Monty Hall shows you three doors, and explains that behind one of the doors is a billion dollars. Pick the door with the money behind it and you get to keep the money (tax-free!). So you pick a door. Monty then opens one of the doors that you did not pick, and shows you that nothing is behind it. He then offers you the chance of sticking with your original door, or switching to the other unopened door.
Are your odds better if you switch doors, stick with your original choice, or does it not matter?
No fair Googling this one! :)
31 comments:
I pick Carol Merrill. Which one is she behind?
(you're probably too young to remember Carol Merill).
I say your odds are better the second time, since you now have a 1:2 chance instead of the initial 1:3.
Thank you for thinking I'm that young ;)
The question was, are your odds better to stick or switch, or does it not matter? If you are saying the odds after being shown an empty door are 1:2, then I take it you are saying it doesn't matter if you stick or switch?
Hmm... sampling without replacement... My guess is that the odds get better if you switch. Is digging out the prob&stats text cheating, too?? ;)
Go ahead :)
I'd say it's even odds between switching and sticking with the door you first chose. Hey, you know it's one or the other, right? But I didn't pay too much attention to probability in math class...
You don't gain any knowledge from knowing that one of the doors you didn't chose has nothing behind it, since that's the case always. Yeah. I'm going with staying and switching being the same odds.
It depends on how you interpret the phrase "are your odds better if..." Because the real question is, better than what? If you're comparing this time to the last time, then yes, your odds are better because now it's 1:2 instead of 1:3.
But if you're comparing your chance for success ONLY for this round, then it doesn't matter which door you choose because you have an even chance for success regardless.
I'm asking what you should do: stick or switch, based on the odds of this second choice (not compared to your first choice, but just based on your options for your second choice). Or if it makes no difference (because it's even odds at that point).
One point I will clarify that does affect the probability of this situation (and this is a clue!), is that the billion dollars is not moved during the entire course of the event.
I'll also tell you that when I first encountered this problem, and heard the solution, I was so convinced that they were wrong, and my solution was correct, that I wrote a program to simulate this puzzle and ran it through thousands and millions of iterations -- and they were right! Then I finally figured out how to explain it so that it made sense to me.
So far only one of you is close to the correct answer ;)
Switch, switch, I'm sticking with switch.
(lol)
I'll go for loyalty. I don't understand what the advantage would be if you switched. It seems to me you'd have a 50/50 chance. Wouldn't the law of averages apply?
OK, OK, this is my final answer.
If you did the experiment a thousand times, the probability of finding the right door would average out to 1/3.
However, if you only do it once, you have a higher probability of getting the right door if you switch your answer.
?? Come on, Kiki, when do we find out LOL!!
Here's the answer :)
It's best to switch. Your odds are 66% (2/3) that you'll get the right door if you switch.
Like I said earlier, I was so convinced when I heard this that the real answer was 50% (it makes no difference what you do) that I wrote a computer program to simulate the problem, convinced I would then prove my answer. I was shocked when the results showed I was wrong and that it was indeed 66%.
Here's why.
Let's call the doors A, B, and C. You pick A. At that point you have a 33% chance of being right and a 66% change of being wrong.
So, there's a 66% chance that the prize is behind B or C. Remember this. This is the key point.
You are shown that B is empty. This does not change the odds. There is still a 66% chance that you were wrong. There is still a 66% chance that the prize is behind B or C, and since you now know that B is empty, this means there is a 66% chance that the prize is behind C.
Since there's a 33% chance that it's behind A, and a 66% chance that it's behind C, it's best to switch to C.
Note that if the prize could be moved between your two choices, then the odds would be 50% for your second choice, and it wouldn't make any difference what you did.
So... to summarize... (for simple people like me)
When you selected Door A, you really only have a 33% chance of being right and a 66% chance of being wrong.
The fact that door B was open to show it was the wrong door, does not change the fact that you still had a chance of being 66% wrong (and still only a 33% chance of being right).
