# What is fundamentally wrong with this?

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?

• 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. – André Nicolas Jun 17 '14 at 0:22

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.