Mrs R claimed an answer of 1/40, which I agreed with, so it seems that we all have a consensus.
However, young Mr R (who is not a mathematician and has no interest in football) challenged the on-field decision and gave an alternative answer of 2/15.
He has sent me to the VAR screen to review my decision.
Mrs R and I are heading to Manchester today (for the football), so it might take me a while before I respond. In the meantime, the crowd will be kept in suspense without being told a thing. Maybe I’m taking the VAR analogy a bit too literally!
When I first gave thought to this teaser, I had in mined something like AvB; AvC; BvC
ie, three possible combinations. And for some reason, thought that having drawn the first team (say) A, there would be 15 teams left from which to draw the second team. 2 of these 15 would be B and C. Hence probability of two Welsh teams drawn together would be 2/15.
I then began to consider the probability of drawing a Welsh team from 16. This would be 3/16
The probability of drawing the next team as Welsh would be 2/15.
The probability of both these events would be (3/16)x(2/15) = 1/40.
I reinforced this view (always a dangerous move !) when I settled back at home, by drawing out on squared paper, all the possible combinations of the draw … AB, AC, AD … AN, AO, AP thru OP and back thru PO to BA.
All 240 of them !!
I randomly selected three teams to represent Wales eg D, J, M. and sure enough there were 6 pairings of DJ; DM; JM; MJ; MD and JD. Hence prob 1/40.
I tried a few more allocations of A ->> P as Welsh teams, just to be a bit more confident.
So, for the moment, I’m (apprehensively) sticking with 1/40.
If there were four Welsh teams in the last 16, what’s the probability that -
there would be a single Welsh v Welsh fixture ?
there would be two Welsh v Welsh fixtures ?
there would be at least one Welsh v Welsh fixture ?
Or worse still …
If there were five Welsh teams in the last 16, what’s the probability that -
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 teaser is quite old, hence the antiquated UK currency …
For those who are unfamiliar with £:s:d … £1 = 20s … 1s = 12d
Smith, Jones and Robinson are the Captain, Co-Pilot and First-Class Flight Attendant on an aircraft, but not necessarily in that order. The aircraft carries three first-class passengers, co-incidentally with the same surnames, but identified with a ‘Mr’ : Mr Smith, Mr Jones and Mr Robinson.
Mr Robinson lives in Nottingham
The First-Class Flight Attendant lives halfway between Nottingham and Sheffield
Mr Jones’s salary is £1,000 : 2s : 1d (One thousand pounds, two shillings and a penny)
Smith can beat the Co-Pilot at billiards
The First-Class Flight Attendant’s nearest neighbour (one of the passengers) earns exactly three times as much as the First-Class Flight Attendant.
The First-Class Flight Attendant’s namesake lives in Sheffield
As a hint, but also to help non-UK citizens and also UK citizens who are younger than 70, you can change Mr Jones’s salary to £1000.21 and the answer will be the same !
Mr Jones’s salary is not EXACTLY divisible by 3 so he can’t be the Flight Attendant’s nearest neighbour – the Flight Attendant’s nearest neighbour must be either Mr Robinson or Mr Smith
However, Mr Robinson lives in Nottingham (given)
Mr Smith must therefore be the Flight Attendant’s nearest neighbour and live in Halfway
The Flight Attendant’s namesake lives in Sheffield (given). Because we have already accounted for Mr Robinson (Nottingham) and Mr Smith (Halfway), the Flight Attendant’s namesake must be Mr Jones - and hence …
… well, at least we now know where most of them live !!
Perhaps the rest wasn’t as straight forward as I thought, or was so obvious that no further explanation was required
Here is my assessment of the situation
The Flight Attendant lives in Halfway (given)
Mr Jones’s salary is not EXACTLY divisible by 3 so he can’t be the Flight Attendant’s nearest neighbour – the Flight Attendant’s nearest neighbour must be either Mr Robinson or Mr Smith
However, Mr Robinson lives in Nottingham (given)
Mr Smith must therefore be the Flight Attendant’s nearest neighbour and live in Halfway
The Flight Attendant’s namesake lives in Sheffield (given). Because we have already accounted for Mr Robinson (Nottingham) and Mr Smith (Halfway), the Flight Attendant’s namesake must be Mr Jones - and hence … the Flight Attendant must be Jones.
This leaves Smith and Robinson as Captain/Co-Pilot
Since Smith can beat the Co-Pilot at billiards, the Co-pilot must be Robinson and
Nothing wrong with the wording Don, just a rush of blood FROM the head leading to some dodgy maths! As ever, Ravvie’s equation demonstrates the neatest method.
