3 Doors, 1 Car

[RobertT]

I'm still not 100% convinced, and by trying it a lot of times I got less than 2/3 when switching, closer to 1/2.  However, by staying it was closer to 1/3 than 1/2.

You tried it out? With actual doors and a car? You must have got some funny looks at work.

It kept me out of trouble for a bit.

[Simon Pieman]

I thought this was going to be a 3 door car, not the Monty Hall problem.

[timmyg]

It's easier to think about 100 doors.

If you pick a door at the start, you've got a 1% chance of picking the right one. If I then remove 98 of the other doors and leave you with the one that you picked first, and one other one then you'd be a fool not to change.

There's a 1% chance that you picked right in the first choice, so a 99% chance that it's NOT in that door. If the 99% chance then is whittled down to one door it's 1% that you've got it and 99% that it's in the other door.

I'm not sure what happens in Deal or No Deal, but if you pick a box at the start and there are 22 boxes split into 4 categories (For the rubbish prizes to the better ones, and better ones and the best ones) There's only a 25% chance that you'll have chosen one of the top prizes. If you're playing just once you'd be better to change.

[Gelbfüßler]

Changing or not makes no difference. Basically it's a 50/50 chance you've picked the right door at the time where you get the chance to change.

If you pick the right door to start with, then afterwards it leaves you 2 doors, 1 with the car, and 1 without

If you pick one of the 2 wrong doors to start with, then afterwards it leaves you 2 doors, 1 with the car, and 1 without

So the first choice doesn't affect the chances.

1/3 chance you pick the right door first time, then a door is removed, so it doesn't matter if you've picked the right door to start with or not, the chances of the car being behind 'door 1' is equal to the chances of it being behind 'door 2' : 50%

[timmyg]

But it's not.

If you pick door A, the chances of being right are 1/3. so the chance of being wrong are 2/3.

with the other door removed your chances are exactly the same as you picked the door before the other one was removed. It's 50/50 if you pick the door AFTERWARDS. However its 33/66 if youpick it BEFOREHAND.

[Gelbfüßler]

But it's not.

If you pick door A, the chances of being right are 1/3. so the chance of being wrong are 2/3.

with the other door removed your chances are exactly the same as you picked the door before the other one was removed. It's 50/50 if you pick the door AFTERWARDS. However its 33/66 if youpick it BEFOREHAND.

Yeah, exactly what I was trying to say. At the end of the day changing your choice when there's 2 doors doesn't improve your chances, it's still exactly the same.

[timmyg]

No it does, because you've picked the door BEFORE.

Think of 100 doors. The chance of you picking the right one is just 1/100.

If I then open 98 of the other doors and leave you with the one that you chose and one other one, it's not a 50/50 split, it's a 1/99 split with the chance of it being the other door being 99% and the chance of it being the door that you chose before only 1%.

If you have 2 doors to choose from then it's 50/50, but if you have 2 doors to chose from KNOWING THAT YOU CHOSE ONE BEFORE THE OTHER DOORS WERE ELIMINATED then the chances of success from changing are bigger.

[Gelbfüßler]

Just did some maths, seems you are right. Feel free to feel smug. Basically if you change once there are 2 doors, there is a 2/3 chance you will find the car.

[Ben Wah Balls]

The key is surely that the door that was opened definitely didn't have the car behind. If you picked a door randomly it would make no difference. Which is what they do on deal or no deal so on deal or no deal it doesn't matter whether you switch or not but in this case you should switch.