Clue to Puzzle #14: 52 Cards Win a Dollar

14. You have 52 playing cards (26 red, 26 black). You draw cards one by one. A red card pays you a dollar. A black one fines you a dollar. You can stop any time you want. Cards are not returned to the deck after being drawn. What is the optimal stopping rule in terms of maximizing expected payoff?

Also, what is the expected payoff following this optimal rule?

This puzzle is really very hard. It's a puzzle though that you can make decent progress with long before getting a full answer. I cover this at length in the main answer page but for this hints page I'm going to breeze over that and go straight to giving clues to the final solution. Quant experts read on.

If you're going to solve this the following might be considered prerequisite.
  • We can't lose. We can always play to zero.
  • I'll quit when I get 'x' dollars pays the best return for x=4
  • Consider the case where I am one dollar up with one card left to draw. I should always quit here.
  • I'll quit when i get x dollars ahead, where x is static, is not an adequate rule.
Having established that, it is helpful to work on a smaller deck, a six card deck with three red cards and three black. The game can only be in 16 positions in total. (3 to 0 of either remaining.) At any one of those positions we can easily calculate our current cash position and the probability of harming or improving our position in the game.

At this point I'm dangerously close to giving too much away so feel free to stop. We essentially now have a cash position for each point in the game and the probabilities of moving to a better or worse position. We need to convert that into an expected return and compare the expected return from gambling to our current cash position.

The final clue, and feel free not to read on.... ....The expected return should be calculate as a probability weighting of the expected returns from the possible outcomes of the gamble. It's helpful to start at the end and work backwards.


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