My last post dealt with the probability curiosity referred to as The Monty Hall Problem. I though it would be useful to call out two ways of conceptualizing its highly counterintuitive solution. Neither of these concepts are mine. I simply offer them to enhance appreciation of the problem’s solution.

Concept 1

Step throught the scenarios:

Behind one door is the car, behind the other two are goats (Nanny and Fanny).

Scenario 1: The contestant initially selects the door with the car. The host then reveals Nanny. If the contestant switches doors, hu will select Fanny (lose).

Scenario 2: The contestant initially selects the door with Nanny. The host then reveals Fanny. If the contestant switches doors, hu will select the car (win).

Scenario 3: The contestant initially selects the door with Fanny. The host then reveals Nanny. If the contestant switches doors, hu will select the car (win).

Hence, switching results in a winning outcome in two out the three possible scenarios.

Concept 2

Increase the number of doors to 100. The player picks a door, then the host opens 98 of the other doors, revealing goats (they have names, but not as good as Nanny and Fanny). The host offers the contestant the opportunity to switch doors. The original odds that the contestant had of picking the door with the car behind it remain 1 in 100. Only one other door remains unopened. The odds of the car being behind that door are therefore 99 in 100.

A friend of mine recently explained the Monty Hall Problem to me in a bar (what do you talk about in bars?) and while it is utterly counterintuitive, the math totally works.

Marilyn vos Savant, the person with the highest IQ ever recorded, was posed the following question in her 1990 Parade Magazine column:

Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you: ‘Do you want to pick door #2?’ Is it to your advantage to switch your choice of doors?

—Craig F. Whitaker, Columbia, Maryland

She responded that the contestant should switch, owing to the fact that hu had a 2/3 chance of winning by switching doors, and only a 1/3 chance of winning by staying fast.

Her response generated thousands of letters, many of them from Ph.Ds in mathematics, telling her that she was wrong.

She is not wrong.

More for fun than proof, you might enjoy playing with Steven R. Costenoble’s simulation (bottom of page, requires Java). The problem with real time simulations like these is that probability seldom bears out using small samples. For instance, everyone knows that with a fair coin and a fair toss, the probability of the coin landing heads is 50%. Toss a coin ten times and see if you get five heads and five tails. Do it a million times, however, and the results will converge on 50%.

Curious, I wrote a simulation of my own, setting it to step through the scenario a million times. The contestant switching doors resulted in winning the car 678,042 times out of a million (67.8% of the time).

Neat.