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Spoiler for try my luck
I picked p in my draw. Lets estimate n = p. What would have been the likelihood that I got the max number ? : 1/p
If say n = p+1. What would have been the likelihood I picked p in my draw ? : 1/(p+1)
1/p > 1/(p+1).
So given that I picked p , the estimate of n that would give the best probability of me picking p is p itself.
So best estimate of n is p
If say n = p+1. What would have been the likelihood I picked p in my draw ? : 1/(p+1)
1/p > 1/(p+1).
So given that I picked p , the estimate of n that would give the best probability of me picking p is p itself.
So best estimate of n is p
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