> 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
The key to the problem is that the solution treats the two coins in each box as separate scenarios (Choosing the coin on the right side of box 1 is not the same as choosing the coin on the left side of the same box).
Let’s look at the three possibilities:
1. We have chosen the only gold coin in box 3 (SILVER, GOLD*).
2. We have chosen the left drawer’s coin in box 1, which contains (GOLD*, GOLD).
3. We have taken the right drawer’s coin in box 1 (GOLD, GOLD*).
In case (A), the remaining coin in the box is silver.
In case (B), the other coin is gold.
In case (C), the remaining coin is also gold.
As you can see, out of the three possible cases, knowing that the first coin is gold, two are favorable. The probability of the other coin being gold is then twice the probability of it being silver. Therefore, the requested probability is 2/3, which is 66.67%, greater than the intuitively calculated probability of 1/2 or 50%.
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