Complicated Conditional Probability Three people role a die starting with person 1, then 2, then 3. First to role a 6 is eliminated.
Find the probability that B is elimination first.
I think this is some summation of probabilities to get a general form for how many times they have rolled a dice. I get $\frac {1}{6}*(\frac{5}{6})^n$ Such that n=1, 4, 7, 10,... (I cant think of how to write a sequence for this). Is there a way to write this because I get lost on how to write out the probability.
After the first one is gone, the remaining two play until the next one wins. Find prob A won the game. I said this was Prob A won given B lost 1st and then C lost 2nd plus the Prob A won given B lost 2nd and then C lost 1st. But since it matters when they each lost to get to this part. How do I go about this?
 A: You have the right idea, but you don’t want to multiply $\frac56$ by those values of $n$: they should be exponents on $\frac56$. It might be easier to think first about A’s probability of being eliminated first. With probability $\frac16$ A is eliminated right away. He’s also eliminated first if all three of them roll non-sixes in the first round, and then he rolls a six on his second roll; that happens with probability $\left(\frac56\right)^3\frac16$. Continuing in this fashion we see that A’s probability of being eliminated first is
$$\frac16+\left(\frac56\right)^3\frac16+\left(\frac56\right)^6\frac16+\ldots=\frac16\sum_{n\ge 0}\left(\frac56\right)^{3n}\;,\tag{1}$$
and since that’s just a geometric series, you can finish off the calculation to get a numerical result. Call this probability $p_A$.
What changes if we look at B? A has to roll a non-six first, and if he does, B is now in A’s position: he’s effectively rolling first from now on. Thus, his probability of being eliminated first is $p_B=\frac56p_A$. 
But if we extend this analysis just a little further, we can actually avoid having to evaluate the summation in $(1)$: the same reasoning shows that C is eliminated first if and only if A and B both survive their first rolls, making C effectively the first player, and C is then eliminated first. That is, C’s probability of being eliminated first is $p_C=\frac56\cdot\frac56p_A=\frac{25}{36}p_A$. One of them has to be eliminated first, so
$$1=p_A+p_B+p_C=p_A\left(1+\frac56+\frac{25}{36}\right)\;,$$
and you have a simple linear equation to solve for $p_A$.
For the second part notice that if B is eliminated first, A is the second player in the resulting two-person game, while if C is eliminated first, A is the first player in the resulting two-person game. Use the ideas above to work out the probabilities of winning for the two players in the two-person game, and them combine them appropriately with the probabilities of B and C being eliminated first in the original game.
