Extended Bertrand's Boxes Problem - 3 boxes (2 coins per box)

Question

Apologies for the long text, but it is to give context to the exercise.

I am having some trouble understanding an extended version of Bertrand's paradox. It comes in a few questions I have answered most, but the last one, I couldn't determine if my answer is the correct one.

Basically, you have 3 boxes, each containing 2 coins, the first one contains 2 gold coins, the second contains one gold coin and one silver coin, and the third one contains 2 silver coins.

Question 1 : you pick one box at random, and take a coin (and DON'T LOOK) what is the probability that the remaining coin is a gold coin ? Answer : all events have the same probability so here it is 1/2

Question 2 : you open your palm, you see that you have a gold coin, what is the probability that the remaining coin is a gold coin ? Answer : 2/3 (out of the 3 gold coins you could get in your palm only 2 are in a box where the remaining would be a gold coin).

Question 3 : you put the gold coin back, and IN THE SAME BOX, you pick a coin, and don't look again, what is the probability that the remaining coin is a gold coin ? Answer : Here you operate under the condition that you have picked a gold coin before, so 2/3 probability of having chosen the first box, and 1/3 probability of having chosen the second box, so the probability that the remaining coin is a gold coin is : $ 1* \frac{2}{3} +\frac{1}{2}*\frac{1}{3} = \frac{5}{6} $

Question 4: you open your palm and you see that you had a gold coin, what is the probability that the remaining coin is a gold coin ? Answer : Here you operate under the condition that you have picked a gold coin before, AND that you picked in the same box a gold coin, in the question 3, having a gold coin remaining, or picking a gold coin are equivalent, so here there can only be a gold coin remaining if you picked from the first box, so the answer is : $ \frac{\frac{2}{3}}{\frac{5}{6}} = \frac{4}{5} $

Question 5 : you get rid of the coin that you picked (there only remains 5 coins now), and pick a box (out of the 3) at random, pick a coin, what is the probability that the remaining coin is a gold coin ?

My Answer is : $ \frac{3}{5} $, basically, I compute conditional on the fact that we eliminated a gold coin from the first box, and conditional on the fact that we eliminated a gold coin from the second box.

First box case, the probability of eliminating that coin is $ \frac{4}{5} $, which as computed before, is the probability of having picked the "eliminated" gold coin from box #1. That said, let $B_{i} , i = 1,2,3$ be the event of picking a coin from box #$i$, we call $O$ the event of having a gold coin remaining in the chosen box, and $G_{i}, i = 1,2$ the event that the eliminated coin is coming from box #$i$ , we have , $P(G_{1}) = \frac{4}{5} $, and $P(G_{2}) = \frac{1}{5} $ and :

$ P(O) = P(O,G_{1}) + P(O,G_{2}) $, with : $P(O,G_{1}) = P(O|G_{1}) P(G_{1}) = P(O,B1|G_{1}) P(G_{1}) + P(O,B2|G_{1}) P(G_{1}) + P(O,B3|G_{1}) P(G_{1}) = 0 * \frac{4}{5} + \frac{1}{2} * \frac{4}{5} + 0* \frac{4}{5} = \frac{2 }{5}$

And : $P(O,G_{2}) = P(O|G_{2}) P(G_{2}) = P(O,B1|G_{2}) P(G_{2}) + P(O,B2|G_{2}) P(G_{2}) + P(O,B3|G_{2}) P(G_{2}) = 1 * \frac{1}{5} +0*\frac{1}{5} + 0* \frac{1}{5} = \frac{1 }{5}$

Total is $\frac{3}{5}$, but the solution says $\frac{2}{5}$, because there are 2 gold coins out of 5, am I missing something ?

Thank you, and sorry again for the long text.

Answer 1

Your first 4 answers look right to me.

In the last question, you have two possible scenarios to begin with. The first is

 1     2     3 
G X   G S   S S  

with probability 4/5. In this case the probability that the remaining coin in the box is gold is the average over the three boxes:

$$ \frac{1}{3}(0 + \frac12 + 0)=\frac16$$

The second scenario, with probability 1/5 is

 1     2     3 
G G   X S   S S  

In this case the probability is

$$ \frac{1}{3}(1 + 0 + 0)=\frac13$$

Then the global probability is

$$ \frac45 \frac16 + \frac15 \frac13 = \frac15$$

So, now we have three answers :-)

Answer 2

Ah Indeed you are right, I see my mistake, I forgot to compute conditional to $B_{i},G_{i}$ instead of what I did, and since $B_{i}$ and $G_{i}$ are independent, $P(B_{i}|G_{i}) = \frac{1}{3}$, so dividing by 3 in my calculation I get the same result as you

Thanks