In a game of Risk, the rules are: The rules for determining how many dice a player may roll: 1. The attacker may roll as many dice as the number of attacking pieces he is using, to a maximum of 3. If A>3, the attacker continues to roll 3 dice per turn. 2. The defender may roll as many dice as the number of armies on his country, to a maximum of 2. If D>2, the defender continues to roll 2 dice per turn. The rules for deciding the outcome of a particular throw of the dice are as follows: 1. The highest attacker die is compared against the highest defender die. Whoever has the lower number loses one army. Ties go to the defender. 2. The procedure is repeated for the second-highest dice, where both attackers and defenders have a second die. If you attack a territory defended by D armies with A armies, what is the probability that you will capture this territory? In such a scenario, how many armies should you expect to lose (whichever side that will prevail)? Previously I had thought we needed a recursive solution, and reached one with the help of some members on this forum. But now I wonder if a closed-form solution is possible. Take a look at this report: http://web.archive.org/web/20060919204627/http://www4.stat.ncsu.edu/~jaosborn/research/RISK.pdf. As noted on page 2, A is the number of attacking armies and D is the number of defending armies. First of all, the probabilities of each outcome on a certain turn are given on page 5. These are in exact form. We don't need to worry about how they were obtained, we can just use them as they are now. What really interests us is the section entitled "The Probability of Winning a Battle", and what comes after that (pages 4 onwards). How do we express the function f(n)ij, found at the bottom of page 4, in closed form, in terms of the values of n, i and j? If we can figure this out, then, as noted half-way down page 5, fij is simply the sum of f(n)ij from n=1 to n=∞. And then, at the bottom of page 5, we perform a summation across all values of j. i=A*D apparently, or perhaps I am interpreting that wrong; it seems strange that a transition probability will be the same for the 4v5 case as for the 10v2 case. Anyway, if someone can give me help on how to reach the final 2 expressions on the bottom of page 5 in closed-form, in terms of A, D and the dummy variables, that would be a great start.