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Can someone help me with this one?

A fair coin is tossed 5 times, what is the probability of a sequence of 3 heads? I can see that there are 2*2*2*2*2 possible outcomes, but how many of these include 3 heads in a sequence and why?

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    $\begingroup$ There are only 32 combinations possible; you could write them all out and just count up the ones that have three heads in them. You could save some effort by noting that all combinations with a tail in the third place cannot have a sequence of three heads, so you actually only have to write out 16 combinations (the ones with a head in the third place) and remember that the other 16 don't have any sequences of three heads. $\endgroup$
    – jbowman
    Mar 2, 2012 at 18:57
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    $\begingroup$ Please count: FFFFF FFFFT FFFTF FFFTT FFTFF FFTFT FFTTF FFTTT FTFFF FTFFT FTFTF FTFTT FTTFF FTTFT FTTTF FTTTT TFFFF TFFFT TFFTF TFFTT TFTFF TFTFT TFTTF TFTTT TTFFF TTFFT TTFTF TTFTT TTTFF TTTFT TTTTF TTTTT $\endgroup$
    – Elvis
    Mar 2, 2012 at 19:06
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    $\begingroup$ Do you mean exactly three successive Heads, or three or more successive Heads? The answer are different in these two cases. $\endgroup$ Mar 2, 2012 at 20:12
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    $\begingroup$ A general analysis of the problem of computing the chance of getting $k$ heads in a row out of a sequence of $n$ independent trials when each head has a chance of $p$ occurring is given in my reply at stats.stackexchange.com/a/23762. The approach given there gives $(3-2p)p^3$ = $1/4$ when $p=1/2$, $k=3$, and $n=5$. $\endgroup$
    – whuber
    Mar 2, 2012 at 21:11

1 Answer 1

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Total number of possible events = 2^5 = 32

Frequency of exactly 3 heads (HHHT*, THHHT, *THHH) = 2+1+2 = 5

Frequency of exactly four consecutive heads (HHHHT, THHHH) = 2

Frequency of five consecutive heads = 1

Frequency of required events = 5+2+1 = 8

Required probability = 8/32 = 1/4

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  • $\begingroup$ Thanks to all those that have provided insights, I could indeed list all possible outcomes and count the one with at least 3 heads but I like the reasoning proposed by Stat-R. $\endgroup$
    – Luca
    Mar 6, 2012 at 13:24

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