# Probability of odd one out at a dinner table

Suppose $n$ friends are trying to decide who will pay for everyone's dinner. They decide to choose the person ( the 'loser' ) by playing 'odd man out'. Each person is given a coin that has probability $p$ of falling heads. Each person tosses his coin. A person is said to be a loser if his toss outcome is different from that of other $n-1$ person's tosses.

Suppose the friends repeat the game until there is a loser. What is the probability that the game would be repeated exactly $k$ times.

I know how to find the probability that there is exactly one loser ( using binomial distributions). By the k'th toss there would be one H versus $n-1$ T or vice versa. However the tosses before the kth toss could be randomly distributed. It could be 2 H and $n-2$ T, or 8 H and $n-8$ T. I'm not able to find a solution for this since the possibilities seem so random.

• Compute the probability that there is exactly one loser for a particular round, call it $m$.
• The number of rounds required would follow Geometric distribution, Geo($m$).
• yes, $q^{k-1}m$ is right. – Siong Thye Goh Nov 22 '17 at 4:15