Place Probability Problem

[mwilding1981]

Been doing a bit of research using the discounted Harville formula for place odds and am stumped. Using these odds:

Horse 1 1.36 Horse 2 5.5 Horse 3 9 Horse 4 29 Horse 5 41 Horse 6 126
I am getting these probabilities:

Horse 1 1.186804 Horse 2 0.667996 Horse 3 0.462834 Horse 4 0.184595 Horse 5 0.139941 Horse 6 0.056887
This obviously cannot be correct. I know I am making an error somewhere in my calculations as I programmed a calculator a while back and it is giving me different figures but I can't for the life of me find where it is. Can anybody help?

[mwilding1981]

Figured it!

[podonne]

Not to be "that guy", but I'd caution you against using this formula, unless its the best at your disposal. The probabilities hold very well for win bets, but win odds are much worse at predicting place bets.

Harville's formula makes sense logically, but only if you accept that the removal of any given horse from a contest means that the odds on the remaining horses scale equally.

Consider an expected speed duel and you can see why that might not be the case. If Horse A and B (out of a field of 8) are speedy types and you expect them to duel it out, they might have a 45-45 shot at winning (with the rest of the field at 10%), but if you know which one won (A), i don't think the other horse (B) has a 81% chance of placing, but that is what Harville's formula asks that you believe.