At the semi finals, given the probability of each team coming 1st and 2nd, what is the probability of each team beating each other in the finals?

Question

We are at the semi finals of a tournament, there are the following games remaining.

• Semi Finals:
  • A vs B
  • C vs D
• Finals
  • (Winner of A vs B) vs (Winner of C vs D)

We are given the following probabilities:

+------+------+------+----------+-----------+
| Team | 1st  | 2nd  | Win Semi | Lose Semi |
+------+------+------+----------+-----------+
| A    | 0.53 | 0.18 | 0.71     | 0.29      |
| B    | 0.17 | 0.12 | 0.29     | 0.71      |
| C    | 0.22 | 0.49 | 0.71     | 0.29      |
| D    | 0.08 | 0.21 | 0.29     | 0.71      |
+------+------+------+----------+-----------+

• 1st means they win in their semi finals match and win their finals match
• 2nd means they win their semi finals match, but lose their finals match

Can we infer from this the following probabilities? (these are the matches that could take place in the finals)

• A wins in a match against C
• A wins in a match against D
• B wins in a match against C
• B wins in a match against D

Maybe some useful information:

• The probability of each team winning their finals match (against either of the two teams they could face) is P(1st | Win Semi) = P(1st) / P(Win Semi)
• And similarly, the probability that they lose their finals match (against either of the two teams they could face) is P(2nd | Win Semi) = P(2nd) / P(Win Semi)

These are calculated here:

+------+-------------+-------------+
| Team |  Win Final  | Lose Final  |
+------+-------------+-------------+
| A    | 0.746478873 | 0.253521127 |
| B    | 0.586206897 | 0.413793103 |
| C    | 0.309859155 | 0.690140845 |
| D    | 0.275862069 | 0.724137931 |
+------+-------------+-------------+

I have been trying to figure this out, but haven't managed to come up with a solution. I've also tried using simultaneous equations, but I think they're linearly dependent (infinite solutions).

Answer 1

You are right that the relevant equations are linearly dependent. I will write $X>Y$ as shorthand for $X$ beats $Y$. Then:

$$P(A=1st) = P(A > B) \times ( P(A>C) P(C>D) + P(A>D) P(D> C) )$$

$$P(A=2nd) = P(A>B) \times ( P(A<C) P(C>D) + P(A<D) P(D>C) )$$

and (importantly) these are the only equations involving unknowns $x= P(A>C)$ and $y= P(A>D)$. Rearranging a bit, the first equation becomes:

$${P(A=1st) \over P(A > B)} = P(C>D) x + P(D> C) y$$

while the second equation becomes:

$$ \begin{array}{rl} {P(A=2nd) \over P(A>B)} &= P(C>D) (1-x) + P(D>C) (1-y )\\ &=P(C>D) + P(D>C) - (P(C>D)x + P(D>C)y)\\ P(C>D)x + P(D>C)y &= P(C>D) + P(D>C) - {P(A=2nd) \over P(A>B)}\\ &= 1 - {P(A=2nd) \over P(A>B)} = {P(A=1st) \over P(A > B)} \end{array} $$

so the two equations are the same (i.e. linearly dependent and consistent) and there are infinitely many solutions.

In fact the only thing you can calculate (from the given info) is this quantity:

$$P(C>D) x + P(D> C) y=P(A > \text{whoever wins the other semi-final})$$

However for any $(x,y)$ that is a solution, if $x$ is slightly higher and $y$ is slightly lower, and the changes are in the right ratio s.t. they balance out s.t. the quantity above remains the same, then the new $(x', y')$ would also be a solution.