I found this question on http://www.cs.uwm.edu/classes/cs317/ I tried to solve it because I have a quals exam on the theory courses next month. Please check my answer if it is correct or not. I assumed that $X$ ( random variable) is equal to the number of rounds coin-flipping.
Suppose $n$ people, $n \ge 3$ play the odd person out game to decide who will buy refreshments. The game works as follows. Everyone flip a fair coin simultaneously. If all the coins but one come up the same, the person whose coin comes up different buys the refreshments.
Otherwise, the people flip the coins again and continue until just one coin comes up dierent from all the others. We would like to know the expected number of rounds of coin- flipping needed to decide the odd person out with $n$ people. Notice that we can view a round of coin-flipping as a "success" if an odd person out is chosen on that round; otherwise, it is a "failure". The game then is interested in reaching a successful round.
a. What is the probability of success? of failure?
My answer is : probability of success is $p$, probability of failure is $(1-p)$.
b. What is the probability that $k$ rounds are needed for a success to show up?
My answer is: $(1-p)^{k-1}p$
c. What is the expected number of rounds needed for the game to end? $E(x) = 1/p$.