Horse Racing A v B

How do you calculate the odds of any horse finishing before another horse (no matter where they finish in the race, just that one beats the other)?

e.g Given Horse A has a winning probability of 20% and Horse B has a winning probability of 12.5%, what is the probability that Horse A will beat Horse B (effectively making it a two horse race)? Assume 5 runners ( other 3 horse probabilities of winning are 35%, 25% and 7.5%).

Here is what we know:
Of all of the results where both A and B beat all three of the other horses, A wins 8/13 of the time.

Whether or not that translates to "A will beat B 8/13 of the time," I am not confident of saying one way or the other.

Problem with this equation is that When horse A loses, if he was favorite due to being only early speed, B + C chances may improve dramatically if either inherits the lead. If B+c have a speed duel, D a closer may go from virtually no chance if A got an easy lead, to 50% if B+C duel for the lead. Theses are horses, not dice or cards. Old track saying " 1 way to win a race, 100 ways to lose it "

Quote: jedijon

How do you calculate the odds of any horse finishing before another horse (no matter where they finish in the race, just that one beats the other)?

e.g Given Horse A has a winning probability of 20% and Horse B has a winning probability of 12.5%, what is the probability that Horse A will beat Horse B (effectively making it a two horse race)? Assume 5 runners ( other 3 horse probabilities of winning are 35%, 25% and 7.5%).

How will you use this information? One problem I can envision towards the back is jockeys not caring about the finish once they're out of the money and there is no real motivation.

Quote: onenickelmiracle

How will you use this information?...

In the UK bookmakers often offer match bets - one horse against the other. Typically they might pick the 2nd and 3rd favourites and the odds offered are usually consistent to the win odds. I can guess the question is where the odds for the horses are fairly different (e.g. 5/1 8/1) and say you can get 8/13 or 5/4, whether the true odds are near 6 to 4 or if there's a bargain to be had.

Exactly - Betfair offer a market called A v B market as per Charlie mentions.
You can take the implied probability of each horse winning from the Win market (bookies' estimates granted).
As per my original post, I just want to know if it's mathematically possible from the information given if you are able to apply any given horse's probability of beating another from these odds?

Quote: jedijon

How do you calculate the odds of any horse finishing before another horse (no matter where they finish in the race, just that one beats the other)?

e.g Given Horse A has a winning probability of 20% and Horse B has a winning probability of 12.5%, what is the probability that Horse A will beat Horse B (effectively making it a two horse race)? Assume 5 runners ( other 3 horse probabilities of winning are 35%, 25% and 7.5%).

This is how I would do it assuming you are not taking into consideration the running styles of other horses. I would estimate that the other horses do not exist and also that the implied odds you have stated also correlate strongly with the likelihood that if one horse finishes 3rd that the other horse would finish worse than 3rd. In the real world of course the running styles of these 2 horses and the other 5 horses is a large factor and also whether or not horse A or B tends to run well even if he does not win is a large factor. Some horses will either win or be way out of the money. But I would speculate that in making a great many bets these factors may become insignificant after a large sample. So, then I would say that horse A will win 20% of the time that B does not win which is 87.5% of the time which is 17.5 times. And horse B will win 12.5% of the time that horse A does not win which is 80% of the time which is 10 times. All of the rest of the races are considered as pushes, they do not count since we have decided not to consider other factors that might determine the winner. So, in the 27.5 races that did count horse A won 17.5 and horse B won 10. Horse A wins just 63.6 of the time and horse B wins 36.3% of the time. So, without calculating the cost of the bets at Betfair, which you would know, fair odds on the A horse would be right around 3/5 and on horse B right around 8/5. I think this is roughly correct. If not, the Mathletes on this site will correct me. I'm sure Betfair has a very accurate method of calculating this and I think it would be unlikely that you would find an advantage by doing a similiar calculation. But, if you do find a horse who either wins or is way out of the money there is a fair chance that Befair is not considering this and you may have an edge betting against this type of horse.