This is fun.

The “Monty Hall Problem” is a logical paradox. It fall into the category of “veridical paradoxes,” which evidently means one that seems patently false at first, but is demonstrably, and ultimately un-controversially true. As such, it is a paradox that reveals not the limits of human reason, but the patterns of human reason that are prone to error.

It is also a paradox, in that it became most famous through a 1990 Parade Magazine column by a woman known as Marilyn Vos Savant, who claims to hold the Guinness Record for having the world’s “highest IQ.” This is paradoxical on two counts: 1) Parade Magazine 2) According to Wikipedia this is her real name.

Here’s the problem for readers who have forgotten, or have never seen it:

A game show host offers a guest a chance to win a prize (car). The prize is behind one of three doors. Behind the other two doors are booby prizes (goats). The guest chooses a door – say door 1. The host proceeds to open one of the other two doors (say door 2) to reveal a goat. The host then offers to let the guest switch his pick from door 1 to door 3. [Assume here that the host knows the correct door, and - to heighten the drama - will always open a goat-concealing door prior to allowing the switch].

The question: should he switch?

The answer: Yes! The player’s odds of owning a Buick will go from 1/3 to 2/3!

I show the reason in the comment section below.

The Wikipedia entry is also quite good, for interested readers: http://en.wikipedia.org/wiki/Monty_Hall_problem. I will note, paradoxically, that this particular Wikipedia entry is significantly longer than the entry for “Zeno.” It is nearly as long as the entry for “Plato,” for that matter.

Plato may not be know for paradoxes, but is also fun, and is arguably more important than Monty Hall, Buicks, and Ms. Vos Savant. Perhaps this reveals not the limits of human reason, but the patterns that are prone to error. Now, back to watching videos of cats on the Internet.

So, to most people it seems that both door 1 and door 3 have a one in three chance of concealing the car, and that opening door 2 cannot affect the odds for doors 1 or 3. At most, by knowing that door 2 conceals a goat, the odds for both doors 1 and 3 go to 50%. OK, then, here’s the correct logic: Door 1 did indeed have a chance of 33%. The host opening an incorrect door (there will always be at least one goat left) can not affect those odds. Perhaps, then one should switch because door 3 now has a 50% chance as opposed to door 1’s 33% chance.? This is incorrect, as well. The odds are actually 67% in favor of door 3. Why? Because the odds at the time of choosing door 1 were 1/3 door 1 and 2/3 not door 1. The host will never open a door concealing the car, so now the odds are 1/3 door 1 and 2/3 door 3 (the only “not door 1” option left). Host behavior is key to understanding the problem and its solution. For further details follow the Wikipedia link, above.

ReplyDeleteI should also point out that the host, in addition to always opening a "goat" door, and always offering a switch, must choose between the possible "goat" doors at random (if there are two). Here is another good summary: http://probability.ca/jeff/writing/montyfall.pdf

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