Finishing postion in a race

How to compare finishing positions across races with a variable number of participants.

Consider the following example:

Race #1: Finished 1 out of 14 participants.
Race #2: Finished 3 out of 9 participants.
Race #3: Finished 5 out of 16 participants.
Race #4: Finished 2 out of 11 participants.

If we assume a negligible difference in racing ability, then I think it would be safe to assume a first place in a 16 participant race is more "valuable" than a first place win in a race with n < 16 participants.

So, for the above example I would express runner i's mean normalized finish position the following way:

1- (FINISH * 1 / n)

(1 / 1) * (1 / 14) = 0.0714. 1 - 0.0714 = 0.9285
(3 / 1) * (1 / 9) = 0.3333. 1 - 0.333 = 0.666
(5 / 1) * (1 / 16) = 0.3125. 1 - 0.3125 = 0.6875
(2 / 1) * (1 / 11) = 0.1818. 1 - 0.1818 =0.8181

The above results average to 0.6161.

Is this the right way to do it?

Thanks.

Neda_Kalantar said:

How to compare finishing positions across races with a variable number of participants.

Consider the following example:

Race #1: Finished 1 out of 14 participants.
Race #2: Finished 3 out of 9 participants.
Race #3: Finished 5 out of 16 participants.
Race #4: Finished 2 out of 11 participants.

If we assume a negligible difference in racing ability, then I think it would be safe to assume a first place in a 16 participant race is more "valuable" than a first place win in a race with n < 16 participants.

So, for the above example I would express runner i's mean normalized finish position the following way:

1- (FINISH * 1 / n)

(1 / 1) * (1 / 14) = 0.0714. 1 - 0.0714 = 0.9285
(3 / 1) * (1 / 9) = 0.3333. 1 - 0.333 = 0.666
(5 / 1) * (1 / 16) = 0.3125. 1 - 0.3125 = 0.6875
(2 / 1) * (1 / 11) = 0.1818. 1 - 0.1818 =0.8181

The above results average to 0.6161.

Is this the right way to do it?

Thanks.

Click to expand...

I believe the average is actually 0.77523. Although what you have done is certainly a measure, the answer to your question depends partially on what yu are trying to do with it.

For example, if you are trying to predict the finish of the horse in a race, you have a problem. Turning your equation around
FINISH = n (1 - WF)
where the WF (weighted finish) would be your average (or one of the other numbers making up the average), i.e.
WF = 1- (FINISH * 1 / n)
If we use that (corrected) average, then
FINISH = 0.22477 n
so that if you have 5 horses or more, you will finish better than first.

Not that I recommend using it to actually bet with, but a 'better' formula might be
WF = FINISH * number of runners
N = Total number of runners in all races
PLACE = (the sum of WF) / N
It at least has the advantage(?) that if the horse had always FINISHed first (or second or third or ...), it would predict that it would PLACE first (or second or third or ...). In this particular case the formula would predict a 2.86 PLACE

There are a multitude of other formulas one might use.

Thanks for the prompt reply.

I understand. I don't intend to wager on any actual races. It's more of an issue of mathematical curiosity. In any case, a competitive model would have to account for dozens, if not hundreds of different values. This is obviously beyond the scope of my ability.