This is the Monty Hall Problem. This Wikipedia page gives a lot of different ways to look at it and understand it. Here is the graphic for the decision tree approach. By always switching, the constestant gets the car 2 out of 3 times.
Ok well i was watching this show on telly the other night and on of the guys did a maths problem. im trying to show this to my mates because it was pretty amazing. the problem is i can't remember how he explained it works. it wasn't so much a problem as a fact. it was...
if there are 3 doors, behind 1 is a car, behind the other 2 is a dud prize...obviously you want the car. so you pick the door you think it is behind. then one of the other doors is opened and it is a dud prize. So the car is either behind your door or the other. Do you change your answer? on the show it turned out that there is more probability of getting the car if the answer is change but i can't seem to remember or figure out how this is possible.
So could someone please explain this to me or is not true?
Thanks.
This is the Monty Hall Problem. This Wikipedia page gives a lot of different ways to look at it and understand it. Here is the graphic for the decision tree approach. By always switching, the constestant gets the car 2 out of 3 times.
One thing that makes me angry about this problem is that people thing that there is a debate and some people say the probability is 1/2.
Anyway, here is my explanation I posted on this forum some time ago.....
You are using the strategy that you will switch....
If you pick a bad door, then you win!! How? If you pick a bad door, two doors remain, good and bad. Certainly the bad is open and the good remains. So when you switch you get the good door. What is the probability of picking a bad door? Simple, 2/3 so the answer is 2/3.
Look at it in reverse....
If you pick the good door, then you lose!! How? If you pick the good door, two doors remain, bad and bad. Certainly the bad is open and the other bad remains. So when you switch you get the bad door and lose. What is the probability of picking the good door? Simple, 1/3 so the probability that you will lose on this strategy is 1/3, that is the probability that you will win is 2/3.
Just think about it....
Let us assume you are using the strategy of staying on the door that you picked. Then the situation is that you end up with two doors and some people think the probability is 1/2and they stay and what they selected. Opening and closing doors has no effect on your initial choice, so how can it be 1/2? Meaning if you pick a door then the probability is 1/2 from the very beginning since you ended up with two doors, but that is not true.
Exagerrate the case....
There are 1,000,002 doors. You pick 1 door. 1,000,000 doors are open and shown to be bad one door remains. Do you still say the probability is 1/2!! And remain! I hope not.
Probability exaplantation....
The sum of the probabilities is 1/3 + 1/3 + 1/3 = 1
For each door.
One is open thus the probability is 0,
Your door has probability of 1/3,
Thus,
0 + 1/3 + 2/3 = 1
So by switching you get the 2/3 probability door.