<s>[INST] Two people A and B throwing dice A begins by throwing a dice until he gets $6$, then B does the same thing.
What is the probability that A throws more times than B?
I try to solve it, but I got 2 different answers:
1. We mark: $X=$How many times A throws, $Y$=How many times B throws.
$$P(X>n|Y=n)=\frac{P(X>n)\cdot P(Y=n)}{P(Y=n)}=P(X>n)\\ \sum_{k=n}^\infty\left(\frac56\right)^{k-1}\cdot \left(\frac16\right)=\left(\frac65\right)^{1-n}=\left(\frac56\right)^{n-1}$$
2. The second way is: We can say that "B throw less times than A", so let's assume that A throws the dice $n$ times. So:
$$P(Y<n)=\sum_{k=1}^{n-1}\left(\frac56\right)^{k-1}\cdot \left(\frac16\right)=1-\left(\frac56\right)^{n}$$
They are opposite, and I don't know which one of them is correct or both of them are incorrect..
Please help with this.
Thank you!
 [/INST] To find the probability that A throws more times than B, we need to sum the probabilities that A throws two or more times given that B throws once, A throws three or more times given that B throws twice and so on.
So we need to evaluate 
$\sum_{k=1}^{\infty}P(A_{throws} > k,B_{throws}=k)$ 
Now for B throwing once, and A throwing two or more times, we have as probability  (using the sum to infinity of a geometric progression)
$\large \frac{1}{6}\times\sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^k\left(\frac{1}{6}\right)=\left(\frac{1}{6}\right)^2\left(\frac{\frac{5}{6}}{1-\frac{5}{6}}\right)=\frac{1}{6}\left(\frac{5}{6}\right)$
For B throwing twice, and A throwing three or more times, we have as probability (using the sum to infinity of a geometric progression as before)
$\large\left(\frac{5}{6}\right) \frac{1}{6}\times\sum_{k=2}^{\infty}\left(\frac{5}{6}\right)^k\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)^2\left(\frac{\left(\frac{5}{6}\right)^2}{1-\frac{5}{6}}\right)=\frac{1}{6}\left(\frac{5}{6}\right)^3$
For B throwing three times, and A four or more times we have
$\large\left(\frac{5}{6}\right)^2 \frac{1}{6}\times\sum_{k=3}^{\infty}\left(\frac{5}{6}\right)^k\left(\frac{1}{6}\right)=\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)^2\left(\frac{\left(\frac{5}{6}\right)^3}{1-\frac{5}{6}}\right)=\frac{1}{6}\left(\frac{5}{6}\right)^5$
Generalising for B throwing $m$ times and A throwing $m+1$ or more times, the probability is given by
$\large \frac{1}{6}\left(\frac{5}{6}\right)^{2m-1}$ 
Thus, the probability that A throws more times than B is given by the following scaled sum to infinity of a geometric progression with ratio $\left(\frac{5}{6}\right)^2$
$\large \sum_{k=1}^{\infty}P(A_{throws}>k,B_{throws}=k)=\frac{1}{6}\sum_{k=1}^{\infty}\left(\frac{5}{6}\right)^{2k-1}=\frac{1}{6}\frac{\frac{5}{6}}{1-\left(\frac{5}{6}\right)^2}=\frac{5}{11}$
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