- #1
Balfo
- 7
- 0
Hi, I have a question about the famous Bertrand's box paradox.
In this puzzle there are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each. After choosing a box at random and withdrawing one coin at random that happens to be a gold coin, the question is what is the probability that the other coin is gold. As in the Monty Hall problem the probability is 2/3.
So my question is, what would the answer to the same question be if we eliminated the box with the two silver coins, or if we added two more boxes with mixed coins, or if for example we added 7 more boxes with two silver coins, so now we had 10 boxes altogether?
I think it would be easier to understand the paradox with answers to a few variations like these, thanks in advance,
Balfo
In this puzzle there are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each. After choosing a box at random and withdrawing one coin at random that happens to be a gold coin, the question is what is the probability that the other coin is gold. As in the Monty Hall problem the probability is 2/3.
So my question is, what would the answer to the same question be if we eliminated the box with the two silver coins, or if we added two more boxes with mixed coins, or if for example we added 7 more boxes with two silver coins, so now we had 10 boxes altogether?
I think it would be easier to understand the paradox with answers to a few variations like these, thanks in advance,
Balfo