Originally Posted by
Ganchrow
If the good runner's fair
odds are , then that would imply a
1 17 ≈ 5.88235% of his winning the race.
Along with the above assumptions, this further implies a
29√5.88235% ≈ 90.69236% probability of of the "good runner" finishing higher than any given "bad runner".
My apologies but I completely dropped the ball on this one.
Let me start over.
If the "good runner" has a 5.88235% of winning the race, then each of the other runners have a (1-5.88235%)/29 ≈ 3.24544% probability of winning the race.
The good runner's probability of placing 2nd conditioned on his not placing 1st would then be 5.88235% / (1 - 3.24544%) ≈ 6.0797%, which is simply his winning probability adjusted for the reduced pool of runners.
To get his absolute probability of finishing 2nd we simply multiply by his own probability of not finishing 1st.
In other words:
P(2nd place finish) = P(2nd place finish | not 1st place finish) * P(Not 1st place finish)
P(2nd place finish) = 6.0797% * (1 - 5.88235%) ≈ 5.72204%
So in general the good runner's probability of placing in position k ≤ 30, given that he did not finish in positions 1 ... k-1 would be:
P(kth place finish | not 1st-(k-1)th place finish) = 5.88235% / ( 1- k * 3.24544%)
And the absolute probability of a kth place finish would then be:
P(kth place finish) = P(kth place finish | not 1st-(k-1)th place finish) * P(not 1st-(k-1)th place finish)
[nbtable][tr][td]= 5.88235% / ( 1- (k-1) * 3.24544%) * (1-[/td] [td] [/td] [td] P(i-1)th place finish) ) [/td] [/tr] [/nbtable]
So:
So:
Edge(+2000 bet) = 5.88235% * 21 - 1 ≈ 23.5294%
Edge(+520 bet) = 17.16507% * 6.2 - 1 ≈ 6.42343%
Edge(+310 bet) = 27.79795% * 4.1 - 1 ≈ 13.97160%
Which yields full Kelly stakes of ~ 0.56555% on the win bet, 0% on the top 3 bet, and 3.94141% on the top 5 bet.
I should also note that the probability of the good runner beating any any other given runner would be:
5.88235% 5.88235% + 3.24544% ≈ 64.44444%
Apologies once again. I just completely spaced.
Originally Posted by
Optional
Is this logic correct? Runner A has a 92.94% implied chance of beating any other, Runner B has a 91.73% chance. Meaning A has 1.21% more chance in a 1 on 1 matchup than B? Therefore ((1-0.0121)/(0.0121+1))+1 = $1.98 is the implied fair odds on the underdog? (or should I be looking at the ratio of their probabilities?)
Assuming regularity conditions hold, runner A's probability of beating runner B would be given by:
Pr(A/B) =
Pr(A/Field)*(1-Pr(B/Field)) Pr(A/Field)*(1-Pr(B/Field)) + (1-Pr(A/Field)) * Pr(B/Field)
Or more succinctly put using the logit function (defined as lg(x) = log(x) - log(1-x)):
lg(
Pr(A/B)) = lg(
Pr(A/Field)) - lg(
Pr(B/Field))
Either way, Pr(A/B) ≈ 54.27191%
To get the probabilities of winning the entire race, we have:
Pr(A) =
Pr(A/Field)*
Pr(A/B)/(
Pr(A/Field)+28*
Pr(A/B)*(1-
Pr(A/Field)))
Pr(A) ≈ 25.19185%
Pr(B) =
Pr(B/Field)*(1-
Pr(A/B))/(
Pr(B/Field)+28*(1-
Pr(A/B))*(1-
Pr(B/Field)))
Pr(B) ≈ 21.22599%
Pr(Other) = (1-
Pr(A)-
Pr(B))/28
Pr(Other) ≈ 1.91365%
Originally Posted by
Optional
I'd very much like to know how to calculate this out when I don't "judge them all to be equally matched each with the same probability of losing" too, if you have the time and patience to set that out as well.
As you can probably tell, these kinds of combinatoric problems can get pretty messy pretty quickly. The best way to tackle it, provided you had the know-how, would just be to write a program to traverse the different combinations.
You just need to remember that at each conditional state all the probabilities need to sum to unity and then keep applying either Bayes' theorem to convert to absolute probabilities and the logit function to compare probabilities between contestants for whom no direct heads-up probability is given.