# Expected value of sum of different geometric random variables

$$n$$ players play a game where they flip a fair coin simultaneously. If a player has a different result than the others, that player is out, and then the remaining $$n - 1$$ players continue until there are two players left and they are the winners. For example, for $$n=3$$, a result $$(H,T,T)$$ makes the first player lose and the other two to win, and $$(H,H,H)$$ will make them toss again.

I'll define the variable: $$Y =\text{number of rounds until there are two players left out of } n$$

I'm looking for $$E(Y), VAR(Y)$$. What I did was:

Define a random variable $$X_i = \text{number of rounds until one player out of } i \text{ is out}$$

so is follows that: $$X_i \sim{\mathrm{Geo}(2\cdot\frac{i}{2^{i}} = \frac{i}{2^{i-1}})}$$ , since we have to choose a player and a value for the coin $$Y =\sum_{i=3}^{n}X_i$$ $$E(Y) = E(\sum_{i=3}^{n}X_i)=\sum_{i=3}^{n}E(X_i)=\sum_{i=3}^{n}\frac{2^{i}}{i}$$ $$VAR(Y) = \sum_{i=3}^{n}VAR(X_i) = \sum_{i=3}^{n} \dfrac{\dfrac{2^{i-1}-i}{2^{i-1}}}{\dfrac{i^2}{2^{2(i-1)}}} = \sum_{i=3}^{n} \dfrac{2^{i-1}(2^{i-1}-i)}{i^2}$$

Is there a closed form solution to this problem?

• @joriki huh, right. Thanks! – Theoretical Economist May 20 '20 at 10:11
• The approach seems valid. Are you hoping to simplify the summation into a closed form? I am not sure that's possible (and but would be happy to be proved wrong). BTW you can add the variances too since the $X_i$'s are independent. – antkam May 20 '20 at 18:53
• "If a player has a different result than the others": What does this mean? On the first round, if half of the players flip heads and the other half flip tails, are they all out? – Michael May 21 '20 at 2:32
• I have updated the question – CodeHoarder May 21 '20 at 6:03