The Monty Hall problem

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Wednesday 23 June 2004 7.12pm
I spent 2 hours on the phone to NY last night discussing this problem and we still disagreed on whether your chances increased from 50/50 by switching. I could have bought a goat. It's doing my 'ead in.

http://www.grand-illusions.com/monty.htm

http://www.cut-the-knot.org/hall.shtml <--good maths site

http://math.ucsd.edu/~crypto/Monty/monty.html
Thursday 24 June 2004 9.33am
When the host opens a door the probability of the person being right *changes* from 1 in 3 to 1 in 2 - so there is no conflict between their first choice and their second.

Yes they *did* have a 1 in 3 chance but now once the host reveals the goat, they have a 1 in 2 chance regardless of whether they change their choice of door.

Any bookie does the same - e.g. the probability of England winning changes as different options are discounted.



Edited 1 times. Last edit at 24 June 2004 9.33am by red bus.
Thursday 24 June 2004 10.00am
Strangely and counter-intuitively, red bus, that's not right, and if you think about it like that, you wont grasp why the probability of winning if they change their choice is so high. Its a great problem because in spite of not being really difficult, its so counter-intuitive.

Think of it as so:
Most of the time (2/3 of the time) what happens is the contestant chooses (unknowingly) a goat. The host, by revealing the other goat, shows them where the car is! So, 100% of 2/3 of the time, they win by changing!

Its easy when you think about it the right way, but finding the right way to think about it is not always easy. Bit like most maths and logic really.
Thursday 24 June 2004 11.46am
Phoney - you'll win the competition anyway, so why worry?

It's not related to the England winning scenario, Red Bus. I liked the similar example on the first website which must explain it better than anything else:

'Some of you will still be shaking your heads at this, and saying we are wrong. So here is another way of thinking about it. Imagine there are 100 doors, with a car behind only 1 of them. You choose a door. Your chance of being right is 1 in 100, right? Now the host opens 98 of the remaining 99 doors, in each case revealing a goat. You can now either stick with your original choice, or you can switch to the one remaining door that is closed. We say if you stick with your original choice, you still have a 1 in 100 chance of being right. And if you switch, you have a 99 in 100 chance of being right. '


Alternatively think about it like this:

1a. You pick one door, so have a 1-in-3 chance of being right. OK?
1b. The TV show is left with the two other doors. Yes? Therefore it has a 2-in-3 chance of being right.

2a. Opening one of the doors doesn't change where the car is. Does it? NO THE CAR IS UNMOVED!
2b. So we can open one of the doors without changing the odds, cannot we?Yes!

3. So the TV show still has a 2-in-3 chance of being right (see 1b), and it then opens one of the doors - without a car.

4a. We know that the door we are standing in front of has a 1-in-3 chance of being right. It always has had; opening another door doesn't change the original probability of our picking the right door.
4b. We know that the door that has been opened has a 0-in-3 chance of having the car. Doh!
4c. So what is the probability of the other door having the car? Well, probabilities must always add up to 100%, i.e. 3-in-3. So our door has 1-in-3 (which has never changed, remember), so the remaining door must have [(3-in-3) minus (1-in-3)] = 2-in-3.

5. So to double our chances of winning the car, we should move doors.


If anybody doesn't believe us, then roll-up roll-up to the TUMJ & Mapmaker gambling saloon, where we will happily take loadsamoney off you, at a tenner a gamble. After 100 rounds of this, you'll be well down & we'll buy you a commiseration pint of zider.

Thursday 24 June 2004 11.51am
Different way of thinking about it.

1. Pick a door - you have a 1-in-3 chance of the car.

2. TV has a 2-in-3 of having the car behind his door.

3a. Imagine you're standing before the car. (1-in-3). Then TV can open either of his doors for you.

3b. Imagine you're standing before a goat (2-in-3). Then TV has no choice which door to open. He has to open his goat, otherwise he reveals his car. So 2 out of every 3 times you play this game, the TV will have a car hiding behind his shut door - not yours.


Therefore you should move for the second round, as 2 in every 3 games you play TV will have the car behind his remaining shut door.

(Apologies that 'TV' appears to have acquired a persona!)

[Edited to add clarity]



Edited 1 times. Last edit at 24 June 2004 12.50pm by Mapmaker.
Thursday 24 June 2004 4.18pm
I like that idea of the gambling saloon, Mapmaker. Shall we set up outside the Tate Modern? Could be very lucrative. I can borrow goats from the City Farm.
Thursday 24 June 2004 4.59pm
I wouldn't mind winning a goat. I don't need a car, as I can't drive and I've nowhere to park it, but I could really go places with a goat.
Thursday 24 June 2004 5.39pm

nooo, you will end up taking all my money !
Thursday 24 June 2004 5.43pm
You could really go places and make goat cheese too. Though you could of course sell the car and buy hundreds of goats. We'd probably need hundreds of goats to get all the apples back from the West Country. Mind you they'd probably eat the apples en route.
Friday 25 June 2004 10.28am
>>nooo, you will end up taking all my money !

Only if we're right. You might be right, in which case you would end up taking all our money.

Are you game for a gamble?
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