Isn’t the Monty Hall Problem two separate games?

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Game One: You are given a choice of three doors. You pick number one. The host opens one of the other two doors, having been given instructions that, if you pick the car, the host is to open one of the other doors, and if you pick a goat, the host opens the other door with a goat. Stalemate. It is a predetermined outcome.

Game Two: The prior game’s outcome stands. The new choice you have is do you keep door number one, or do you switch?

How do you have a 2/3 chance of winning if you switch?

In: 107

34 Answers

Anonymous 0 Comments

Imagine 100 doors instead of 3.

You pick one door which means you have 1% chance of picking the correct door. The host decides to even the field by opening 98 doors. Now there are 2 doors left, that other door has a 99% chance of being the correct door. Basically your original % chances don’t change, it’s only through the act of leveling the field that makes changing doors a higher %

Think of it backwards instead. Remember the host knows which door has the price but you don’t. So the host let’s you open a total of 99 doors, so you pretty much have a 99% chance of nailing the right door. Another way to see this is knowing the fact that the host will never open the correct door, so everytime he opens 1 door it adds 1% to the door he didn’t pick. The only way to lose this is if you stick to your original door [1% chance] or if you were extremely un/lucky in having picked the correct door at 1%.

Basically it’s best to switch doors either way as that gives you a higher %

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