## Tossing a biased coin — Part 2

If John has to select one of three alternatives, he tosses the coin in batches of three until he comes up with a mixed group. Here’s one way he can assign options to coin patterns.

THH or HTT — Option 1
HTH or THT — Option 2
HHT or TTH — Option 3

Each set of patterns has the same probability

$p^2(1-p) + p(1-p)^2 = p(1-p)$

The same strategy lets him choose between n alternatives, although things do get messy for n > 5.

The binomial calculation helps. The fact that

$\binom{5}{2} = 10$

means there are 10 ways in which 2 heads and 3 tails can be arranged. So if John has 10 options he needs only batches of five tosses.

Cryptic? Yes. I leave it to you to expand the idea, although it’s drying up.

Enough.