Originally Posted by

**HallsofIvy** The probability of "one head, two tails" in **that** order is (1/2)(1/2)(1/2)= 1/8. It is not difficult to see that that is the probability for any specific order. How many orders are there?

It is easy to write them: HTT, THT, TTH. Three possible orders. That is the same as $\displaystyle \begin{pmatrix}3 \\ 2\end{pmatrix}= \begin{pmatrix} 3\\ 1\end{pmatrix}= \frac{3!}{(1!)(2!)}= 3$.

So the probability of "one head, two tails" is $\displaystyle \frac{3}{8}$. In general if you flip n times, the probability or 'h heads, t tails", where h+ t= n, is $\displaystyle \frac{\begin{pmatrix}n \\ h \end{pmatrix}}{2^n}= \frac{n!}{h!t!2^n}$.

Even more generally if you do something that has probability p of outcome A and probability q of outcome B, where p+ q= 1, N times, the probability of outcome A happening n times and outcome B happening m times, with m+ n= N, is $\displaystyle \begin{pmatrix} N \\ n\end{pmatrix}p^nq^m= \frac{N!}{n!m!}p^nq^m$.