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Alice and Bob play chess, if one of them can win three times then the other will be lost the game.
How many different results are possible between them on the chess game?

I guess there are 3 different conditions,
1. one of them achieve a triple-win : 2
2. one of them win 3 times and lost 1 time : 2 * C(3,1)=6
3. one of them win 3 times and lost 2 times : 2 * C(3,2)=6

So there are 14(2+6+6) possible results on the chess game.
Is it right? I hope someone can check it.thanks

[UPDATE]
Suppose draws don't happen.
Also, AABBA is the DIFFERENT result as ABABA.

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In chess, there are, all too often, draws. This gives infinitely many possibilities. But if we don't allow draws, it depends what you mean by result. Is AABBA the same result as ABABA? –  André Nicolas Apr 16 '12 at 16:02
    
I don't understand what the premise means. "If one of themc an win three times then the other lost the game". Are they playing a single game? Are they playing a series of matches? I'm just confused, I guess. –  Arturo Magidin Apr 16 '12 at 16:03

1 Answer 1

A draw is always possible in chess. So there are infinitely many possibilities. But let us assume, as presumably we are intended to, that a game always ends in a win by one of the players. We count the number of sequences of wins/losses that end with Alice winning the match, and multiply by $2$. For concreteness, it can be helpful to write A for "Alice wins" and B for "Bob wins."

There is only $1$ such $3$-game sequence, namely AAA. For reasons that will be clear later, I prefer to write $\binom{2}{2}$ instead of $1$.

Now let's count the $4$-game sequences that result in a win by Alice. So the $4$-th game must be won by Alice, the $4$-th and final letter of the sequence is an A. Of the first $3$ games, exactly $2$ were won by Alice. The location of these $2$ games can be chosen in $\binom{3}{2}$ ways.

Finally, let's count the $5$-game sequences that result in a win by Alice. She won the $5$-th game, so the $5$-th and final letter of the sequence was an A. Of the first $4$ letters, exactly $2$ were A's. The location of these $2$ A's can be chosen in $\binom{4}{2}$ ways.

Thus the total number of possibilities is $$2\left(\binom{2}{2}+\binom{3}{2}+\binom{4}{2}\right).$$

Remark: In the baseball World Series, and in the hockey Stanley Cup finals, the first team to win $4$ games wins the series. By reasoning very similar to the one above, if A and B are playing, there are $$2\left(\binom{3}{3}+\binom{4}{3}+\binom{5}{3}+\binom{6}{3}\right)$$ possible outcomes.

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