott Nov 17 '12 at 8:26
pls explain in mathematical analysis –  mohanasundaram Nov 17 '12 at 8:34
There really isn’t much to explain, but give me a few minutes, and I’ll write up an answer. –  Brian M. Scott Nov 17 '12 at 8:44

1 Answer 1

This game is essentially equivalent to betting on the toss of a fair coin: each of the possible outcomes, inside and outside, occurs with probability $\frac12$ unless the dealer is cheating somehow. To see this, suppose that the dealer deals first to the inside pile and then to the outside pile. Then the cut card will end up on the inside pile if its position in the deck is an odd number, and it will end up on the outside pile if its position in the deck is an even number. (Cards $1,3,5,\dots$ go on the inside pile; cards $2,4,6,\dots$ go on the outside pile.) Half of the cards are in odd-numbered positions and half are in even-numbered positions, and all positions in the deck are (in theory) equally likely to be cut, so the probability of cutting a card in an odd-numbered position is $\frac12$; the same goes for cards in even-numbered positions.

Each cut is independent of any earlier cuts: the deck has no ‘memory’ of the earlier cuts, just as a tossed coin has no ‘memory’ of earlier tosses. Thus, each game is a fresh start, just as if it were the first game. Consequently there is no strategy: no matter which way you bet, your chance of winning is $\frac12$.

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