Monty Hall Problem: Solution


Yesterday’s Problem, Monty Hall Paradox

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice and why? Give a proper mathematical explanation.

Now here is the solution given by my elder brother

Probability of winning when we don’t switch = (1/2) * (2/3) = 1/3 (probability that car is behind gate 1 and gate 3 has a goat)

Probability of winning when we switch = 1-(1/2)*(2/3) = 2/3.

But here is another solution if you did not understood the above solution. This is from good old Wikipedia.

Car hidden behind Door 3 Car hidden behind Door 1 Car hidden behind Door 2
Player initially picks Door 1
Player has picked Door 1 and the car is behind Door 3 Player has picked Door 1 and the car is behind it Player has picked Door 1 and the car is behind Door 2
Host must open Door 2 Host randomly opens either goat door Host must open Door 3
Host must open Door 2 if the player picks Door 1 and the car is behind Door 3 Host opens Door 2 half the time if the player picks Door 1 and the car is behind it Host opens Door 3 half the time if the player picks Door 1 and the car is behind it Host must open Door 3 if the player picks Door 1 and the car is behind Door 2
Probability 1/3 Probability 1/6 Probability 1/6 Probability 1/3
Switching wins Switching loses Switching loses Switching wins
If the host has opened Door 2, switching wins twice as often as staying If the host has opened Door 3, switching wins twice as often as staying
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