Nov 2014
I'm having trouble with a question, and need some help:
A race track betting board has the following odds against:
Horse A 9:2
B 6:1
C 11:3
D 19:2
E 10:1
F 13:2
A bet called a "triactor" is when you name the exact order of the first 3 horses in the race. What is the probability of winning the triactor?

I have been told that odds at the race track doesn't translate from straight probabilities, but factors in the number of betters etc.

I'm treating it more as a question on probabilities.

I assumed that the question is assuming that I'll pick the horses with the top 3 probabilities to win: so C, A, and B, in that order. What I'm also assuming is that the odds/probabilities listed are for finishing first. So, I thought I'd take 3/14 and multiply by (2/11)/(1-(3/14)) since we want to know the probability that A finishes first given C is out and then do the same for B. Is this reasoning flawed?


Forum Staff
May 2007
The probability of winning the triactor depends on which horses you pick!
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Reactions: 1 person
Nov 2014
Yes, but if I pick C, A, and B in that order (as mentioned) bc they have the highest probabilities, did I do the rest of it correctly?