Question

Consider a deck with 2626 black and 2626 red cards. You draw one card at a time and you can choose either guess on whether it is red beforehand or simply observe the result. If the card is red you get \$1$1 and the game ends whenever you decide to guess. What is your strategy to play this game and the expected earnings

Answer 1

The maximum earning is v(r,b)

Let v(r,b) be the expected value of the game for the player, assuming optimal play, if the remaining deck has r red cards and b black cards.

Then v(r,b) satisfies the recursion

and

The stopping rule is simple: Stop when v(r,b)=0.

To explain the recursion . . .

If r,b>0, and the player elects to play a card, then:

• The revealed card is red with probability $\frac{r}{r+b}$, and in that case, the player gets a score of +1, and the new value is V(r-1,b)

• The revealed card is black with probability $\frac{b}{r+b}$, and in that case, the player gets a score of −1, and the new value is V(r,b-1)

Thus, if r,b>0, electing to play a card yields the value f(r,b).

But the player always has the option to quit, hence, if r,b>0, we get v(r,b)=max(0,f(r,b)).

Implementing the recursion in Maple, the value of the game is

v(26,26)=41984711742427/15997372030584

v(26,26)  ≈2.624475549

and the optimal stopping strategy is as follows . . .

• If 24≤b≤26, play while r≥b−5.
• If 17≤b≤23, play while r≥b−4.
• If 11≤b≤16, play while r≥b−3.
• If 6≤b≤10, play while r≥b−2.
• If 3≤b≤5, play while r≥b−1.
• If 1≤b≤2, play while r≥b.
• If b=0, play while r>0.  

So, The maximum earning is v(r,b)

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