Monday, 23 January 2012

Tabletop Games and the Odds

I have been thinking for a while about probability in tabletop role-playing games. Mechanics that determine actions influenced by chance make for interesting times, but a few things recently have had me thinking seriously about what the odds are of things happening in games. The first thing that has taken my thoughts in this direction are the influences of the OU module that I took on games (mostly on computer games, but there was a good chunk of time spent on the philosophy of games). The second was playing Apocalypse World.

In Apocalypse World, virtually every action that has a chance element is determined by rolling 2d6. Without any modifier, one needs a 7 to get a "hit" (at least some kind of success), and a 10 or better to get a high hit. There are 11 outcomes (2, 3, ..., 11, 12) for rolling 2d6. The probability of getting a particular outcome is not evenly distributed though - you have to get 1 on both dice to get an outcome of 2, but there are six ways of making your required 7 minimum.

Without modifiers, any chance situation has a 21 in 36 chance of working out (at least somewhat) in the player's favour. More than half, actually, 7 in 12. This makes for quite a balanced sort of game mechanic (I think). But by the end of the campaign my character had a +3 modifier for whenever he tried to do something involving aggression or violence (my character's class was "Gunlugger"). So then from a 4 onwards he was going to get a "hit" - maybe not totally get his whole way on a situation, but he would be successful. What does that mean? Avoid a 2 and a 3 and I'm home free! (well, OK, more or less). Suddenly I have an 11 out of 12 chance of some success. Quite a jump. And the chance of me getting a high hit is more than 50/50.

So OK, enough of all that: this has lead me to thinking that there must be a simple way of qualifying the odds of getting X or more on 2d6. Or Y on 3d8. Or Z on AdN (arbitrary integer A, N (and X, Y, Z)). It is simple enough to draw some tables, noodle some numbers down and do it case-by-case. But my gut says that there is a formula. If this screams "Obviously!" then for now please don't tell me, I want to see if I can get this from working some examples and then generalise. More to follow on this, when I have time to work it out.

1 comment:

NathanRyder said...

OK, for those in the know, who have read what I've said, a thought occurs: counting the number of permutations of Z of length A such that the parts of the permutations are at most length N. That has to be a starting point. Now to dig out my old combinatorics notes...