## Joseph Bertrand and the Monty Hall Problem

The death a few days ago of Monty Hall reminded me of something I was going to write about the *Monty Hall Problem*, as it did with another blogger I follow, namely that (unsrurprisingly) Stigler’s Law of Eponymy applies to this problem.

The earliest version of the problem now called the Monty Hall Problem dates from a book, first published in 1889, called *Calcul des probabilités* written by Joseph Bertrand. It’s a very interesting book, containing much of specific interest to astronomers as well as general things for other scientists. Ypu can read it all online here, if you can read French.

As it happens, I have a copy of the book and here is the relevant problem. If you click on the image it should be legible.

It’s actually Problem 2 of Chapter 1, suggesting that it’s one of the easier, introductory questions. Interesting that it has endured so long, even if it has evolved slightly!

I won’t attempt a full translation into English, but the problem is worth describing as it is actually more interesting than the Monty Hall Problem (with the three doors). In the Bertrand version there are three apparently identical boxes (*coffrets*) each of which has two drawers (*tiroirs*). In each drawer of each box there is a medal. In the first box there are two gold medals. The second box contains two silver medals. The third box contains one gold and one silver.

The boxes are shuffled, and you pick a box `at random’ and open one drawer `randomly chosen’ from the two. What is the probability that the other drawer of the same box contains a medal that differs from the first?

Now the probability that you select a box with two different medals in the first place is just 1/3, as it has to be the third box: the other two contain identical medals.

However, once you open one drawer and find (say) a silver medal then the probability of the other one being different (i.e. gold) changes because the knowledge gained by opening the drawer eliminates (in this case) the possibility that you selected the first box (which has only gold medals in it). The probability of the two medals being different is therefore 1/2.

That’s a very rough translation of the part of Bertrand’s discussion on the first page. I leave it as an exercise for the reader to translate the second part!

I just remembered that this is actually the same as the three-card problem I posted about here.

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October 4, 2017 at 10:30 pm

I think you’ve got this wrong. If you find, say, a silver medal, then there are three ways that this could have happened, each equally probable: one of the silver/gold combination, one of the silver/silver combination, or the other of the silver/silver combination. The probability of the silver/gold combination is therefore 1/3. Applying Bayes theorem works equally well.

October 4, 2017 at 10:48 pm

I was translating the text from the first part of Bertrand’s statement of the problem…

The answer of 1/3 is the probability that the you picked the silver/gold box in the first place. However, opening one drawer and finding that it is silver allows you to update the probability that it is the silver/gold box by eliminating the possibility that it it gold/gold.

October 4, 2017 at 11:21 pm

I agree with Andrew above. It is easy to see if you consider 300 trials of this game, ignoring shot noise:

100 times you will pick the GG box and see a gold, 100 times you will pick the SG box and see silver 50 times and gold 50 times; and 100 times you will pick the SS box, and see 100 silver. So, if the first one opened is silver, then there is a 2/3 probability that you are holding the SS box, and 1/3 SG.

(NB the above assumes the SG box is randomised top/bottom:

if the silver in SG were always on top and you see silver on top, then it would be 50/50; or if the silver in SG is always on the bottom and you see silver on top then it’s certainly SS. )

You can also see 1/2 is wrong because the problem is symmetric between S & G: the initial probability is 2/3 of getting the same colour in both drawers, and since it is 50/50 whether you see S or G first, opening one drawer doesn’t actually change this.

October 4, 2017 at 11:48 pm

As I tried to explain, I was just translating the first bit of Bertrand’s text (on the first page)….The next page criticises the argument he just presented.

See also here for a similar problem..

https://telescoper.wordpress.com/2016/02/29/the-three-card-puzzle/

October 5, 2017 at 11:21 pm

The answer is 1/3, even after opening one drawer, because you have three possible drawers left and only one of them contains a unique color.

October 6, 2017 at 3:02 pm

But if there are an up drawer and a down drawer and you know that the box containing the two different types of medals has the up drawer containing a gold medal, what are the odds then?

October 7, 2017 at 2:40 am

I meant top drawer and bottom drawer, not up and down…

October 7, 2017 at 6:47 am

Or strange and charmed?