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I'm studying probability and came across the following problem:

Alice and Bob are playing a game in which each of them chooses a 4-letter string of $\{H,T\}$ (heads/tails), following which a coin is tossed repeatedly until one of the players' chosen strings appears. When that happens, that player has won.

If Alice chooses the string $THTH$, is there any string Bob can choose with which his probability of winning is greater than $50\%$?

I know that if I'm given a specific string for Bob, I can calculate each player's probability of winning by doing recursive conditional probability, but short of checking each of Bob's 15 possible choices for a string this way I'm not sure how to prove or disprove the assertion that Bob's chance to win can be more than $50\%$.

(I can't find a problem like this anywhere on SE so if this is a duplicate I apologize in advance).

Or Bairey-Sehayek
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  • Perhaps relevant: https://math.stackexchange.com/a/1859864, https://plus.maths.org/content/os/issue55/features/nishiyama/index, https://math.stackexchange.com/questions/3381462, https://math.stackexchange.com/questions/2862698, https://math.stackexchange.com/questions/3390215. – joriki Nov 29 '19 at 15:26
  • Google "Penney's game", you will find all the detail you desire. –  Nov 29 '19 at 16:54

2 Answers2

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If they play the game many times, and the winner scores each game based on how long it takes their opponent to finish after them, and they add together their scores from each game, then scores will, in the long run, be even.

The trick here is for Bob to pick a sequence such that if he wins, he wins just before Alice finishes her sequence, and if Alice wins, it's a long time until Bob finishes his. That way, since Bob often scores less points than Alice, he must win more often.

With that in mind, I suggest Bob goes for TTHT.

Arthur
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Turns out that HTTH gives player B a larger chance of winning — I ascertained this by looking for strings that have no prefixes in common with THTH but could potentially capitalize on player A’s losses.

It was mostly guess and check... Solving the system of equations given by following the conditional probability through to A’s win and B’s win gives us $P(B)=\frac{9}{16}$.

Or Bairey-Sehayek
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