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View Full Version : If Monty Hall were more devious... (puzzle)

Half Man Half Wit
08-28-2010, 06:25 AM
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:

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!):
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.

Palooka
08-28-2010, 07:35 AM
Unless I'm misreading this, it's faulty because the second box is independent from the first. So the probability is some unknown of whatever the producer felt like doing.

How I've read it is you're saying there is a green box and it does contain a prize and you've the option of taking the second box. To win, it needs to be a second green box.

Half Man Half Wit
08-28-2010, 07:49 AM
Unless I'm misreading this, it's faulty because the second box is independent from the first. So the probability is some unknown of whatever the producer felt like doing.

How I've read it is you're saying there is a green box and it does contain a prize and you've the option of taking the second box. To win, it needs to be a second green box.

No, the other box doesn't need to be green. The second box is independent from the first only in the case of infinitely many colours, were the probability of both boxes being the same colour becomes infinitesimal.

However, I should probably be more explicit that absent all other information, a box either contains a prize or not with equal probability; just think of boxes in general as things that can be in either of two states, one with a prize and one without.

Little Nemo
08-28-2010, 08:26 AM
However, I should probably be more explicit that absent all other information, a box either contains a prize or not with equal probability; just think of boxes in general as things that can be in either of two states, one with a prize and one without.The equal probability is a pretty important point. Otherwise this situation falls into the common fallacy of thinking that "there's two possibilities - yes or no" equals a 50/50 chance.

As written the OP is essentially asking "You have a choice between taking a box, which may or may not contain a prize, or picking a door, which has a one third chance of having a prize." Now my general observation of the world is that the vast majority of boxes do not contain prizes. (On a side note, I found out last week that Crackerjack and breakfast cereal boxes no longer contain prizes inside - you have to send away for the prizes. Another childhood memory lost to time.) So I would be better off picking the door with a one third probability.