Monty Hall Problem - random variation
For those who don't know, the Monty Hall problem involves a game show where there are three doors. One of them contains a car, the other two contain a goat. You get to choose any door you want. You also get to keep whatever is behind that door. So if you pick a goat, you get a goat. And if you pick the car, you get the car. You want to pick the car because it's worth more.
You don't know what is in what door, but you choose to pick door 1. The game show host knows which doors contain the goats, and opens door 2 to reveal a goat, and asks, "Do you want to stay with door 1, or do you want to switch to door 3?". Most people would stay with door 1 because they don't want to change their choice, and they also believe that chance of there being a car in door 1 is now 50/50. But mathematically, the contestant should switch to door 3 because it's 2/3 more likely that door has the car. That, in a nutshell, is the Monty Hall problem.
Now suppose you happen to have a program that can simulate the Monty Hall game show. You run the program, and the graphics show three doors, two of which has a goat behind them, one of which has a car. The program simulates the game show, but with a variation, the program doesn't know behind which door the goats are, nor the car. At the beginning of the game, like the original Monty hall show, the user is given a choice to click on one of the three doors. However, unlike the original Monty Hall show, the computer picks one of the other two doors at random to open. This time, it may contain a goat, or it may contain the car.
And here's the crucial part. If the computer, who randomly opens one of the other two doors you didn't choose, reveals the car, the program terminates right there, and starts again. If on the other hand, the computer opens one of the other doors, and by chance it is a goat, the program continues and you are given a choice to switch or not.
My question is, in this simulation of the Monty Hall show, with this extra random variation where the computer doesn't "know" which door contains the goat or car, but reveals a goat in one of the other doors by chance, is the probability still 2/3 that the other unopened door contains the car? Or does the probability change because of the extra random process?
You don't know what is in what door, but you choose to pick door 1. The game show host knows which doors contain the goats, and opens door 2 to reveal a goat, and asks, "Do you want to stay with door 1, or do you want to switch to door 3?". Most people would stay with door 1 because they don't want to change their choice, and they also believe that chance of there being a car in door 1 is now 50/50. But mathematically, the contestant should switch to door 3 because it's 2/3 more likely that door has the car. That, in a nutshell, is the Monty Hall problem.
Now suppose you happen to have a program that can simulate the Monty Hall game show. You run the program, and the graphics show three doors, two of which has a goat behind them, one of which has a car. The program simulates the game show, but with a variation, the program doesn't know behind which door the goats are, nor the car. At the beginning of the game, like the original Monty hall show, the user is given a choice to click on one of the three doors. However, unlike the original Monty Hall show, the computer picks one of the other two doors at random to open. This time, it may contain a goat, or it may contain the car.
And here's the crucial part. If the computer, who randomly opens one of the other two doors you didn't choose, reveals the car, the program terminates right there, and starts again. If on the other hand, the computer opens one of the other doors, and by chance it is a goat, the program continues and you are given a choice to switch or not.
My question is, in this simulation of the Monty Hall show, with this extra random variation where the computer doesn't "know" which door contains the goat or car, but reveals a goat in one of the other doors by chance, is the probability still 2/3 that the other unopened door contains the car? Or does the probability change because of the extra random process?
Comments (35)
I'm not saying this is incorrect, but some people try to explain the Monty Hall problem in terms of what the host knows and has to reveal.
Here's an example:
And that is relevant, because the reason you are getting additional information out of the host's decision is that this decision follows a predetermined rule.
You have recreated that rule though, by having the game terminate. If the game does not terminate, the player gains information.
Edit: the Monty Hall problem can always be solved with a decision tree. There is still a 2/3 chance to initially select the wrong door, and if you did, then you should switch if one wrong door is subsequently eliminated.
Compare this with the same game where Monty knows where the car is: Monty always picks the door with the car to leave, except the 1 in 100 where you have picked it.
Still 2/3. The door you chose retains the initial 1/3 chance and the other door must have a 1 - 1/3 chance, i.e. 2/3.
https://en.wikipedia.org/wiki/Monty_Hall_problem
The goat likewise raises the probability of door 1.
Switch because the computer didn't shut down???!!! :chin:
If we play the game 99 times with 3 doors then we'd expect our initial selection to be right 33 times because the car has an equal chance of being behind each of doors 1, 2, and 3, so we can't win more than 33 games (out of 99) by not switching.
I've changed my opinion on this after running some simulations. Unless I'm doing it wrong, there's a 43% chance of winning by not switching and a 57% chance of winning by switching.
There is a difference between the OP's version of the game and the original. In the original you would expect your initial door to contain the car 33 times out of every 99 games you're asked to switch. However, in the OP's version of the game you're not asked to switch every game; some games are ended early. My simulations suggest that 22% of games end early.
So it's still better to switch, just not as better.
Door 1 has the car. Program doesn't restart.
Door 1 has a goat, Program opens door 2, and it's a goat. Program doesn't restart.
Door 1 has a goat, Program opens door 3, and it's the car. Program restarts.
So 1/2 times you pick the goat in the door you choose, the computer invalidates that decision, and restarts. It's like you are given an extra chance.
Where's the indication that one door is more likely, assumptions that one chose wrong aside?