The odds don't change if everything else remains constant.
Yes, that's a good summary. SunGrooveTheory's "sampling without replacement" sounds like a good term. I have no idea what it means, but it sounds like maybe it applies here!
Someone I work with who actually remembers probability class from college (and apparently did well in it) said that all you're receiving from the open door is "free information", which doesn't affect the probability space (or something like that).
My head hurts.
It's all coming back to me of why I was an english lit major.
I don't buy it. Once door B was opened and shown that nothing is behind it, the calculations change. B is not longer a factor do everything has to be recalculated. As such, back to 50% chance. It's either behind one door or the other. let's do it this way. Let's say you had two chances. you choose door B (33% chance u r right) but it is not there. Now there are two 33% chances left. two 33% chances would be an even chance that either is correct. if it's even, then it becomes 50%. just my humble opinion.
lol James
Ailyn, I had trouble buying it for a long time, too. Maybe this wording will help:
You pick a door. There's a 33% chance you will win, and a 66% chance you won't. Now you are allowed to either stick with the door you picked, or choose both other doors, and if the prize is behind either one of those other doors, you win. In this scenario it's clearly best to switch, because you now have a 66% chance to win, right? Well the original wording is no different than what I just described, because you will always be shown an empty door, which is the one you didn't pick that is empty (and there will always be such a door). So whether you are shown a door is empty or not, you'll always get the other two doors combined (in effect) (33% + 33% = 66%). Whether you are shown the empty door before or after your second choice (to stick or switch to the other doors, one of which will be shown to be empty at some point) makes no difference. Does that help?
So what you are really saying is that... X = 9?
Yes, if it doesn't rain.
I always forget to factor in the rain... must be why my picnics are not very succesful, either...
That was an excellent explanation, Kikki, thank you so much =)
"sampling without replacement" means that it is one of those problems similar to this standard P&S problem:
You have 4 blue balls and 3 red balls in a box. You will draw one ball at a time, taking them out of the box and not replacing them. If you know that you have drawn BBRRB, what is the probability that the 6th ball that you draw will be a red one? Of course the similarity is that you are not replacing the balls , so you are sure how many balls of each color you have left, and like the door problem, you know which colors the balls were that you have already drawn. Being given that information also affects the problem.
Thanks again for the brain-tickler =) This has been the most fun!!!!
What I meant by the last statement was, whether you are given the information or not, affects the problem..
Thanks for that explanation of sampling and replacement, SunGrooveTheory :) You know, I made it all the way through Calc 3 in college, and found it easy. But then probability and statistics gave me trouble. It felt like I was jumping into an entirely different field mid-stream, and I struggled to answer the basic problems.
Do you like games? It is just a matter of how you think about it... Try to put it into words of something you like. Chess, maybe? Or computer programs... Given the letters A-Z to choose from, "sampling with replacement," and telling the computer to randomize the its choices for output, repeat loop 26 times, what is the probability to get one repeat letter? or the probability to have no repeat letters? or to have the letter 'p' repeated twice? See, that is something you could get into, right?
I can understand the rational and the chance going along with the wording. I just have to disagree with the real world possibility. Ok, maybe I'm just being stubborn.
SunGroove, yes, I think I could -- I think I wanted more of a solid grounding in the fundamentals. The class I took went at a very fast pace.
Ailyn, I was very stubborn with this problem for quite awhile, so I know exactly where you're coming from. That's why I wrote the simulation program. And it was only after being confronted with the evidence of the program results which produced 66% for switching every time (and meticulously verifying that the program was written correctly!) that I soon saw how to think of it differently.
Dude, I am NOT reading all of those comments. But you switch, of course. A 50-50 chance is much better than a 1 in 3.
I'd have said something fluffy, like 'follow your intuition, stick to the first choice".. but then, I'm not a billionaire.. yet.
PS, that's better than Schroedinger's cat, nobody dies..
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