The Monty Hall Problem
Probability is complex. It can make or break your game by making you fall into the abyss of its statistical illusion. The comparison to illusion is apt. Even though you know that your eyes are deceived but you tend to rely on your brain’s assessment. Such probabilities emphasize, how crucial it is to check that you’re satisfying the assumptions of statistical analysis before trusting the results. A famous case of it is ‘The Monty Hall Problem’.
This is a probability puzzle named after Monty Hall, the original host of the TV show ‘Let’s Make A Deal’. It’s a famous paradox that has a solution that is so absurd, that most people refuse to believe it’s true.
Suppose you are on a game show, and you’re given the choice of three doors: Behind one door is the prize; behind the others, goats. The prize is behind one of the closed doors, but you don’t know which one. You select a door and state it out loud. Now Monty, who knows all the items behind the door opens one of two unselected doors which obviously has a goat behind it. Now you that the prize is either behind the door you chose or the one left. Thus comes the main question from Monty, “Would you like to switch the door or stick with your decision?”.
The majority of the people assume that both doors are equally like to have the prize. It appears like the door you chose has a 50/50 chance. Because there is no perceived reason to change, most stick with their choice. But as I said earlier, the probability is complex, and nor would this question be so debatable if it was so simple. The truth is that you double your chances of winning by switching the doors.
Believe it or not, it’s actually to your benefit to switch. If you switch you have roughly a 2/3 chance of winning the prize. If you stick with your original choice you have roughly a 1/3 chance of winning the prize. This fact has been approved over and over again with a plethora of mathematical simulations. Still not able to grasp it. Don’t fret. Even a great mathematician like Paul Erdös was baffled and not convinced until he saw the practical results of this problem. A more logical approach to this problem could be seen in the table.
It turns out that there are only 19 possible combinations of choices and outcomes. Not switching results in a 33% chance of winning whereas switching doubles the chance to 66%. The first row shows the scenario where you pick door 1 and it turns out to be the one with the prize behind it.
In this case, you switch, you lose, or stick with your original decision and win. For the second row, you pick door 1 and the prize is behind door 2. Here you switch from door 1 to door 2 you win or you stay you lose. The table shows all the potential situations. We just need to count up the number of wins for each door strategy.
The mathematical approach to this problem is through ‘The Bayes’ Theorem’, which is a way to figure out the conditional probability. Let’s assume you pick door 1 and then Monty shows you the goat behind door 2. In order to use Bayes’ theorem, we need to first assign an event to A and B. Let event A be that the prize is behind door 1 Let event B be that Monty opens up door 2 to show the goat
The probability of event A is pretty simple to figure out that is 1/3 and the remaining 2 have the probability of 1/2. Here’s when conditional probability comes into the picture, which makes getting the probability of B in the denominator a little trickier. Analyzing the situation:
- You choose door 1. Monty shows you the goat behind door 2
- If the prize is behind the door Monty will not choose it. He’ll open door 2 and show a goat 1/2 of the time
- If the prize is behind door 2, Monty will open door 3 as he never reveals the prize
- If the car is behind door 3, well then you know the drill.
As Monty has opened door 2, you know the prize is either behind door 1, which is your selection, or behind door 3. The probability of the prize being behind door 1 is 1/3. This means that by the law of probability, the chances of the prize being behind door 3 is 1–1/3 which is equal to 2/3. And that is why you switch.
Another reason why this problem baffles a large number of people is the small numbers. Let’s look at the same problem with 100 doors instead of 3. You pick a random door. Instead of 1 door, Monty eliminates 98 doors, which he knows are the ones with no prize behind them. You’re now being presented with a filtered choice in comparison with the wild guess you made. It should be clear that now your odds are much better if you switch.
The reason why this problem gained attention is due to media furore. Vos Savant, one of the early pioneers in decoding this problem, wrote in her first column on the Monty Hall problem that the player should switch. She received innumerable letters from her readers — the vast majority of which, disagreed with her. But eventually, she proved to be right.
In his book The Power of Logical Thinking, cognitive psychologist Massimo Piattelli Palmarini writes:
“No other statistical puzzle comes so close to fooling all the people all the time [and] even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer.”
Another version could be that Monty reveals the goat, and he suddenly has a seizure. He does close the door and mixes all the prizes, including your door. Does switching help? The answer to this is No. Monty started to filter but never completed it — you have 3 random choices, just like the beginning. Hence this version could be ruled out of the statistical circle.
The solution to Monty Hall’s problem seems weird because our mental assumptions for solving the problem do not match the actual process. Therefore it can definitely be termed a ‘Statistical Illusion’. The fatal flaw in the Monty Hall paradox is not taking Monty’s filtering into account, thinking the chances are the same before and after — it’s to realize how subsequent actions and information challenge previous decisions. This again brings us to our old statement. Probability Is Complex.
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