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This is one of the questions of a set of exam review questions that don't have solutions to them. I can't get my head around this but it seems so simple.

By flipping a fair coin repeatedly and independently, we obtain a sequence of H's and T's. We stop flipping the coin as soon as the sequence contains either HH or TH. Two players play a game, in which Player 1 wins if the last two symbols in the sequence are HH. Otherwise, the last two symbols in the sequence are TH, in which case Player 2 wins.

A = "Player 1 wins" and B = "Player 2 wins."

Determine Pr(A) and Pr(B)

Caleb Stanford
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Roman
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3 Answers3

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First flip is either heads or tails. If the second flip is heads we have a winner no matter what. If the second flip is tails we have no winner, but it follows that Player 2 must win. Why?

Flip three is either heads or tails. If it is heads, player 2 wins. Tails, no one wins. Flip four and each afterward either results in a heads and player 2 wins or a tails and no one wins. Given the last flip was tails, HH will never occur before TH.

Using this information, we have a $\frac{1}{2}$ chance of the game ending on the second flip. Assuming it ends on the second flip, each player wins $\frac{1}{2}$ of the time (HH or TH). If the game does not end on the second flip, Player 2 wins.

This must mean $\operatorname{Pr}(A) = \frac{1}{4}$ and $\operatorname{Pr}(B) = \frac{3}{4}$.

Brad
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Hint: Note, the only way player 1 wins is if the sequence starts "HH". Once a "T" appears, the second player is guaranteed a win eventually.

Macavity
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The above answers are right.

The key point to understand here is that Player B wins for TH.

So the only chance for Player A to win is through HH in the first two flips.

Once you get a T it'll nullify Player A's chances of winning because you can never get 2 H's again. The game will become whether Player B wins or not.

So if we consider the unrolling of the probability tree for the first two levels,
we see that:
P(HH occurring before TH) = P(HH in first two flips) = P(A winning) = 1/4
P(TH occuring before HH) = P(B winning) = 3/4

I too was initially confused with this problem but the above answers helped me articulate the solution.

sdasara
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