There are 3 unnumbered doors. I pick one at random. That door is then numbered "1". The left-most of the remaining doors is numbered "2" and the other numbered "3". A door is selected at random and a car placed behind it. If we play 99 games then 33 times the door is behind door "1" (my door), 33 times the door is behind door "2", and 33 times the door is behind door "3".
If I don't switch then I win 33 times. If I switch then I win 66 times (if we're playing the standard game; as my post above explains, the situation is slightly different with the OP's variation, although it's still better to switch).
In theory the chances are 50/50 from beginning to end as the revealed dud is nonfactor; it only appears to be 2/3 if you focus on the dud with the assumption that you chose wrong, if you assume you chose right vice versa.
In practice, you might not win one game by switching.
Forget about Monty opening a door and just look at the initial selection where I have to pick from doors 1, 2, or 3. If we play 99 games then that first selection will be right 33 times, so I can't win more than 33 games by not switching. Monty opening a door can't make it that I can win 50 games by not switching.
And you might not win one game by not switching. But we're discussing probability theory here.
Without the assumption that you chose wrong, the three doors are equiprobable - and staying or switching you would win 33 games anyway, by the posited logic.
If we simply choose one and reveal them all it's 1/3.
If we remove the dud from the equation, it should be obvious you're limited to two options and it's 1/2.
If we by condition of the dud, assume we chose wrong, then yes, you can skew the chances to 2/3.
You can win all 99 by switching as well.
Quoting Shamshir
That's not right. Probability dictates that if we play 99 games then I am likely to pick the correct door 33 times. That means that I will win 33 games and lose 66 games by not switching, and so therefore win 66 games and lose 33 games by switching.
You got me there.
In any case, if you're feeling up to it we can continue this in private, lest this thread spiral out like the other two. I made my case in the addendum to my earlier post.
33 times the car is behind "my door".
33 times the car is behind "other door 1".
33 times the car is behind "other door 2".
If I don't switch then I will win 33 times when the car is behind "my door", 0 times when the car is behind "other door 1", and 0 times when the car is behind "other door 2". My win rate is 33/99 = 1/3.
If I switch then I will win 0 times when the car is behind "my door", 33 times when the car is behind "other door 1", and 33 times when the car is behind "other door 2". My win rate is 66/99 = 2/3.
66 times out of 99 games we're being asked if we want to switch to the winning door.
Doesn't "initial probability" mean irrespective of all later factors? In which case I don't see how it could be anything but 1/3.
Monty's G_G___C_[s]G[/s]___C_[s]G[/s]
Six permutations, but the combination G+G is eliminated by the exposure of the car, as indicated above by the invisible strikes. One game in three my door wins, one game in three Monty's door wins, and one game in three is abandoned. So it's even odds.
In the original problem, the combination G+G is avoided by Monty's choice; no games are abandoned Monty's odds are doubled.
Not in this case because the program can loop.
[s]I don't think that's right.
There's a 1/2 chance that he picks door 2 and a 1/2 chance that he picks door 3.
There's a 1/3 chance that the car is behind door 2 and a 1/3 chance that the car is behind door 3.
What is the chance that the car is behind the door he picks? 1/2 * 1/3 = 1/6. So of 99 games, ~17 are abandoned. Therefore 82 games are played to completion.
In 33 of those games the car is behind door 1 (my door). If I don't switch then I win 33/82 = 40% and if I do switch then I win 49/82 = 60%. This is pretty close to what my simulations showed (I'll look into why they're not exact).[/s]
Actually, you're right. It's 50% for the OP's modified version.
I think I get it too now. If I get the option to switch, that means the game didn't restart. It's less likely I picked the right door initially. But, given that the game did not restart, it becomes more likely that I picked the right door. Not sure exactly how the mathematical operation looks though.
Monty hall problems are apparently difficult even if you know the basics.
If we're playing 99 games then in 33 games the car is behind my door, in 33 games the car is behind the door the computer picks, and in 33 games the car is behind the door the computer doesn't pick. The 33 games where the car is behind the door the computer picks are dismissed so I'm only ever asked to switch in 66 games; 33 in which the car is behind my door and 33 in which the car is behind the door the computer doesn't pick.
As I see it, you should be indifferent to keeping or switching in the modified game. My reasoning:
If you initially correctly pick the door with the car, then the computer will continue the game. However there is a 2/3 probability that you are initially incorrect. In those games, there is a 1/2 probability that the computer will open the door with the car and the game will be terminated. Then the same logic applies to the rerun games.
So if you play a game where you are asked to keep or switch, then that is one of the non-terminated games. Out of 300 games, you will (on average) be correct on 100 games that therefore continue, incorrect on 100 games that continue, and the remaining 100 games will be terminated and replayed. Of those replayed 100 games, you will be correct on 33, incorrect on 33, and 33 will be replayed. In this way, you will approach being correct on 150 games and incorrect on 150 games.
For the millionaire scenario, the computer randomly removes two wrong answers from the set of three wrong answers. That selection is statistically independent of your (secret) guess, so the remaining two answers are equally probable.
millionaire answers, Right and Wrong.
my guess:- R_____W_____W_____W
Computer
leaves:- WWW____R______R_____R
Exactly. :up: