Conditional variance problem Three people - Alice, Bob and Eve play a game. Every round, each of them tosses a fair coin. The one of the three who gets a different side of the coin wins the game. If three of them got the same side (for example three heads or tails), the game continues to the next round. There are up to 4 rounds. The game ends in a tie if there is no winner after the fourth round.
Probability for a person to win the game is $\cfrac{85}{{256}}$. A person can't lose the game, but it can end as a tie for everyone. They play the game 10 times (each game takes up to 4 rounds). If it is known that Alice won the game 4 times, what is the variance of games won by Bob?
My attempt was to use the conditional variance:
$Var(X|Y=4)=E[X^2|Y=4]-(E[X|Y=4])^2$
Where $X$ represents the games won by Bob, and $Y=4$ represents the 4 games Alice won (out of 10). So:
$E[X|Y=4]=\sum_{x=1}^{10}  x⋅P(x|y=4)$
Now I can sum over the iterations of x and calculate the probability of each $P(X=x|y=4)=P(X=x∩y=4)/P(y=4)$. I think that they are not independent, so I need to calculate the probability of $P(X=x∩y=4)$ for sure.
Then, I must also calculate $E[X^2|Y=4]=\sum_{x=1}^{10}  x^2⋅P(x|y=4)$
All that looks like an enormous work. I think that it is possible to solve it like that, but I'm sure there is a better way. May somebody give me a hint?
 A: I will use the letters $A,B,C$ for the three players Alice, Bob and Eve. One can model the given game as follows using the space $\Omega=\Omega_0^{\times 10}$, where $\Omega_0$ is the probability space with four elements $A,B,C$ and tie, where $A,B,C$ (corresponding to the win of the corresponding player in a "simple tossing game" with at most four rounds) have each the probability $85/256$, and where the tie has the remained probability $1/256=1/4^4$. It is convenient to denote by $B'$ the event in $\Omega_0$ consisting of $C$ and the tie. It has probability $(85+1)/256$.
An element of $\Omega$ is of the shape $\omega=(\omega_1,\omega_2,\dots,\omega_{10})$, where the components $\omega_j$ are from $\Omega_0$.
Now we restrict inside this space to the subspace $\Omega'$ consisting of $10$-tuples $\omega$ with exactly four components equal to $A$. Let $S_\omega$ be the subset of $\{1,2,\dots,10\}$ of the places with the component $A$. Then we can split $\Omega'$ in a natural manner in disjoint subsets $\Omega_S'$, indexed by subsets $S$ of $\{1,2,\dots,10\}$. (So $\omega\in\Omega'_S$ iff $S_\omega=S$.)
Now we fix such an $S$, and consider the given problem on $\Omega'_S$. By permutation symmetry we may and do assume $S=\{7,8,9,10\}$. Then on the remained places $1,2,3,4,5,6$
we can put independently either $B$ or $B'$. The two choices have weight $p=85/(85+86)$ and $q=1-p=86/(85+86)$. So we are dealing with a binomial distribution with parameters $n=6$ and $p=85/(85+86)$. The needed variance - restricted to $\Omega'_S$ - is
$$
n\;pq =6\cdot\frac{85\cdot 86}{(85+86)^2}\approx 1.4999487021647\dots\ .
$$
Since this does not depend on $S$, this is also the variance on $\Omega'$.
