Putting the Pro in Probability
The Monty Hall Problem
The door to Riches or Rags?
For all the Brooklyn Nine-Nine fans out there, you’re finally about to understand why Kevin was right, and what Amy was trying to explain to Holt all along. In S04E08, “Skyfire Cycle”, Capt. Raymond Holt and his husband, Kevin, had a row over a notorious math puzzle: the Monty Hall problem. They decide to consult the detectives of the precinct, Rosa and Amy, to settle their tiff. Kevin gives the premise: there are three doors, behind one of which is a car, and behind the other two are goats. You pick a door. The host, who knows where the car is, opens a different door with the goat behind it. Now the host asks if you’d like to choose the other unopened door. Should you do it? Kevin says the probability of winning the car is ⅔ if you switch, ⅓ if you don’t. But Holt thinks that switching is pointless — out of the two doors left, both have an equal probability of having the car, i.e., a 50–50 chance case.
Let’s say you choose door 1. Now, the probability that the car is behind door 1 is ⅓, and for doors 2 and 3, it is ⅔. When Monty reveals that the goat is behind door 2, this ⅔ probability gets concentrated to door 3. Hence, door 3 has double the probability of having the car behind it as compared to door 1, which implies you should switch. Here’s a more in-depth explanation of the Monty Hall problem, along with some history:
“The Monty Hall Problem” was based on the famous American television game show “Let’s Make a Deal” and was named after its original host, Monty Hall.
The game was pretty simple :
- You are on a Game Show, standing in the spotlight.
- Three doors stand in front of you.
- Behind one of the doors is a Prize Car, luxurious and comfortable enough for you.
- Behind the rest of the two doors stands a Goat each, which we assume you wouldn’t want.
- You are given one chance to choose your door.
Once you choose your door, the host, our clever Monty, will not reveal what lies beyond the door you chose.
Rather, Monty will open the door, which has a Goat standing behind it.
Now Monty will present you an opportunity to change your choice for the last time between the two still closed doors.
What will you do??
I know what you think. You think that since Monty revealed the door containing the Goat, you have a 50/50 chance of winning irrespective of whether you choose the third door or not.
Well, it isn’t as simple as it seems to be. So bear with me for some more time.
Now let’s divide our problem into 2 cases.
Case 1: You never change your choice. (The Cowboy in you doesn’t prefer to reload his guns)
Now, what’s the probability that you will select the right door?
It’s ⅓. Yes, you heard us right.
Now, since you won’t change your choice when presented the opportunity, the probability of you winning the game will remain unaltered, that is 1/3.
Case 2: You always change your choice when presented with the opportunity.
Well, give this a thought. If you choose the wrong door, Monty will open the other wrong door containing the Goat.
And since you always change your choice if allowed to do so, you change your choice and select the third door. And of course, you get your Car!
This indicates that to win the car, in this case, you have to always choose the wrong door.
The probability of choosing the wrong door = 2/3, which is also the probability of winning the game!
Hence, we conclude that the best chance of you winning the car of your dreams is to change your choice every time!
Most of us think like Holt: if door 2 does not have the car behind it, it comes down to door 1 and 3, and thus, the chances of finding the car are 50–50. This is because both the doors look the same, and we don’t know any more about one than the other. They seem symmetrical, and hence, the 50–50 probability argument arises. The argument can be resolved using the following table, wherein we assume that you initially pick door 1:
Hence, in 2 out of 3 cases where you switch, you get the car.
VARIATIONS:
In order to understand the problem better, let’s take a variation: a million doors instead of 3. This time, you have a one-in-a-million chance of choosing the right door in the beginning. You choose a door, and Monty opens 999,998 out of the 999,999 doors that are left, revealing that all those doors had goats behind them. Now you’re left with the door you chose initially and the one Monty did not open.
By opening 999,998 doors, Monty essentially filters out the *one* door from 999,999 and drastically improves your chances of winning. The probability gets concentrated to the one door Monty did not open. So what do you do, stick with your initial choice (which has a 1/1,000,000 probability of being the right door) or switch to the door that has been filtered out for you (which has a probability of 999,999/1,000,000)?
Another variation, called the Lazy Monty Hall Problem, takes an interesting take on the original puzzle. In this case, Monty has a preference for a specific door, so much so that he opens it 3 out of 4 times (only because he’s standing close to it — pretty lazy indeed). Can this give the contestant a clue as to whether they should switch or not?
Let’s assume you choose door 1. There is a ⅓ chance that it is the car door. Now, Monty opens door 2 ¾ of the time. In this case, you lose ¼ of the times if you switch ((1/3)×(3/4)=1/4). So, if we run the game 60 times, door 1 will be the car door 20 times, and 15 of those times, door 2 will have a goat, and you’ll lose by switching.
There is a ⅔ chance that door 1 is the goat door. This time, Monty is forced to reveal the other goat door. We assume both doors have an equal probability of having the goat. Monty opens door 2, revealing the goat, and we win by switching. So, switching wins (2/3)×(1/2) = 1/3 of the time door 2 reveals a goat. Again, if we run the game 60 times, door 1 will be the goat door 40 times, and 20 of those times, door 2 will have a goat, and you’ll win by switching.
There are endless variations of the problem, but you’ll be baffled to know that Monty Hall himself did not offer the contestant an opportunity to switch doors! Marilyn Vos Savant, an American columnist, author, and lecturer with the highest recorded IQ of 228, was the one who actually proposed the problem. You can check out the interview here.
GAME THEORY:
From the game theory aspect, the Monty Hall problem is treated as a finite two-stage two-person zero-sum game. The two players are the host and the contestant, and they both want to keep the car. According to von Neumann’s Minimax theorem (check it out here), they both have a minimax strategy, and the game has a value p (here, 2/3), such that 2 cases arise:
- Host uses his minimax strategy of hiding the car and opening either door with equal chance. In this case, no matter what strategy is used by the contestant, he will go home with the car with probability at most p.
- Contestant uses his minimax strategy of choosing a random door and switching after the host opens a door. In this case, no matter what strategy is used by the host, the contestant will go home with the car with a probability of at least p.
With his minimax strategy, the contestant wins the car with a probability of 2/3 exactly, no matter what strategy the host uses. With the host’s minimax strategy, the contestant can’t do better than 2/3 (random initial choice and thereafter switch).
The Monty Hall problem has left many scratching their heads in confusion. The solution seems weird because it is a far cry from what we had mentally assumed it would’ve been. But fret not! You’re not alone. Many expert mathematicians have fallen prey to its statistical illusion. So maybe, brush up on your college-to-kindergarten statistics. Or you could take a hint from Rosa’s advice.
