Tuesday, July 22, 2008

learning statistics and computer games - recharging (energy) is same as (negative) randomness (entropy)

This is a well known problem that illustrates a nice area of proability/statistics (recently showed up inthe so so movie with Kevin Spacey, 21):

In a game show, a contestent has to guess which of three doors
the prize lies behind (e.g. a fast car) - the other two doors lead to nothing.
The game show host knows the right answer.

Now, lets say the contentst guesses door A. The game show host
now opens one of the other two doors (obviously showing
nothing), and asks the contestent
if they want to stay with their choice, or change to the door that
they didn't chose and the host didn't open.

What should the contestent do, and why?

So the answer is change (always - since you have 2/3 chance of the prize being behind the two doors you didn't pick, and in the second go, you get to know it is 50/50 between the door you did pick (with 1/3 chance) and the one you didnt that didnt get opened, so either way, you are twice as likely to win by changing as by sticking.

The more interesting problem is:
how do you convey this (teach it) to people?
I asked 4 random kids - 2 got it (using the 1/3 v. 2/3 or the 50/50 v 1/3 argument above) . two refused to believe me after explanation

Soome said: cast the change in information as like recharging in a game, then maybe they'd get it - so information increase is entropy decrease entropy is just negative heat. maybe there is something in this - could we devise a game whch illustrates this idea generally?

derek says: why not do the 100 (or million) door version and in 2nd go, host opens 999,998 doors...


Richard G. Clegg said...

Ah, the famous Monty Hall problem -- but if Marilyn vos Savant cannot successfully explain it to PhD students and professors do you really have a chance of explaining it well to children?

Actually, despite being familiar with the problem, I was quite confused by your explanation. The 50/50 is surely misleading -- no probability of 1/2 shows up anywhere in the calculations (although many people think it does). If the car is behind the door you picked (1/3 chance) you lose by switching. If the car is not (2/3 chance), you win by switching.


jon crowcroft said...

so the 50/50 is the chance if you had not had any prior goes, when what is left in front of you is 2 doors and 1 prize. the point is that you got here
first with no information, and were then given more info...yes its confusing:)

by the way, the trick of chaning the problem to 100 (or 1Million) doors
did not work to help the two kids who didn't get it, understand it any better:)

jon crowcroft said...

...more specifically (about kids learning)
the hope was
a) they had fewer preconceptions
b) the way they misunderstood it (viz 50/50 explanation) might reveal a better choice of learning strategy

so far, no joy:)

Richard G. Clegg said...

Ah, thanks. I understand the origin of your 50/50 now. To my shame, I got the answer wrong the first time I came across the question (although I was asked a few pints into the evening so I have something of an excuse).