{"id":1286,"date":"2024-05-20T14:14:21","date_gmt":"2024-05-20T14:14:21","guid":{"rendered":"https:\/\/www.allendowney.com\/blog\/?p=1286"},"modified":"2024-05-20T14:14:21","modified_gmt":"2024-05-20T14:14:21","slug":"bertrands-boxes","status":"publish","type":"post","link":"https:\/\/www.allendowney.com\/blog\/2024\/05\/20\/bertrands-boxes\/","title":{"rendered":"Bertrand\u2019s Boxes"},"content":{"rendered":"\n

An early draft of Probably Overthinking It included two chapters about probability. I still think they are interesting, but the other chapters are really about data, and the examples in these chapters are more like brain teasers — so I’ve saved them for another book. Here’s an excerpt from the chapter on Bayes theorem.<\/em><\/p>\n\n\n\n

In 1889 Joseph Bertrand posed and solved one of the oldest paradoxes in probability. But his solution is not quite correct \u2013 it is right for the wrong reason.<\/p>\n\n\n\n

The original statement of the problem is in his Calcul des probabilit\u00e9s<\/em> (Gauthier-Villars, 1889). As a testament to the availability of information in the 21st century, I found a scanned copy of the book online<\/a> and pasted a screenshot into an online OCR server. Then I pasted the French text into an online translation service. Here is the result, which I edited lightly for clarity:<\/p>\n\n\n\n

\n

Three boxes are identical in appearance. Each has two drawers, each drawer contains a medal. The medals in the first box are gold; those in the second box, silver; the third box contains a gold medal and a silver medal.<\/p>\n\n\n\n

We choose a box; what is the probability of finding, in its drawers, a gold coin and a silver coin?<\/p>\n\n\n\n

Three cases are possible and they are equally likely because the three chests are identical in appearance. Only one case is favorable. The probability is 1\/3.<\/p>\n\n\n\n

Having chosen a box, we open a drawer. Whatever medal one finds there, only two cases are possible. The drawer that remains closed may contain a medal whose metal may or may not differ from that of the first. Of these two cases, only one is in favor of the box whose parts are different. The probability of having got hold of this set is therefore 1\/2.<\/p>\n\n\n\n

How can it be, however, that it will be enough to open a drawer to change the probability and raise it from 1\/3 to 1\/2? The reasoning cannot be correct. Indeed, it is not.<\/p>\n\n\n\n

After opening the first drawer, two cases remain possible. Of these two cases, only one is favorable, this is true, but the two cases do not have the same likelihood.<\/p>\n\n\n\n

If the coin we saw is gold, the other may be silver, but we would be better off betting that it is gold.<\/p>\n\n\n\n

Suppose, to show the obvious, that instead of three boxes we have three hundred. One hundred contain two gold medals, one hundred and two silver medals and one hundred one gold and one silver. In each box we open a drawer, we see therefore three hundred medals. A hundred of them are in gold and a hundred in silver, that is certain; the hundred others are doubtful, they belong to boxes whose parts are not alike: chance will regulate the number.<\/p>\n\n\n\n

We must expect, when opening the three hundred drawers, to see less than two hundred gold coins the probability that the first that appears belongs to one of the hundred boxes of which the other coin is in gold is therefore greater than 1\/2.<\/p>\n<\/blockquote>\n\n\n\n

Now let me translate the paradox one more time to make the apparent contradiction clearer, and then we will resolve it.<\/p>\n\n\n\n

Suppose we choose a random box, open a random drawer, and find a gold medal. What is the probability that the other drawer contains a silver medal? Bertrand offers two answers, and an argument for each:<\/p>\n\n\n\n