Bertrand's box problem

I recently wrote an article dissecting the famous and particularly deceptive Monty Hall problem from a mathematical perspective (which might come in handy if you happen to participate in any game shows in the future).
When researching the historic origins of this paradoxical brain teaser I encountered several other interesting problems which can be considered to be somewhat related to the former. One of these and the first documented prototype of this kind is Bertrand's box paradox. While it may be not as treacherous as Monty Hall, it reveals very similar flaws in our intuitive understanding of (conditional) probability.

Pixabay by hamiltonleen

The problem

Suppose there are 3 boxes with 2 separate compartments each. Each of the 6 total compartments contains exactly one precious metal bar, either silver or gold, for a total of 3 silver and 3 gold bars distributed in the following way:

• Box 1: gold and gold (GG)
• Box 2: gold and silver (GS)
• Box 3: silver and silver (SS)

Now someone will show you one randomly chosen out of the 6 compartment revealing its content but none of the others. (Maybe this person is trying to sell one of the boxes to you but does not want you to see its complete content.)
Suppose you are seeing a gold bar, what is the probability of the other compartment of that box containing another gold bar?

The intuition

Let's take the intuitive approach and say that there are 3 boxes in total. You are seeing one gold bar so obviously you are not presented box number 3 (SS) because it does not contain any gold. So it could be either box 1 (GG) or 2 (GS) and because the initial choice was uniformly distributed the probability should be 1/2.

The math

But unfortunately the deceptive nature of riddles strikes again: The answer is not 1/2 but rather 2/3.
We are experiencing the Monty Hall situation all over again, the human brain does not understand conditional probability very well.
You see, just because the initial choices of the compartment were uniformly probable our conditional probabilities of seeing a second gold bar or not do not have to be.

With a somewhat refined intuition this result can be made plausible:
There are 6 equally probable initial choices for the compartment, 3 for a silver and 3 for a gold one. We are seeing gold so we condition this on 3 again equally probable choices for gold. But 2 of the 3 choices represent one of the 2 gold bars in (GG) and 1 choice the one gold bar in (GS).
So with a probability of 2/3 we have box (GG) and thus a second gold bar, and with 1/3 we have (GS) and no second one.

This can be made rigorous with another result from probability theory, Bayes' theorem (Wiki):
Let us denote the event of seeing gold by A and the events of having box GG,GS,SS in front of you also by GG,GS,SS respectively. We are interested in the conditional probability of having GG (thus a second gold bar) under the condition A which is usually written as P(GG|A). Bayes' law then states:

P(GG|A) = P(A|GG)xP(GG)/P(A)

where:
P(A|GG) = 1 (probability of seeing gold when you know GG is in front of you),
P(GG) =1/3 (probability of choosing a compartment belonging to GG, so 2 in 6 cases) and
P(A)=1/2 (probability of seeing gold in the compartment alltogether so without knowing which box it belongs to, so 3 in 6 cases)

Plug everything in and you get your desired result:

P(GG|A) = 1x(1/3)/(1/2) = 2/3

which corresponds exactly to our result and approach of refined intuition.

I hope you have outsmarted your intuition and wish you a nice day!

Source: Bertrand's box paradox

https://steemit.com/science/@galotta/the-monty-hall-problem-if-you-get-this-wrong-you-are-in-excellent-company