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?

(In reply to Wrong by Tristan)

Under the assumption that all that is known is the number from each trial (rather than which ones where blue)

I get 148 red-only coins (452 red-blue) for part 2 via this method:

we want the number m of red/blue coins with the highest likelihood of both observations, i.e. the product of the likelihoods.

obviously 250 is a minimum and the likelihood will increase from there for a while.

so, l(m) = ((m choose 200) / 2^m) * ((m choose 250)/2^m =

to find a max, lets look at l(m)/l(m-1).  When this hits 1 the likelihood will start to decrease.
l(m)/l(m-1) = m^2/(m-200)(m-250)*4 = 1 (point it changes)
solving the quadratic gives: 452.75...
So, at 452 l(m) is greater than l(m-1) and at 453 it is smaller.

Thus, m=452 is the most likely.

Posted by Joel on 2006-12-23 01:13:04