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Think about it that after one goat door is open, the remaining doors contain a goat and a car and what behind your door can | Think about it that after one goat door is open, the remaining doors contain a goat and a car and what behind your door can | ||
either be a goat or a car. Now in order to win by swapping, you have to have a goat behind your door (b/c if you have a car now,you | either be a goat or a car. Now in order to win by swapping, you have to have a goat behind your door (b/c if you have a car now,you | ||
| − | + | will end up with a goat after swapping). so we can samplely calculate the possibility of getting goat in the first place, because after swapping you will end up with a car. | |
| − | will end up with a goat after swapping). so we can samplely calculate the possibility of getting goat in the first place, because | + | |
| − | after swapping you will end up with a car. | + | |
P(car after swap) = P(not a car in first pike) = P(goat) = 2/3 | P(car after swap) = P(not a car in first pike) = P(goat) = 2/3 | ||
Revision as of 11:13, 16 September 2008
before we start, ask yourself, how are you going to win a car. the answer is that there are two ways.
Case 1. win by not swapping
in this case you have to chose the car at first place and stick with it by now swapping. so the possibility will samplely
P(car) = 1/3
case 2. win by swapping
Think about it that after one goat door is open, the remaining doors contain a goat and a car and what behind your door can either be a goat or a car. Now in order to win by swapping, you have to have a goat behind your door (b/c if you have a car now,you
will end up with a goat after swapping). so we can samplely calculate the possibility of getting goat in the first place, because after swapping you will end up with a car.
P(car after swap) = P(not a car in first pike) = P(goat) = 2/3
