# The Probability one Player will have more Kills than another based on a distribution of Kills? [closed]

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.

## closed as off-topic by Harish Chandra Rajpoot, Seyhmus Güngören, Did, Claude Leibovici, user91500Jul 30 '15 at 8:43

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• What ends a game: finally making all 10 kills or getting timed out? Makes a huge difference in the approach. If it's total kills before timed out, perhaps the distribution of kills for any one player is (roughly) by a Poisson distribution. Then it would be easy to compare their likely number of kills in various situations. I would need to see the entire distribution for several players in order to see whether that approach is feasible. Failing that, how would you feel about just comparing the total number of kills for each player gets? – BruceET Jul 30 '15 at 3:08
• There is a time limit so 10 kills might not be reached. The percentages represent the % the player ends the game with 1 kill, 2 kills, etc. – Martavis Griffin Jul 30 '15 at 3:09
• OK then. I'd need to know total number of games, and entire distribution (no ...) for a few players. Just for verification, can you also give avg number of kills per game for each player? – BruceET Jul 30 '15 at 3:16

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.

• I think you mean and rather than given that. The probability that P1 gets $1$ kill or more given that P2 got exactly $0$ kills is in fact $0.9$ (assuming the kills are independent from each other). Unless I'm misunderstanding your explanation. – Ian Jul 30 '15 at 3:17
• You're absolutely right. Edited. Good catch! – John Jul 30 '15 at 3:38

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.