Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
1  
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

2 Answers 2

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.

share|improve this 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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.