When I first heard about the Monty Hall problem was probably when I read about it in The Man Who Loved Only Numbers, the biography of Paul ErdΕ‘s, by Paul Hoffman. I was reading it nervously on the Megabus(which advertised 1\textdollar rides) on my way to New York City from Towson, MD, to meet with Dr. Peter Lax. At the time I was still in Junior College and only had experience with differential, integral, and maybe multivariable Calculus, and Linear Algebra, and I was totally nervous and curious about what was to come ahead in Math. Much like closing your eyes in a scary movie when the scary parts are about to happen, leaving it up to your mind to fill in the blanks somehow makes it worse because you tend to assume the worst. The Monty Hall problem proved to me that whatever intuition I had about statistics or probability, was as good as a boat with a hole in it.

If you’re not familiar with the problem:

“Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?”

The correct answer is to switchΒ because you would have \frac{2}{3} chance versus only the \frac{1}{3} chance of your initial choice. I have to admit, my first intuition was switching would not make a difference because I figured it was 50/50, much like the many math PhDs that wrote letters to Marilyn vos Savant declaring her answer to switch was wrong, but then again I was skeptical about my first guess, and I wasn’t expecting to get it right the first time anyways, and when I read that the right answer was to switch I was truly stumped, I had no intuition to understand what was going on, how to get that answer, or where \frac{2}{3} came from.

The explanation that “What if Monty had 100 doors instead of 3 (a car was behind one door, the other 99 doors had a goat behind them), you choose a door, and then from the other 99 doors he opens 98 of them that have a goat behind them, and asks you if you want to switch or not? Of course you should switch because he obviously knows something about that one door he didn’t open” didn’t satisfy me either at the time, although I thought it was clever.

The adventure began to try to understand this problem. Looking back, I think now the most important line in the problem is probably “and the host, who knows what’s behind the doors.” It took years, and some specific events and reasoning about those events wrongly to really understand what was really going on. There were two key events that helped me understand this problem, one was actually taking the mandatory Probability 1 and Probability 2 courses in my Pure & Applied Mathematics program at my University, and the other was playing a game of ‘3-card Monte.’

I played the game of 3-card Monte before taking the Probability courses, but I’ll start with the knowledge gained from actually learning Probability. First we start by assigning probabilities to each situation:

🐐 for goat, πŸš— for car

πŸš— 🐐 🐐  Β  Β  🐐 πŸš— 🐐  Β  Β  🐐 🐐 πŸš—
πŸšͺ πŸšͺ πŸšͺΒ  Β  Β  πŸšͺ πŸšͺ πŸšͺΒ  Β  Β  πŸšͺ πŸšͺ πŸšͺ

those are the only possibilities, the car is either behind the first, second or third door, and picking the right door, or the door with the car behind it, would mean guessing right out of 3 possible situations, thus regardless of the door you choose you’d have a \frac{1}{3} chance of guessing the right situation, or the car is behind the door you chose. For simplicity, the problem says you pick door no.1, so let’s stick with door no.1, but the logic is the same regardless of the door chosen.

By virtue of your choice, there is a \frac{1}{3} chance of guessing right, and correspondingly a 1-\frac{1}{3}=\frac{2}{3} chance of guessing wrong because you can either guess right or wrong, and the probabilities must sum to 1=\frac{1}{3}+\frac{2}{3}. Thus you have a higher chance of guessing wrong,Β if Monty opened the doors you didn’t choose, right after you chose yours, you’d see a car behind either of the two unchosen doors \frac{2}{3} of the time. This was the key insight learned from those courses. I knew more or less how to theorize probabilities about a situation, but how to theorize about the whole situation (that all the probabilities must sum to 1) and then deduce the probabilities about the anti-situation was formally introduced to me through those textbooks on Probability I read.

So if you were a contestant and were wondering “Wow! This sucks! Thanks Monty, for nothing. I have less than a 50%(33.33%) of guessing right, and a 66.66% chance of guessing wrong,Β if only I could open the other two doors I didn’t choose at the same time, I could then leave this place in style…” Monty suddenly throws you a bone, and opens one of the two doors you didn’t choose with a goat behind it and offers you the chance to switch your choice. Suddenly your wish came true! You couldn’t open both doors at the same time because you can only open one door, but Monty helped you out with that problem by opening from the doors you didn’t choose, the door with a goat behind it.

More importantly, because when the game was set up, because each door has a \frac{1}{3} chance of having a car behind it, whichever door you choose, there is a \frac{2}{3} chance that it is behind the other two doors, and Monty opening one of the two doors that has a goat behind it doesn’t change the probabilities, some new information is gained about the original system we set up, not suddenly you have a system with two doors to choose from with equal chance of having a car behind them. The system always has three doors, you just happen to know after choosing a door that one of them has a goat behind it.

