# Efficient Method for Calculating the Probability of a Set of Outcomes?

Let's say I'm playing N different independent "games". For each game, I know the probability of winning, the probability of tying, and the probability of losing.

From these values, I've also calculated the probability of winning exactly X games, the probability of tying exactly X games, and the probability of losing exactly X games (for X = 0 to N).

I'm just trying to figure out the probability of each outcome after playing all N games. For example, if N = 10, what is the probability of winning 7, losing 2, and tying 1?

Any ideas, or a proof that this is impossible to compute efficiently?

-
Have you read the faq? I don't think this is a question of interest to research mathematicians. –  Gerry Myerson Oct 3 '10 at 3:51
you might try posting your question to stats.stackexchange.com they seem to deal with more basic stuff as well. [altho i'm not sure how "basic" you question actually is. i suppose with statistical software the answer for any particular outcome can be computed. maybe you want to ask about appropriate software for doing the required calculations. –  ronaf Oct 3 '10 at 4:02
Maybe you could edit the question so it asks for an efficient means of calculating the probability of each outcome (or a proof that there is no efficient way), if that's what you meant. –  Bjørn Kjos-Hanssen Oct 3 '10 at 6:16
@Bjørn: updated, thanks for the feedback! –  Kenny Oct 3 '10 at 6:22

-

Excel (or any spreadsheet) is your friend for this calculation. In rows 1-3 of each column you put the probabilities of a win, tie, and loss in that game. To ease the following, in rows 4-6 put in the probability of a nonwin, nontie, and nonloss in that game. Then in rows 7-7+11^3=1338 you put the probability of having won/lost/tied a number of games (0-10) so far. If you are clever in the organization, you can just copy down and right a lot. Certainly once you have the second column you can just copy right. Look in column J, row 712+6=718 for the answer.

-
Can you explain a little bit more about what would go in rows 7-1338? You can assume there are only 2 games to keep in simple. Therefore the possible outcomes are (W-L-T): (2-0-0), (1-1-0). (1-0-1-). (0-1-1). (0-2-0). (0-0-2) –  Kenny Oct 3 '10 at 4:49
Each row represents a number of games won, lost and tied. So I confused you because it should be 3^11=177477 rows instead of 11^3=1331. And Excel won't do that many. At the start, the probability of (0-0-0) is 1. Let pn(win) be the probability of winning game n. pn(loss) is the probability of losing game n. Then the probability of (1-0-0) is p(0-0-0)*p1(win). The probability of (1-1-0) is p(1-0-0)*p2(loss)+p(0-1-0)*p2(win). The message in problems like this is to lose the info you don't need. After game 5, you don't need to know which games were won,lost,drawn-just how many –  Ross Millikan Oct 4 '10 at 3:20