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The question appeared in one of the assignments I am using to self study Probability and Statistics:

I begin turning a deck of cards one at a time. At some point (before I have turned over all $52$ cards) you can bet the next card is red to receive one dollar; otherwise you receive nothing. Design a strategy to maximize the probability that you will receive the dollar. When should you bet?

$E_k$ is the event that bet is placed on card number $k$ and that it is red. Obviously, $E_1=E_2=\frac{1}{2}$ which led me to believe it is immaterial when one bets on the cards. I tried to derive the same using a case for $E_{k} \text{ when }k-1$ cards have already been drawn.

$P(E_k)=\frac{26}{52-(k-1)}\frac{26\choose{k-1}}{52\choose{k-1}}+\frac{26-1}{52-(k-1)}\frac{{26\choose{k-2}}{26\choose{1}}}{52\choose{k-1}}...\frac{26-(k-1)}{52-(k-1)}\frac{{26\choose{0}}{26\choose{k-1}}}{52\choose{k-1}}$
$P(E_k)=\frac{1}{(53-k){52\choose{k-1}}}\sum_{i=0}^{i=k-1}(26-i){26\choose{k-1-i}}{26\choose{i}}$

How do I evaluate this value? Also I realize for my particular solution, $k\le 26$. What would be a more general solution?

PythonSage
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  • Expectation, $\mathbb{E}(\cdot)$, is defined for random variables $X$, not events like ${X=k}$ so while you may or may not be able to compute the written RHS, the LHS has no meaning. Compare to the computations $\mathbb{E}(\mathbb{1}_{X=k})=\mathbb{P}(X=k)$, the expected value of the indicator of the event ${X=k}$ and $\mathbb{E}(X|X=k)=k$ the conditional expectation of $X$ conditional on $X=k$... – Nap D. Lover Dec 12 '19 at 19:15
  • @NapD.Lover I will edit that. I'm sorry I didn't put that correctly but can you please elaborate your comment? I don't think I quite understand it. – PythonSage Dec 12 '19 at 19:20
  • without knowing specifically what you want elaborated, let me just remark that for any constant real number $c$, we have $\mathbb{E}(c)=c$. For a discrete valued RV $X$, $\mathbb{E}(X)=\sum_k k \mathbb{P}(X=k)$ (where the sum is over the countable range of $k$). The expression $\mathbb{E}(X=k)$ is not even defined. And definitely $\mathbb{E}(1)\neq 1/2$ by the first property I mentioned... – Nap D. Lover Dec 12 '19 at 19:26
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    If your "Strategy" is to simply wait until a particular number of cards has been drawn and just bet on that particular turn without regard to what has been drawn earlier, then of course you always have a probability of $\frac{1}{2}$ of winning. Suppose, however that if I were to delay the game long enough to have seen every black card be drawn, leaving only red cards. Then of course at that point, having known that all black cards had already been drawn, I am guaranteed to win. – JMoravitz Dec 12 '19 at 19:32
  • @NapD.Lover. I edited it. Can you check if it is correct? Also could you help with the solution? – PythonSage Dec 12 '19 at 19:32
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    The point is to try to determine a strategy (if one exists) that performs better than one that succeeds merely $\frac{1}{2}$ of the time, and if possible, to show that it is optimal. For instance, I could try waiting until there are at least two more reds left in the deck than blacks, and bet then, unless if it so happens that there are about to be no reds left. How often would I win then? – JMoravitz Dec 12 '19 at 19:33
  • @JMoravitz I thought about it too but it is possible the last card is black, say. And I wait until all cards have been drawn and I loose. Say, for example, all red cards finish before all black cards. Then in that case, what do I do?

    Is there a better strategy? Or is it indifferent?

     – PythonSage Dec 12 '19 at 19:34
  • @JMoravitz Thanks. It does. Just so I know, am I supposed to delete this post? I guess this is a duplicate right? – PythonSage Dec 12 '19 at 19:37

1 Answers1

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This is an old puzzle, and the answer is that there's no strategy that gives better than $\frac12$ chance of winning. (Most of the ideas presented so far give ways of having a larger winning probability if certain situations arise, but no indication of what to do if those situations don't arise.) Here's a pair of observations that I find convincing:

  • The game is exactly the same as: any time you want, you can stop looking at cards bet on whether the last card (not the next card) is red or black.
  • The probability of guessing the last card correct is $\frac12$.
Greg Martin
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