Briefly discuss the Monty Hall problem and share your insights and solution to this brain teaser problem.
The
Monty Hall problem is a probability puzzle named after the late Monty Hall, the original host of the TV show Let’s Make a Deal. In this problem, there are
3 doors. Behind 1 door is
a car; the grand prize that the contestant can win, while behind the other 2 doors contain
goats. The contestant chooses a door at random which he or she believes is the door that contains the grand prize. The game host then opens one of the 2 remaining doors which reveals one of the goats inside. Now the contestant has the option to stick with his or her chosen door or switch to the other or remaining door to win the prize. The majority of people assume that both doors have the same probability to win, where the door you chose and the door that was left unopened have a 50/50 chance. Because there is no perceived reason to change, most stick with their initial choice. However, in terms of chances, that is not the case.
The chances of actually winning the car is, at first, distributed equally among each door, meaning that each door will have a ⅓ chance of containing the grand prize. If a contestant chooses 1 door, he or she will have a ⅓ chance of having the grand prize, and it would stay that way if he or she sticks with his or her chosen door. That would mean that there must be a ⅔ chance that the car is on the other 2 doors. However, even though the host had already revealed the contents of the second door, the contestant would still have a ⅔ chance of winning the grand prize if he or she switches. The odds are better if you switch because the host curates the remaining choices (Statistics How to, 2021).
To better understand the concept of probability in this game, let us look into the answer of columnist Marilyn vos Savant in her magazine Parade when she was asked about the same problem. According to her, there is about 66% chance (roughly ⅔) of winning the prize by switching (Frost, 2021). This is because there are only nine different combinations of choices and outcomes. Let us refer to this table:
## row1 row2 row3 row4
## 1 1 1 Win Lose
## 2 1 2 Lose Win
## 3 1 3 Lose Win
## 4 2 1 Lose Win
## 5 2 2 Win Lose
## 6 2 3 Lose Win
## 7 3 1 Lose Win
## 8 3 2 Lose Win
## 9 3 3 Win Lose
## 10 3 Wins (About 33%) 6 Wins (About 66%)
| 1 |
1 |
Win |
Lose |
| 1 |
2 |
Lose |
Win |
| 1 |
3 |
Lose |
Win |
| 2 |
1 |
Lose |
Win |
| 2 |
2 |
Win |
Lose |
| 2 |
3 |
Lose |
Win |
| 3 |
1 |
Lose |
Win |
| 3 |
2 |
Lose |
Win |
| 3 |
3 |
Win |
Lose |
|
|
3 Wins (About 33%) |
6 Wins (About 66%) |
The first column contains the numbers 1, 2 and 3 which represent each of the doors or labels each door. Meanwhile, the second column represents the door that has the prize. The third and fourth column tells whether the contestant wins or loses the game if he or she either sticks with his or her first choice or switches to the other door respectively. By enumerating these nine different combinations of choices and outcomes, we now have different potential situations where we could analyze the probability of winning between switching and not switching the contestant’s first choice. In the third column which represents the outcome if the contestant didn’t switch, most of the result would indicate that the contestant would have a 33% chance of winning if he or she sticks with his or her first choice; having a total of 3 wins and 6 losses considering all 9 scenarios. Meanwhile, if the contestant switches his or her door for the other unopened door, then he or she would have a bigger chance to win; having a total of 6 wins and 3 losses considering all 9 scenarios, which gives him or her a 66% chance of winning. If we compare the probabilities of winning between switching and not switching the door, then from the probability percentages we could conclude that the contestant would usually have a better chance of winning the prize if he or she switches his or her choice.
Still not convinced that switching would give us a better chance of winning? Then let’s try increasing the number of doors, from a measly 3 to a hundred. Each door would now have 1/100 chance of containing the prize. The contestant still picks one door, but the host opens 98 doors so that the contestant would have the option to either stick with the door he or she initially chose or switch to the other. If we always keep our original door, then our probability of getting the car is the probability we choose the door on the first try, which is 1/100. If we choose to always switch, then our probability of ultimately getting the car is the probability we choose a goat times the probability we switch from a goat to a car. Our probability of picking a goat initially is clearly 99/100. Then, once we pick a goat and one goat door is opened, there are 98 other doors, one of which contains the prize. This would mean that our chance of switching from a goat door to a car door, or the chance of guessing correctly after switching is 1/98, which also demonstrates sampling without replacement.
Here is the solution for solving the probability of winning the prize by switching:
Let
P(A) be the
Probability of getting a goat
Let
P(B|A) be the
Probability of guessing correctly after switching
Let
P(A intersection B) be the
Probability of getting the prize or winning by switching
\[ P(A)= \frac{99}{100} \] \[ P(B|A) = \frac{1}{98}\] \[ P(A \bigcap B) = P(A)*P(B|A) = \frac{99}{100} * \frac{1}{98} \approx 0.0101 \]
Now let’s compare that to the probability of getting the prize on the first try, which is 1/100 or approximately 0.01. You’ll notice that the Probability of getting the prize or winning by switching is
slightly greater than the probability of getting the prize on the first try of sticking with the initially chosen door (0.0101 > 0.1). This would still indicate that we would usually have a higher chance of getting the prize if we ultimately switch from our first choice rather than sticking with our first choice.
Names: Justine Sison and Sun Phil Zablan
References:
Frost, J. (2021, February 01). The monty hall problem: A statistical illusion. Retrieved July 15, 2021, from https://statisticsbyjim.com/fun/monty-hall-problem/
Statistics How to (2021, June 13). Monty Hall problem: Solution explained simply. Retrieved July 15, 2021, from https://www.statisticshowto.com/probability-and-statistics/monty-hall-problem/
