Hence, given that the strategy does not guess the first card, the probability of winning is But this latter conditional probability is equal to the probability of winning when using an n-1 card deck containing a single ace of spades it is thus, by the induction hypothesis, 1/(n-1). ** On the other hand if the strategy does not guess that first card is ace of spades, then the probability that the player wins is the probability that the first card is not the ace of spades, namely (n-1)/n, multiplied by the conditional probability of winning given that the first card is not the ace of spades. Given that it does then the players probability of winning is 1/n. Fix any strategy and let p denote the probability that this strategy guesses that the first card is ace of spades. Since this is clearly true for n = 1, assume it to be true for an n-1 card deck, and now consider an n card deck. To show this, we will use induction to prove the stronger result that for an n card deck, one of whose cards is the ace of spades, the probability of winning is 1/n, no matter what strategy is employed. What is a good strategy what is a bad strategy?Įvery strategy has probability 1/52 of winning. In addition, the player is said to win if the ace of spades has not yet appeared when only 1 card remains and no guess has yet been made. At any time the player can guess that the next card to be turned over will be the Ace of spades if it is, then the player wins. The cards are shuffled and then turned over one at a time. Consider the following game played with an ordinary deck of 52 playing cards.
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