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I simply cannot wrap my head around this fact:

"A fair coin is no more likely to produce any specific 10-toss sequence than any specific other one, but it is about 250 times as likely to produce one with exactly 5 heads as it is to produce one with 10 heads."

The probability of getting a specified coin toss sequence of e.g. 10 tosses is 1/10, so getting a sequence like HHHHHHHHHH has the same probability like getting HTTTHTHHTH. BUT according to the text snippet above, the probability of getting a sequence with exactly 10 heads is smaller than getting a sequence with 5 heads.

Can someone please explain how this is possible? My intuition is failing me at this point, I understand both statements separately, but I have a hard time seeing how they are both true at the same time because for me P('HHHHHHHHHH') = P('sequence with exactly 10 heads') holds.

Maybe it helps if someone puts it in different words or mathematical terms? Thank you!

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If you throw a coin 10 times, there are $2^{10}=1024$ different possible sequences. If your coin is fair, each sequence is equally likely. And there is exactly one sequence HHHHHHHHHH. So there is a chance of $\frac{1}{1024}$ to produce HHHHHHHHHH. Or, for that matter, any other specific sequence. (Not 1/10, as you write.)

However, there are many different sequences that contain exactly five heads, e.g., HHHHHTTTTT, TTTTTHHHHH, HTHTHTHTHT, HHTHTTTHHT and so forth. We could try to count them, or we could simply notice that the probability of getting exactly five heads is a binomial probability for 5 successes out of 10 trials with a success probability of $\frac{1}{2}$, or

> dbinom(5,10,1/2)
[1] 0.2460938

Multiplying this by 1024 gives us that there are 252 sequences with exactly 5 heads.

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  • $\begingroup$ Oh, I think now I see where the problem was, there is always only 1 occurrence of a specific sequence, but there can be many sequences with some property, e.g. 5 heads, got it! However, I just tried the binomial pdf function in matlab binopdf(5,10,.5)*1024 and it gave me 252, which seems to be the "almost 250" that the author is describing? $\endgroup$ – Maria Jul 18 '19 at 12:22
  • $\begingroup$ You are right, I picked the wrong function. Sorry. Updated the answer. $\endgroup$ – Stephan Kolassa Jul 18 '19 at 12:25

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