Old puzzle for the statisticians

[Cohiba]

I know this isn't magic, but because of the ACAAN posting that's digging into probabilities, here's a little puzzle for you. Some of you may have heard this problem posed before (it was famous years ago):

Like the old "Let's Make a Deal", a game show contains 3 doors, designated A, B, and C. The game show host tells you that there is $1 million behind one of the doors, and the other two doors contain goats. You get to take home whatever lies behind the door that you open.

You are offered a free selection of the three doors, and let's assume you choose door A. The host then opens door C, showing you a goat. He then gives you the opportunity to change your original choice, and select door B - only if you would like to.

The Question: Are you better off
1 - sticking with your original choice (door A),
2 - changing to door B, or
3 - does it not make a difference?

[Joe M. Turner]

Change to another door. Didn't Bob Farmer have an article on this?

[Keith Raygor]

Statistics and my brain sometimes occupy different worlds, so I wasn't until I watched this explanation that the answer was clear to me:
http://www.youtube.com/watch?v=mhlc7peGlGg&e

Wikipedia has an excellent chart that demonstrates the results of each choice.
http://en.wikipedia.org/wiki/Monty_hall_problem

[RogueMD]

If the host knows which door the money is hidden behind...switch doors to B.(This increases your odds of winning)
If the host DOESNT know...it doesnt matter if you switch or not.
(Odds are the same)

RogueMD

[erlandish]

Heh, that blew my mind. Totally counter-intuitive.

It's like playing three card monte in a random mix where you have 2 guesses to figure out where the winning card is, and it doesn't matter if you get the first one wrong. Your chances improve to 2/3 in that format rather than in the format of one guess only. This problem simulates having two guesses, as opposed to one.

If it was a random mix, then there's a %33 chance that it's behind A, and a %66 chance that it's behind B or C. He was always going to be able to show you a goat in the set of B or C anyway, so that part is irrelevant. Very sneaky. The goat and the car would have to be remixed again in order to get to a %50 chance.

[erlandish]

Sorry, maybe that's a poor explanation.

Think if the game was in the reverse. You get to choose two doors instead of one. Your odds of winning are 2/3. It doesn't matter if one of those doors shows a goat because there's a %100 chance that one of the two was a goat anyway (the combinations are goat-car, car-goat, and goat-goat), but only a 1 in 3 chance that BOTH of them are goats.

So, if you started out with two guesses instead of one, you'd do better to stick with your two guesses even after a goat is shown. Similarly, if you started out with one guess instead of two, you're stuck with a 1 in 3 chance even when one of the goats is revealed.

[Richard Hatch]

Here's an easy (I think) way to understand why you should switch. Assume there are 100 doors rather than 3 and that after you pick a door, the host (who knows where the prizes are) will open all the doors that have goats behind them except yours and one other one. Before he opened the doors, your odds were 1 in 100. Once you know where 98 of the 99 goats are, the odds change dramatically. Switch!

[Cohiba]

You guys are good! I thought there would be at least some disagreement on this. Maybe the problem is more well known than I thought.

Nice job!

[erlandish]

A quick question, though... Conceptually I get it, but I was wondering, has anybody sat down and written out a computer program to see if the concept works out? Say, with 10,000 random games played?

[Adam Brooks]

It's been done. From Ian Stewart's wonderful book "The Magical Maze" (a book on recreational mathematics), in the chapter "Marilyn and the Goats":

"I ran 100,000 trials on a computer. The contestant's strategy [don't switch] gave the correct answer in 33,498 trials, but the wrong answer in 66,502. Withe the host's recommended strategy [always switch], the numbers were exactly the other way around. The corresponding probabilities of 0.33498 and 0.66502 are convincingly close to the host's claimed values of 1/3 and 2/3."

So there ya go.

[Nick]

The mathematical reasoning for the "switch" strategy is correct, and computer trials are great, but I never really "believed" the solution at a gut level until I coerced a friend into sitting down with me and playing the game several times with three cups and a rolled-up dollar bill. When you're the game show host, 2/3 of the time you actually have *no choice* about which cup you show (or door you open) after the initial choice, because the player has chosen an empty cup 2/3 of the time. It was that knowledge of having no choice 2/3 of the time that moved my belief in the correctness of the switch strategy from my head to my gut.

[Larry Barnowsky]

The Monte Hall Problem was well discussed in a Parade Magazine column many years ago where Marilyn vos Savant correctly stated the answer (always switch). She then received letters from mathematicians and other experts who said she was wrong. Other experts concluded she was right including I believe Perci Diaconis. The problem is seductively sneaky. We have to assume however that Monte Hall always allows you to switch after choosing. If he knows you are using the switch stragegy he could offer the switch only when it would make you lose. However, for the purpose of the problem Monte always offers the switch.

http://en.wikipedia.org/wiki/Marilyn_vos_Savant

[Bob Farmer]

I have written about this extensively in both MAGIC and GENII. However, the defintive article, with a complete psychological analysis of WHY this is so confusing, will appear in my soon-to-be-forthcoming masterpiece, THE BAMMO FLIM-FLAM CONgomeration.

It doesn't matter if Monty knows or doesn't know where the prize is -- what's important is that he opens a door and shows you where it isn't. You then switch and you'll win 2/3rds of the time.

Of course, if he doesn't know where the prize is, then he will not open a door to show you where it isn't -- and in those circumstances it doesn't matter if you switch.

The manner in which this problem is explained causes as many arguments as its solution -- but I think I've come up with the best way of doing that, a way that no one else has used. It's in the book!

[Michael Kamen]

Thanks for the refs Larry. Very interesting -- and fooled me. To paraphrase the solution and check my own understanding, we have a 2/3 chance of choosing the wrong door in the first place. So, our first choice is probably wrong (2 out of 3 times). Learning where one of the goats resides makes switching the obvious choice, since we were probably wrong in our first choice and we know where the other wrong choice is now.

[Mark Ratekin]

Hi, Bob -

Any update on when your book will be published?

Thanks -

Mark

[Bob Farmer]

I'm aiming for the summer and having it available for the IBM convention.

[Larry Barnowsky]

Monte Hall strikes again!!!
http://www.nytimes.com/2008/04/08/scien ... .html?8dpc
It appears that some fundamental reserach done studying Cognitive Dissonance is flawed due to the researchers not accounting for the "Monte Hall Effect". The statistical basis of their research is called into question.

[Jonathan Townsend]

Thanks Larry, wonderful when folks who do the rationalzing forget that readers can tell.

[Michael Kamen]

That is too diabolical Larry. Kind of scary that we can fool ourselves that easily.

[Jonathan Townsend]

pretty much goes back to the joke about the scientist who claimed spiders hear through their feet.