Winning an asymmetric dice toss over multiple rounds

Question

I'm working on a (computer) game where some events will be resolved by dice rolls. Two players, A and B, will roll against each other, whomever rolls highest wins. In the case of a tie B always wins, because life is not fair.

I've reformulated the problem as follows:

• Player A rolls $k$ number of dice, each numbered $1$ through $x_k$. The score of A is the sum of each roll plus a multiplier $m$. The score is strictly positive.
• Player B does about the same, with $j$ dice, numbered $1$ through $x_j$, and multiplier $l$.
• Player A wins a round if their roll is strictly higher than B's (a tie means B wins).
• Player A wins the game if they win $n+1$ out of $2n+1$ rounds
• $x_k$ and $x_j$ values don't vary between rounds.

I'm looking for the calculation for:

1. The probability $p$ of A winning a single round, as a function of all the $x_k$ and $x_j$
2. The probability $q$ of A winning a game, as a function of $x_k$ and $x_j$ and/or $n$ and/or $p$

Answer 1

I don't think there's any reasonable closed-form function that will solve this problem, but there are algorithms that will calculate the probabilities for any number of dice of any number of sides.

For a single round, we can simplify the problem a little bit by reducing the results of all the rolls to just one variable. That is, starting from zero, add the dice rolled by A and the "multiplier" $m,$ then subtract the dice rolled by B and the "multiplier" $l.$ If the result is positive, A wins.

For example suppose A rolls four dice whose largest numbers are $4, 4, 6,$ and $13$ (where you get a fair $13$-sided die is another question, but let's suppose you can) with "multiplier" $6,$ and suppose B rolls three dice whose largest numbers are $4, 12,$ and $12,$ with "multiplier" $5.$ Then you can go to http://anydice.com/ and put this "program" in the input text window:

output 2d4+d6+d13+6-d4-2d12-5>0

Click "Calculate" and select View "Export" and Data "At Least". Look in the output window where you see the line

1,52.8669482283

That tells us that A has approximately a $52.8669482283\%$ chance of winning each round.

You can get a few more digits in the answer by omitting >0 from the "program." The output can get quite long in that case and you may have to scroll down to find the line that starts with $1.$ But the last few digits in that format are not quite correct, anyway; you can get an idea of how accurate this calculator is by selecting Data "At Most" and scrolling to the bottom of the output window. The probability there should be $100\%$ but it typically is not exactly $100$ for the examples I have tried.

If "multiplier" actually meant something to multiply the outcome by rather than something to add to it, the "program" is

output 6*(2d4+d6+d13)-5*(d4+2d12)>0

Another on-line calculator is http://topps.diku.dk/torbenm/troll.msp, but it works a little bit differently and I have not experimented as much with the programs there.

Once you have your probability for a single round, you can formulate the game as a binomial random variable $Y$ with parameters $2n+1$ and $p$ (where $p$ is the probability that A wins one round). For example, suppose there will be $25$ rounds ($n=12$). Then the probability that A wins is $P(Y \geq 13),$ which we can also calculate on-line:

https://www.wolframalpha.com/input/?i=prob+x%3E%3D13+for+x+binomial(25,0.528669482283)

Wolfram Alpha gives the answer $0.614021,$ that is, A has a $61.4021\%$ chance to win the game.

If you want to set up all these calculations yourself rather than use the on-line resources, a reasonably simple program will do it, or even a computer spreadsheet with appropriate formulas. I could go into more detail on that if that's desired.