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Greetings.

I have been trying to figure out a formula that would give the probability of getting at least h heads after f coin flips. I have that:

(2^f - 1) / 2^f gives the odds of getting at least one heads for f flips and

(2^f - 1 - f) / 2^f gives the odds of getting at least two heads for f flips.

I couldn't make it to three. I also understand that f!/(h!(f-h)!2^f) gives the odds of getting exactly h heads for f flips and that summing the "exacts" formulae for h through f heads would give the odds of getting at least h heads for f flips if I actually knew what f was; but, since I don't know f, I don't think I can sum the "exacts" together.

Could you tell me please what formula would give the probability of getting at least h heads after f coin flips?

Thank you.

You do need to sum the number of getting exactly h' flips for all h'≤h. You can "sum the 'exacts' together" -- that is precisely how you compute this. See here. For example, to compute the probability of getting 5 heads out of f flips, you sum:

5

Σ f! / (h'!(f-h')!2^f) = ( f^5 - 5f^4 + 25f^3 + 5f^2 + 94f + 120 ) / ( 15 × 2^(f+3) )

h'=0

That's all you can do. For a fixed number of heads (like this one, five or less) you can simplify the expression (as above) into a polynomial in f divided by the power 2^f. But if you want the number of heads to be variable, the best you can do is the summation from h'=0 to h'=h, where h has to be specified before you can simplify it.

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