Bertrand’s Box of Paradox

What we think at first is not always true, and we are often tricked by our own myopic thinking. There are 3 boxes: one which has two gold bars, one which has a gold and silver bar, and one which has two silver bars. If you pick up a bar randomly from a randomly chosen box, give that it is a gold bar, what is the likelihood that the other bar is silver? Look at the picture carefully and think about it, and then scroll below the picture if you want to see what the answer is.

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The immediate thought is that the chances are ½, because one of the boxes with gold has two gold bars and the other has a gold and silver. Let’s look at the cases of which bar you could pick up, and what the other bar is.
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Upon thinking about the different options, you realize that there are two ways to choose a gold bar for the other to be gold, while there is only one way to choose a gold so that the other is silver. Which results in a chance of 1/3. Sound familiar? Yep, this is essentially a different form of the Monty Hall Problem, which I’m sure you’ve heard of (otherwise be sure to look it up). To be honest, I’m quite sure most people will fall for this one, and found it to be an interesting variation of a widely known question (people who have already seen the Monty Hall problem will likely be fooled as well). An easy way to think about this is to imagine that there are 10 gold bars in one box, and one gold bar and 9 silvers in another (and obviusly the last box with 10 silvers). The question would now be given that you picked a gold, what is the likelihood that the other bars are silver. One could quickly realize that there would be a large likelihood of there being gold seeing the large number of gold bars in that box. This idea is also applicable for the Monty Hall problem, where you imagine 10 doors and only one door with a car behind it. Thinking of the same problem in a different light (or the same light in a different magnitude) often gives unexpected results, and I never fail to be amazed by it.

 

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