How can you decide the winning percentage of a player in a game based on PMFs for each player?

In a two player game, each player is trying to score the higher number of points. We can assume that the probability that a player scores $$n$$ points is independent of how many points his opponent scores. For example, if this were the setup for player one and player two:

$$P_1(4) = 0.8$$

$$P_1(5) = 0.1$$

$$P_1(6) = 0.1$$

$$P_2(6) = 0.1$$

$$P_2(7) = 0.9$$

In this example, the two players will tie $$10\%$$ of the time, and player two will win 90% of the time. For more complicated PMFs, how can we solve this problem in general?

• If the scores are independent, then it appears they tie $1\%$ of the time and player 2 wins $99\%$ of the time. May 23 '19 at 14:10
• It doesn´t seem that you have found the pmf, did you? May 23 '19 at 14:20
• @callculus The $P_1$ and $P_2$ above seem to be valid pmfs to me, they respectively sum to $1$ over their supports. May 23 '19 at 14:24
• @JackCrawford I was thinking of the joint pmf. May 23 '19 at 14:26
• @JackCrawford The OP wants a solution for a general case although she/he hasn´t found the joint pmf in this specific case. Such postings don´t encourage me to answer. May 23 '19 at 14:35

You have the idea. Assuming independence $$P_{12}(a,b)=P_1(a)P_2(b)$$. If the distributions are discrete you can just sum over all the cases where $$b \gt a$$ to get the chance $$2$$ wins. If they are continuous you integrate. In your example the joint pmf has six pairs of values where the probability is greater than $$0$$, so you can just add them up.
As @InterstellarProbe commented above, in the example you've given a tie only occurs $$1\%$$ of the time and player $$2$$ wins the other $$99\%$$ of the time, since the only score they overlap on is $$6$$ and so a probability of getting a tie is $$P_1(6)\cdot P_2(6) = 1\%$$. For all other states, player $$2$$ wins.
In order to solve this problem in general, let's first consider that to find the probability of getting a tie, you must simply find all of the scores that the players can possibly get in common, multiply together all of the pairs, and sum them. For example, if $$P_1(5) = 0.3$$, $$P_1(6) = 0.2$$, $$P_2(5) = 0.1$$, and $$P_2(6) = 0.4$$ (not exhausting their entire pmfs) where $$5$$ and $$6$$ are the only potential scores they have in common, then the probability of yielding a tie is $$P_1(5)\cdot P_2(5) + P_1(6)\cdot P_2(6)$$.
Likewise, you can find all the pairs of numbers where player $$2$$ is greater than player $$1$$, or vice-versa, and take the sum of the products of the pairs to find those probabilities, too.