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We can find out how many trails we need on average to get n consecutive heads with the equation below:

$$A =\dfrac { p^{−n} − 1 } {1 − p}$$

So, for $n=6$ with an unbiased coin, we'd require 62 trials.

My question is, how to deal with a biased coin.

Putting a coin with a .9 probability of landing heads, says we need -8 trials.

user577215664
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phantom234234
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  • This is hard to follow. What's $p$ meant to be in that equation? Also, I think you meant $n=5$ if you want to get $62$. And, in your last sentence, how many Heads are you seeking such that you get $-8$ trials? If $n$ is a positive integer then $.9^{-n}>1$ so $A$ should be $>0$. – lulu Oct 11 '20 at 13:11
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    As to the question, see this question and adapt the recursion to the weighted case. – lulu Oct 11 '20 at 13:15
  • you need $62$ expected tosses for $5$ consecutive heads and not $6$. Also, you can substitute $p$ with a different probability than $0.5$ if it is not a fair coin. So I am not sure what you are seeking. Is it the proof of a generic case that you are seeking? – Math Lover Oct 11 '20 at 14:12

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