A random distribution of rugby players.

Question

Prompted by a recent news item, it was suggested, anecdotally, that when 'randomly' allocated seats on a recent flight, only $2$ out of $30$ rugby players wound up sitting together (by which I assume is meant side-by-side in the same three seat cluster).

Assuming this was a $186$-seat aircraft ($31$ rows of six seats with a central aisle), what is the probability of this occurring, and, from that, can we say with any confidence the probability of this being 'random'?

Edit: conceptually, I think $54$ rows of $3$ seats each is a fairer interpretation of the plane layout; https://www.seatguru.com/airlines/EasyJet_Airlines/EasyJet_Airlines_Airbus_A320.php

Also, I think we should assume that the rugby players were first to check in, so there were no other seat allocations.

Edit 2: I think another way of formulating the problem is to distribute 3 decks of cards (a red deck, an orange deck, and yellow deck, shuffled together) amongst 30 players (one card each) , and then to ask what is the probability of only two players receiving the same card, one of which is orange. (OK, that's 52 instead of 54, but that seems like a small technicality)

Answer 1

It is not exactly what you were asking, but we can calculate the expected number of pairs sitting together. Consider one set of three seats. You have three players seated there with probability $\frac {30 \choose 3}{186 \choose 3}=\frac {203}{52762}\approx 0.00385$ which would give you two pairs. You have two seated next to each other with probability $\frac {2{30\choose 2}156}{3{186 \choose 3}}=\frac {1131}{26381}\approx 0.04287$ which gives you one pair. The numerator is the number of ways to choose two rugby players and one other and the $\frac 23$ accounts for the fact that two of six seating configurations have the players together. By the linearity of expectation, the expected number of pairs sitting together is then about $54(2\cdot 0.00385+.04287)\approx 2.87$ Having two pairs together looks reasonably likely.