Originally Posted by
TennTradition
There may be a more clever formula for this, but the fact that each game can have a different probability takes away the greatest opportunities for really simplifying it. Regardless, it isn't hard, just a little cumbersome.
Basically to know the odds of winning x number of games out of y total games, you would multiply two quantities:
1) The odds of winning all games
by
2) The sum of the likelihood of losing all the games you need to lose divided by the likelihood of winning those games for each combination that satisfies your criterion of losing y - x games.
That second quantity might not be totally clear so I'll go into more detail. Basically if you have four games and you want to know the probability of winning three of them, you first enumerate the easy you van win three. In this case that is easy. You either lose game 1, 2, 3, or 4. So that second term would be the prob of losing game 1 divided by the prob of winning game 1 plus the prob of losing game 2 divided by the prob of winning game two plus the same for game3 plus the same for game 4. If it were the probability of winning two games out of four, now the games you need to lose come in combinations. There are 6 of them. For example, you could lose games 1 and 2. So the contribution to the sum in part 2 above for that particular combination would be (prob of losing 1*prob of losing 2)/(prob of winning 1*prob of winning 2). There would be a term in the sum for each of the six combinations of games you must lose to satisfy the condition of winning 2 of four games.
To write out the math, lets say we have a 4 game season and we want to know the probability of winning three games.
First, how many combinations is that. The answer is 4 choose 3, which is 4.
If the probability of winning game 1 is w1 and the probability of losing is l1, then the probability of winning three of four games is:
P3(4) = w1w2w3w4 * ( l1/w1 + l2/w2 + l3/w3 + l4/w4 )
On the other hand if we want to know the probability of winning 2 of 4 games, then the number of combinations is 4 choose 2, or 6. We can enumerate those as losing the following combinations, 1,2 ; 1,3 ; 1,4 ; 2,3 ; 2,4 ; 3,4. The probability becomes:
P2(4) = w1w2w3w4 * ( l1l2/w1w2 + l1l3/w1w3 + l1l4/w1w4 + l2l3/w2w3 + l2l4/w2w4 + l3l4/w3w4 )
Now, if your question is what is the probability of winning at least x games of y games, then you would just use:
P of winning at least x games of y is equal to: Sum from i = x to i = y of Pi(y)
Or, using the previous example, the probability of winning at least 2 is: P2(4) + P3(4) + P4(4).
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To give actual numbers, if there are four games and the the probabilities of winning are .2, .8, .3, and .9, then :
The prob of winning two games is 50.2%.
The prob of winning three games is 28.9%.
Note that this form should work for all numbers of games.
If it is the prob of winning one game then the terms in the sum look like l1l2l3/w1w2w3 ... Three terms per combination because you would have to have three losses.
The prob of winning zero games becomes w1w2w3w4 * (l1l2l3l4/w1w2w3w4), which reduces to the the obvious l1l2l3l4.
The prob of winning 4 games is just w1w2w3w4 since there are no losing combinations to sum over in the second portion of the probability equation.
This should be fairly straightforward in a spreadsheet but I don't like that you have to enumerate all combinations. To make sure you get gem all use e following equation.
N choose r is n!/(r!(n-r)!).
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