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.