- #1

cosmicminer

- 20

- 1

- TL;DR Summary
- multiparticipant event probabilities from the head to head probabilities

Is there a formula for this ?

Consider the following simple looking problem.

We have three contestants A, B and C, there is a competition between them and the best wins.

For example a race, discus throw, javelin throw ...

We know that A beats B with a probability 0.6, A beats C with a probability 0.8 and B beats C with a probability 0.7.

That is we know the head to head probabilities but what about the probabilities when they are all together competing ?

My approach was this:

Suppose C is left out and A fights against B and then the winner fights against C.

Or B is initially left out, or A is initially left out.

There may be psychological factors which make this different from the real event but we neglect them. We assume independence.

So if C is initially left out then A goes forward with prob. 0.6, then it's 0.8 against C, equals 0.48. For B it is 0.4 then 0.7, equals 0.28. For C it is the remainder, 0.24.

If B is left out then similarly A has probability 0.48, C has 0.06 and B has 0.46.

If A is left out then B has 0.28, C has 0.06 and A has 66%

The three possibilities are equivalent so for the real situation we take the averages and finally:

P(A) = 0.54, P(B) = 0.34 and P(C) = 0.12

Now the real problem is when there are N participants and we know the head to head probabilities (Pij).

It becomes tedious, how to write a computer code ?

Is there a general formula ?

Consider the following simple looking problem.

We have three contestants A, B and C, there is a competition between them and the best wins.

For example a race, discus throw, javelin throw ...

We know that A beats B with a probability 0.6, A beats C with a probability 0.8 and B beats C with a probability 0.7.

That is we know the head to head probabilities but what about the probabilities when they are all together competing ?

My approach was this:

Suppose C is left out and A fights against B and then the winner fights against C.

Or B is initially left out, or A is initially left out.

There may be psychological factors which make this different from the real event but we neglect them. We assume independence.

So if C is initially left out then A goes forward with prob. 0.6, then it's 0.8 against C, equals 0.48. For B it is 0.4 then 0.7, equals 0.28. For C it is the remainder, 0.24.

If B is left out then similarly A has probability 0.48, C has 0.06 and B has 0.46.

If A is left out then B has 0.28, C has 0.06 and A has 66%

The three possibilities are equivalent so for the real situation we take the averages and finally:

P(A) = 0.54, P(B) = 0.34 and P(C) = 0.12

Now the real problem is when there are N participants and we know the head to head probabilities (Pij).

It becomes tedious, how to write a computer code ?

Is there a general formula ?