Monty Hall died Saturday, a fact I learned after it was posted on the Facebook page of my brother-in-law, a retired math professor.
And if you're wondering the connection between the host and creator of "Let's Make a Deal" and mathematics, then you are not familiar with The Monty Hall Problem, a simple yet vexing statistical puzzle.
Simple, if you see it, and vexing, if you don't, particularly for those trying to explain it.
Remember the show: Would-be contestants in the audience came to see the show in outrageous costumes, waving signs and props, hoping to catch Monty's attention and be selected to compete. They would swap what they had brought, bartering their way higher and higher, hoping to get to the final challenge, which involved three doors, cleverly named Door No. 1, Door No. 2, and Door No. 3. Behind one of the door was a new car.
The lucky contestant would be asked to pick one of the three doors—let's say she picked Door No. 1.
Then Monty would announce that he would let that contestant stick with the door she picked, or choose another door. But first, he'd say, let's see what is behind one of the doors you didn't pick. He would open, let's say, Door No. 3, and reveal a burro, or a goat, or some such thing.
Now, Monty would say, do you want to stick with the door you picked, Door No. 1? Or would you like to take your chances with Door No. 2?
What should the contestant do?
The average person trying to puzzle through the problem might say, "Okay, the chances of getting the new car are one in three at the beginning and, tossing one door out, become one in two, but that holds for either door, so it doesn't matter whether you stick or switch.
That's wrong.
According to statistics, the contestant should always switch. The odds are better with the other door, and not just a little better, but twice as good.
How can that be?
The quick trick to understanding is to imagine two types of strategies: Always Stick, and Always Switch.
In order to get the car, what must Mr. Always Stick pick in order to end up with the new car? He must pick the door hiding the car because later, given the chance to switch, he won't, and is left with his initial choice. To get the car, he has to pick the car right off of the bat.
And what are the chances of picking the car first thing? One door in three, or 1/3.
Now thinks about Always Switch. How does Miss Always Switch get the car? By picking the car? No, because if she picks the car, she'll later switch away from it. So to get the car, she has to pick one of the donkeys. And what are the chances of picking a donkey? Two in three, or 2/3.
So Always Stick's chances are 1/3, and Always Switch 2/3, or twice as good.
Do you see it? No? Don't feel bad. My experience is that people often don't. A lot of people. Twenty-five years ago, after "Ask Marilyn" columnist Marilyn vos Savant presented the problem, up to then the stuff of obscure mathematical journals, in her popular column in Parade Magazine, readers deluged her with mail claiming she was wrong.
Many people I've tried to explain the problem to just can't get it, to my eternal frustration. What throws them off is the final choice, between two doors, seems like it should be 50-5o, because they know one door has a car and one a burro. They are forgetting the winnowing process, the throwing out of the third door, which affects the odds for switching. The odds for sticking remain what they were at the beginning—one in three; but tossing out one remaining door changes the odds considerable for taking the door that isn't thrown out.
Let's put it another way. I am going to give you a choice between two wallets, and your goal is to find the wallet stuffed with cash. Wallet A is picked from a pair of wallets, one containing cash, one empty. Wallet B is selected from a group of a hundred wallets, only one of which contains cash. Of the two wallets I offer, which wallet should you pick? Even though you are being presented with two wallets, you should always pick A, because the odds of B containing any money are very small. Thus can the odds of picking between two things not be 50-50.
Very confusing, I know. And I probably should have addressed the slaughter in Las Vegas instead. But honestly, at least this problem has an actual solution that can, with the application of brainpower, be solved. The other one seems merely impossible, at least for now and the foreseeable future.