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.

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 audience came 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

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 beause 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 switch away. So to get the car, she has to pick on 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.

I remember seeing this in the Parade magazine a long time ago, and after thinking on it I still believe there's a flaw in Ms. von Savant's reasoning.

ReplyDeleteThe nagging problem I have with this is: The original odds were 1 chance in 3 to get the car. Say I picked Door #1, then find out Door #2 has a donkey behind it. The fact that Door #2 has a donkey behind it doesn't PHYSICALLY move the car to another door. So, tell me again why I should switch? The odds after the donkey is revealed in Door #2 are still 50/50 between my original pick (Door #1) and the other door (Door #3). Was this theory proven to be correct on the show?

There are three doors. The thing to remember is that Monty always opens a donkey door. The result is that if you switch, you are picking TWO doors--the one that you end up with and the donkey Monty reveals. Thus your chance of a car is 2/3. If you don't switch, you only get the door you actually selected.

DeleteThat's interesting. In other words, odds are pretty good that your initial pick was wrong. Why get stuck with what was probably the wrong choice.

ReplyDeleteWhat I don't understand is what any of this has to do with the picture at the top of the Vietnam Women's Memorial.

I'm just wondering, that when Monty Hall's friends & family got to the undertakers, were they told to pick from chapel #1, chapel #2 or chapel #3 & did they get a zonk if they picked the wrong chapel?

ReplyDeleteI understand the two out of three (you're given two choices), but that final choice is still a coin toss. Like Sandy, I'd like to see the data supporting this theory.

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ReplyDeleteI knew it, within the great unknown there are normal people. When I read Marilyn's column in Parade it didn't make sense. So I started writing a Pascal program to simulate the problem using the Monty Carlo method. The same solution you provide became obvious as correct, Marilyn was right. Now we can hypothesize an alternate universe with an Evil Monty Hall. He would only make the offer when the contestant made the right choice. In fact he would have all manner of tricks and traps to turn all contestants into losers, no matter what choices they made. Soon the ratings would grow as he gained an audience of sadistic fans. And similar behavior would spread all over. Not a happy place, thank goodness we live in a good universe, at least I hope so.

DeleteFascinating. To many people the correct answer is absurd and no matter how many ways you approach it (a pack of cards whittled down to 2 is my favorite gambit), they fail to grasp the concept. Just like those who refuse to see that the very presence of so many guns in our society is pernicious and needs no evil-minded maniacal villain to blame.

ReplyDeletejohn

Sorry, been thinking about this for several minutes (all I have time for this morning) and still not seeing it.

ReplyDeleteBTW, speaking of unfavorable odds, that Sanskrit online gambling spam has popped up in the previous thread.

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ReplyDeleteHmmmm...OK, now I'm starting to see it. Thanks, Unknown.

Delete(So I snuck back here from work. Sue me.)

Did you use a program to get this or just write it down? Not disagreeing just wondering how you got it.

DeleteAnd where is the Burro?

I'm getting a headache.

ReplyDeleteOK, OK, I surrender to the math gods. That Monty Hall was a tricky little devil.

ReplyDeleteOK, I played 20 hands of three-card Monty after lunch, and FWIW, here were my results:

ReplyDeleteNo switch: Car 7, donkey 3

Switch: Car 4, donkey 6

That's remarkable. You should play the lottery today.

DeleteI see Eric Zorn has written a column on this subject:

ReplyDeletehttp://www.chicagotribune.com/news/opinion/zorn/ct-perspec-zorn-monty-hall-problem-1004-20171003-story.html

I talked about your math and had a very cognoscente remark from a friend of mine. He said he wanted the Burro. We laughed a bit and then I thought of the taxes, insurance and every thing else involved with the car. I realize there are costs with the burro too, but I already have a car and 2 motorcycles, but I have no burro. I think I would stick.

ReplyDeleteNow... what were the odds that Neil and I would both eschew feckless pontificating about Las Vegas to ruminate on the Monty Hall problem? I gave him a shoutout in my newsletter http://e.chicagotribune.com/a/hBZ1RO4B8hLfLB831YEAAA9U7Kn/chi3-2 -- and while I'm at it, here's something I just wrote to a guy who boasted an IQ of 155 and told me I was full of shi on this matter:

ReplyDeleteimagine a five-card game. There's one black card and four red cards, all face down. The black card is the winner. You pick one of the cards but don't look at it.

What are the odds that you have selected the winning card? 1 in 5, right?

And what are the odds that one of the other four cards is black ? 4 in 5, right?

I randomly and with no knowledge of the outcome turn over just one of the other four cards, which remain in front of me. It's red. Have your odds of winning the game improved? Are they now one in four?

If you think so, then you need a basic course in probability. Of course they have not!!! They are 1 in 5. Always, always 1 in 5.

I turn over another card. It's also red. Have your odds of winning the game NOW improved?

No, they remain 1 in 5, always, always 1 in 5.

It doesn't matter how or why I'm turning over the red cards. Maybe I know where the black card is, maybe I don't. But even if I turn over one more card and it's red, the odds that your card is the black card REMAIN 1 in 5.

So should it come down to one of my original four cards and the one card you selected, what are the odds that your card is black? Repeat after me... 1 in 5... always, always 1 in 5.

And the odds that the remaining card from my original set is black? 4 in 5. Definitely, obviously, provably, always 1 in 5.

Play this game yourself if you're not satisfied. Over time, you will lose 80 percent of the time if you don't switch at the end when it comes down to two cards.

155? Wow. One tenth of one percent of the population is over 145. You'd think he wouldn't have cognitive deficiencies. Maybe he meant 115. That'd put him at the high end of normal.

DeleteEric -- Exactly. Showing what's behind the third door doesn't affect the odds of your initial pick being correct, which remain one in three. What they do is double the odds of your second pick being correct, by eliminating possibilities. As for the odds of us both picking a relatively obscure subject, those numbers I can't fathom.

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