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I am trying to solve this problem.

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.

(a) How many possible outcomes for the individual games are there such that the player ends up with 3 wins, 2 draws and 2 losses?

(b) 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?

My attempt:

(a)This is the same as the number of permutations of the WWWDDLL. There are $7\choose{3}$ ways to decide where to put the W, then $4\choose2$ ways to decide where to put the D and $2\choose2$ ways to decide where to put the L.

This gives ${7\choose3}\cdot{4\choose2}\cdot{2\choose2}=\frac{7!}{3!2!2!}=210$ possibilities.

(b) A must end up with 4 points, and by symmetry B must end up with three points.

We are then interested to find the number of non-negative solutions to the equation

$1\cdot{x_{1}}+0.5\cdot{x_{2}}+0\cdot{x_{3}}=4$

This amounts to having 3 distinguishable boxes, into which 7 indistinguishable particles can be put. Each particle can be either a win, a draw or a loss.

For example,

 W  D  L

|***|**|**|

There are 7+3-1=9 slots and we need to select 2 out of the 9 slots, where two |'s can go. But, a draw counts only as 0.5 points. So, I don't think this yields the correct answer.

It would be nice, if someone could verify part (a) and help with part(b).

Edit 1

There is a also a part (c) to the question.

(c) Now assume that they are playing a best-of-7 match, where the match will end when either player has 4 points or when 7 games have been played, which is first. For example, if after the 6 games, the score is 4 to 2 in favor of A, then A wins the match and they do not play a 7th game. How many possible outcomes for the individual games are there, such that the match lasts for 7 games and A wins by a score of 4 to 3?

My attempt:

c) If the match is to last till the 7th game and A to win by a score of 4 to 3, the 7th game should not be a loss for A. Thus, we have

  • 4 wins, 3 loss (WWWLLW) - ${7\choose4}\cdot{3\choose3}-{6\choose4}\cdot{2\choose2}$
  • 3 wins, 2 draws, 2 loss (WWWLLDD) - ${7\choose3}\cdot{3\choose2}-{2\choose2}\cdot{2\choose2}$
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    $\begingroup$ The first is fine. For the second, let's work with B, since $3$ is smaller than $4$. Could be $6$ draws or $4$ or $2$ or $0$. Count all separately. For example, $4$ draws is for reasons you are familiar with $\binom{7}{4}\binom{3}{1}$. $\endgroup$ – André Nicolas Jul 17 '16 at 10:21
  • $\begingroup$ @AndréNicolas , so its the sum of 1) DDDDDDW - (7 Choose 1) x (1 Choose 1) 2) DDDDWWL- (7 Choose 4) x (3 Choose 1) x (1 Choose 1) 3) DDWWWLL - (7 Choose 2) x (5 Choose 3) x (2 Choose 2) 4) WWWWLLL - (7 Choose 4) x (3 Choose 3) $\endgroup$ – Quasar Jul 17 '16 at 11:28
  • $\begingroup$ Yes, the case numbers are right, then we compute and add up. The $\binom{1}{1}$, $\binom{2}{2}$, and so on are all $1$, so unnecessary though harmless. There are other more "general" kinds of approaches, for example using generating functions, but for small numbers like these the approach you carried out is efficient. The answer by Shagnik uses essentially the same sort of case analysis. In the answer you wrote out, you forgot to add up at the end, $\endgroup$ – André Nicolas Jul 17 '16 at 12:57
  • $\begingroup$ @AndréNicolas, is the answer to part(c) also correct? $\endgroup$ – Quasar Jul 18 '16 at 1:43
  • $\begingroup$ I had not seen c). There are a lot more cases than you indicate. Sure we can divide into last is win and last is draw. But last is win can happen in several ways: (i) Your $3$ losses and $3$ wins in first $6$. There are $\binom{6}{3}$ of these. (ii) $2$ draws, $2$ wins in first $6$. Number of ways $\binom{6}{2}\binom{4}{2}$. (iii) $4$ draws, $1$ win in first $6$, $\binom{6}{4}\binom{2}{1}. (iv) $6$ draws in first $6$, $1$ way. Then there are the ones that end in a draw, again many cases. Add up. $\endgroup$ – André Nicolas Jul 18 '16 at 2:16
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Part $(a)$ is correct

For part $(b)$ easier to count for $B$ (only $3$ points), and count only wins and draws

$3W \;or\; 2W,2D\;or\;1W,4D\;or\;6D \to \binom73 + \binom7{2,2} + \binom7{1,4}+\binom76$

For part $(c)$, parallel approach to $(b)$ considering $3$ points in six games ($7_{th}$ is a must win)

$\binom63 + \binom6{2,2} + \binom6{1,4}+\binom66$,

and also $3.5$ points in $6$ games with the $7_{th}$ a must draw:

$3W,1D \;or\; 2W,3D\;or\;1W,5D \to \binom6{3,1} + \binom6{2,3} + \binom6{1,5}$

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  • $\begingroup$ blue nail, 7th could also be a draw, so I also need to add those up, right? $\endgroup$ – Quasar Jul 18 '16 at 7:06
  • $\begingroup$ Quite right, I have added ! Btw, true blue anil. $\endgroup$ – true blue anil Jul 18 '16 at 7:31
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Your solution to part (a) is correct.

For part (b), there are two equations that need to be satisfied: $x_1 + x_2 + x_3 = 7$ and $1 \cdot x_1 + 0.5 \cdot x_2 + 0 \cdot x_3 = 4$. I am not sure if the bars-and-stars argument can be adapted to this setting.

However, note that if you fix the value $x_1$, then the two equations determine $x_2$ and $x_3$ uniquely. Hence you can run a case analysis: how many outcomes are there where A wins $4$ games? $3$ games? $2$ games? and so on. Do not forget to show that your case analysis is complete - you should argue that you have not overlooked any cases.

Finally, for part (b), when considering a single case with $x_1$ wins, $x_2$ draws and $x_3$ losses, it looks like you should be counting how many ways these results can be obtained, as in part (a). Once you have determined this for each case, how should you combine the results?

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The correct answer to part (b) would be as follows then,

B could get 3 points by either

  • 1 Loss, 6 Draws - ${7\choose6}\cdot{1\choose1}$
  • 2 Loss, 4 Draws, 1 Win - ${7\choose2}\cdot{5\choose4}\cdot{1\choose1}$
  • 3 Loss, 2 Draws, 2 Win - ${7\choose3}\cdot{4\choose2}\cdot{2\choose2}$
  • 4 Loss, 3 Wins - ${7\choose4}\cdot{3\choose3}$
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