# Prove the probability to even number of “Heads” is $\frac{1}{2}$.

Let $n$ coins, where at least one of them is a fair coin. Each one of the $n$ coins is tossed - Prove the probability to get even number of "Heads" is $\frac{1}{2}$.

I'd be glad for a direction.

Thanks.

Lat $A$ be one of the fair coins. Let $p$ be the probability that the number of "heads" among the rest (i.e., the coins $\ne A$) is even. Then the probability for a total number of even "heads" is $$P(A\text{ tails})P(\text{rest even})+P(A\text{ heads})P(\text{rest odd})=\frac12p+\frac12(1-p)=\frac12.$$
• Can you explain $\frac12p+\frac12(1-p)$? – AlonAlon Nov 27 '14 at 17:59
• @AlonAlon: This is Bayes's formula, conditioning on whether $A$ turned up heads (the second term) or tails (the first term). – Ted Shifrin Nov 27 '14 at 18:03