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I tried for a while to find a quick and elegant solution but failed. Finally, I made a brute-force count using excel. First, I should be clear about what I counted, which is the number of distinct results of the form "Player A got x1 points, B got x2, C got x3, D got x4." The number I came up with is 1285.
My method was as follows. There are 6 pairs of players. For each pair there are 5 possible results: 4-0, 3-1, 2-2, 1-3, 0-4. This makes for 5^6 = 15625 outcomes. For each outcome I computed the final tournament result, and then I counted the number of distinct results.
I feel pretty confident of my plan, but wouldn't be surprised if I made a mistake in the execution.
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... okay, I've already found a mistake in my execution. With that fixed I now get 1289, which is the same number that Blutigeroo came up with. Blu, can you elaborate a little on how you came up with that total?
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phystutor,
One possible error in your reasoning is here:
"Each player has 6 possible outcomes in his 3 games as white." If I understand what you mean by this, there are 7 possible outcomes: 0, 1, 2, 3, 4, 5, or 6 points.
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So OK guys, what exactly are we talking about here. So go over this one more time for us slow guys. Thanks
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phystutordotcom wrote:
| The problem is not well defined. |
Yes, but not the way you think. equus defined it as, "The different results that can occur in a group, for example: A - 12, B - 6, C - 4, D - 2." Thus, it just matters how many points each player gets and not, for example, how many they score with black and how many with white.
The only ambiguity is whether A stands for the winner of the group or just some player. In other words, are we just looking for sequences like 12-8-4-0, 12-8-3-1, 12-8-2-2, ... down to 6-6-6-6, or are all the permutations of which player gets which score also counted as separate solutions? I'm going to assume the latter.
| If winning with white and drawing with black is different than vice versa the answer is 3^12 = 513,441 |
Agreed but this isn't what's being asked for. (And it's actually 531,441 but that's just a typo.)
| If an outcome is determined only by the number of points each player has we get 6^4 = 1296 but some outcomes are more likely than others. |
No. Your method is incorrect and you've implemented it incorrectly, too. You say that each of the four players can score either 0, 1, ..., or 6 as white. That's seven options, so your answer should be 7^4=2,401, not 6^4. But there's a huge amount of double-counting here because a player's score with white does not uniquely determine his total score. In particular, as I said above, a player gets the same score if he beats an opponent with white and loses with black as he does if he draws both games or wins with black and loses with white. So you can't count outcomes just by looking at scores with white.
If you don't believe me, consider the case where there are only two players in the group. By your argument, each player can score 0, 1 or 2 points as white, so the number of possible outcomes would be 3^2=9. But there are clearly only five possible outcomes: 4-0, 3-1 (which your method counts twice), 2-2 (three times), 1-3 (twice) and 0-4.
I guess that, since two people have independently calculated the answer to be 1289, they're probably right. 
Dave.
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