The Monty Hall Problem

9

February 17, 2009 by esarsea

monte

Imagine you are a contestant on the 70’s game show Let’s Make A Deal with host Monty Hall. You’ve made it to the final round, and have a chance to win a brand new car!

Monty asks you if you want what is behind Door #1, Door #2 or Door #3. Before you choose, Monty explains that there is a new car behind one of the doors, and a goat behind each of the other two doors.

You select Door #1. Before revealing your prize, Monty opens Door #3 and reveals a goat. The new car is behind Door #1 or Door #2! Then Monty asks you if you want to change your selection from Door #1 to Door #2.

Assuming you would prefer to win the new car instead of a goat, is it to your advantage to change your selection?

You might be surprised to learn that the answer is yes! While at first it might appear as a 50/50 proposition, it is not.

When you choose Door #1, you had a 1 in 3 chance of being correct, or inversely, you would pick a goat 2 out of 3 times. After Monty opens Door #3 to reveal one of the goats, you still have a 1 in 3 chance of winning a new car if you retain your original selection of Door #1. In other words, your original choice has a 2 out of 3 chance of being incorrect.

If you change your choice to Door #2, you have the inverse of keeping your original selection, or a 2 out of 3 chance of being correct! You will win the new car 2/3rds of the time by changing your choice in this example.

This is an example of what is commonly referred to as the Monty Hall Problem,  which according to Wikipedia is a probability puzzle with a counterintuitive solution. It’s the mathematical equivalent to the Three Prisoners Problem and the Bertrand’s Box Paradox.

This is something that has been around for quite a while, but I just discoved it after reading an article in Card Player Magazine which I subscribe to (I enjoy playing poker, especially No Limit Texas Hold’em and Omaha).

Here is a webpage where you can play the Monty Hall game yourself, and track your own results.

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9 thoughts on “The Monty Hall Problem

  1. Bill says:

    VIZZINI: Where’s the poison? But it’s so simple! All I have to do is divide from what I know of you – are you the sort of man who would put the poison into his own goblet or his enemy’s? Now, a clever man would put the poison into his own goblet knowing that only a great fool would reach for what he was given. I am not a great fool, so I can clearly not choose the wine in front of you. But you must’ve known I was not a great fool, you would’ve counted on it, so I can clearly not choose the wine in front of me. I haven’t made my decision yet, though. Because Iocane comes from Australia, as everyone knows. And Australia is entirely people with criminals. And criminals are used to having people not trust them, as you are not trusted by me, so I can clearly not choose the wine in front of you. And you must’ve suspected I would’ve known the powder’s origin, so I can clearly not choose the wine in front of me. You’ve beaten my giant, which means your exceptionally strong, so you could’ve put the poison in your own goblet trusting on your strength to save you, so I can clearly not choose the wine in front of you. But, you’ve also bested my Spaniard, which means you must’ve studied, and in studying you must’ve learned that man is mortal, so you would’ve put the poison as far from yourself as possible, so I can clearly not choose the wine in front of me. Ha, it’s worked; you’ve given everything away. I know where the poison is. And I choose- (points behind the Pirate) what in the world can that be?! (switches goblets) Oh, I could’ve sworn I saw something. Well, no matter. Let’s drink – me from my glass, and you from yours. (drinks, then laughs) You think I guessed wrong, that’s what’s so funny! I switched glasses when your back was turned! Ha ha, you fool! You’ve fallen victim to one of the classic blunders! The most famous is “Never get involved in a land war in Asia”, but only slightly less well known is this – “Never go in against a Sicilian, when death is on the line!” Ha ha ha ha ha ha ha ha h-!

    (falls over dead)

    Princess Bride

  2. rsr348 says:

    sigh…. Stu’s making me do math again….I sort of get it. I suppose I’ll check out the Monty Hall links.

  3. rsr348 says:

    Ok. I read the article and the specific paragraph about the inverse selection several times. I don’t get it, but neither do most people, according to the article.

  4. Stu says:

    Think of it this way. Suppose the 3rd door was never opened to show you that it was a goat.

    Then you’re offered the choice to switch Door #1 for both Door #2 and Door #3, not knowing what was behind either door. That’s a no brainer, right? You’re changing a 1-in-3 chance for a 2-in-3 chance.

    So then after you agree to swap Door #1 for both Door #2 and #3, they open Door #3 and it is a goat. Well, at least ONE of those doors MUST be a goat, right? There is only one car.

    So opening Door #3 before or after you agree to change matters not, you’re still exchanging a 1-in-3 chance for a 2-in-3 chance.

    Does that help any?

  5. rsr348 says:

    Well, yeah, I get that switching 1 door for 2 doors increases my chances. And I see that knowing what’s behind door #3 doesn’t change my odds, but if Monty shows me one goat, I still don’t get why the other 2 doors aren’t a 50/50 chance. Why do I have a better chance if I open door 2 than door 1? (Now is where my math teacher husband would get impatient.) You don’t have to try to help me any more. I’ll be ok not getting it.

    I can, however, follow the Princess Bride script. Midevil fantasy….much easier.

  6. Stu says:

    I know it’s “counterintuitive” as they say.

    If your initial chance of being correct is 1 in 3, and you have the opportunity to make another selection, and there is only one other selection available, and one of the choices must win, the remaining choice MUST be the inverse of the 1st, (a 2 in 3 chance)….because the chances must add up to 3 in 3.

    Another way to look at it is this. If you retain your original 1 in 3 chance pick, over repeated trials you win approx 33% of the time. So what must the remaining choice represent?

    It’s hard to get one’s mind around the concept, but repeated trials do bear this out. It works.

    It’s fun stuff to think about. It opens new channels in one’s brain and forces us to look at things differently – not just in this example, but how one might see and/or use this concept in other aspects of our being! It fascinates me how adding and then removing a variable (that was a losing variable anyway) can impact what otherwise appears to be a 50/50 chance. It lends itself to all kinds of speculation as to how other seemingly insignificant and/or unrelated variables in our own lives might be impacting us in ways we are not aware of.

  7. rsr348 says:

    Are you saying that if I exchange my original choice for the second choice, I get to open door 2 AND 3? Then I get it. Otherwise I cannot wrap my mind around it, but I will keep trying, dang it all!

  8. rsr348 says:

    I take that back. I only get it if door number 3 is not opened, which goes back to the original scenario–If I know door 3 is a goat, I still believe either of the other two doors is a 50/50 chance. Whatever… I gotta go to work. Have a nice weekend!

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