One of the books that I'm reading made mention of "the three box puzzle" that I've encountered once or twice in the past. It seemed like a fine subject for a post on this blog, and so here we are.
For those not familiar with it, the scenario is as follows:
You are presented with three identical boxes, one of which contains a prize of some sort. You aren't told which box has the prize, but the PuzzleMaster knows, and he invites you to pick one of the three boxes as your guess. Once you've made your selection, he doesn't tell you whether you're right or wrong in your choice. Instead, he chooses one of the other two boxes and shows you that that box is, in fact, empty. He then asks, "Would you like to change your pick now?" What should you do?
From a probability point of view, there's a right answer to this puzzle (in other words, an answer that will give you a greater likelihood of ending up with the prize). I'm going to turn Comment Moderation on for a day or two so that people can post their responses, if they like, and not influence, or be influenced by, each other. Then I'll add the correct answer as a Comment for all of us, and future generations of blog readers, to enjoy...
Subscribe to:
Post Comments (Atom)
8 comments:
Definitely you should switch.
Originally you had a 1/3 chance of being right. Which means the chance of the prize being in one of the two unselected boxes is 2/3.
When the host eliminates the othe empty box switching your choice means you have a 2/3 chance of getting the prize.
You should change your pick. You have better odds if you do. I could try to explain the math, but then my head would explode.
This is the classic monty hall problem. You choose the third box. Unfortunately, I don't remember how the probabilities break down off-hand, but let's see if I can derive it now. The initial choice is a 1-in-3 probability - all three have equal likelihood of being the correct one. As soon as one of the remaining boxes is shown to not be the prize, the probability breakdown changes.
This is the hard part and I really can't get past it at this time of night, so I'll think about it and post again later. But choose the untouched box.
I'd switch to the other box. So when you first choose you have a 1 in 3 chance. Once he's eliminated one....then isn't it 50/50 for the other box but still 1/3 for the box in your hand?
Ah, the Monty Hall problem. Classic!
Always choose the other door and increase your probability of a win from 1/3 to 2/3.
Personally, I've always thought it would be neat to have a goat for my backyard. I'm just sayin'...
Looks like those who are going to chime in on this brain teaser have now done so, and to a person they've made the right choice! Good job, everyone! It warms my heart to know that I'm chumming around with such a smart set of folks!
Anyway, the right answer is to switch boxes when given the chance. Several respondents included explanations for why that is so, and I'll add my own now (which overlaps somewhat with what others said).
I don't have one of those brains that can immediately solve puzzles like this one as soon as I hear it, but I can usually work through it logically. In this case, I looked at it a couple of ways:
1) Suppose we call the 3 boxes A, B and C, and suppose that you picked A originally. At that point, each of A, B and C had equal probability (1 in 3) of holding the prize. More specifically, though, A had a 1/3 chance whereas B & C had a combined 2/3 chance. By eliminating one of B or C, the PuzzleMaster left you with a 2/3 chance that the remaining, unopened B or C box now holds the prize. There's still a 1 in 3 probability that A holds the prize, in which case you'll kick yourself for switching away from it, but the odds are in your favour by a 2-to-1 margin if you abandon A and go for whichever of B and C hasn't yet been opened.
2) A similar but slightly different perspective is this: Since the PuzzleMaster knows where the prize lies (inside A, B or C), he's either randomly opening B or C (because he knows that the prize is in A) or he's intentionally opening the one box (among B and C) that he knows is empty. There's only one scenario for the former case (the one in which you picked the right box) but there are two for the latter one (you've wrongly picked A and the prize is in B, so he opens C; or you've wrongly picked A and the prize is in C, so he opens B). In other words, you have one shot if you stick with your original choice but two if you switch (even though it seems like you only have one choice in the latter, as well, because there's only one box left to choose).
In both views, the odds are twice as good if you switch as they would be if you stood pat. Fun, huh?
Somebody else asked this one last night ... a different variation. I cannot get my head to what everyone says is the right answer. I tried all day before reading this tonight.
OK with the 33% chance for box A initially. But the only reason there is a 66% chance for Band C is because there are 2 of them. The minute one is gone, A and B are left. one has it. One doesn't. I think probability changes, but I think at that point it changes for both. So now it is 50/50. Somebody on 'deal or no deal' must know the answer to this, I'm sure.
Anyway my other friend got it from a math text. But I have been mulling it over all day and can't get there from here.
Based on Sue G's comment, I'll try one more approach to explaining this, which occurred to me only after I'd posted my original comment.
Imagine that the scenario is run 3 times in succession. In each case, you'll make the same choice (either to stick with box A or to switch to the one of B or C that hasn't been opened yet). And the PuzzleMaster will put the prize in a different box each time.
So, if you stick with box A, you'll be right 1 time out of 3. If you use the "switch to B or C" approach, then you'll be right the other 2 times. Therefore, statistically speaking, you're always better off switching.
Does that help, Sue?
Post a Comment