Thursday, December 20, 2007

Marilyn vos Savant Puzzle

I suppose this is a bit old, but I was reminded of it by 'The Curious Incident of the Dog in the Night-Time' by Mark Haddon. It's not about church or theology . . .

This is the puzzle which came up on a programme hosted by Marilyn vos Savant (purported to have the highest IQ in the world!)

In a game show a contestant is shown three doors. He is told that behind two of the doors there is a goat. Behind the other door there is a car. He must guess the correct door in order to take home the car. After the contestant makes a guess, the host opens one of the other doors to reveal a goat. The contestant must now choose whether to stick with his original choice or to choose the other unopened door.
What is the sensible thing to do?

I'll put the answer in a comment if I get any other comments to this!


Unknown said...

Well if you change then you may have a mathematical 2/3 change to win nd a 1/3 change to loose. Answer is apply maths it.

bugs said...

True, however, statistics shows that your first answer is usually the correct one.

Jenny Hillebrand said...

Thanks for your comments! It's interesting that most people think that there is a 50/50 chance when you are left with two closed doors - and both comments were unhappy with that.
As well as applying maths to the problem I took three playing card (an ace and two eights) and played the game 20 times with my husband as the contestant. He quite quickly picked up that he was winning when he changed his choice. Out of 20 times he would have won 14 times if he changed and only 6 times if he had stuck with his original answer. So the experiment showed 70% chance of winning if he changed. The maths shows 66% chance of winning if he changes, so that was a pretty good correlation.
To see the truth of the maths the best things is to draw a 'choice picture'. Draw three blocks across the page that represent your first choice. Write goat in two of them and car in the third. Now draw two boxes under each of these - label one stick and one change. In the stick box write the original choice, in the change box write the other. Now you have six boxes. You should see that of the three 'stick boxes' two have goat in them. Of the three 'change boxes' two have car in them. Try it - it's hard to describe without the picture.
Or you can think of it mathematically using numbers as Herman was describing. I'll leave that as an exercise for the student!

I love numbers!!!!

Anonymous said...

Solution to this probability statement assumes, as the 'brainteaser' does, that the behaviour of the showhost is constrained by simple logical rules. This being the case , the behaviour of the host is a 100% given, and the outcome is fixed and determined entirely by the contestants initial choice. Thereafter the outcome is certain.

Assuming the contestant takes the switch, which MUST be offered:-

If the contestant initially chooses the concealed prize, a 1 in 3 chance , he loses.
If the contestant chooses a concealed goat, a 2 in 3 chance, he wins

Why? If he chooses the prize, (1 in 3) the host can't open this door, (against the 'rules' and defeats the object) He can only open one or other remaining doors, both with goats, it does not matter which. He then has to offer the switch to the only remaining door not yet selected, concealing the remaining goat, and the contestant loses.

If the contestant chooses a concealed goat (2 in 3) the host cannot open the door with the prize( again, against the 'rules' and defeats the object) and must open the door with the other goat. The only remaining door has the prize, and since the host must offer it, the contestant wins, 2 out of three times.

If the contestant refuses the switch, whatever happens after his initial choice is irrelevant....and he has a simple 2 in 3 chance of losing based on his initial choice. 1 prize, 2 goats!

All the wrangling about this problem confuses the brainteaser, in which the hosts behaviour is fixed by logic. with a real show.

In a real show the host may:

Decide to open the door revealing the prize, which rather defeats the object, and has nothing to do with the probabilites contained in the brainteaser.

Not offer the switch or convince the contestant not to accept it, which is not part of the brain teaser and has to do with 'cheating' or psychology, not the problem being posed.

Not be aware of which door hides the prize, which was also not the problem being posed.

The brain teaser makes the offer to switch part ot the game 'rules'

idlingdove said...

Comment to Jenny Hillebrand: interestingly, I made exactly the same experiment as you did with the cards, only I chose an ace and two sevens, and did it 100 times. Exactly 67 times it turned out to be better to switch versus 33 times better to stay. Probabilities work!!!
Once I had satisfactorily proved to myself that Marilyn was right, I set about thinking why this was so, and quickly realised that it was quite obvious...

Unknown said...

I have seen this before, but was struck today that it can apply to the Howie Mandel game show, "Deal or No Deal."

So I will be playing with numbers today. :)