I have some concerns with the original answer to the Monty Hall problem. The assumptions made in the probability distribution, as many others have raised in many other forums and threads, require further consideration.
The original premise: you are a contestant on a game show, and when you come out on stage you see three doors. The host (Monty) explains that there is a brand new car behind one of the doors, but there is nothing (or goats) behind the other two. You are asked to choose one door, then Monty will open one of the other doors to reveal nothing (or a goat), and lastly he asks if you want to switch your answer before the final reveal occurs and you go home with a new car (or nothing).
The counterintuitive answer is that you should switch, essentially arguing that your first guess only had a 33.3% chance of success, and with the reveal Monty has raised the likelihood of success of the third door. I would like to challenge the assumptions used.
Your first guess is random, but Monty’s subsequent door reveal is not. However, you are not in a position to understand why his action isn’t random. Monty’s action, assuming you first guess the right door, is not purely random, but it does represent two distinct possible scenarios that should be acknowledged separately, creating 12 scenarios instead of 9. If you guessed wrong on the first try, then Monty’s action is limited to one choice, but accounting for the two possibilities that you guess wrong and the two possibilities of Monty’s reveal if you originally guessed right balances out the likelihood of success if you switch doors or not.
One of the original extended examples considers 100 doors, with Monty opening 98 of them once you’ve made your first guess. This is a flawed comparison as the probability is now wildly different than the first example (i.e., in the first scenario, Monty opens 33.3% of the doors, in the second scenario he opens 98% of them). Another concern is the argument that Monty is adding information that only raises the value of the unchosen door. If you choose door 1, and he opens door 2, the argument is that he has raised the value of door 3 but not door 1. I raise this concern because if you chose door 3, and he opened door 2, now door 1 has added value but not door 3. In a model elaborating all 12 scenarios, if you stick with your first choice you should win exactly 6 times and lose exactly 6 times. Likewise, if you switch in all 12 scenarios, you still win and lose 6 times each.
To expand upon this position, I suggest a revised Monty Hall problem where there are 6 doors, and Monty will open two of them once you’ve made your first guess. Again, there is only 1 car, but now the number of possible choices expands greatly, although the likelihood of success if you switch your choice does not.
Assuming the car is behind door 1, there are now 40 possible permutations of outcomes specifying which door you guess, and which two Monty opens. If you do not switch, there are 10 possible outcomes in which you win (i.e., you correctly guessed the door on the first try and make no changes). This creates a probability of 0.25, after Monty has opened two doors at random. From the contestant’s perspective it is random – or perhaps a different viewing, it isn’t random but you as the contestant do not have enough information to intuit the non-random nature of Monty’s actions.
In total, assuming you do not switch your choice after he opens two doors, there are 240 total possible outcomes, with 6 initial possible positions for the car, 6 original guesses, and the total number of variations of doors that Monty could open once you’ve made your guess. Without switching answers after the first guess, you will win the car 60 times (60/240 = 0.25).
Assuming you do switch, now we have to explore the total number of possible outcomes that you might choose, arguing that you will drop your first guess and use only the three remaining doors, regardless of which doors Monty (seemingly) opened at random. Now there are 720 total possible outcomes, and out of all random guesses that you can make, statistically you will be correct in your second guess only 180 times (180/720 = 0.25).
The original premise: you are a contestant on a game show, and when you come out on stage you see three doors. The host (Monty) explains that there is a brand new car behind one of the doors, but there is nothing (or goats) behind the other two. You are asked to choose one door, then Monty will open one of the other doors to reveal nothing (or a goat), and lastly he asks if you want to switch your answer before the final reveal occurs and you go home with a new car (or nothing).
The counterintuitive answer is that you should switch, essentially arguing that your first guess only had a 33.3% chance of success, and with the reveal Monty has raised the likelihood of success of the third door. I would like to challenge the assumptions used.
Your first guess is random, but Monty’s subsequent door reveal is not. However, you are not in a position to understand why his action isn’t random. Monty’s action, assuming you first guess the right door, is not purely random, but it does represent two distinct possible scenarios that should be acknowledged separately, creating 12 scenarios instead of 9. If you guessed wrong on the first try, then Monty’s action is limited to one choice, but accounting for the two possibilities that you guess wrong and the two possibilities of Monty’s reveal if you originally guessed right balances out the likelihood of success if you switch doors or not.
One of the original extended examples considers 100 doors, with Monty opening 98 of them once you’ve made your first guess. This is a flawed comparison as the probability is now wildly different than the first example (i.e., in the first scenario, Monty opens 33.3% of the doors, in the second scenario he opens 98% of them). Another concern is the argument that Monty is adding information that only raises the value of the unchosen door. If you choose door 1, and he opens door 2, the argument is that he has raised the value of door 3 but not door 1. I raise this concern because if you chose door 3, and he opened door 2, now door 1 has added value but not door 3. In a model elaborating all 12 scenarios, if you stick with your first choice you should win exactly 6 times and lose exactly 6 times. Likewise, if you switch in all 12 scenarios, you still win and lose 6 times each.
| Car Position | Contestant Guess | Host Reveal | Switch? | Correct? |
1 | 1 | 2 | Yes | No |
1 | 1 | 3 | Yes | No |
1 | 2 | 3 | Yes | Yes |
1 | 3 | 2 | Yes | Yes |
2 | 1 | 3 | Yes | Yes |
2 | 2 | 1 | Yes | No |
2 | 2 | 3 | Yes | No |
2 | 3 | 1 | Yes | Yes |
3 | 1 | 2 | Yes | Yes |
3 | 2 | 1 | Yes | Yes |
3 | 3 | 1 | Yes | No |
3 | 3 | 2 | Yes | No |
1 | 1 | 2 | No | Yes |
1 | 1 | 3 | No | Yes |
1 | 2 | 3 | No | No |
1 | 3 | 2 | No | No |
2 | 1 | 3 | No | No |
2 | 2 | 1 | No | Yes |
2 | 2 | 3 | No | Yes |
2 | 3 | 1 | No | No |
3 | 1 | 2 | No | No |
3 | 2 | 1 | No | No |
3 | 3 | 1 | No | Yes |
3 | 3 | 2 | No | Yes |
To expand upon this position, I suggest a revised Monty Hall problem where there are 6 doors, and Monty will open two of them once you’ve made your first guess. Again, there is only 1 car, but now the number of possible choices expands greatly, although the likelihood of success if you switch your choice does not.
Assuming the car is behind door 1, there are now 40 possible permutations of outcomes specifying which door you guess, and which two Monty opens. If you do not switch, there are 10 possible outcomes in which you win (i.e., you correctly guessed the door on the first try and make no changes). This creates a probability of 0.25, after Monty has opened two doors at random. From the contestant’s perspective it is random – or perhaps a different viewing, it isn’t random but you as the contestant do not have enough information to intuit the non-random nature of Monty’s actions.
In total, assuming you do not switch your choice after he opens two doors, there are 240 total possible outcomes, with 6 initial possible positions for the car, 6 original guesses, and the total number of variations of doors that Monty could open once you’ve made your guess. Without switching answers after the first guess, you will win the car 60 times (60/240 = 0.25).
Assuming you do switch, now we have to explore the total number of possible outcomes that you might choose, arguing that you will drop your first guess and use only the three remaining doors, regardless of which doors Monty (seemingly) opened at random. Now there are 720 total possible outcomes, and out of all random guesses that you can make, statistically you will be correct in your second guess only 180 times (180/720 = 0.25).
