Two people A and B throwing dice

A and B throwing dice are two independent events, so you need to multiply them and evaluate an infinite sum of geometric series.

$\displaystyle P(X>n)=\left(\dfrac{5}{6}\right)^n\\\\ P(Y=n)=\left(\dfrac{5}{6}\right)^{n-1}\cdot \left(\dfrac{1}{6}\right) $

and the required probability is:

\begin{align*} P(X>Y) &=\sum_{n\ge 1} P(X>n)\cdot P(Y=n)\\ &=\sum_{n= 1}^{\infty} \left(\dfrac{5}{6}\right)^{\!n}\cdot \left(\dfrac{5}{6}\right)^{\!n-1}\cdot\left(\dfrac{1}{6}\right)\\ &=\frac{6}{5}\cdot\sum_{n= 1}^{\infty} \left(\dfrac{5}{6}\right)^{\!2n}\cdot\left(\dfrac{1}{6}\right)\\ &=\frac{1}{5}\cdot\sum_{n= 1}^{\infty} \left(\dfrac{25}{36}\right)^{\!n}\\ &=\frac{1}{5}\cdot\frac{25/36}{11/36}\\ &=\dfrac{5}{11}\\ &= 0.45\overline{45}. \end{align*}

In order to avoid the double sum, note that the problem is symmetric. The probability that A throws more times than B is equal to the probability that B throws more times than A; call this probability $p$.

Now let's calculate the probability that both A and B throw the same number of times; this will be $1-2p$. This is

$P(A=B) = \sum_k P(A=k, B=k) = \sum_k P(A=k) P(B=k) = \sum_k P(A=k)^2$

and we can write out the sum explicilty as

$\sum_{k=1}^\infty ((5/6)^{k-1} (1/6))^2$

or, after some simplification,

$(1/36) \sum_{j=0}^\infty (25/36)^j$.

Summing the series, this is $1/11$. So $1-2p = 1/11$ and therefore $p = 5/11$.

I am a new user and I really need 5 points. Hope this helps.