# Rationale behind the Monty Hall problem

Yesterday, I was asked for the rationale behind the Monty Hall problem, which I originally referred to back in August 2004. I thought it worth sharing. The original problem involves three doors; I also worked out the logic for four.

The original problem goes like this. There are three doors, behind one of which is a car, behind the other two of which are goats. (Not sure why goats, but that’s the version I heard.) The assumption is that you’d like to go home with a car as opposed to a goat.

You’re invited (by a game show host, naturally) to pick a door, which you do. Irrespective of which door you pick, the game-show host opens one of the non-picked doors and reveals a goat. He then asks you whether you want to stick with your original choice, or switch to the other unopened door.

The answer is that you should always switch, as this doubles your chances of driving home as opposed to attracting bemused stares while walking home accompanied by a goat.

Let’s refer to the doors as A, B and C. For the sake of argument, let’s assume you choose door A. (Choosing each of the three doors is equally likely, and therefore the odds you see below can be divided by three and then multiplied by three at the end, with no overall impact. Doing so confuses and adds no value, so I won’t.)

There are three potential cases:

– Case 1: car is behind door A

– Case 2: car is behind door B

– Case 3: car is behind door C

If you don’t switch doors, then your chances of being correct are 1 in 3, as you ignoring any extra information being given. If you select door A, then in case 1 above, you’ll win; in cases 2 and 3, you’ll lose. The odds of each case occurring are equal, so the odds of winning the car if you don’t switch are 1/3.

In case 1 above, the game show host will open either B or C. Either way, switching will result in a goat.

In case 2, he will open door C (he can’t open A because you chose it, and he can’t open B, because it hides a car). Switching will give you door B, which will result in a car.

In case 3 he will open door B (he can’t open A because you chose it, and he can’t open C, because it hides a car). Switching will give you door C, which will result in a car.

So, in equally likely scenarios, (1, 2 and 3), scenarios 2 and 3 give you a car; scenario 1 gives you a goat.

So if you don’t switch, you have a 1/3 chance of winning. If you switch, the probability of winning goes up to 2/3 – double.

Now, let’s move this up to four doors and see what it does to the odds.

The doors are A, B, C, D. You choose A. Cases 1 through 4 are that the car is behind A, B, C, D respectively.

The chances of winning if you don’t switch are 1/4. Now, let’s assume you switch.

In case A, if you switch, you’ll lose, as you chose the correct door in the first place. In cases B through D, if you switch, there’s a 1/2 chance that you’ll win. This is because the car is not behind the door originally selected, and there are only two other doors to go for, one of which has a car.

So, your odds of winning are (1/4 * 1/2 * 3) = 3/8. This is:

[a 1/4 chance of one of a specific potentially-successful scenario

happening] *

[a 1/2 chance of success from switching] *

[3 scenarios]

The 3/8 chance of winning having switched is 50% greater than the 1/4 chance you would have had if you’d stuck.

### Comments

**8 Responses to “Rationale behind the Monty Hall problem”**

**Leave a Reply**

Surprised at you Dan, I stopped reading at the explanation of the 3 doors as I think your logic is flawed. You’ve confused “switching” with making a brand-new selection.

Once you’ve picked a door, the host opens a door, leaving two doors. Changing your selection is not the same as making a brand new selection. So I think what you mean is “you should always re-chose”, as opposed to “you should always switch”. Re-chosing opens the fair possibility of picking the correct door(1 chance in 2) in the event you guessed right first time (1/3).

Switching I don’t think is the same.

Ok, I take it back. 2/3 > 1/2 – You should switch. Damn, this has messed with my sleep deprived mind.

The beauty of comments is that I can choose to keep them, or delete them. I think I’ll keep these for posterity 😉

And you’re already blaming your son for things. Shame on you…

Not blaming my son, blaming the earth for being round and creating different time-zones as a result system owners that I have calls with on the other side of the globe are awake when I should be asleep. Is it really the earth’s fault though? Maybe I should blame gravity and the curvature of space for squashing the earth into a spheroid. Damn gravity. This just lends credence to my theory that all Gravity is evil.

The earth’s round? I’m not convinced.

It’s really the fault of the people like Columbus (or that Chinese fella) for going beyond their natural boundaries. If everyone inhabited a strip of the earth 15 degrees wide stretching from the Arctic to Antarctica (I’d suggest Europe/Africa to maximise land; N and S America are deceptively not on top of one another latitudinally), then we’d all be in the same time zone, and no other time zones would effectively exist.

Here’s the maths to support the argument:

– World radius: 6,378km

– Surface area: 511m sq km

– Surface area in 15 degrees: 21m sq km

– Percent of land in 15 degrees: 60% (wet finger based on 20-35 degrees East)

– Surface area of land in 15 degrees: 12.8m sq km

– Population density required to house the world’s 6bn people: 469 people per sq km

– Current population density of the UK: 383 people per sq km

It’s quite a big project to make this happen, but I think I’ve presented enough of a business case for someone to start running with this. High-level plan anyone?

BTW, here are the formulae, where n is the number of doors:

P(winning without switching): 1/n

P(winning if you switch): (n-1)/(n(n-2))

% uplift from switching: (n-1)/(n-2) – 1

Here’s the % uplift in your chances of winning for various values of n:

n % uplift

=================

3 100%

4 50%

5 33%

6 25%

7 20%

8 17%

9 14%

10 13%

15 8%

20 6%

30 4%

40 3%

50 2%

75 1.4%

100 1.0%

Dear, oh dear.

The chances are not improved when you switch, what’s wrong with you all.

You have a 1 in 3 chance of getting it right with 3 doors. Now, Monty (or whoever) knows where the car is so shows you where it isn’t (this helps a lot when extrapolating to more doors. He opens all but your door and only one other, not he opens just one more).

Does this mean there is less of a chance it’s behind the door you originally picked? No, of course not, ther’s still a 1/3 chance it’s behind your door. Odds do not change once an outcome is shown. The logic that you should switch is as dumb as that of you should pick lottery numbers on what has passed.

Read this if you’re not convinced:

http://math.ucsd.edu/~crypto/Monty/montybg.html

Or google it and read any number of other explanations.

Or, Uncle Rob, you could do your own homework 😉

http://exploringdata.cqu.edu.au/montyexp.htm

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

Dan.