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N tosses of a fair coin. There are $\binom{2+N-1}{N} = N+1$ ways to choose with replacement from {h, t}. So the probability of N heads is $\frac 1{N+1}$. Obviously not all ways are equivalent but is there a better explanation for the absurd result?

I got this playing/trying to derive the binomial distribution formula using a "combinations with replacement" approach. Is it possible?

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    $\begingroup$ You have identified the problem: If $X$ is the number of heads, then $\Pr(X=k)$ is highly variable, small for $k$ small or large, and big in the middle. To get the right answer, note that all sequences of length $N$ made up of the letters H and/or T are equally likely, and that there are $\binom{N}{k}$ such sequences with exactly $k$ H. $\endgroup$ – André Nicolas Jun 17 '14 at 0:22
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There may be $N+1$ different possible counts of heads in $N$ tosses (0 heads, 1 head, ..., $N$ heads), but those outcomes are not equally likely.

Consider tossing two distinguishable* coins
enter image description here
(or consider tossing one coin twice). The outcomes when you consider each toss in order are HH, HT, TH and TT. With fair coins, those four outcomes are equally likely.

When you just count the heads in two tosses, you combine HT and TH into "one head"; this means that "one head" from two tosses is twice as probable as either of the outcomes "0 heads" and "2 heads".


You could always try it and see - toss two coins at a time, and keep track of the number of times 0,1 and 2 heads come up - you'll soon see that 1 head is more likely than either of the other two outcomes - and if the two coins are readily distinguishable (such as different denominations), and you keep track of HT and TH separately, it's soon clear why.

*(original coin images are Wikimedia commons images stated to be public domain)

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