# Math Question - 3 Outcome Probability **URL:** \ **Category:** Factual Questions\ **Created:** [December 6, 2008, 11:09am UTC](https://boards.straightdope.com/t/math-question-3-outcome-probability/475986 "2008-12-06T11:09:34Z")\ **Posts on this page:** 11\ **Page:** 1
**Author:** ![gks1](https://avatars.discourse-cdn.com/v4/letter/g/22d042/32.png) [@gks1](https://boards.straightdope.com/u/gks1)\ **Post date:** [December 6, 2008, 11:09am UTC](https://boards.straightdope.com/t/math-question-3-outcome-probability/475986/1 "2008-12-06T11:09:34Z")
Hi, I was wondering which formula to use for an event with 3 possible outcomes. The probability of each outcome for a single event can be expressed this way: A=47.5% B=47.5% C=5% For instance, a sporting event with two evenly matched teams where Team A wins 47.5% of the time, Team B wins 47.5% and they tie 5% of the time. As an example, what is the probability of Team A winning 2 games out of 16 (no ties)? Thanks for the help! ---
**Author:** ![Student\_Driver](https://avatars.discourse-cdn.com/v4/letter/s/3da27b/32.png) [@Student\_Driver](https://boards.straightdope.com/u/Student_Driver)\ **Post date:** [December 6, 2008, 11:38am UTC](https://boards.straightdope.com/t/math-question-3-outcome-probability/475986/2 "2008-12-06T11:38:39Z")
The easiest method I can think of is a binomial probability distribution; your 3-outcome trial can be treated as 2-outcome for your example, since you’re interested in one of two things-- whether A wins, or whether A does not win. (You’ll have to look up the distribution and other stuff on your own, though-- I’m doing statistics homework myself and getting ready for a final, so I find myself wary at the possibility of doing someone else’s homework!) ---
**Author:** ![Outpits](https://avatars.discourse-cdn.com/v4/letter/o/4bbf92/32.png) [@Outpits](https://boards.straightdope.com/u/Outpits)\ **Post date:** [December 6, 2008, 1:04pm UTC](https://boards.straightdope.com/t/math-question-3-outcome-probability/475986/3 "2008-12-06T13:04:43Z")
The question of the OP is a little ambiguous. Are you stating as a fact that there are no ties in the 16 tries, in which cases, its a simple 50:50 situation? Or are you saying, as the previous poster said, that it is a 47.5/52.5 situation in which A wins or does not win (losses plus ties)? ---
**Author:** ![Student\_Driver](https://avatars.discourse-cdn.com/v4/letter/s/3da27b/32.png) [@Student\_Driver](https://boards.straightdope.com/u/Student_Driver)\ **Post date:** [December 6, 2008, 1:14pm UTC](https://boards.straightdope.com/t/math-question-3-outcome-probability/475986/4 "2008-12-06T13:14:51Z")
Oh, didn’t notice that quirk of the OP’s question, and just assumed “we don’t count ties as times that A wins”-- you’re right, that wording is ambiguous. Reminds me of a poorly-worded problem in my current stat class where the teacher and I went around for almost an hour… it boiled down to a single instance of the word “the” in the problem, and how one interpreted it. ---
**Author:** ![gks1](https://avatars.discourse-cdn.com/v4/letter/g/22d042/32.png) [@gks1](https://boards.straightdope.com/u/gks1)\ **Post date:** [December 6, 2008, 1:29pm UTC](https://boards.straightdope.com/t/math-question-3-outcome-probability/475986/5 "2008-12-06T13:29:28Z")
Thanks for your reply. SD. I should have mentioned it but I’m not a student, I finished college over 20 years ago 🙂 I tried the Binomial Probability Formula [http://www.mathwords.com/b/binomial\_probability\_formula.htm](http://www.mathwords.com/b/binomial_probability_formula.htm) and it seems that whatever numbers I plug in I get the same result. For instance, with my above example of 2 wins and 14 losses (let’s ignore ties for the moment) the equation works out to (.5)^2 x (.5)^14 = 0.000015292 or 0.001529% chance. This makes sense to me as one of the two evenly matched teams shouldn’t dominate the other so. If I use 8 wins and 8 losses I get the same result. If I’m not mistaken there are 17 possible outcomes of the 16 game series (ignoring ties): Team A wins the series 16-0 15-1 14-2 13-3 12-4 11-5 10-6 9-7 for 8 combinations (or Team A loses for another 8 combinations) plus an 8-8 split. If there are only 17 combinations how can each result only have a 0.001529% chance? It just doesn’t seem right that the odds of a series ending 14-2 is the same as the odds of it ending 8-8 between two evenly matched teams. What am I doing wrong? I’m sure I’m having a brain fart somewhere but I can’t see it. ---
