Thetownend.com

So there are 3 doors, lets say A, B and C.

Behind one of the doors there is a car.

Once you have chosen your door (but not opened it) ONE of the other doors that definately doesn't have the car behind it is removed.

You now have the option to STAY WITH YOUR ORIGINAL SELECTION or CHANGE TO THE OTHER REMAINING DOOR.

What do you do?

May as well stick, the removal of the door without makes no odds to your original choice.

That's the thing though it does.  I can't remember why or how though.

This is like deal or no deal. I'd stick. I'd rather have made the wrong choice all along, that have made the right one and thrown it away.

I saw Deal or No Deal when I was back home at Easter. I missed the start though. Is it the same as this? Do they choose one box at the start?

Surely you would always swap If you chose one at the start then you've only got a 1/3 chance of it bein right (in the doors I mean).

Yeah, they choose one at the start. I know what you're saying about odds, but my head wouldn't work like that...

But you've got a little chance of getting it right at the start. 1/3 in fact. So that means that there is 2/3 chance that you've got it wrong. So you should swap.

How many boxes are there in Deal or No Deal? 20? You've got a 1/20 chance of getting it right and you'd stick? there's a 19/20 chance that it's NOTin your box. I understand that you'd rahter be wrong than be right and throw it away (and that could happen) but statistically you've got a very small chance of being right...

22 boxes. As DB says, i'd stick with my original choice

Got my head around the maths, and statistically speak you have better odds by changing your choice.  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.

However, in the original example of this questions (a game show) there would have been more to influence the decision than straight maths.

Yeah you should change. Can't remember the maths off the top of my head. Sure if you look about on tinternet then there are plently of example.

Never really thought about how it would affect deal or no deal.

As I understand it, the answer for Deal or No Deal, was to switch off and watch something a little less shit

Ok let's look at a diagram
Right ---- Stay Right
1/3
    \_____ Change Wrong

Wrong ----- Stay wrong
2/3
    \______ Change right


The only way you can be right if you can stay is if you pick the right one at the start (1/3), however if you're wrong at the start because (2/3) because the other wrong one will be removed you will always be right if you change.

So because you're more likely to have been wrong than right at the start you should change. Meh.

This is termed the 'Monty Hall Problem' in maths circles; I did it in my first year of uni.

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.

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

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.

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%

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.

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.

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.

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.