Combinatorics with Blitzstein and Hwang

Also from Blitzstein and Hwang, seven chess matches are played between players A and B.  How many ways can the final result be 4-3 to A?

In doing this question, I am reminded about how almost all of the difficult work in maths questions comes in unpacking the question.  This unpacking is also the act of modelling.  And it isn't easy when you're doing it from scratch.

So my initial thinking was to transform the problem into a counting problem.  We don't need to model B at all here - it can all be done in modelling how A gets those 4 points.  This is because each game has precisely one point to distribute, and that will certainly happen at the end of each game.  So B disappears from the analysis.  This is like the 'degrees of freedom' idea in statistics 

With only seven games, there's no way A could have finished with 4 points without at least one clear victory.  At the other end of chess playing competence is the possibility that A won 4 (and therefore lost 3).  Note also that the number of draws must always be 0 or some multiple of 2.

Now that B has been dispatched, we look at the triplet which is a result.  In general, there are the following four ways that 4-3 can result, were W=A win, D=draw, L=A loss:

• 1W 6D 0L
• 2W 4D 1L
• 3W 2D 2L
• 4W 0D 3L

Now we can use the 'degrees of freedom' trick one more time.  If you know two elements of this triplet, then you'll know for sure what the third element is.  So we don't need to model three distinct values for D,W and L. We only need two, but which two?

I always prefer to go for smaller numbers, and on that basis, I'd go for W + L.  We shall henceforth model only A's wins and losses, since A's draws are implied.

Now we have a sum of 4 possibilities.  Let's take them one by one. ${7} \choose {1}$  represents all 7 locations where that one win would happen.  We don't have losses since by implication all the six other slots were draws.  So for this sub-case, we're done.  Once we place the  win, we have no more variations to deal with.

  ${{7} \choose {2}} { {5} \choose {1}}$ represents first how we can distribute two wins among the seven games, then when that's done we have 5 unfilled slots which we need to drop one loss into.  We multiply those two together (choices multiply).

Similarly,   ${{7} \choose {3}} { {4} \choose {2}}$ represents 3W + 2L.  And finally   ${{7} \choose {4}} { {3} \choose {3}}$ represents the case where we drop in four wins, then all remaining slots contain the full set of 3 losses.

The final answer to the question is: ${{7} \choose {1}} + {{7} \choose {2}} { {5} \choose {1}}  + {{7} \choose {3}} { {4} \choose {2}} + {{7} \choose {4}} { {3} \choose {3}}$

R thinks this amounts to 357 different possible ways to arrive at a 4-3 result for A.