Probabilities in three way contest: A beats B w/ prob 0.6, A beats C w/ prob 0.8, B beats C w/ prob. 0.7

[kajamix]

Three athletes A,B,C compete in the same event, such as 100 meters sprint, discus throw etc so the best wins.
We know from previous data that A beats B with probability 0.6 in head to head, A beats C with probability 0.8 and B beats C with probability 0.7.
When it's all three of them competing what are the win probabilities for each ?

 

[BigBeachBanana]

Three athletes A,B,C compete in the same event, such as 100 meters sprint, discus throw etc so the best wins.
We know from previous data that A beats B with probability 0.6 in head to head, A beats C with probability 0.8 and B beats C with probability 0.7.
When it's all three of them competing what are the win probabilities for each ?

What have you tried?

I would start by listing out all the different scenarios where each player wins.

 

[kajamix]

What have you tried?

I would start by listing out all the different scenarios where each player wins.

I say this:
Consider the equivalent contest in which C stays out, A confronts B and then the winner meets C in a "final" and also the other two equivalent contests in which B initially stays out, A initially stays out.
In this way I find P(A) = 0.54, P(B) = 0.34, P(C) = 0.12.
That looks like the correct answer.
But the real problem is what if we have N contestants (N= 5, 6, 7, .... 30 ...). Is there a general formula, given N and the pairwise probabilitie Qij to compute P(A), P(B), P(C) ... ?

 

[BigBeachBanana]

I say this:
Consider the equivalent contest in which C stays out, A confronts B and then the winner meets C in a "final" and also the other two equivalent contests in which B initially stays out, A initially stays out.
In this way I find P(A) = 0.54, P(B) = 0.34, P(C) = 0.12.
That looks like the correct answer.
But the real problem is what if we have N contestants (N= 5, 6, 7, .... 30 ...). Is there a general formula, given N and the pairwise probabilitie Qij to compute P(A), P(B), P(C) ... ?

My answer doesn't agree with yours.
Can you show me how you got your answers?
Are we to assume the events are independent and there are no ties?

 

[kajamix]

My answer doesn't agree with yours.
Can you show me how you got your answers?
Are we to assume the events are independent and there are no ties?

No ties.
But my solution is a good approximation really, checked. Or it depends rather. If they are drawing lots of some kind then it's like I said. But if it is a physical contest, such as racing, marathon ... then it's different and mathematicians use other probability models.
My real problem was how to use this method of mine to more than three contestants. It works out one has to make pair eliimnations stagewise and if N gets big, say N = 30, it is difficult - long time to compute.

 

[BigBeachBanana]

No ties.
But my solution is a good approximation really, checked. Or it depends rather. If they are drawing lots of some kind then it's like I said. But if it is a physical contest, such as racing, marathon ... then it's different and mathematicians use other probability models.
My real problem was how to use this method of mine to more than three contestants. It works out one has to make pair eliimnations stagewise and if N gets big, say N = 30, it is difficult - long time to compute.

Approximated how? Check against what? Depends on what?
What is your method?

Your question lacks a lot of details.

 

[kajamix]

Approximated how? Check against what? Depends on what?
What is your method?

Your question lacks a lot of details.

Well look.
Consider a three horse race like this:

horse 1 completes the race with standard time 60 ± 0.5 seconds
horse 2 completes the race with standard time 60.3 ± 0.6 seconds
horse 3 completes the race with standard time 60.7 ± 1 seconds

There is an integral that computes the probabilities.
I solved the integral for each pair (1 v 2), (1 v 3), (2 v 3) and computed certain values for the head to head probabilities and certain values for the win probabilities when all three compete.
Then I solved the same integral for all three together and computed slightly different win probabilities, but those are the correct ones.

This demonstrates that in a real Physics problem the simple approach is not good.
But the simple approacch is an approximation.

 

[Steven G]

Well look.
Consider a three horse race like this:

horse 1 completes the race with standard time 60 ± 0.5 seconds
horse 2 completes the race with standard time 60.3 ± 0.6 seconds
horse 3 completes the race with standard time 60.7 ± 1 seconds

There is an integral that computes the probabilities.
I solved the integral for each pair (1 v 2), (1 v 3), (2 v 3) and computed certain values for the head to head probabilities and certain values for the win probabilities when all three compete.
Then I solved the same integral for all three together and computed slightly different win probabilities, but those are the correct ones.

This demonstrates that in a real Physics problem the simple approach is not good.
But the simple approach is an approximation.

Can we see your work?