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### Course: Statistics and probability>Unit 7

Lesson 1: Basic theoretical probability

# The Monty Hall problem

Here we have a presentation and analysis of the famous thought experiment: the "Monty Hall" problem! This is fun. Created by Sal Khan.

## Want to join the conversation?

• Isn't it always going to be a 50/50 chance? If #3 is eliminated you are just choosing between #1 and #2. Switch or no switch it is still a 1/2 chance. I don't understand how your odds increase by always switching. Thanks for anyone's help
• That's the initial trick. You are tricked into believing here that the 3rd curtain/door doesn't matter, since it's seemingly "eliminated" from the equation. This is an illusion and is meant to fool you into thinking the 3rd curtain doesn't matter. You HAVE to consider the 3rd curtain as well. Why? Because you made your pick BEFORE the 3rd curtain got eliminated. It would only be a true 50/50 chance if this 3rd curtain was eliminated BEFORE you made your pick and not after.

To see this more easily, you just have to exaggerate the numbers to the extreme. Suppose you had 1000 doors and only one door had the prize. The chances of picking the right door would be extremely small, right? Ok, you pick a door and afterwards, the host eliminates 998 doors, leaving the door you picked and a second door. Now,you tell me if you think it likely that your initial pick would just HAPPEN to be the right door chosen out of 1000 doors. It sure as hell wouldn't be 50%.
• Could this question also be understood by multiplying them as dependent variables? Given that you will always switch. P(winning if you first choose wrong)= 2/3 * 1/2 = 1/3. P(winning if you first choose right)= 1/3 * 1/2 = 1/6. The 1/2 probability comes from the two remaining doors, either you have the correct one or the wrong one. It does make sense to switch because you are more likely to choose the incorrect door at the beginning.
• What makes the Monte Hal Problem interesting in that the host knows, and will always chose the goat. So I tried to break it down by probability of winning the car behind doors A, B or C...

A. Initial Conditions:
1) P(A) = P(B) = P(C) = 1/3
2) P(A)+P(B)+P(C) = 1 (or 100%). This is just a way of stating that the prize can be anywhere.

B. You Choose 'A' -- can be any door, but we can call any door you choose 'A'
...This leads to the following probabilities:
1) P(A) = 1/3
2) P(not A) = 2/3.
* Also note that P(not A) = P(B)+P(C)

C Monte Opens 'B' -- can be any door, but we can call any door he chooses 'B'
...Monte will never open the door with the prize, or he will be fired. So we have the following certain probability:
1) P(B) = 0/3 = 0

D. Monte says, "Do you want to switch?"
...So, let's look at our certain probabilities:
1) P(A)=1/3 (From section A)
2) P(not A)=2/3 (From section B)
3) P(B) = 0 (From section C)

E. Using the last 2 certainties, we end up with the following
1) P(not A) = P(B)+P(C). So, substituting, we end up with
2) P(B)+P(C) = 2/3
3) Since P(B) = 0, Then we can substitute '0' for P(B), which gives us
... 0+P(C) = 2/3.
...This simplifies to
*4) P(C) = 2/3

F. So it makes sense to switch...

Notice that Monte's new information tells you nothing about A. Nor does it change the initial odds. It only gives you information about * P(not A)*.

