[thylacine]

There is a lot going on here. You suggest the following strategy for the Picker (P) when the Writer (W) chooses N distinct numbers. (I'll assume distinct.) For k about N/e, the picker looks at the first k numbers and then stop as soon as there is a number bigger than those first k. (P certanly loses if the biggest was in the first k.) Let's call this 'Basic Strategy' with N and k.

I figure P's probability of success is

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

and I believe it is best to choose k about N/e, so that probability of success is about 1/e for big N.

Now suppose that as the numbers are pulled from the hat, that P is not told the actual numbers, but is only told where each new number slots into the ordering of the numbers drawn so far, (e.g. 'the 10th number drawn is the 7th biggest of the first 10 numbers drawn). Then I could easily believe that the above strategy is the best possible.

But if P is told the actual numbers, then he may be able to do better than Basic Strategy. Let's suppose that W selects from some probability distribution of sets of N numbers, and that P knows this probability distribution. Then maybe P has a lot more information and can do better. It could be that after looking at, say, the first m numbers, that P can figure the conditional probability distribution of N-tuples containing those numbers, and can take advantage of that.

In the simplest case N=2, If using Basic Strategy then P could go with k=0 or k=1, since f(2,0)=f(2,1)=1/2, so P might as well flip a coin. But it could be that if the first number drawn is, say, x, then P knows that in the conditional probability distribution of pairs of numbers including x has the other number in the pair being bigger than s say 53% of the time, and so P could go for the 2nd number. W can only thwart this if he can come up with a probability distribution where for every x, the conditional probability that a pair containing x, has x being the smaller, is exactly 50%. BUT THERE IS NO PROBABILITY DISTRIBUTION!! (Do you believe me?) ON THE OTHER HAND, AT LEAST W CAN MAKE P's SUCCESS AS CLOSE TO 50% AS DESIRED, JUST NOT EXACTLY 50% !! (Do you believe me?)

Now consider N=4. For Basic Strategy use k=1. In other words look at the first number drawn, and then stop at the next number that is bigger. This Basic Strategy has success rate f(4,1)=11/24. But now P will always see at least 2 numbers, and it may be that after seeing, say, x then y, that in some cases even when y>x, that P may modify Basic Strategy by not stopping at y, in cases where the conditional probability that y is the biggest of the 4 numbers, given that the 4 numbers include x and y, is less than 1/2.

Can W come up with any probability distribution such that P can have success no better than 11/24?

Can W even come up with any probability distributions such that P can have success no better than something arbitrarily close to 11/24, similarly to the N=2 case? Or is it instead that P can always have some success rate S>11/24, regardless of W's chosen probability distribution?

I will start a related thread 'Another interesting proposition' on a related problem.