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This is a famous demonstration of how hard we find it to work out probabilities. When it was published in Parade magazine in 1990, the magazine got around 10,000 letters in response92% of which said that their columnist, Marilyn vos Savant, had reached the wrong conclusion.2 Despite the weight of correspondence, vos Savant had reached the correct conclusion, and here's the confusing problem she put forward, based roughly on the workings of the old quiz show Let's Make a Deal presented by Monty Hall.

Imagine you're a participant on a game show, hoping to win the big prize. The final hoop to jump through is to select the right door from a choice of three. Behind each door is either a prize (one of the three doors) or a booby prize (two of the doors). In this case, the booby prizes are goats.

You choose a door.

To raise the tension, the game-show host, Monty, looks behind the other doors and throws one open (not yours) to reveal a goat. He then gives you the choice of sticking with your choice or switching to the remaining unopened door.

Two doors are left. One must have a goat behind it, one must have a prize. Should you stick, or should you switch? Or doesn't it matter?

Most people get this wrongeven those with formal mathematics training. Many of the thousands who wrote to Marilyn vos Savant at Parade were university professors who were convinced that she had got it wrong and insisted she was misleading the nation. Even the famous Paul Erdos, years before the Parade magazine incident, had got the answer wrong and he was one of the most talented mathematicians of the century (and inspiration for Erdos numbers, which you may have heard of3).

The answer is that you should switch you are twice as likely to win the prize if you switch doors than if you stick with your original door. Don't worry if you can't see why this is the right answer; the problem is famous precisely because it is so hard to get your head around. If you did get this right, try telling it to someone else and then explaining why switching is the right answer. You'll soon see just how difficult the concepts are to get across.

The chance you got it right on the first guess is 1 in 3. Since by the time it comes to sticking or switching, the big prize (often a car) must be behind one of the two remaining doors, there must be a 2 in 3 chance that the car is behind the other door (i.e., a 2 in 3 chance your first guess was wrong).

Our intuition seems compelled to ignore the prior probabilities and the effect that the game show host's actions have. Instead, we look at the situation as it is when we come to make the choice. Two doors, one prize. 50-50 chance, right? Wrong. The host's actions make switching a better bet. By throwing away one dud door from the two you didn't choose initially, he's essentially making it so that switching is like choosing between two doors and you win if the prize is behind either of them.

Another way to make the switching answer seem intuitive is to imagine the situation with 1000 doors, 999 goats, and still just one prize. You choose a door (1 in 1000 chance it's the right door) and your host opens all the doors you didn't choose, which have goats behind them (998 goats). Stick or switch? Obviously you have a 999 in 1000 chance of winning if you switch, even though as you make the choice there are two doors, one prize, and one goat like before. This variant highlights one of the key distractions in the original problemthe host knows where the prize is and acts accordingly to eliminate dud doors. You choose without knowing where the prize is, but given that the host acts knowing where the prize is, your decision to stick or switch should take that into account.

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# Strange counterintuitive probability problem

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