*ahem*
Very small. Infinitesimal has a precise meaning, and unless we're going into trans-finite arithmetic (clue - we're not) it's not appropriate.
I haven't been following the detail (I'd need a cold compress and a spreadsheet of my own, and not to have drunk some rather nice wine), but I suspect that what you collectively have shown is that the expected winnings are the same whether all five ducks are in one race or whether they're spread out across multiple races. But if your simplified example is indicative (the way you've constructed it makes me think it is) suggests that the variability of winnings is bigger if the ducks are spread out across multiple races. So if the target is to maximise your maximum possible prize pot then you should spread the ducks.
No. I think I have shown, after a false start, that for greatest expected winnings, where there is a prize for a heat win as well as an overall prize, you should put all the ducks in a single heat. It doesn't matter the relative size of the prizes
As
@srw implies, we need to be very careful about the difference between winning exactly one heat and winning at least one heat.
Suppose you enter n ducks. Suppose there are y heats, each of x ducks. n<x,y
If you enter all your ducks in the same heat:
P(one of your ducks wins that heat)= n/x
P(one of your ducks wins overall)=n/x . 1/y = n/xy
If you enter one duck each in n different heats:
P(at least one of your ducks wins a heat)=1 – P(none of your ducks wins a heat)
=1 – ((x-1)/x)^n
=1-(1 – 1/x)^n
If x is large we can do a Taylor expansion
=1-(1 – n/x+ O(1/x^2))
=n/x- O(1/x^2)
So the chances of winning at least one heat are smaller if you enter one duck in each heat than if you enter all the ducks in the same heat, but by a term of order 1/x^2. So as x gets larger, the chances of winning at least one heat converge on the same value whichever strategy you adopt.
Consider now the expected winnings E, where the prize H for each heat = H.
If you enter all your ducks in the same heat:
E = P(one of your ducks wins that heat).H =H n/x
If you enter one duck per heat in n different heats:
E = P(one of your duck wins).H + P(two of your ducks win).2H + P(three of your ducks win).3H…
= n. 1/x. ((x-1)/x)^(n-1). H + n.(n-1)/2. 1/x^2. ((x-1)/x)^(n-2).2H + n.(n-1)(n-2)/6. 1/x3. ((x-1)/x)^(n-3).3H…
This is just a binomial distribution with n experiments and p, the probability of success in each trial, of 1/x. The mean of a binomial distribution is np or in this case n/x so the expected winning is Hn/x.
So the expected or average (mean) winnings are exactly the same whichever strategy you adopt. But if you enter one duck each in multiple heats, you have a slightly higher risk of not winning any heats, compensated by a non-zero chance of winning two or more heats; if you enter them all in the same heat, you have a higher chance of winning one heat, but, obviously, no chance of winning more than one heat.
No spreadsheets were harmed in the making of this public information post.
