Updated

I was doing Risk calculations and got a probability greater than 1, so something was wrong but I don't feel like messing with it now. Also a pretty poor lecture episode of SVU. On the bright side, that country song is almost done.

Update: I got the prob. for 2-2 dice rolls. At least I think so. I think the problem was with the summations I was using, as it involved a double summation, which is notoriously easy to mess up, so I did the summation more or less by hand and got some results that seemed reasonable to me. I could theoretically check my work experimentally, but since the probabilities I found unsurprisingly have denominators of 6^4 (there are four dice involved, after all), checking would be rather tedious.

The real problem is moving on to 3-2. The method can be generalized to any number of dice, which should come as no surprise, but it would be much easier if I could get the summations to work out. Generalization isn't even very hard once you have the mechanism for "choosing" the higher dice, especially if you stick to only the top two dice mattering. Actually, in trying to work out how to do it, I was thinking about dice with fewer sides so that I could check my work along the way. I got to thinking what a two-sided die would look like, and then I realized it is a coin.

So, some notions about that. When I say that the probabilities I found seemed reasonable, I mean that they met with the intuitive (and experimental by playing) notion that the defense has a slight edge (due to ties going to the defender) and that it is most likely to end with winning one and losing one as opposed to winning or losing two, but that none of the probabilities are especially high or low (nothing like 90%).

If we look at our theoretical two-sided dice, we see that each person, defender and attacker, can throw one of four possibilities, equally likely; 1,1; 1,2; 2,1; 2,2. Let's take those to be the defender's dice and look at the likelihood given each of the attacker losing two [this is my method].

1,1. If the attacker throws 1,1, he loses two by tying, but other cases yield at least one win.

1,2. If the attacker throws 1,1; 1,2;or 2,1 he loses two since we pair the highest dice and ties go to the defender.

2,1. This is the same as the previous case, 3/4 cases mean a loss of 2.

2,2. In all cases, the attacker loses two.

As I mentioned, each of the defender's throws are equally likely, so we can say the probability of losing two is

P(lose 2) = 1/4 (1/4 + 3/4 + 3/4 + 1) = 11/16.

An astute reader might note that since P(lose 2|(n,m)) = P(lose 2|(m,n)). One might also note that then we can simplify the process by combining the cases (n,m) and (m,n) by assuming n is greater than or equal to m and double its probability of occurring. This only works for when n != m since, for example, there are two ways of throwing (5,3), that is, a five and then a three or a three and then a five, but only one way of throwing a (4,4). Those probabilities work out to 2/N^2 and 1/N^2 respectively if N is the number of sides on the die.

Anyway, what I wanted to talk about was that in the case of N = 2, we see the defense has a huge advantage. You can work out that P(win 2) = 1/16, so P(win 1) = 1/4. The probabilities aren't so skewed if you use standard six-sided dice. It would seem that if you increase N, you lessen the defender's advantage, which makes sense if you think about it, because the advantage only comes into play given a tie. If people are throwing 20 sided dice, for example, the chance of a tie is pretty low, so the advantage is basically negated. We wouldn't expect it to change much though, that the most likely event is a "tie" of win 1, lose 1 because even if we stretch out the number of possible rolls, the probability of both being high or low is low. You can think of it like an area problem, I suppose. It's kind of hard to explain the mental picture here.