SOLUTION TO PROBLEM H-13

Every strategy has probability $\frac{1}{52}$ of winning!!! To show this, we will use induction to prove a stronger result that for an n card deck, one of whose cards is the ace of spades, the probability of winning is $\frac{1}{n}$, no matter what strategy is employed. Since this is clearly true for n = 1, assume it is true for an n-1 card deck, and now consider an n card deck. Fix any strategy, and let p denote the probability that this strategy guesses that the first card is the ace of spades. Given that it does, then the player' s probability of winning is $\frac{1}{n}$. On the other hand, if the strategy does not guess that the first card is the ace of spades, then the probability that the player wins is the probability that the first card is not the ace of spades, namely $\frac{(n-1)}{n}$, multiplied by the conditional probability of winning given that the first card is not the ace of spades. 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, $\frac{1}{(n-1)}$. Hence, given that the strategy does not guess the first card, the probability of winning is

$$\frac{n-1}{n}\cdot \frac{1}{n-1} = \frac{1}{n}$$

Thus, letting G be the event that the first card is guessed, we obtain that

$$P(\hbox{win}) = P(\hbox{win}\vert G)P(G) + P(\hbox{win}\vert G^c)(1-P(G))$$

(where P(A|B) denotes the conditional probability of the event A given the event B, and A^c is the complement of the event A)

We get

$$P(\hbox{win}) = \frac{1}{n}p + \frac{1}{n}(1-p) = \frac{1}{n}$$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Apr 19 08:46:05 CDT 2002