The champion beat the winner of the 3rd place playoff in the semifinal: 15 times

1938 1954 1958 1962 1966 1974 (1978) 1982 1994 1998 2002 2006 2010 2018 2022

The champion beat the loser of the 3rd place playoff: 5 times

1934 1970 1986 1990 2014

The condition specified in the problem is met if 1 gets to play 2 or 3, and not met if 1 gets to play 4. Therefore 2/3 is the answer. ]]>

This leads to the three possible cases:

A>B>C>D

A>C>B>D

C>A>B>D

A wins with probability 2/3 and C with probability 1/3

]]>It’s better to start at the semifinal stage. At that point we have no info about their relative ranks, so the 24 orderings of A,B,C,D are equally likely. Adding conditions A>B>D and C>D rules out all but 3 cases, leaving A>C with probability 2/3.

Since the Monty Hall problem also has a probability 2/3 (that switching choices is better), I wonder if there’s a way to map this problem to Monty Hall that would let reasoning from one illuminate the other.

]]>where P(A = 1 | B > D) + P(C = 1 | B > D) = 1.

First, let’s be letter agnostic. For the 4 semifinalists, we know each will fit into one of the following ranges:

– Finalist 1: 1

– Finalist 2: 2 through 5 (5 occurs when 1 through 4 are all in one bracket quarter together)

– Finalist 3: 3 through 9 (same reasoning as finalist 2)

– Finalist 4: 4 through 13.

Bringing letters into it, we can further state that:

D must be in the finalist 4 group, since it must be worse than all other finalists (worse than A indirectly through transitivity).

B must be either in the finalist 2 or 3 group, since it must be better than D but worse than A.

C must be in one of the finalist 1,2, or 3 groups, since its only condition is that it is better than D.

A must be in either the finalist 1 or 2 groups, since it must be better than both D and B.

This means that A is between 1 and 5, and if it is not 1, C must be 1.

So if A is 1, A > C, but is A is 2,3,4,5, C > A.

P(A in (1,2,3,4,5) | B > D) = P(A = 1 | B > D) + P(A in (2,3,4,5) | B > D)

P(A in (2,3,4,5) | B > D) = P(A = 2 | B > D) + P(A = 3 | B > D) + P(A = 4 | B > D) + P(A = 5 | B > D)

We can translate slightly to make easier to calculate!

P(A = 1 | B > D) = P(1 in A “cluster”) = 4/16

P(A = 2 | B > D) = P(2 in A “cluster” & 1 in C “cluster”) = 4/16 * 4/15

P(A = 3 | B > D) = P(3 in A “cluster” & (1,2) in C “cluster”) = 4/16 * 4/15 * 3/14

P(A = 4 | B > D) = P(4 in A “cluster” & (1,2,3) in C “cluster”) = 4/16 * 4/15 * 3/14 * 2/13

P(A = 5 | B > D) = P(5 in A “cluster” & (1,2,3,4) in C “cluster”) = 4/16 * 4/15 * 3/14 * 2/13 * 1/12

Now for our final answer we take P(A = 1 | B > D) / P(A in (1,2,3,4,5) | B > D)

First, we can multiply all terms by 16/4

So we now have (conditionality assumed): P(A = 1) / [P(A = 1) + P(A in (2,3,4,5))] = 1/(1 + 1/3) = 3/4 or 75%.