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here is a probability question. Can someone explain the flaw in my reasoning?

Two players, A and B, alternatively toss a fair coin (A tosses the coin first, then B tosses the coin, then A , then B .. .). The sequence o f heads and tails is recorded. I f there is a head followed by a tail (HT subsequence), the game ends and the person who tosses the tail wins. What is the probability that A wins the game?

P(A) = 1 - P(B)

Then, if A tosses H and B tosses T (probability 1/4), B wins. Also, if A tosses T (probability 1/2), the game effectively restarts and B has the same probability of winning as A. Finally, if A tosses H and B tosses H (probability 1/4) then the game effective restarts with the same chances of B winning. Hence,

P(B)= 1/4 + 1/2P(A) + 1/4P(B)

Solving this gives me P(A) = 2/5 when the answer is 4/9. Can someone explain?

Anon
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1 Answers1

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If A tosses H and B tosses H, that's not an effective restart, since A can win immediately with a T on their second turn.

Basically, once the first H is tossed, then the game becomes "the next T wins".

Greg Martin
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