This cannot be right -- something is not getting defined properly. Let's look at Fred's first question: having drawn m coupons with k distinct, what is the probability that there are n available values? Suppose m=k=1! If we follow Erich Friedman's suggestion, we get a probability of 1 for all values of n! A little less flippantly: suppose m=k=2. We have drawn two distinct coupons; In Friedman's method, the probability of this outcome in a 1-coupon universe is 0; in a 2-coupon universe it's 1/2. In a 3-coupon universe it's 2/3. In a universe with n coupons, the probability that the first two draws are distinct is (n-1)/n -- so Friedman's method prefers larger values of n, without limit. If I draw 2 distinct coupons, Friedman says it is more likely that there are two million coupons than that there are six. Something is wrong here; I apologize that I have not been able to point out the problem more coherently. On Wed, Feb 3, 2016 at 1:55 PM, Erich Friedman <erichfriedman68@gmail.com> wrote:
I don't know in advance how many distinct coupons are available, but have collected m among which there are just k distinct. What is the most likely value of n ?
i give a similar problem in my intro prob class. i would approach this with a maximal likelihood approach.
assuming each coupon is equally likely to appear. for a given value of n, calculate the combinatorial probability of getting what you have collected. then choose the value of n that maximizes this probability.
but i don't know how to approach the asymptotics.
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