The Monty Hall Problem

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- Qtian
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The Monty Hall Problem
OP - September 14, 2014, 09:10 PM

The Monty Hall problem is good example of the counterintuitive nature of probability.

My mind runs, I can never catch it even if I get a head start.
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The Monty Hall Problem
Reply #1 - September 14, 2014, 09:10 PM

My mind runs, I can never catch it even if I get a head start.
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The Monty Hall Problem
Reply #2 - September 14, 2014, 09:11 PM

My mind runs, I can never catch it even if I get a head start.
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Reply #3 - September 14, 2014, 09:13 PM

My mind runs, I can never catch it even if I get a head start.
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Reply #4 - September 14, 2014, 09:14 PM

My mind runs, I can never catch it even if I get a head start.
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Reply #5 - September 14, 2014, 09:15 PM

My mind runs, I can never catch it even if I get a head start.
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Reply #6 - September 14, 2014, 09:16 PM

My mind runs, I can never catch it even if I get a head start.
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Reply #7 - September 14, 2014, 09:17 PM

Some solutions

Quote from: curiouser.co.uk
Consider this: in what cases will switching doors lead to a goat? If one chooses a goat door originally, then switching doors can only lead to a car, as there is only one other goat door and that has been opened. Therefore, the other door must reveal the car. Conversely, if one selects the car door originally, switching doors must lead a goat.

One can therefore see that the only time when switching doors can be wrong is when one selects the car door originally. If one selects a goat door originally, switching must lead to the car. The probability of selecting the car door originally is 1/3. The probability of selecting a goat door originally is 2/3. Therefore the probability of winning the car by changing doors is 2/3. And the probability of losing the car by changing doors is 1/3. Therefore, the rational decision, as Ms. Savant stated, is to change doors in order to double the probability of winning the car.

My mind runs, I can never catch it even if I get a head start.
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Reply #8 - September 14, 2014, 09:22 PM

Quote from: curiouser.co.uk
it is worth considering what would happen it there were more than 3 doors. Let us assume there are 10 doors, with 9 goats and 1 car. The contestant selects one door. The host, Monty Hall decides to open 8 doors that he knows will all reveal goats. What is the probability now of winning the car if the contestant decides to change his/her selection? Again, there is only one way in which changing the selected door can lead to a goat, and that is if the contestant selected the car door originally. If the contestant didn’t select the car door originally, then the other door must reveal the car. The chances of the contestant selecting the car door originally are just 1/10. Therefore the probability of winning the car by changing selection in this instance is 9/10.

In general, with n doors, and (n-2) goat doors knowingly opened by the host, the probability of winning the car by changing from the original selection is (n-1)/n.

My mind runs, I can never catch it even if I get a head start.
- Quod Sum Eris
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Reply #9 - September 14, 2014, 09:23 PM

`But I don't want to go among mad people,' Alice remarked.
`Oh, you can't help that,' said the Cat: `we're all mad here. I'm mad. You're mad.'
`How do you know I'm mad?' said Alice.
`You must be,' said the Cat, `or you wouldn't have come here.'
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Reply #10 - September 14, 2014, 09:24 PM

Quote from: curiouser.co.uk
If all else fails, try some mathematics:

Let the 3 doors be A, B and C.

Let your selection be A.

Let the door the host reveal be B.

The a priori probability that the prize is behind door X = P(X) = 1/3

The probability that Monty Hall opens door B if the prize were behind A is:
P(Monty opens B|A) = 1/2

The probability that Monty Hall opens door B if the prize were behind B is:
P(Monty opens B|B) = 0

The probability that Monty Hall opens door B if the prize were behind C is:
P(Monty opens B|C) = 1

Therefore:

The probability that Monty Hall opens door B is:
p(Monty opens B) = p(A) * p(Monty opens B|A) + p(B) * p(Monty opens B|B) + p(C) * p(Monty opens B|C) = 1/6 + 0 + 1/3 = 1/2

Using Bayes's Theorem:

P(A|Monty opens B)    = p(A) * p(Monty opens B|A)/p(Monty opens B)

   = (1/6)/(1/2)

   = 1/3

and
P(C|Monty opens B)    = p(C) * p(Monty opens B|C)/p(Monty opens B)

   = (1/3)/(1/2)

   = 2/3

In other words, if you choose A originally and Monty opens door B to reveal a goat, the probability of there being a car behind door C is 2/3. Therefore you should always switch your selection

My mind runs, I can never catch it even if I get a head start.
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Reply #11 - September 14, 2014, 09:24 PM

Quote from: Quod Sum Eris on September 14, 2014, 09:23 PM

?

My mind runs, I can never catch it even if I get a head start.
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Reply #12 - September 14, 2014, 09:29 PM

That's me sitting here sipping my drink as I read the thread. It's interesting. Maybe this would work better.

`But I don't want to go among mad people,' Alice remarked.
`Oh, you can't help that,' said the Cat: `we're all mad here. I'm mad. You're mad.'
`How do you know I'm mad?' said Alice.
`You must be,' said the Cat, `or you wouldn't have come here.'
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Reply #13 - September 14, 2014, 09:30 PM

Ahh, fair enough.

My mind runs, I can never catch it even if I get a head start.
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Reply #14 - September 14, 2014, 11:11 PM

Quote
Imagine that instead of 3 doors, there are 100. All of them have goats except one, which has the car. You choose a door, say, door number 23. At this point, Monty Hall opens all of the other doors except one and gives you the offer to switch to the other door. Would you switch? Now you may arrogantly think, “Well, maybe I actually picked the correct door on my first guess.” But what’s the probability that that happened? 1/100. There’s a 99% chance that the car isn’t behind the door that you picked. And if it’s not behind the door that you picked, it must be behind the last door that Monty left for you. In other words, Monty has helped you by leaving one door for you to switch to, that has a 99% chance of having the car behind it. So in this case, if you were to switch, you would have a 99% chance of winning the car.

My mind runs, I can never catch it even if I get a head start.
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Reply #15 - September 14, 2014, 11:16 PM

A good visual explanation: http://www.youtube.com/watch?feature=player_embedded&v=BigmlbUOxrc

My mind runs, I can never catch it even if I get a head start.
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