[thylacine]

See my post above. If you look at the first k out of the N numbers, your success rate is


f(N,k):=(1/C(N,k))\sum_{i=1}^{n-k} C(n-i-1,k-1)/i

For given N you want to choose the k that maximises this.

When N is big, then setting p=k/N this is about

-p ln(p) (natural log)

which is maximised when p=1/e, so k is about N/e.

Actually your 25% is a lower bound for

-0.5 ln(0.5) =approx 0.34657359

but this is still smaller than

-(1/e) ln(1/e) = (1/e) =approx 0.36789441.

My post above asks if a different type of strategy can do even better!