A way to picture it is (remember you chose door no.1, the leftmost door):

πŸš—| 🐐 🐐  Β  Β  🐐 |πŸš— 🐐  Β  Β  🐐| 🐐 πŸš—
πŸšͺ| πŸšͺ πŸšͺΒ  Β  Β  πŸšͺ |πŸšͺ πŸšͺΒ  Β  Β  πŸšͺ| πŸšͺ πŸšͺ

The doors you didn’t choose are to the right of the ‘|’, note that the car is only to the left of the ‘|’ once out of the three situations, and in two situations there is a car to the right of the ‘|’.

Now Monty shows you a door with a goat behind it:

πŸš—| 🐐 🐐  Β  Β  🐐 |πŸš— 🐐  Β  Β  🐐| 🐐 πŸš—
πŸšͺ| πŸšͺΒ πŸšͺΒ  Β  Β  πŸšͺ |πŸšͺ πŸšͺΒ  Β  Β Β πŸšͺ| πŸšͺΒ πŸšͺ
❔| ☝ ❔  Β  Β  ❔ |❔ ☝  Β  Β  ❔| ☝ ❔

Monty is doing the dirty work for you by opening the door that has a goat behind it to the right of the ‘|’ because if you switch, Monty’s actions together with yours amount to opening the two doors you didn’t choose, and because Monty got rid of a door that is a ‘dud’, the door you’d choose by switching would have a car behind it \frac{2}{3} of the time.

Now I will describe a situation where the probability actually is 50/50. Let’s say you were playing a game with three cards, two Queens, and a Joker. The dealer has two cards in one hand, a Queen and the Joker, and the other Queen in the other. The dealer shows you the Joker first with the Queen behind it, and says follow this card (the Joker). Here the dealer can perform a sleight of hand by making it look like the Joker is being placed down on the table first, but really, the dealer is putting down the Queen behind it, and then the dealer places down the other cards and begins to shuffle them flat on the table. If the dealer didn’t perform the sleight of hand and you had digital cameras for eyes that were able to keep track perfectly of the first card he placed down, then the card you would choose would indeed be the Joker. Otherwise, if the dealer managed to fool your perfect eyes, and perform the sleight of hand successfully then the card you would choose would be a Queen. You’re at the whims of the dealer.

But say you didn’t listen to the dealer, you don’t take kindly to being told what to do, and decide to rebel by following the lone Queen in the dealer’s other hand. With infallible eyes that can follow a card no matter what, you follow the lone Queen as the dealer shuffles the cards flat on the table. Note that the dealer isn’t perfect either, the dealer doesn’t know where the Joker is, the cards have been shuffled too many times and the dealer has lost track long ago. So now you can disregard a card, because you know it is the lone Queen, and you’re left with the choice between two cards. Say you naively go through the motions as if it were the Monty Hall problem. You choose a card (not the lone Queen), and in your head you act it out, “Monty shows me where the lone Queen is, and now offers me a chance to switch.” Is switching useful in this case? No, we don’t have the same system as the Monty Hall problem at the onset. First, the dealer showed you each card before shuffling, the shuffling being an attempt to randomize the situation, but you’re able to follow the shuffling of one card so the chances of guessing right where the Joker is isn’t necessarily \frac{1}{3}. Second, the dealer doesn’t know where the Joker is, all the dealer is doing is showing you the cards as he has them in his hands before potentially performing a sleight of hand and shuffling, and can’t offer you any extra information. Monty knows what is behind each door. Furthermore, you already know where a Queen is as well, the dealer or Monty isn’t showing you. Because you were able to follow perfectly the lone Queen, the system starts as a 50/50 chance of guessing where the Joker is, and still is a 50/50 chance after the three cards are shuffled flat on the table. Therefore it is not enough to have a situation where one has to choose the right card/door out of 3, to be able to ‘get rid of a card/door’, and that switching improves the probability of guessing right. One needs the full formulation of the Monty Hall problem for it work. Note that if the dealer performs the sleight of hand 50% of the time, then just following perfectly whichever card is puts down first on the table will give you a 50/50 chance of guessing right where the Joker is. Also note that if you know that the dealer will always perform a sleight of hand (the dealer is a real defector), the dealer always puts a Queen down first on the table, and if you know where the dealer places the lone Queen, to the left or right of the Queen already on the table, then you would know where the Joker is immediately, and since your eyes can keep track flawlessly, you’d know where the Joker was after the shuffling. I guess the lesson here is: pay attentionπŸ‘πŸ‘.

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