Here is a simple strategy: Assume there are N floors (in this case N = 10). We pick some M
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After the M-th floor, pick the first diamond which is larger than each of the first M diamonds.
To find the optimal M, we do the following calculation: let X_M be random variable of the rank of the largest diamond among all. (here we rank diamonds from small to large, so the smallest diamond gets rank 1 and the largest gets rank N), the following is easy to see:
p(X_M
p(X_M
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and from these, we get the distribution of X_M:
p(X_M=K) = p(X_M
Now, the probability that the aferomentioned strategy picks the largest diamond is:
The idea is that, if X_M = K, then there will be N-K diamonds that are larger than any of the first M diamonds, and in order for the strategy to work, the largest diamonds has to apear first among these N-K diamonds.
Then, we only have to find M that maximize this probability -- I guess it will be hard to get a close form formula, but clearly solvable for a N as small as 10.
How about this: find a strategy, so that you will get the largest expected rank.
For example, you pick one diamond out of the 10, and it happened to be the largest, then you get a score of 10. If it is the smallest, you get a score of 1. Your goal is to maximize the expected score.