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Consider tossing a fair coin; you win only if you get k+1 consecutive heads after k successive tails. What is the probability you win in finite trials?

I encountered this problem in an interview. My initial thought was to calculate the probability that the game never stops: $P=1-\frac{1}{2}-\frac{1}{2}^2-.....$.

However, I feel my solution is wrong. Could anyone suggest?

user334639
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    Does consecutive mean the same as successive? Also, does the run of $k$ successive tails have to be exactly of length $k$ or would $k+1$ consecutive Heads after a gazillion $\gg k$ successive tails be acceptable for a win? – Dilip Sarwate Aug 21 '22 at 22:23
  • @DilipSarwate Consecutive = successive in this case. The $k$ consecutive tails have to be exactly of length $k$. – user334639 Aug 22 '22 at 01:42
  • The second link, "How to prove that an event occurs infinitely often," isn't truly a duplicate, but the methods used to answer that question apply obviously and directly to answering this question. Another approach is to make it harder to win by insisting that the $k+1$ consecutive heads begin after exactly $2(k+1)n+1$ trials for $n=0,1,2,$ etc. Your chance of not winning the harder game by this time is is $(1- 2^{2k+2})^n,$ which becomes arbitrarily small; whence your chance of not winning the original game must be smaller still. – whuber Aug 22 '22 at 15:45

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