Duck race probability

[swansonj]

*Like everyone, I'm assuming that the ducks are IID (independent, identically distributed) variables - there's nothing special about any of the individual ducks, and they don't form teams to help the team GC duck win. Add that complexity in and you're in a whole different game.

The other rather key assumption is that all heats are arranged to be equal. If some heats are bigger than others, that changes things too.

 

[Custom24]

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.

Thank you. When I calculated the expected winnings, I forgot the rather obvious consideration that if 2 of the ducks win, then you win two prizes, 3 = 3 prizes, etc. When I take that into account, it does indeed come out at 5% of the little prize for the expected winnings.