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"I have three boxes, each with two compartments.

  • One has two gold bars

  • One has two silver bars

  • One has one gold bar and one silver bar"

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  • You choose a box at random, then open a compartment at random.

  • If that bar is gold, what is the probability that the other bar in the box is also gold?

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This is just an old problem that I found in an old newspaper.

So I simply looked at the question and I said "$\frac 12$" easy peasy.

My first thoughts: $1$ in $2$ chance

  • Since there are only two boxes with a gold bar in it, I reasoned, I must have picked one of these.

  • Since one has a gold bar and the other has a silver bar on the other side, the probability that I have another gold bar is $\frac 12$.

$$Right?$$

Well anyway, I checked the solutions at the bottom of the newspaper page and it said I was wrong!!! According to the newspaper, the actual answer is $\frac 23$ chance. However, the newspaper didn't offer any explanation so I was left dumbfounded. So I went back to the drawing board and racked my head to see if I could somehow conjure the number $\frac 23$ from the problem. No such luck.

So here is my question: could someone please explain the answer?

Any guidance hints or help would be truly greatly appreciated. Thanks in advance :)

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    $\begingroup$ Have you heard of bayes theorem? Or similarly look at the monty hall problem? $\endgroup$
    – Chinny84
    Commented Jun 22, 2015 at 10:26
  • $\begingroup$ @Chinny84 No I have not. $\endgroup$
    – anonymous
    Commented Jun 22, 2015 at 10:28
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    $\begingroup$ You haven't chosen a box at random, you've chosen a compartment. There are in total three compartments that have a gold bar in them. Two of those have gold in their partner compartment. $\endgroup$
    – Arthur
    Commented Jun 22, 2015 at 10:38
  • $\begingroup$ @Arthur Not particularly well understood. $\endgroup$
    – anonymous
    Commented Jun 22, 2015 at 10:39
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    $\begingroup$ @anonymous This is a shame, because Arthur's comment states everything that one needs to neatly solve the problem. Maybe try harder? $\endgroup$
    – Did
    Commented Jun 22, 2015 at 10:47

5 Answers 5

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Assume each of the gold bars have the number 1,2 or 3 stamped respectively on their underside where you cannot see it and the gold bars stamped 1 and 2 are in the same box.
You randomly pick a box and after looking in one compartment you see a gold bar.
Knowing that you have picked a box which contains a gold bar, there are three possibilities.

1) The gold bar has a 1 stamped on it therefore you have chosen the box containing the two gold bars.
2) The gold bar has a 2 stamped on it therefore you have chosen the box containing the two gold bars.
3) The gold bar has a 3 stamped on it therefore you have chosen the box containing one gold bar and one silver bar.

The above three events are all equally probable therefore the probability that the other bar in the box is also gold is $\frac{2}{3}$.

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If you picked a box, and without opening any compartments were told that "There is gold inside this box", then yes, there's a $1/2$ chance that it's double gold. But that's not what's happening.

You pick a box, and instead of asking "Does this box have gold in it?" you open a compartment and check. That means that if you picked the box with one of each, there's a $1/2$ chance that you will open a silver and reset the whole trial. That means that where there originally were probability $1/2$ of success, you throw away half the failures. That makes the probability of winning $2/3$.

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the probability describing this event $P(A)$ is to choose a box that has two gold bars in them. We have three ways to choose a compartment that has a gold bar in it, and two ways to choose a compartment such that the second compartment also has gold bar in it.

The reason why there is two ways, and not one, is because this problem of course assumes that the choice of the compartment is independent. I like to think that there is two ways to open the box with the two gold bars, where the solution, $P(A)=\frac{2}{3}$ arises.

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The mechanical solution

In this case the set of possible elementary events is $$\Omega=\{(f,u),(f,l),(s,u),(s,l),(th,u),(th,l)\},$$

where $u$ refers to "upper", "l" refers to "lower" and $f,s,th$ refer to "first", "second", and "third", respectively. For example $(f,u)$ means that you open the upper compartment of the first box.

The probabilities of these elementary events equal, all are $\frac{1}{6}$. The conditional probability sought is

$$P(\{(f,u),(f,l)\} \mid \{(f,u),(f,l),(th,u)\})=$$ $$=\frac{P(\{(f,u),(f,l)\}\cap \{(f,u),(f,l),(th,u)\})}{P(\{(f,u),(f,l),(th,u)\})}=$$ $$=\frac{P(\{(f,u),(f,l)\})}{P(\{(f,u),(f,l),(th,u)\})}$$ $$=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}.$$

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The only way in which you can get the second bar gold as well is if you have selected the first box (the one with two golds). Notice that since both gold bars are considered different, there are actually 2 ways in which you can get both bars gold. And the third instance is where you have originally chosen the box with a gold and a silver. Hence it is 2/3. The whole thing revolves around the fact, the box having two golds can produce two different results where both bars selected are gold.

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