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    <p>This can be done with dynamic programming, I'm not sure if there is a better method as the games are independent.</p> <p>Have a 4-D array, of wins, losses, ties, and games. You can limit wins/losses/ties to the number you want (let these be W, L, T, W+L+T=G) , time complexity will be O(W*L*T*G), which is bounded by O(G⁴).</p> <p>The algorithm is basically:</p> <pre><code>A[][][][] = new double[G+1][W][T][L] // A[g][w][t][l] is the probability of have w wins, t ties, l losses // after g games. This can be computed from A[g-1]. // Let P[g][o] be the probability of outcome o for game g //everything else is initially 0. A[0][0][0][0] = 1 for g=1..G for w=0..W for t=0..T for l=0..L A[g][w][t][l] = A[g-1][w-1][t][l]*P[g][win] // assume out of bounds +A[g-1][w][t-1][l]*P[g][tie] // reference returns 0 +A[g-1][w][t][l-1]*P[g][lose] return A[G][W][T][L] </code></pre> <p><strong>edit)</strong></p> <p>We can do this in O(W*L*T*G/max(W,L,T)), i.e. O(G³). Note that if we have W wins and T ties after G games, then we must have L losses.</p> <pre><code>// we should pick the conditions we loop to be the smallest two. // here we just use wins and ties. A[][][][] = new double[G+1][W][T] A[0][0][0] = 1 for g=1..G for w=0..W for t=0..T A[g][w][t] = A[g-1][w-1][t]*P[g][win] // assume out of bounds +A[g-1][w][t-1]*P[g][tie] // reference returns 0 +A[g-1][w][t]*P[g][lose] return A[G][W][T] </code></pre> <p>Maybe it's possible to do this significantly faster, by computing the probabilities of x wins/ties/losses separately (O(G)), and then adding/subtracting them intelligently, but I haven't found a way to do this.</p>
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    1. POEfficient Method for Calculating the Probability of a Set of Outcomes?
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    1. POEfficient Method for Calculating the Probability of a Set of Outcomes?
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    1. COThis sounds promising, but can you elaborate a little bit more on your algorithm? I'm not sure I follow how to implement it. Thanks! :-)
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    2. CObeautiful, this is exactly what I was looking for! thanks :-)
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    3. COI don't see a reason for doing it in O(N^3). See my response. Am I missing something?
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