The Mathematics of Chess Pairings

There was a recent article is ChessBase regarding a problem that had arisen last year involving the world's highest ranked woman player, Hou Hifan. The article began:

The Gibraltar Masters wrapped up Thursday, with Levon Aronian in first place. This year Round Ten passed without incident, in contrast to 2017 when, on February 2nd, the story of the day was a rare scandal involving women's World Champion Hou Yifan deliberately losing a game in protest of the high number of women she was paired against. She was further confounded when a similarly unlikely string of pairings happened in October at the Isle of Man Open. Johannes Meijer looks at the odds in detail. Hou did not return to Gibralter in 2018, but instead competed in the Tata Steel Chess Masters.

Imagine, you are at a tournament with 255 players of which 43 are female. You are to play ten rounds. How many female opponents would you expect to face? Three? Five? I am pretty sure you wouldn't say seven. Yet, this was exactly the number of female players Hou Yifan faced at the Gibraltar Open 2017 when, a year ago today, she threw her last game in protest of these seemingly odd pairings.

The article goes on to ask the question: How probable is such a pairing? Could it have happened by chance at all? Well, the approach to solving this problem involves the hypergeometric distribution, a discrete probability distribution commonly covered in high school probability and statistics courses. Wikipedia describes it thus:

Thus to find how likely, or unlikely, Hou's pairings were we only have to replace "green marbles" with women and "red marbles" with men. So k=7, K=43, N=255, n-10, n-k=3 and N-K=212. Substituting into the formula we get:$$P(X=7)=\frac{^KC_k \text{ . } ^{N-K}C_{n-k}}{^NC_n}=\frac{^{43}C_7 \text{ . } ^{212}C_{3}}{^{255}C_{10}}\approx 0.000188 $$Thus it seen that the likelihood is very small that this could happen and yet the pairings were allegedly arranged using a computer draw. The quoted article carries out similar calculations but follow a somewhat different approach.

To be strictly accurate, since possible pairings with Hou Hifan are under consideration, we should make K=42 and thus N=254. This gives a slightly lower probability of 0.000164 and so even more unlikely. It means that out of 10,000 random pairings of a woman with ten competitors, the result of being paired with another woman in seven out of the ten rounds would be expected to occur less than twice. On the other hand, the probability of not being paired with any women is nearly 16% and is given by:$$P(X=0)=\frac{^KC_k \text{ . } ^{N-K}C_{n-k}}{^NC_n}=\frac{^{42}C_0 \text{ . } ^{212}C_{10}}{^{254}C_{10}}\approx 0.158251 $$