Conditional probability
Posted Friday, January 8th, 2010 at 12:15
This is a quiz show meme, but worth remembering.
You are shown 3 doors. Behind one door is a car, behind the others are goats. Choose one door to win the car.
You make your choice. One of the other two doors is then opened and reveals a goat.
You are now left with two closed doors. You KNOW that behind one of them is a car; behind the other is a goat.
Should you change your mind and choose the other door? Most people would say no, it’s 50/50.
But the original choice was 1 out of 3. So now the balance of probability is 2/3rds in favour of the other closed door.
So you are more likely to win the car if you change your mind and choose the other door.
This probably shows why I would never have been a successful gambler.
January 11th, 2010 at 16:32
I am no expert on probability but surely as soon as one door has been opened and shown not to contain the car, the odds become 50/50. It would be exactly the same situation as if the other door had not existed and you had been given the choice of the two doors to begin. Each of the two doors would be equally likely to contain the car.
January 11th, 2010 at 17:04
Not at all, Dave. You have a 33% chance of picking the car first time round. Whether you switch your subsequent choice or not, you will always end up with a goat.
But if you pick one of the two goats the first time around — a 66% chance — and get shown the other goat, and you switch, you will inevitably get the car.
So the odds are 2/3rds in your favour by switching.