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        riddles >> easy >> Each Way
        
(Message started by: EdwardSmith on Feb 2^nd, 2015, 2:14am)

Title: Each Way
Post by EdwardSmith on Feb 2^nd, 2015, 2:14am

I dont know if this belongs in easy, medium or hard.
I just know that I dont know the answer so I'm asking you.

There is a horse race with nine horses.

Horse A has a 1 in 4 chance of winning the race.
B has a 1 in 5 chance.
C,D,E,F,G and H each have a 1 in 12 chance.
I has a 1 in 20 chance.

What chance does horse A have of being placed in the first 3 ?.

I'm basically trying to figure out how each way odds are worked out. Any help is appreciated.

Title: Re: Each Way
Post by rmsgrey on Feb 2^nd, 2015, 6:16am

We have insufficient information.

We'd need to know what the chance of horse A coming second or third is.

For example, it could be that horse A is always either unbeatable (on a good day) or comes in way at the back (on a bad day) - in which case, it's a 1/4 chance of him placing in the top 3; or it could be that he always manages to produce a photo finish for first place, in which case he'd be certain to finish in the top 3.

Or it could be anything in between - so anything from 1/4 to 1/1.

To get a single answer, you need to make assumptions about how the horses' first-place performance reflects their chances of coming second when they don't win, and third when they don't come in the top two.

An obvious possibility is to assume that if you remove one or more horses from the race before it starts, the remaining horses will have the same relative chances of winning the modified race. So if you ignore horse A, B would beat the rest 4/15 of the time, C-H 1/9 each and I 1/15.

The calculations for this model get a little involved - assuming I've not messed up any figures, the chances of A coming second are 709/3344, or 1545/3344 of finishing in the top two places - I arrived at that figure by calculating the chances of finishing second to any given horse and summing them.

So, A's chances of finishing second to B are the (1/4) / (4/5) chances of his being fastest apart from B times the 1/5 chance of B being the fastest for 1/16 overall. Finishing second to C is (1/12) * (1/4) / (11/12) = 1/44 and the same for each of D-H. Finishing second to I is (1/20) * (1/4) / (19/20) = 1/76

1/16 + 6*(1/44) + 1/76 = 709/3344

Performing the same calculations for the other horses would give you their chances of finishing second to any given horse, as well as their chances of finishing second overall. Using those intermediate figures for every pair that doesn't involve A will let you work out A's chances of coming third to each pair, summing those will give you A's overall chance of coming third, and adding that to A's chance of coming first or second will tell you A's chance of placing in the top three.

If horses X, Y, Z (all different) have probabilities p, q, r of winning, respectively, then, under this model, the chances of Y coming second to X are:
q*p/(1-p)
and the chances of coming in X first, Y second, Z third are:
r*q*p / ( (1-p)(1-p-q) )

Title: Re: Each Way
Post by EdwardSmith on Feb 2^nd, 2015, 4:20pm

Thank you for your assessment.
Taking the winning horse out of the race then recalculating the odds seems a good method.

I may try to write a program to do this. Studying form is a bit of a hobby of mine.

Most bookmakers offer either 1/4 or 1/5 the odds for a place, but this does not reflect the actual chance and bookmakers are out to make a profit.