2 player dice game probability For some homework in one of my classes, we are given this problem:

In a certain dice game, player $A$ rolls six six-sided dice vs. player $B$ who rolls nine four-sided dice. Each player rolls exactly once, and $A$ wins provided that the sum of his dice is strictly greater than $B$'s, otherwise $B$ wins. What is $A$'s probability of winning? Solve this analytically.

So, seeing how the highest number either one can get is $36$, I calculated each player's probabilities of making a number from within $1\dots 36$. However, I am stuck in terms of how to figure out $A$ probability of winning the dice roll. Can anyone explain to me the steps to figure this out?
Thank you kindly.
 A: Once you have the probabilities for each player to roll each number you are almost there.  For each roll of $A$, calculate the probability that $B$ gets a lower number and add them up.  For example $A$ has a chance of $\frac 1{6^6}$ to roll $36$.  He wins unless $B$ also rolls $36$, which is $\frac 1{4^9}$.  So this contributes $\frac 1{6^6}(1-\frac 1{4^9})$ to $A$'s chance to win.  Since $B$'s average is $9 \cdot 2.5=22.5$, $A$ will win precisely $50\%$ of the time he rolls $23$.  If his chance to roll $23$ is $p$ (which you have calculated), it contributes a chance to win of $\frac p2$.  Add them all up.
A: Well, if you have all the individual probabilities figured out, your answer would be:
$$
P(winning)=P(A>36\mid B=36)+P(A>35\mid B=35)+\cdots+P(A>4\mid B=4)
$$
Since your events are independent, we have
$$
P(winning)=P(A>36)P(B=36)+P(A>35)P(B=35)+\cdots+P(A>4)P(B=4)
$$
Then $P(A>n)=\sum_{k=n+1}^{36} P(A=k)$ so
$$
P(winning)=0\cdot P(B=36)+P(A=36)P(B=35)+\cdots+[P(A=36)+\cdots+P(A=4)]P(B=4)\\
$$
Finally,
$$
P(winning)=\sum_{n=4}^{36}\left(P(B=n)\sum_{k=n+1}^{36} P(A=k)\right).
$$
A: A probability generating function can solve this:
$
G(x)={\left(\dfrac{x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x}{6}\right)}^{6} {\left(\dfrac{\dfrac{1}{x} + \dfrac{1}{x^{2}} + \dfrac{1}{x^{3}} + \dfrac{1}{x^{4}}}{4}\right)}^{9}
$
Sum all the coefficients of x which are greater than 1.
E.g. In Sage, it can be done like this:
fx = expand((sum(x^i for i in range(1,7))/6)^6*(sum(1/x^i for i in range(1,5))/4)^9)
sum([fx.coefficient(x,i) for i in range(1,36-8)])

which is:
$
\dfrac{725864657}{2038431744}\approx 0.356089753378566
$
which also agrees with a simulation (done in J)
sim=: 3 : '(+/1+?6#6)>(+/1+?9#4)'
(+/%#)(sim"0)100000#0

$\approx 0.35643$
