tag:blogger.com,1999:blog-3317748811909502538.post7656412920499775424..comments2019-06-29T14:01:15.876+01:00Comments on Intropy: Probability and sexismDaveoftheNewCityhttp://www.blogger.com/profile/04140446220455064332noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-3317748811909502538.post-27163684671138365692013-04-27T22:44:30.217+01:002013-04-27T22:44:30.217+01:00Say you have N identical boxes, N gold coins, and ...Say you have N identical boxes, N gold coins, and N silver coins. Put two coins into each box in such a way that M boxes have one of each kind. That means that (N-M) have just one kind: (N-M)/2 have two gold coins, and (N-M)/2 will have two silver coins.<br /><br />Pick a box at random. The chances that it has two coins of the same kind are (N-M)/N. But what if, without looking, you reach into a box and pull out a coin in your closed fist so that you still can’t see it. Whatever color it is, there were [M+(N-M)/2]=(N+M)/2 boxes that had a coin of that color, and (N-M)/2 boxes that had two of them. So it would seem the probability this box has two coins of the same color is now (N-M)/(N+M). The probability changed simply because you took a coin out without looking at it. But it would change the same way if you did look, because it can’t depend on what you would see. How can that be?<br /><br />This apparent paradox is known as Bertrand’s Box Paradox, named for Joseph Bertrand who first described it in 1889. He used N=3 and M=1, but the basic idea holds no matter what the values are. His point was that when you count cases for this type of problem, you shouldn’t count all of the cases that fit the observation. You should only count the proportion of each case that, when you allow all observations to be made equally, would result in what you observed. So in the above problem, of the M boxes that have a gold coin and a silver coin, you should only count M/2 of them because there are two different kinds of coins that could have been withdrawn.<br /><br />If Gary Foshee had not mentioned Tuesday, the problem is a Bertrand’s Box Problem with N=4 and M=2. The wrong answer is (N-M)/(N+M)=(4-2)/(4+2)=1/3, It is wrong because it assumes the statement “Gary Foshee has a boy” is logically equivalent with “Gray Foshee tells you he has a boy.” They are not equivalent, because it is possible for Gary Foshee to tell you he has a girl, when he has a boy and a girl. The correct answer is (N-M)/N=(4-2)/4=1/2. And when you add in Tuesday, the problem is a little more complicated, but the result is the same. The answer is not 13/27, it is still ½, and your intuition that “Tuesday” shouldn’t matter is confirmed.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.com