The VAR referee has surprisingly (controversially?) came up with a ruling that was neither 1/40 nor 2/15.
Mrs R’s method was similar to Don’s. She envisaged a league schedule of 16x16 teams, giving 256 possible matches, of which the diagonal can be removed. She said there were 120 matches in each of two triangles, each of which had 3 all-Welsh matches. Hence 1/40 chance.
Young Mr R said if I’m a Welsh team there is a 2/15 chance that I will play another Welsh team. He also indicated that this doesn’t consider all possibilities and felt the answer could be higher.
One way to resolve it is to consider all permutations of the draw, but as there are 16! of them (about 21 trillion), that may be a challenge too far.
Let’s solve a simpler problem, say there are two Welsh teams in the semi-finals. There are only 4! = 24 permutations. Mrs R’s method would give 1/6 and Young Mr R gets 1/3. There are 8 permutations with two Welsh teams: 1234, 1243, 2134, 2143, 3412, 4312, 3421, 4321. This rules out Mrs R’s method (which initially I went along with). It doesn’t prove Young Mr R’s method though, nor answer his doubts about it being higher with a third team.
Let’s now try three Welsh teams out of six. There are 6! = 720 permutations which is a bit too many to list, but if we consider two teams drawn in any match, there are 4! = 24 ways of drawing the rest. In the first match there are 6 possible draws for two Welsh teams. There are three matches so the answer is 6x3x24/720 = 3/5. Young Mr R’s method gives 2/5. Assuming he is team 1, he had only considered 1 vs 2 and 1 vs 3 but there is also 2 vs 3.
This method does extend to 16 teams, giving 3/15.
I now await boos from the crowd!