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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'll 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. | 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'll 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 | ||
| + | |||
| + | That's the youtube or orginal monte hall probelm solution, but '''doesn't apply to this 3.1B''' | ||
| + | in this problem, two people are picking! | ||
| + | |||
| + | case 1. win by not swapping | ||
| + | in this case you have to chose | ||
Latest revision as of 17:15, 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'll 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
That's the youtube or orginal monte hall probelm solution, but doesn't apply to this 3.1B in this problem, two people are picking!
case 1. win by not swapping
in this case you have to chose
