Another card puzzle

I seem to have accumulated a backlog of subject matter that I have found interesting recently.  I’ll get back to it soon… but right now, in the spirit of not getting back to work just yet, following is another puzzle involving a card game.  I found this problem in Peter Winkler’s Mathematical Puzzles: A Connoisseur’s Collection.  This is an excellent book, with more than just the “same old” problems, but also including some interesting twists that were new to me.

For example, recall the card puzzle motivating the discussion of other players’ strategy (not) affecting your expected return in blackjack.  Winkler discusses this problem, but only as a prelude to the following more challenging variant:

Problem: Consider a standard, shuffled poker deck of 52 playing cards, of which 26 are red and 26 are black.  You start with one dollar.  For each of 52 turns, I will deal one card at a time face up on the table between us.  At each turn, you may bet any fraction of your current bankroll (from “passing” to betting everything you have) on the color of the next card to be dealt, paying even odds if your guess is correct.  For example, a conservative strategy would be to wait until the last card, whose color you will know with certainty, and bet your whole dollar, guaranteeing a net return of one dollar.

Can you do better than this?  Like the earlier problem, it is interesting to consider maximizing both the expected (i.e., average) return, as well as the worst case return.  And also like the earlier problem, this is a great combination of “write a short program to compute optimal strategy” and “use pencil and paper to show why the program’s output makes sense.”

 

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3 Responses to Another card puzzle

  1. mendel says:

    Hmm, a back-of-the-envelope analysis of the endgame (up to 4 cards left) seems to indicate that the average return doesn’t actually depend on the betting strategy in dubious positions, so I could just wait until the remaining cards are all the same and start doubling my money then to realize a good average. It also seems that there is always a way to guarantee the average return as the worst-case if betting properly.

  2. Tom Breton says:

    This seems like a case for the Kelly betting criterion, http://en.wikipedia.org/wiki/Kelly_criterion Whenever you have some advantage, you bet a fraction of your bankroll, specifically

    (p (b + 1) – 1) / b

    where p is the probability of winning and b is the odds vs 1. Here since b = 1, it’s just:

    p – 1/2

    Eg on card 2, you have 26/51 odds of winning, so you bet 0.98 cents – well, you round it up to 1 cent, or the guy lets you run a tab in arbitrary-precision. It’s useful to be able to bet mils instead of cents.

    And so on, always referring to your current odds and current bankroll. Obviously you go all in on the last card (then the guy reveals that it’s the joker!)

    • This is an interesting game for at least one of the reasons you mention. Actually, you can do better than Kelly here. But even when actually maximizing expected return, your point about “arbitrary precision” bets is important– the answer actually changes significantly if you cannot make “real-valued” bets, but must instead bet in, say, cents.

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