The Probability one Player will have more Kills than another based on a distribution of Kills? Alright, I'm definitely not a math guy so bare with me. I'll make this short and simple. I have a dataset of players and the # of kills (video game) they have per game. For instance, if there are 10 possible kills in a game it looks like...
Number of Kills | %-Player 1 | %-Player 2
0--------------10%-----------5%
1--------------15%-----------12%
2--------------20%-----------15%
etc, etc, etc
What I'd like to do is come up with an equation that says Player 1 has what % probability of having more kills than Player 2 based off this data. 
I can't figure it out. So if player 1 has 2 kills, Player 2 has a 17% chance of having less than 2 and a 68% chance of having more than 2, and a 15% chance of having the same. But how do you quantify that for the overall % chance of having less/same/more? 
Any help will be appreciated. Thanks in advance.
 A: Denote by $X_1$ (respectively $X_2$) the random variable representing the number of kills of player $1$ (respectively player $2$).
We assume that $X_1$ and $X_2$ are independent from each other, i.e. they are playing in different games.
Then
$$
\mathbb P\left(X_1\le X_2\right)
=\sum_{k=0}^{10}\sum_{i=k}^{10}\mathbb P\left(X_1=k\right)\mathbb P\left(X_2=i\right),
$$
and you can also compute
$$
\mathbb P\left(X_1<X_2\right)
=\sum_{k=0}^{9}\sum_{i=k+1}^{10}\mathbb P\left(X_1=k\right)\mathbb P\left(X_2=i\right).
$$
To compute the above quantities, simply plug in the values in your table, i.e. $\mathbb P\left(X_1=1\right)=0.1$, $\mathbb P\left(X_2=1\right)=0.05$, etc.
A: If P2 has $0$ kills, then P1 needs $1$ kill or more.
The probability of P2 getting (exactly) no kills is $0.05$.  The probability that P1 gets $1$ kill or more is $0.9$.  So the probability that P1 gets $1$ kill or more, and that P2 got exactly $0$ kills, is $0.05 \times 0.9 = 0.045$.
Next, if P2 has (exactly) $1$ kill, then P1 needs $2$ kills or more.
The probability that P1 gets $2$ kills or more, and that P1 got exactly $1$ kill, is $0.12 \times 0.75 = 0.09$.
Repeat this calculation for P1 getting $3$ or more kills and that P2 got exactly $2$ kills, for P1 getting $4$ or more kills and that P2 got exactly $3$ kills, etc., and add them all up.
