The solution to Tuesday’s puzzle is under the fold:
So I have a well-shuffled deck of 52 cards. I turn them over one by one, starting at the top. You can raise your hand whenever you like. After you raise your hand, you win a prize if the next card I turn over is red.
Okay, you’ve just raised your hand. At this point I say to you: “Hey. Would you mind if I turn over the bottom card instead of the top one?”. It clearly makes no difference whether you say yes or no. If, for example, I have 17 cards left, 12 red and 5 black, then you’ve got a 12/17 chance of winning, regardless of whether I turn over the top card, the bottom card, the middle card, or any other randomly chosen card.
In other words, we might as well change the rules of the game so that I deal from the top until you raise your hand, and then I deal one card from the bottom. What’s your chance of winning that game? Regardless of your strategy, you win when the bottom card is red and lose when it’s black — so regardless of your strategy, you win with probability 50%.
- If I deal the last card from the bottom, your strategy doesn’t matter.
- If I deal the last card from the top, your chances of winning — under any strategy — are exactly the same as if I deal the last card from the bottom.
- Therefore, when I deal the last card from the top (i.e. in the original game) your strategy doesn’t matter.
You can plan to raise your hand the first time you’ve seen more blacks than reds, or the first time you’ve seen more reds than blacks, or the first time you’ve seen 20 blacks, or any other damn thing. Your chance to win will still be 50%.
You can find this solution in the comments on the original post. Ramprasad got it in Comment #2(!), and others, including Ben Kenendy (#92 there), spelled it out in nice detail. My own solution was pretty much the same as Mike H.’s (comment #72 on the original post). But I like this one much better.
Hat tip to Richard Stanley of MIT.