I got this problem:
It is required to select a person from n. To do this, each person spear a coin and it is agreed that if one obtains a different result from the others, then that person is the selected one; otherwise coins are thrown again under the same conditions. Determine the expected number of games made until a person is selected.
I tried to define a random variable which tells us the number of games that it takes to select the n-th person Tn, so Tn has a distribution Geometrically with parameter 1/2. So the mean is:
E(Tn) = $\sum_{n=0}^{\infty} (1/2)^{n} = \frac{1}{1/2} \ = 2$
Am I right?
Guide:
First compute the probability of exactly one person get a different outcome. That is either everyone but one person get heads or everyone but one person get tails.
$$\binom{n}{1}\frac1{2^n}+\binom{n}{n-1}\frac1{2^n}=\frac{n}{2^{n-1}}$$
Now, apply this probability value to a geometric distribution.