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    <p>For your example, you need to consider the possible ways that the outcome can occur.</p> <p>For win 7, lose 2, tie 1. There are <code>10! / (2!*7!)</code> or 360 possible ways. So multiply all the outcomes as you did, then multiply by that many permutations of the outcomes.</p> <p>For all wins, you can just multiply because there's exactly one permutation of ten wins. For a mix, you need to consider the permutation.</p> <p>In general for this problem the permutations will be <code>10!/(w!*l!*t!)</code> where w is number of wins, l is number of losses, and t is number of ties.</p> <p><strong>Edit 1</strong> Note that the above only indicates how to count the permutations. The total probability is the number of permutations times (pw^w*pl^l*pt^t) where pw is probability of a win, pl a loss, pt a tie. w, l, and t, are the counts of each.</p> <p><strong>Edit 2</strong> OK, in light of the new information, I don't know of a general way to do this. You'll have to individually computer each outcome by hand and add them together. With your two-game example above. If you want to find the probability of 1 win and 1 tie, you'll have to find every possible way of getting exactly 1 win and exactly one tie (there are only two) and add them up.</p> <p>For ten games with the initial example, you'll have 360 outcomes that meet your criteria. You'll have to do each permutation and add up the probabilities. (wwwwwwwllt, wwwwwwwltl, etc) Unfortunately, I don't know of a better way to do this.</p> <p>Further, in your two-game example, for one win and one tie, you must add the probability of winning the first game and tying the second to the probability of tying first, then winning.</p> <p>So, there are nine independent outcomes:</p> <pre><code>W W W T W L T W T T T L L W L T L L </code></pre>
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    1. POEfficient Method for Calculating the Probability of a Set of Outcomes?
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    1. COHere's the problem. Let's consider the case of 10 wins, 0 losses, and 0 ties This probability of going 10-0-0 SHOULD be the same as the probability of getting 10 wins that I've already calculated (since there is only 1 way to get 10 wins). However, your formula would have me multiply this probability TIMES the probability of getting 0 losses and then multiply again times the probability of getting 0 ties (which will both be extremely small numbers)
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    2. COAh, not quite. My equations here are only to determine the number of permutations. The w, l, and t, are no probabilities. I should expand on that, but my intent is to calculate the probability he did (and you have as O in your first equation), then multiply that number by the total count of permutations. So, you'd take O (exactly as you have it, I agree completely), then multiply by 360. In the case of ten wins, 0 losses, and 0 ties, there is exactly one permutation, so you'd multiply by one (10!/(10!*0!*0!) = 1).
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    3. COright but in that case, wouldn't O be too small? The probability of going 10-0-0 and is just probability of getting 10 wins... not the probability of getting 10 wins TIMES the probability of getting 0 losses TIMES the probability of getting 0 ties.
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