### NOT Metcalfe’s Paradox – A Puzzle

Bob Metcalfe posed this puzzle at an unconference we were recently at. He says, however, he didn't invent it. Most people, including me, get the answer wrong; in fact, I got the answer wrong twice – once by not listening and once by not thinking.

Here's the puzzle:

You are a guest on a game show. There are three closed doors; behind one of them is a car you want to own; behind the other two are goats you don't want despite the fact that they don't burn gas.

You have to pick a door. After you do that, the host will pick a door behind which there is a goat (he knows what's where and has to follow the rules). You then get to decide whether you should be awarded what's behind the door you picked initially or what's behind the door that neither of you picked.

The questions, smart reader, are:

- Does it matter which strategy you pick?
- If so, which strategy is favored?
- What is the quantitative advantage, if any, of the favored strategy?
- For extra credit: why?

Incidentally, this same puzzle was presented to the math whiz by his teacher in the recent movie "21".

Posted by: sprague | August 01, 2008 at 05:29 PM

Most of you are right at least in part. It's 1 in 3 if you stick and 2 in 3 if you switch. see full explanation at http://blog.tomevslin.com/2008/07/answer-to-not-m.html or in wikipedia http://en.wikipedia.org/wiki/Monty_hall_problem

Posted by: Tom Evslin | July 16, 2008 at 02:19 PM

I am also familiar with the "Monty Hall" problem. Here is a real life example. I was away on a trip scanning the TV in my hotel room. Deal or No Deal was on. I wouldn't normally watch it, but I did. A woman was down to a small number of cases. The one she chose and several others. She is hoping that she has chosen the 1 million dollar case. I think there were 20 to start. She gets down to the last two cases, hers and one other. They are 1,000,000 and 750,000, so she had a good deal win or loose. The host gives her the option of switching. I know that the odds of her choosing the wrong case initially were 95% (19 out of 20) so I knew she should switch. Of course, she did not. The 1 million was in the OTHER case.

Posted by: Phil Marshall | July 15, 2008 at 07:52 AM

I ran across this a few years ago and was always bothered I couldn't match my intuition with the statistical explanation. When I changed the numbers though, it made a lot more sense to me. Imagine there were 99 goats, and 1 car, and the gameshow host reveals 98 of the goats after you picked.

That is essentially an exaggerated version of the same scenario, but the odds seem a lot clearer that your first pick has a 1 in 100 chance of being the right one, and so changing is the right thing to do.

It is a fascinating mental hack, I kept thinking about it wondering if there were real-world situations where I was falling into the same trap.

Posted by: Pete Warden | July 14, 2008 at 04:09 PM

As others have commented on, it is a well known problem and in my opinion the reason it remains a gotcha is because most try to solve the problem logically. Probability in general and conditional probability in particular defies intuition and an algebraic solution would produce the correct result as evidenced by the last section in the wikipedia article. To this day I can not follow Martin Gardner's article on Random Walk, even though my doctorate is on stochastic processes. I always stick to formal analysis when it comes to probabilities.

Posted by: Aswath | July 14, 2008 at 06:48 AM

It's the Monty Hall problem. The solution already has a wikipedia page. I remember talking about this back at University, almost caused a fist fight! :)

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

Since the odds of you picking the winning door are 1/3, the odds of the winning door being in the other pair is 2/3.

The host _never_ opens the winning door.

Therefore the probability of winning with a switch in doors is 2/3.

Remember, the odds of you having selected the winning door don't change.

Posted by: Jason | July 14, 2008 at 01:10 AM

Here's how to cheat and make it intuitive -- imagine same scenario with a million doors. The host opens all doors but one. Do you switch?

Posted by: Jon | July 13, 2008 at 09:54 PM

I would switch doors.

Posted by: Eric | July 13, 2008 at 08:43 PM

Actually, this is known as the "Monty Hall Dilemma" and the answer is to pick door 1 and switch to the remaining door after the door has been opened.

The reason is that while the doors initially all have a 1/3 probability of being right, once one door has been removed, the probability changes.

I don't remember the math but I seem to remember that the odds of the remaining door being the one with the car are higher than the odds of your door being the right one.

Posted by: TNLNYC | July 13, 2008 at 08:39 PM

Ok, I'll play...

1) Yes

2) Switching

3) it's too hot to do math...

4) see below.

Assuming that the goats and car are randomly placed, I have a 1 in 3 chance of picking the car the first time. However, this means I have a 2 in 3 chance of picking a door with a goat. These odds are not at all affected by the host later identifying one of the doors with a goat.

Once the host identifies one of the goat doors, my decision is whether to switch or not. If I believe I've picked the car, I should not switch and if I believe my odds are 50/50 there's no point in switching. As noted, the first premise is unlikely since my odds of picking the car the first time were only 1 in 3. The second is false since the odds will not have changed to 50/50 with the elimination of one of the goat doors.

My odds of having picked a goat are 66.6% though... so it's twice as likely that, if I keep the initial guess, I'll own a goat. I would thus switch.

Now, since you've characterized this as a paradox, I'm wondering if that fairly obvious answer is incorrect and, if so, what the reasoning is.

Posted by: rick | July 13, 2008 at 07:28 PM