I created six hundred coins. I tell you that each is red on one side, but may be red or blue on the other side. I flip each coin, and show you the resulting colors. You count 400 red and 200 blue. What is your best estimate of the number of coins that are red on both sides?

I flipped all the same coins again, and you count 350 red and 250 blue. How should you modify your estimate?

Well, both of these questions are interesting to me, and neither strike me as straightforward.

Part I)  While I agree with Oskar's part I answer, I can't help thinking about what I would estimate if the first flip was 299 red and 301 blue.  In the absence of an a priori assumption about the distribution, I guess we can't do better than guessing the distribution which gives the highest chance of producing 301 blues, namely that there are 600 red/blue coins and 0 red/red coins.  But I really find myself wishing (unsuccessfully) that there were some basis for determining an a priori distribution, because I hate to conclude on this evidence that there is not a single Red/Red in the bunch, even if it is the most likely point estimate of the actual distribution.

Part II) And I agree that, at a minimum, we need to average the two results together to form a modified estimate.  But averaging is flawed, and you really want something Bayesian here, I think.  Consider if there were three flips, and the number of blue sides displayed on the three flips were 50, 50, and 203.  It would be very wrong to calculate an everage of 101, double it, and estimate that there were 202 red/blue coins.  Clearly there are at least 203!  And I think that it is also wrong to estimate that there are only 203, since it is wildly against the odds that every one of the 203 came up blue on the third flip.  My intuition is that the correct estimate of the number of read and blue coins is something greater than 203, in this case.

Posted by Steve Herman on 2006-12-22 09:59:11