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!
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!)
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)?
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.
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 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.
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?
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.
I knew I was missing something! Thank you, Ximenean, I understand now.
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.
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.
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.