The Monty Hall Problem


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Problem: “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 door #1. The host, who knows what's behind the doors, opens door #2, which has a goat. He then asks if you’d like to keep door #1, or switch to door #3. Is it to your advantage to switch your choice?”

Simplified Solution: Initially, when door #2 is not open, all three doors have an equal chance of 1/3 of the car being behind either one of them. Let's say we pick door #1. And door #1 has a 1/3 chance of the car being behind it, which in tern means that there is a 2/3 chance of the car being behind either door #2 or #3. But when we remove door #2 from the equation, that means the 2/3 chance gets concentrated on door #3, which means you should always switch. But it has to be stated that, in this specific incident the car may not be behind door #3 because it has a 1/3 chance of the car not being behind it. But in the long run, always switching would be the best strategy.

The key factor that needs to be considered here is, looking at the problem from the Game Show Host's perspective. There is no chance in hell that he is going to open the door which has a car behind it. To consider a scenario where the car is behind door #3, and door #1 and #2 both have goats behind them, and given that we pick door #1, the Host will always open door #2 to keep things interesting. If he opens door #1, just because we picked it, the game wouldn't be fun anymore. And again, he wouldn't dare open door #3 which has the car behind it.

 

July 23, 2019, 2:56 p.m. by - Lesith