this is probably very simple for you guys but i'm not sure how to think about this and just wanted to understand how badly i was being ripped off if at all.

so in an 11 horse race we place a bet on a horse at odds of 3.0, one horse pulls out before the race leaving 10. Now the exchange reduces our odds to 2.5 before the race has started.

so they've shaved off .5 on our price and are saying our chances have increased by 6.7%

I figured that is correct because odds of 3 represent 33.3% chance and 2.5 is 40%.

This is where I get a bit confused because if all the horses have an even chance, then the horses should have 10% chance each. But obviously that isn't the case yet their 6.7% feels very steep considering the horse wouldn't have had much impact on the outcome imo.

So is that reduction in price fair or how do they calculate it on exchanges?

Sorry if that's a bit muddled I'm just a beginner trying to figure out how to think about this mathematically. If anyone could explain this it would be much appreciated

It depends on the odds of the scratched horse. If the scratched horse had probability p of winning, your horse's (1/3) becomes (1/3)/(1-p).

As heehaww stated, it entirely depends on how strong was the scratched horse. Its odds are not distributed by the remaining horses equally; rather, they are distributed proportionally to other horses' odds.

Just to make an example, suppose Federer, Djokovic and Nadal at the beginning of a major. They are the strongest players and bookies gives each of them a 33% chance of winning, while the remaining 1% is shared by the other 125 players. At the first round, unexpectedly, Nadal loses. Odds of weaker players get mostly unaffected by that. On the other hand, of course, Federer and Djokovic odds rise to almost 50% each.

In your case, the scratched horse had about a 1/6 (~16.7%) chance of winning. Your horse, being one of the favourite, gets a good portion of it. Another very weak horse with basically 0% chance isn't as much affected. On the extreme case, if the scratched horse was 2/3 (meaning that all the remaining horses had a 0 chance to win), now your horse had become a lock.

I completely agree with the above two posts. That is typically the correct way to think about things in this situation since that is typically all the information you have.

However, in some situations in which you or the odds-makers have information on the likely order of finishers in an event (or ordered preferences expressed over multiple items -- the topic of my dissertation), it can sometimes be a little tricky.

The probability of one horse (H1) winning a race when another horse (H2) is scratched is not tautologically given by P1/(1-P2).

To see this suppose that the race order between three horses is randomly decided between ABC, BAC, and ACB. Then it is clear that Horse A will win 2/3 of the time, Horse B will win 1/3 of the time, and Horse C will never win.

Now suppose Horse A is scratched. Using the same orderings excluding Horse A, the race will now be randomly decided between BC, BC, and CB. Here Horse B will win 2/3 of the time and Horse C will win 1/3 of the time.

Generally speaking, whenever there is "interaction" between Horses' finishes, the "excluded" probability of the scratched horse winning may be "allocated" to the other horses in non-obvious ways.

The tennis example can illustrate this as well.

Just to add to what whosnext posted:

This wiki page might be of interest: https://en.wikipedia.org/wiki/Indepe...t_alternatives (it explains pretty much what whosnext just said though...)

The book "The Compleat Horseplayer by David Edelman" outlines a different method of re-normalising the odds after a runner is removed (see pages 70, 71 & 72).

Finally, I'm not sure what exchange you are using, but on Betfair you can see the actual "Reduction Factor" that will be used in the event of a removal - just click on the little graph icon on the left of the horse's name.

Juk

Here's a worked example to show the difference between the "naive" method and the "Edelman" method:

Horse A = 60%
Horse B = 30%
Horse C = 10%

If horse A is removed then using the "naive" method you get:

Horse B = 0.3/(0.3+0.1) = 0.75 = 75%
Horse C = 0.1/(0.3+0.1) = 0.25 = 25%

If horse A is removed then using the "Edelman" method you get:

InverseCDF[NormalDistribution[0, 1], 0.3] = -0.524401

InverseCDF[NormalDistribution[0, 1], 0.1] = -1.28155

FindRoot[CDF[NormalDistribution[0, 1], -0.524401 + x] + CDF[NormalDistribution[0, 1], -1.28155 + x] - 1, {x, 0}] => 0.902976

-0.524401 + 0.902976 = 0.378575
-1.28155 + 0.902976 = -0.378574

Horse B = CDF[NormalDistribution[0, 1], 0.378575] = 0.647498 = ~65%
Horse C = CDF[NormalDistribution[0, 1], -0.378574] = 0.352502 = ~35%

Juk

PS: Sorry the "FindRoot" command doesn't actually seem to work using WolframAlpha, but it still shows the method... The book itself suggests using Newton–Raphson to find the required offset.

The above posts explain the theory. However, the actual calculation on the betfair exchange is a bit of a black box and while the reduction will usually not be very far from the theoretical, there can be some unusual reductions in low liquidity markets, usually early in the day. Also the reduction isn't based on the relative prices when you place the bet, but when the horse is withdrawn which can lead to substantial reductions if the odds of the withdrawn horse shorten a lot in the interim.

you have 11 horses all with different percentage chance of winning (combined total of 100%)
1 is scratched and now there are 10 horses

when 1 horse is taken away, the percentage chances of the other 10 are normalized so that total of race entrants is 100% again.

Sometimes the odds are not normalised like this... especially if the same trainer has multiple runners. One horse e.g. might be a pacemaker for another. This is something I've seen in UK/Ireland.

horses aren't coupled (same connections) in uk/ire. (if that is what you are referring to)