Tossing Dice repeatedly, probability that 2nd trial had more tosses than 1st one John repeatedly tosses a die until a six occurs for the first time. Alice then repeats the experiment. What is the probability that Alice made more tosses
than John?
 A: Outline: Let's first find the probability that Alice and John have the same number of tosses. They can either each get their first $6$ on their first toss, or they can each get their first $6$ on their second toss, or each on their third toss, and so on.
The probability it occurs in  the first round is $\frac{1}{6^2}$. The probability it occurs in the second is $\frac{1}{6^2}\frac{5^2}{6^2}$. The probability it occurs in the third is $\frac{1}{6^2}\frac{5^3}{6^3}$. And so on. This is an infinite geometric progression. Calculate the sum, and call it $p$.
Let $a$ be the probability Alice made more tosses. Then $2a+p=1$.  For by symmetry either John made fewer tosses, or Alice made fewer tosses (probabilities equal by symmetry) or there was a tie. 
Added: The infinite geometric sequence
$$\sum_0^\infty \frac{1}{6^2}\left(\frac{5^2}{6^2}\right)^n$$
has sum $\frac{1}{6^2}\frac{1}{1-\frac{25}{36}}$, which simplifies to $\frac{1}{11}$.
It follows that the probability Alice makes more tosses than John is $\frac{5}{11}$. 
