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I've been playing a board game called risk lately, and I'm very interested in the probability behind it. Sorry if this question is it the wrong category, but I assumed that it was largely statistical. Consider the following:

Suppose two people are playing a dice game involving troop exchanges called Risk. Each player (either attacker or defender) begins with a certain number of "troops" and rolls three dice. Depending on how many of one player's dice beats those of another (based on a special scoring system), each player can lose a fraction of their troops. A defending player has a 62% chance for a net win, while an attacking player has only a 37% chance. Specifically, a defending player has a 37.5% chance to take three of the attacking player's troops, while an attacking player has a 15.6% chance to do the opposite. There is a 25.5% chance that the attacking player will lose 2 troops while the defending one loses 1, and there is a 21.4% chance that the attacking player loses 1 troop while the defending player loses 2.

I've gotten this far, through a very long program, but I can't figure this out: How many troops should the attacking player have in comparison to the defending player to make it fair, or so that there should be 0 players left on both sides in theory?

Thanks!

bifflet
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    Possibly related? [The “Risk” game dice problem](http://stats.stackexchange.com/questions/1424/the-risk-game-dice-problem?rq=1) ... if your question is substantively different please make those differences clear. – Glen_b Jul 27 '15 at 04:42
  • The question about a closely related Risk-like game at http://stats.stackexchange.com/questions/91814 includes working code to answer questions like this and many related ones. – whuber Jul 27 '15 at 14:04

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