# Question about conditionalization and probability

I am working on a problem set and I am not sure if I am heading in the right direction. The scenario: "Suppose three identical boxes are presented to you, and you are told that one box contains two gold coins, one box contains two silver coins, and the remaining box contains one gold and one silver coin. You pick a box at random and pull out one coin (without observing the other), and you observe that it is gold.

The question is: What is the probability that the other coin in the box that you selected is also gold? What would you conditionalize upon and how would that change your beliefs.

My initial credence (C1) before knowing that I pull a gold coin is that all scenarios (drawing gold & gold; gold & silver; silver & gold; silver & silver) have a 1/3 chance of occurring. Since I have pulled a gold coin, it is certain that I did not pull out of the silver box. Taking that into consideration, I am thinking that the probability that the other coin in the box is also gold is 1/2. You already have the gold coin in your hand, so that leaves you with two options: either a gold coin or a silver coin in the box, so that gives you 1/2 chance that the other coin is gold.

• This question is more suitable for Math SE. – Conifold Dec 17 '19 at 0:34
• In addition to en.wikipedia.org/wiki/Bertrand%27s_box_paradox given by @AdamSharpe, which directly answers your specific problem, you might also want to look at the similar probability-brain-teaser Monty Hall Problem, en.wikipedia.org/wiki/Monty_Hall_problem which likewise confounds your intuitive reasoning. – user19423 Dec 17 '19 at 5:18
• By the way, I'm not googling it, but the paradox seems to also work, and even more easily, with just two boxes -- {gold,silver} and {gold,gold}. If you first happen to randomly select a gold coin from a random box, then you might think your odds are 50-50 that the other coin in that same box is gold. But your odds are again two-thirds. – user19423 Dec 18 '19 at 5:54

This is called Bertrand's box paradox. It's a "paradox" in the sense that the correct answer, P(other_coin_is_gold) = 2/3, is unintuitive to most people at first.

To get a feel for the correct answer, break it down into cases. Let's call the box with 2 gold coins "G", the box with two silver coins "S", and the box with one of each "M" (for "mixed"). Further, imagine each coin in each box being labeled "1" or "2".

Then, one of six things can happen when you pick the first coin out.

Case 1: You have box G, and you pick coin number 1; the coin is gold.

Case 2: You have box G, and you pick coin number 2; the coin is gold.

Case 3: You have box S, and you pick coin number 1; the coin is silver.

Case 4: You have box S, and you pick coin number 2; the coin is silver.

Case 5: You have box M, and you pick coin number 1; the coin is gold.

Case 6: You have box M, and you pick coin number 2; the coin is silver.

You're interested in the conditional probability of the other coin being gold given that the first coin pulled was gold. Since the first coin you picked is gold, you can eliminate cases 3, 4, and 6. So, case 1, 2, or 5 must be true. And in 2/3 of those cases, the other coin is gold.

If you wanted to approach these and other similar problems more formally, you could use Bayes' theorem:

``````EVENTS
* box_g: you are holding the box that has two gold coins
* first_is_g: the first coin that you drew was gold

We're interested in the conditional probability of us having box G (and hence that that other coin is gold), given that the first coin is gold ("box_g given first_is_g").

BAYES' THEOREM:

P(first_is_g | box_g) * P(box_g)
P(box_g | first_is_g) = --------------------------------
P(first_is_g)

WHAT WE KNOW
* We know that P(first_is_g), the prior probability of the first coin being gold is 1/2.
* We also know that P(box_g), the prior probability of selecting box G is 1/3.
* Finally, we know that P(first_is_g | box_g), the probability of the first coin being gold given we chose box G is 1.

Plugging in all these values yields the correct result of P(box_g | first_is_g) = 2/3
``````

You wrote:

My initial credence (C1) before knowing that I pull a gold coin is that all scenarios (drawing gold & gold; gold & silver; silver & gold; silver & silver) have a 1/3 chance of occurring. Since I have pulled a gold coin, it is certain that I did not pull out of the silver box. Taking that into consideration, I am thinking that the probability that the other coin in the box is also gold is 1/2. You already have the gold coin in your hand, so that leaves you with two options: either a gold coin or a silver coin in the box, so that gives you 1/2 chance that the other coin is gold.

You are correct that there are two possible events, you can either draw a silver coin or a gold coin. But what you did not take into account is that there are more ways for the second coin to be gold, as can be seen in our analysis of the six cases above.

In addition to @AdamSharpe's analysis of "break it down into [six] cases", you can also analyze the situation your intuitive way, i.e., first choose a box, and then choose a coin from that box.

Start with all three of your boxes: let's call them GG, SS and GS, with the obvious meanings. Then, first of all, if you happen to choose the SS box, then your subsequent coin choice will necessarily be S, and you'll just immediately discard that entire trial.

So let's just focus on the remaining two GG and GS boxes. If you choose the GG box then your coin choice will necessarily be G, so you'll never discard any trial where you've chosen the GG box. But if you choose the GS box, then half the time your coin choice will be S, so you'll discard half the trials where you've chosen the GS box.

That means you'll end up with the GG box twice as often as you end up with the GS box. And with the GG box, the other coin is always G. And with the GS box, the other coin is always S. So, since you end up with the GG box twice as often, the other coin will likewise be G twice as often.

So, two times out of three, when the coin you choose is G, the other coin is also G. qed.