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?
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: