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Two chess players, A and B are going to play 7 games. Each game has three possible outcomes: a win for A (which is a loss for B), a draw (tie), and a loss for A (which is a win for B). A win is worth 1 point, a draw is worth 0.5 points, and a loss is worth 0 points.

How many possible outcomes for the individual games are there, such that A ends up with 4 points and B ends up with 3 points?

Source: Introduction to Probability

There are four possible ways how A can get 4 points while B also gets 3 points:

  1. 1x Wins, 6x Draws, 0x Losses
  2. 2x Wins, 4x Draws, 1x Losses
  3. 3x Wins, 2x Draws, 2x Losses
  4. 4x Wins, 0x Draws, 3x Losses

I calculated the following combinations:

  1. ${7 \choose 1}{6 \choose 6}{0 \choose 0} = 7$
  2. ${7 \choose 2}{5 \choose 4}{1 \choose 1} = 105$
  3. ${7 \choose 3}{4 \choose 2}{2 \choose 2} = 210$
  4. ${7 \choose 4}{3 \choose 0}{3 \choose 3} = 35$

From a former question I believe that I am over counting in 2. and 3. so that the correct answer should be:

  1. $\dfrac{{7 \choose 2}{5 \choose 4}{1 \choose 1}}{3!} = \dfrac{7!}{3! 2! 4! 1!} = 17.5$
  2. $\dfrac{{7 \choose 3}{4 \choose 2}{2 \choose 2}}{3!} = \dfrac{7!}{3! 3! 2! 2!} = 35$

However 17.5 does not look like a valid result, as we can't have non-integers for combinations.

What approach is correct and why does dividing with 3! not work out?

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bodokaiser
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  • The method for all four is basically correct, and the numbers $7,105,210,35$ are right. The end terms are in each case $1$ so unnecessary. For 4) your expression should be $\binom{7}{4}\binom{3}{0}$, Or you could use the Multinomial Formula, and write $\frac{1}{7!0!3!}$. There was no overcounting, so the attempted modifications are not correct. – André Nicolas Mar 30 '15 at 16:46
  • @AndréNicolas Why does 3. and 4. not over count? When applying the multinomial formula shouldn't I need to divide by 3!? – bodokaiser Mar 30 '15 at 16:51
  • Let's see why you don't divide by 3!. For 2), we are counting the number of permutations of {win, win, draw, draw, draw, draw, loss}. Then use that method in the problem that you link to and you'll see that you get $\frac{7!}{2! \ 4! \ 1!}$ without the 3!. – D Poole Mar 30 '15 at 16:54
  • The number of losses in 4 is incorrect. It should be $3$. However, the final term does not affect your answer. Once you have selected the wins and draws, the number of ways to select the losses will always be equal to $1$. – N. F. Taussig Mar 30 '15 at 16:55
  • I had a "typo" for 4) Multinomial gives $\frac{7!}{4!0!3!}$. Still $35$. We are counting the words of length $7$ in the alphabet W, D, L that have certain specified numbers of each letter. For $2$ wins and $4$ draws and $1$ loss, there are $\binom{7}{2}$ ways to place the W's, and for each there are $\binom{5}{4}$ ways to place the D's. – André Nicolas Mar 30 '15 at 17:00
  • But for example if we want to form 3 teams from 12 people then we have $\dfrac{12!}{3! \cdot (4!)^3}$ - we also divide by 3! because the order of the teams does not matter – bodokaiser Mar 30 '15 at 17:04
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    What ordering do you think that you are getting rid of by dividing by 3! in this problem? – D Poole Mar 30 '15 at 17:13
  • The case where we would have $(W, W, D, D, D, D, L)$, $(D, D, D, D, W, W, L)$ or $(L, W, W, D, D, D, D)$ and so on – bodokaiser Mar 30 '15 at 17:16
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    But those are different outcomes, we would want to count each of those outcomes once. – D Poole Mar 30 '15 at 17:29
  • Ah okay so in this case the order does matter. I see. Do you want to put this in an answer? – bodokaiser Mar 30 '15 at 17:42
  • As others have noted, your answers are correct. In 2., you are selecting which two of the seven games $A$ wins, then selecting which four of the five other matches result in a draw. Once those matches have selected, the matches that $A$ loses have already been determined. By selecting which matches $A$ wins and which matches result in a draw, you are determining the order of wins, draws, and losses. The number of solutions is the number of ways $A$ can win twice, draw four times, and lose once. – N. F. Taussig Mar 31 '15 at 10:29

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