Probability of having X more heads than tails for N tosses.

Question

Giving a a fair coin, and tossing it N times, in how many possible outcomes would there be a point wherein there were more heads than tails tossed, ie, net heads.

Answer 1

The paths of length $N=2n$ such that this does not happen are the Dyck paths, enumerated by the Catalan number $C_n=(n+1)^{-1}{2n\choose n}$. The total number of paths is $2^{2n}$. Hence the probability you are after is $p_{2n}=1-2^{-2n}C_n$ (and $p_{2n+1}=p_{2n+2}$).

For example, $1-p_N\sim cN^{-3/2}$ when $N\to\infty$, with $c=2\sqrt{2/\pi}$.

Answer 2

If $N$ is odd, there is no possibility of equal heads and tails. Thus, the probability of excess heads is $\frac{1}{2}$

If N is even, the probability of equal heads and tails, $P_{equal}$, is given by:$$P_{equal}=\left(\frac{1}{2}\right)^N\frac{N!}{\left(\frac{N}{2}\right)^2}$$The probability of excess heads is then given by$$P_{Heads}=\frac{1-P_{equal}}{2}$$

Answer 3

Tossing a coin N times is a binomial distribution and you want to solve the for the CDF=0.5.

You can work it out from the formula in the article but as a short cut ...

The chance of more heads is half the chance of unequal results.

If N is odd the chance of equal results is 0, so the chance of more heads is 0.5.

If N is even the probability of getting equal results is ${N \choose \frac{N}{2}}\frac{1}{2^N}$, so the chance of more heads is $\frac{1-{N \choose \frac{N}{2}}\frac{1}{2^N}}{2}$.