**Author:** ![gks1](https://avatars.discourse-cdn.com/v4/letter/g/22d042/32.png) [@gks1](https://boards.straightdope.com/u/gks1)\ **Post date:** [December 6, 2008, 1:36pm UTC](https://boards.straightdope.com/t/math-question-3-outcome-probability/475986/6 "2008-12-06T13:36:30Z")
I apologize for the poorly worded OP. My brain is mush and I’m having a hard time explaining myself. The question is: given the three possible outcomes what are the odds of it _exactly_ coming out as 2-14? ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [December 6, 2008, 2:12pm UTC](https://boards.straightdope.com/t/math-question-3-outcome-probability/475986/7 "2008-12-06T14:12:23Z")
> [@gks1](#): > > If I’m not mistaken there are 17 possible outcomes of the 16 game series (ignoring ties): > > Team A wins the series > 16-0 > 15-1 > 14-2 > 13-3 > 12-4 > 11-5 > 10-6 > 9-7 > > for 8 combinations (or Team A loses for another 8 combinations) plus an 8-8 split. If there are only 17 combinations how can each result only have a 0.001529% chance? Because those 17 possible outcomes are not equally likely. There many ways the series can end up 9-7 – AAAAAAAAABBBBBBB, BBBBBBBAAAAAAAAA, ABABABABABABABAA, etc. etc. – but there is only one way it can end up 16-0. In your binomial formula calculations, you are omitting the number-of-permutaions element (the pair of numbers in parentheses), which accounts for the difference in the number of possible ways to reach an outcome. ---
**Author:** ![gks1](https://avatars.discourse-cdn.com/v4/letter/g/22d042/32.png) [@gks1](https://boards.straightdope.com/u/gks1)\ **Post date:** [December 6, 2008, 2:20pm UTC](https://boards.straightdope.com/t/math-question-3-outcome-probability/475986/8 "2008-12-06T14:20:06Z")
I knew I was missing something! Thank you, Ximenean, I understand now. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [December 6, 2008, 2:21pm UTC](https://boards.straightdope.com/t/math-question-3-outcome-probability/475986/9 "2008-12-06T14:21:59Z")
Let’s go with a five game series, to make things simpler for a starting point. There’s one chance in 32, or around 3 percent, of any particular outcome, but that’s for a specific pattern of wins and loses in each numbered game, for instance: “Player A wins every game but game 5, which B wins” and “Player A wins every game but game 2, which B wins” are different outcomes at this level If all that you’re interested in is the final series score, then as **Ximenean** noted, you need to look at permutations, or conversely, to look at all 32 possible outcomes and count up the wins: 1 sequence where A wins 5 times 5 sequences where A wins 4 times 10 sequences where A wins 3 times 10 sequences where A wins twice 5 sequences where A wins once 1 sequence where A does not win at all. This is actually the 6th row off of Pascal’s triangle, and you should be able to get the relative probabilities for your 16-game series off the 17th row of the triangle. Hope that that helps. ---
**Author:** ![Student\_Driver](https://avatars.discourse-cdn.com/v4/letter/s/3da27b/32.png) [@Student\_Driver](https://boards.straightdope.com/u/Student_Driver)\ **Post date:** [December 6, 2008, 2:40pm UTC](https://boards.straightdope.com/t/math-question-3-outcome-probability/475986/10 "2008-12-06T14:40:19Z")
For the binomial distribution, it looks like you’re leaving out the first portion, which is (n choose y), which looks like a set of parentheses with n above y on the page link you gave. n choose y is calculated using factorials (!)-- 4! is 4 \* 3 \* 2 \* 1, 5! is 5 \* 4 \* 3 \* 2 \* 1, etc.-- and is n! / (y! \* (n-y)!). That will give a similarly small percentage for 2 wins, a more realistically large percentage (19.25%) for 8 wins, but all assuming ties can occur in the other 8 games. For no ties at all, my first thought is this: calculate the probability of a run of 16 games with no A/B ties (0.95 ^ 16) which is about 44%, multiply with a binomial distribution where probability of A or B winning is 0.5 (since, in the event that there are no ties, A and B are evenly matched each game and has 50% likelihood of winning any given game). Still a small percentage (0.08%), though I note that a situation where no ties have occurred in individual games, and A and B each have won 8 games, around 8.6% probability. Then again, my brain is mush too, so I may be wrong. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [December 6, 2008, 4:33pm UTC](https://boards.straightdope.com/t/math-question-3-outcome-probability/475986/11 "2008-12-06T16:33:48Z")
The binomial distribution works here because there’s really only one interesting event, but in general for problems like this, you want the [multinomial distribution](http://en.wikipedia.org/wiki/Multinomial_distribution).