The other door must have the rest of the chances, or 2/3. The best I can do with my original choice is 1 in 3. Monty could add 50 doors, blow the other ones up, do a voodoo rain dance - it doesn’t matter. If I rigidly stick with my first choice no matter what, I can’t improve my chances. My first guess is 1 in 3 - there are 3 random options, right? If I pick a door and hold, I have a 1/3 chance of winning. That’s the hard (but convincing) way of realizing switching works. Just play the game a few dozen times to even it out and reduce the noise. If you had a coin, how many flips would you need to convince yourself it was fair? You might get 2 heads in a row and think it was rigged. There’s a chance the stay-and-hold strategy does decent on a small number of trials (under 20 or so). Is it above 50% Is it closer to 60%? To 66%? Pick a door, Monty reveals a goat (grey door), and you switch to the other. Now reset and play it 20 times, using a “pick and switch” approach.
Just pick door 1 (or 2, or 3) and keep clicking. Try playing the game 50 times, using a “pick and hold” strategy. You’re probably muttering that two doors mean it’s a 50-50 chance. The game is really about re-evaluating your decisions as new information emerges.
Today let’s get an intuition for why a simple game could be so baffling. If you switch doors you’ll win 2/3 of the time! Here’s the game: Do you stick with door A (original guess) or switch to the unopened door? Does it matter? (If both doors have goats, he picks randomly.) Monty Hall, the game show host, examines the other doors (B & C) and opens one with a goat.There are 3 doors, behind which are two goats and a car.The Monty Hall problem is a counter-intuitive statistics puzzle:
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