Would you like to fry your
brain in a completely organic way? You can, simply by trying to understand the
Monty Hall problem.
Maybe you’ve already heard
of it. Until M told me about it, I had not. Also, until M told me about it I
thought I had a reasonably good grasp of basic logic.
The problem is named for
the host of the American game show, “Let’s Make a Deal,” though it doesn’t
actually have anything to do with him or the show. It is a probability puzzle
in which a game show scenario is used to illustrate an apparent paradox.
Here it is.
You are a contestant on a
game show. You stand before three identical, closed doors. The show’s host –
let’s say it’s Monty himself -- tells you that behind one of the doors is a
car, and behind the other two are goats. You are allowed to choose one door and
to claim as a prize whatever is behind it.
You tell Monty,
“I choose door number 1.”
Monty then says that before
he opens your chosen door he is going to give you a chance to change your mind.
He walks over to door number 3 and opens it, revealing a goat. He says to you,
“Do you want to stick with your original choice of door number 1, or do you
want to switch to door number 2?”
And now we arrive at the
brain bursting bit. Do you think you should change your choice to door number
2?
I hope you said that it
doesn’t matter, that both door 1 and 2 are now equally likely to hide the car.
That will make me feel better, because that’s what I said and that’s what I
argued for two hours to M the wretched day he told me about the problem.
But it is wrong. You’re
actually better off switching your choice to door number 2, and here is why.
As you probably already
assumed, the host knows what is behind each door. This is one of two things at
the nub of the riddle. The second thing, and the one that I think most people
pay too little attention to upon initial consideration, is that this is a
probability puzzle. It does not predict what will happen in a single enactment
of the scenario but rather gives chances based on what would happen if you
played the game a whole bunch of times.
So you mustn’t think about
the problem in terms of having a single chance to choose a door and probably
ending up with a goat, which you will name Ethel. Think about playing the game
a hundred times.
Two-thirds of the time you
are going to choose (as your first choice) a door that has a goat behind it. In
these cases, when Monty opens his door – and he’ll always open one with a
goat behind it – he will have ‘shown’ you that the door neither of you
chose hides the car. In these cases he has given you a one hundred percent
chance of finding the car.
But wait, you say. We don’t
know which times I have first picked a goat door. I accept that it
occurs two-thirds of the time, but what is the use of that if I don’t know
which times they are?
Just hold on, dearie, and
focus the old coconut on what I said about it being a probability puzzle.
The remaining one-third of
the times you play the game your first choice will be the car door. In the
excitement of considering this, you might neglect logic. You desperately want
the car and you’re tired of taking Ethel home (she has wrecked your house and
farts in front of company). Why on earth would it be a good idea to switch your
door in such cases, when you’ve landed on the right one the first time? It
wouldn’t, of course. But even if you do switch your choice of door and end up
with a goat, this is still only happening one-third of the overall times you
play the game. The other two-thirds of the time, if you switch doors after
Monty shows you a goat, you’re going to get the car.
This means that if you
always take the chance to switch doors, you will go home with the car
two-thirds of the time. Ethel, alas, will only get the chance to poison the
atmosphere of your dinner parties every third week.
Now do you see? I don’t
mean to patronise. Believe me, I smelled smoke when I was trying to work out
the first time. But then again, I didn’t have a kindly blogger explaining it to
me in words that I could understand. I had only M, sitting on the couch with
his superior knowledge of mathematics and intuitive ability to understand
things like terminal velocity, while my dogged, pack-mule-like brain stumbled
through the underbrush.
If it’s still not making
sense to you, this might help. Think about playing the game if Monty did not
know which door hid what. Of course it would be a silly way to play because
sometimes he would open the door hiding the car and the audience would probably
feel less than hysterically excited to watch you try to decide between the
remaining two doors. But of the times when Monty opened a door onto a goat
there would be no advantage to you changing your choice of door. There would be
exactly a fifty percent chance of the car being behind either of the remaining
closed doors.
The reason the Monty Hall
problem is so perplexing is that it seems to be a simple random chance
situation, but it’s not. Monty knows where the goats are and acts upon that
information. This adds an element of predictability to the results, but only an
element. There is still a random element, too, and the co-existence of the two
in one simple problem seems to jam our mental gears.
My mental gears got jammed
just trying to find a way to say that. So I’m going to go now and unload the
dishwasher and give the ol’ circuits a chance to cool down. As for you, I can
tell that you so thoroughly enjoyed the Monty Hall problem that you are heart-broken
to find yourself at the end of the blog. Don’t cry, I’ve got a treat to leave
you with.
It’s called the birthday
problem, and it shows mathematically that the chances of two people in a group
of 23 sharing the same birthday are fifty percent. Can you explain why?
Ah, come on. You need to hone your mental skills. What if the human race gets nearly wiped out and it's up to a dozen or so of us to re-invent combustion engines and mobile phones? Don't look to me. I'm going to be swamped with my mould-growing operation, hoping one of them turns out to be penicillin.