Probability that one player rolls more more 5s or 6s than the other player when one player rolls $X$ dice and the other player rolls $Y$ dice

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There is a game I play that utilizes a particular mechanic that I am trying to assess mathematically. Me being a layman, I am having some problems figuring out one of the aspects for myself.

The basic mechanic works like this: you roll a number of 6 sided dice. Any die that rolls a 5 or a 6 nets you a success. You count your total number of successes, and the more of them there are the better.

Now, I know how to calculate the chance that I will succeed when I am trying to roll a fixed number of successes. But, sometimes the game has you roll against someone else's roll. For example:

Player A rolls X number of dice with 6 sides. Player B rolls Y number of dice with 6 sides. The person who rolls more results of 5 or 6 succeeds.

How do I calculate the probability that player A will succeed?

Thank you very much in advance!

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On any die, you either "win" by rolling a $5$ or a $6$, or you don't.

The probability of rolling at least $n$ winning dice out of $d$ rolled is

$$P(n,d) = \sum_{k=n}^{d} {d \choose k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{d-k}.$$

Further, if $n > d$, $P(n,d) = 0.$

So if you have $x$ dice and your opponent has $y$ dice, then the probability of you winning is

$$P_{win}(x,y) = \sum_{k=0}^{x} {x \choose k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{x-k} [1-P(k,y)].$$