I really hope this helps. The notation really took work to put together. It is more mathematically rigorous than Sal's demonstration, but I was a mathematics major at university. But I love his explanation for the simplicity...
• Also why in "deal or no deal," with 30 cases, if you could get it down to your case and only one other case, you should switch. You get a 96.67% chance of getting your million dollar case. Of course it would take quite a bit of luck to get that far, but nevertheless the theory behind the situation is identical.
• The difference is that the 28 other cases are not all eliminated for the player by the host. I think that the host of "Deal or No Deal" doesn't even know which cases have the good prizes. Ultimately the situation is different.
• Why is the probability 2/3?Lets take it this way,the hosts shows you the door that doesn't have the prize and then you are left with two possible outcomes.One of them has the prize.Therefore,P(the outcome would be req. prize)=1/2
• No, Syomantak, you're changing the situation in an essential way. In the case you've given, you are only choosing once. In the actual Monty Hall problem, you choose twice, and the probability of the second choice is affected by the first. Dependent probability.
Think of it this way:
if you choose the correct door the first time, switching will always make you lose.
If you choose one of the two wrong doors the first time, switching will always make you win.
What's the probability of choosing the wrong door the first time?
Yes, that's right, 2/3.
Another expansion of the thought that several people have brought up:
instead of 3 doors, think of 10. Monty will do a similar thing once you choose for the first time, but, instead of revealing 1 goat, he will reveal 8.
you have the same scenario:
If you choose the right door the first time, switching will make you lose.
If you choose any of the 9 wrong doors the first time, switching will always make you win.
What's the probability of picking the wrong door the first time? 9/10
So, what's the probability of winning if you always switch? 9/10
And, what's the probability of winning if you never switch? 1/10.
• The 2/3 probability when always switching does not make practical sense. Knowing ahead of time that a wrong answer will be identified for you, that just eliminates one pick from the possibilities, so the probability is always 50% that you will pick correctly, whether you stick to your initial pick or switch. It appears to me that they get the 2/3 probability by mixing the initial probability of picking the first time correctly with the altered probability once you know which one is incorrect.
(1 vote)
• That would be correct if it was independant probability. But, this is dependant probability, because if you should switch or not depends on if you picked the car or one of the goats in the initial pick. This means that the second pick depends on the first pick.
There is a higher chance of picking a goat in the initial pick, so their is a higher probability of winning, if you swith in the second pick.
Thats why there is not a 1/2 probability, but a 2/3 one.
Hope this helps you understand :)
• i'm confused.

"if you pick wrong the first time, you should switch." / "if you picked wrong, and switch, you will always win."

it seems like the condition for these probabilities is "if you picked wrong." but the contestant will NEVER KNOW if they picked wrong - we are under the assumption that the contestant DOESN'T know they picked wrong the first time.

he doesn't know what he picked at all.

and so, with that condition out the window, doesn't this fall apart? if someone could explain, that'd be great =/
• The win isn't guaranteed even if you switch, but more likely.
• I remember having this argument with one of my high school math teachers. I finally convinced him by scaling this problem up to 1 million doors. If you start with 1 million doors, and are asked to pick one, then 999,998 are opened, only leaving your door and one other surely the other door is far more likely to be the one.

When it comes to most math puzzles my initial instinct is to scale it up and see if there's an obvious result, it really helps sometimes.
• i think its easier if you think there are 100 doors and only one of them has the prize. Then you pick one door and Monty Hall opens all the other 98 doors leaving only your door and another one closed. What is the probability that you picked the right door out of 100 doors? It is very unlikely so you will want to switch. The monty hall problem has 3 doors instead of 100. It is still more likely that you pick a goat.
• The probability of winning is 1/3 because there are 3 doors and 2 doors are wrong and 1 door is right so the chance of losing is higher than the chance of winning. You said if a person picks door 2 the Monty Hall will close door 1 and 3. If a person picks door 1 which is wrong the Monty Hall will close door 3 and give you chance to switch to the right answer, so it means they want always people win the prize. What is the point of playing the game and there isn't losing?
• An old puzzle and a good one. I enjoy asking maths teachers and maths graduates this one because they almost always get it wrong! It is deeply counter-intuitive that what happened in the past - your past decision - can change the chances of there being a car behind either of the two doors in front of you in the present. I mean, whatever happened in the past, right now you've got two doors, and if the car is not behind one then it is behind the other. Only two possibilities so a 50/50 chance, right?

You can do the maths and STILL not believe it because it's so counter-intuitive.

The way I got over the counter-intuitive aspect of it was the same as the way Peter Collingridge does below (or above?) which is to do a thought experiment and imagine that there is a huge number of doors, and all but two are opened by the game show host, leaving the same situation as before, namely that there are two doors, with a goat behind one and a car behind the other. In the case of a hundred doors or a thousand doors it is intuitively obvious that, for all practical purposes, you may as well assume that you intially chose a goat, and that you are therefore almost always going to win a car if you switch. Once you see that, you can mentally reduce the number of doors until there are just three, but you will still retain the intuition that you are better off switching.

In Philosophy this method of reasoning is called 'reductio ad absurdum' (reduction to the absurd) and it's a really useful tool in all kinds of situations.

Great video.