M is for . . .

The Monty Hall Problem

Image source: Wikipedia Commons

This math problem is dear to my heart because I distinctly remember where I was and what I was doing when the solution came: driving up one of those tight, cloverleaf, exit ramps off a highway. I couldn't wait to get home to write it down.

This problem gets its name from the original host of the TV game show, Let's Make a Deal, where a player faces three doors, two of which hide booby prizes and one hides something worth winning. Here 'tis:

"Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?" (Wikipedia)

What do you think? You've picked a door. After that, the host shows you that one of the two doors you didn't pick hides a booby prize. You could stick with your first choice or switch. Seems a fifty-fifty chance now, doesn't it? But that's where the math comes in.

The answer is that you should always switch. In fact, you have a 2/3 chance of winning the car if you do switch and only a 1/3 chance of winning if you don't. Surprised?

Here's the key: At the beginning, when you selected one door out of the three, you probably picked the wrong one. Why? With 2 goats and 1 car, random picking means there's a 2/3 chance you picked a goat. Therefore, odds are the car will be yours if you switch.

Cool, huh?

Do you (or did you ever) watch game shows? Would you prefer a goat over a car?

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This week's short story giveaway, The President and the Pea, features a sci-fi tale inspired by The Manchurian Candidate.