Bertrand’s box paradox

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> There are three boxes:
> – a box containing two gold coins,
> – a box containing two silver coins,
> – a box containing one gold coin and one silver coin.
>
> Choose a box at random. From this box, withdraw one coin at random. If that happens to be a gold coin, then what is the probability that the next coin drawn from the same box is also a gold coin?

My thinking is this… Taking a box at random would be 33% for each box. Because you got one gold coin it cannot be the box with TWO silver coins, therefore the box must be either the gold and silver coin or the box with two gold coins. Each of which is equally likely so the chance of a second gold coin is 50%

I understand that this is a veridical paradox and that the answer is counter intuitive. But apparently the real answer is 66% !! I’m having a terrible time understanding how or why. Can anyone explain this like I was 5?

In: Mathematics

11 Answers

Anonymous 0 Comments

The phrase that helped me understand is “all coins are equally likely”. Don’t think of it as selecting a box, then selecting a coin, but just immediately selecting a coin at random. Two of the gold coins will get you into the “2 golds” box, and only one gold coin gets you into the “1 gold” box. Since all coins are equally likely, the percentage 66% follows.

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