So, this gameshow host informs you that, in his possession, he has two boxes, each of which can be either red, green, or blue. He also tells you that at least one of the two is green, and contains a prize. Then, he gives you the following choice: You can either have the other box (which may or may not contain a prize), or you can choose one of three doors, behind one of which there’s again a prize (all prizes are the same).
Which should you choose?
I’ll first give the answer, so those playing along at home can compare, then I’ll give two hints if your answer didn’t match up with mine, and at the end, I’ll give the full solution:
Answer:
You should take the other box. (However, the probability that it contains a prize is not 1/2!)
Hint 1:
The right question to ask is: “What is the probability that both boxes contain a prize?”
Hint 2:
Yes, the colours are important.
Solution (no peeking!):
[spoiler]No, ‘it doesn’t matter’ isn’t the right answer. The probability that the other box contains a prize isn’t 1/3. In fact, it is slightly smaller than 1/2 (5/11), and hence, you should take it. To see this, consider the following:
Box 1
| contains prize | contains no prize|
| | |
---------------------------------------------------------------------
| | |
contains | X | X |
prize | | |
Box 2 ---------------------------------------------------------------------
| | |
contains | X | |
no prize | | |
---------------------------------------------------------------------
We have four possible cases, of which the three cases marked with ‘X’ are those not ruled out by the information that at least one box contains a prize; since only one of those cases amounts to both boxes, and hence the second, containing a prize, the probability for that appears, naively, to be 1/3, right?
But you have to also consider the information about the colours: there are three possible ones, and at least one of the boxes containing a prize has to be green. Thus, the diagram now looks like this:
Box 1
| contains prize | contains no prize|
| R G B | R G B |
---------------------------------------------------------------------
R | X | |
contains G | X X X | X X X |
prize B | X | |
Box 2 ---------------------------------------------------------------------
R | X | |
contains G | X | |
no prize B | X | |
---------------------------------------------------------------------
We have now 6 * 6 = 36 cases in total, of which 11 are not ruled out by the information ‘one box contains a prize and is green’; of those 11, 5 give the desired result of both – and thereby, the other one – containing a prize. The probability that the other box contains a prize as well is then 5/11 = 0.4545…, significantly exceeding the 1/3 for guessing the right door.[/spoiler]