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Hi there, here's another puzzle I've been looking at.

Suppose you are to guess the colour of the next card in an ordinary deck of 52 cards---red or black---one at a time. How many can you expect to get right?

Here's what I think. Let $E(r,b)$ denote the expected number you will get right given that there are $r$ red cards left and $b$ black cards left. Then

  • If $r \geq b$, then you would guess red (WLOG in the case of equality) and thus $$E(r,b) = \frac{r(1+E(r-1,b))+ b E(r,b-1)}{r+b}.$$

  • If $r < b$, then you would guess blue and thus $$E(r,b) = \frac{r E(r-1,b) + b(1+E(r,b-1))}{r+b}.$$

Thus, if we let $$F(r,b) = \frac{r E(r-1,b) + b E(r,b-1)}{r+b},$$ then for all $r,b$ we have $$E(r,b) = F(r,b) + \frac{max(r,b)}{r+b}.$$

How might one solve these equations? Simulation tells me that $E(26,26) = 30.0392...$.

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  • $\begingroup$ You can solve $E(r,b)$ by recursion. The base of the induction is when there is only one color left, and you are sure to be right. Then, your equations provide the value at $E(r,b)$ from smaller values. So the complete table of values can be exactly calculated. $\endgroup$
    – Joel David Hamkins
    Commented Feb 13, 2012 at 1:19
  • $\begingroup$ You already set up a recursion, and the initial cases are not too hard to work out. Please feed this information to a computer algebra system. $\endgroup$
    – S. Carnahan ♦
    Commented Feb 13, 2012 at 1:31
  • $\begingroup$ Sure, so there is no explicit formula for a situation like this? $\endgroup$
    – A Chuh
    Commented Feb 13, 2012 at 1:45
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    $\begingroup$ There is a big simplification possible. If there are $k>0$ more red cards than black cards, you will make $k$ more correct than incorrect guesses by the time the deck is even again. So, the expected number of correct minus incorrect guesses is a sum of $1$ for each time the deck changes from even to uneven. That is the expected number of times the deck will be even before the end, or the sum of the probabilities that the deck will be even at time $0, 2, 4, ... 2n-2$. $\endgroup$
    – Douglas Zare
    Commented Feb 13, 2012 at 7:04
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    $\begingroup$ Perhaps the question was closed too quickly? Based on Douglas Zare's comment there is a clear answer: $E(r,b)$ can be computed as a single sum / recursion involving binomial coefficients (rather than over two variables). In particular it seems that $E(n+1,n+1) = E(n,n) + 1 + (1/2)\cdot (2\cdot 4 \cdots (2n))/(3\cdot 5 \cdots (2n+1))$, and that asymptotically, $E(n,n) = n + \sqrt{\pi n}/2 + o(\sqrt{n})$. $\endgroup$
    – Johan Wästlund
    Commented Feb 13, 2012 at 10:54

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