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Consider the following game played with an ordinary deck of 52 playing cards. The cards are shuffled and then turned over one at a time. 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. 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. What is a good strategy; what is a bad strategy?

Solution

Every strategy has probability 1/52 of winning. 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. 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. Fix any strategy and let p denote the probability that this strategy guesses that the first card is ace of spades. Given that it does then the players probability of winning is 1/n.

** 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. 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). Hence, given that the strategy does not guess the first card, the probability of winning is

$ {{n-1}\over {n}} {{1} \over {n-1}} = {1 \over n }$

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

$P(win) = P(win|G)P(G) + P(win|G^c)(1 - P(G))$

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

$={1 \over n}$

My problem

I fail to understand what they are doing in the second paragraph of the solution (**).

They are essentially saying that $ P(win|G^c) = {{n-1}\over {n}} {{1} \over {n-1}} ... (1)$

But $ P(win|G^c) = {P(win \bigcap G^c) \over P(G^c)} ... (2)$

I fail to see how (2) -> (1)

Would appreciate it if someone can explain by expanding the equation correctly.

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  • $\begingroup$ the first paragraph proves that $P(win|G) = 1/n$. The second paragraph explains that $P(win|G^C) = 1/n$ as well. That doesn't come from (2), it is just basic logic - winning without guessing on the first card requires that that card not be the Ace of spades, followed by winning the subsequent game with $n-1$ cards. $\endgroup$ – Nick Pavlov Oct 13 '17 at 12:13
  • $\begingroup$ Maybe to clarify my question more: how can the second paragraph be translated into an equation such as shown in (2) $\endgroup$ – Johnny M Oct 13 '17 at 13:12
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Here is how you can write out the logic of the second paragraph using the conditional probability identities. Let us call $A$ the event that the first card is the ace of spades, and accordingly, $A^c$ that it is not (I'll also write $W$ instead of "win").

$$ P(W|G^c) = \frac{P(W \cap G^c)}{P(G^c)} = \frac{P(W \cap G^c \cap A) + P(W \cap G^c \cap A^c)}{P(G^c)}, $$ where I have used that $A$ either happens or does not. Note that the first term in the numerator is zero, because you cannot win if you didn't guess, but the ace came up ($G^c \cap A = \rm loss$).
Also note that the other term is $$ P(W \cap G^c \cap A^c) = P(W \cap (G^c \cap A^c)) = P(G^c \cap A^c) \cdot P(W|(G^c \cap A^c)) .$$ [The first step just groups events into a combined event, the second step uses the conditional probability formula in the form $P(A\cap B) = P(B)P(A|B)$.]

So we are left with $$ P(W|G^c) = \frac{P(G^c \cap A^c) \cdot P(W|(G^c \cap A^c))}{P(G^c)} = P(A^c) \cdot P(W|(G^c \cap A^c)), $$ because $A^c$ and $G^c$ are independent, so $P(G^c \cap A^c) = P(G^c)P(A^c)$ and the denominator gets cancelled.

Finally, $P(A^c) = {n-1\over n}$, while $P(W|(G^c \cap A^c)) = {1\over n-1}$ by the inductive assumption (given that you didn't guess, and the ace didn't come up, you just play out the rest of the deck, which is the same as an $(n-1)$-card game with the same rules). Thus you get $(1)$.

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  • $\begingroup$ Beautiful, thank you this is what I was looking for. $\endgroup$ – Johnny M Oct 13 '17 at 14:03
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No real answer to your question, but an alternative way that uses induction.


Assumption: if the game is played with $k<n$ cards then every strategy gives probability $\frac1{k}$ to win.

Now check the strategies when $n$ cards are used.

One of them is guessing that the first card is the ace of spades and evidently it gives probability $\frac1n$ to win.

Now take another strategy and let $A$ denote the event that the first card is ace of space.

Then $$P(\text{win})=P(\text{win}\mid A)P(A)+P(\text{win}\mid A^{\complement})P(A^{\complement})=0\cdot\frac1{n}+\frac1{n-1}\frac{n-1}{n}=\frac1n$$

Here $P(\text{win}\mid A^{\complement})=\frac1{n-1}$ on base of the assumption.

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  • $\begingroup$ Should the $P(win|A)$ not equal to 1? Another issue I have is that I do not follow how the 1/n-1 is obtained (according to the explanation given in the solution). $\endgroup$ – Johnny M Oct 13 '17 at 13:20
  • $\begingroup$ Taking another strategy means here that the first card is not guessed to be the ace, then consequently $P(win\mid A)=0$. $\endgroup$ – drhab Oct 13 '17 at 14:05
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$\def\w{\text{win}}$ Letting $A$ be the event that the first card is the ace of spaces, you want to understand why $P(\w|G^c\cap A^c)=\frac1{n-1}$.

Conditioning on an event means restricting the sample space to the outcomes in that event. In the event $G^c\cap A^c$, the ace of spades cannot be the first card, but it is equally likely to be any of the other cards. Furthermore, the card you guess is not the first card, but it can be any of the others. Therefore, this conditional probability is exactly the same as the probability of successfully guessing the ace of spade's location in a deck of $n-1$ cards. By the induction hypothesis, this probability is $\frac1{n-1}$.

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