Monty Hall Problem - random variation

Wheatley

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?

Hanover

↪Purple Pond

I don't see how your variation changes the odds. The issue isn't the foreknowledge of the host, but of the knowledge of the contestant as revealed by the host. The way you've set this up, if I'm understanding correctly, is that you have a contestant who chose door #1, then another door is revealed to him that does not contain the car, and then you ask whether he should change doors. Just as in the initial case, the answer is yes. My understanding is that you disregard those instances where the computer chooses and shows the car or the goat, so they really aren't being included in your decision making.

Wheatley

The issue isn't the foreknowledge of the host, but of the knowledge of the contestant as revealed by the host. — Hanover

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:

Echarmion

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. — Purple Pond

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.

unenlightened

The best way to make it intuitive is to add more doors. So - 100 doors, 99 goats, you choose a door, then Monty chooses 98. There's going to be a lot of abandoned games in this scenario. But eventually you arrive at the two doors one of which has the car. In effect you have chosen one and Monty the other from a position of equal ignorance. Monty is not more likely to be correct than you.

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.

Michael

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? — Purple Pond

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.

Shamshir

↪Michael

How does it retain the initial chance?

Michael

↪Shamshir

https://en.wikipedia.org/wiki/Monty_Hall_problem

Shamshir

↪Michael

Again - why are you biasing in favour of one door?

The goat likewise raises the probability of door 1.

TheMadFool

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. — Purple Pond

Switch because the computer didn't shut down???!!! :chin:

Michael

Again - why are you biasing in favour of one door?

The goat likewise raises the probability of door 1. — Shamshir

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.

Michael

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. — Michael

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.

Wheatley

I don't think the initial probability of picking a car at a random door is 1/3. Think about it. Consider you pick door 1, there are 3 possibilities:

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.

Shamshir

↪Michael

One door has zero chance and chances are split to 1/2.

Where's the indication that one door is more likely, assumptions that one chose wrong aside?

Michael

One door has zero chance and chances are split to 1/2.

Where's the indication that one door is more likely, assumptions that one chose wrong aside? — Shamshir

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).

Michael

↪Purple Pond

See my post here. You're right that it's no longer 1/3, but it isn't 1/2 either.

Shamshir

↪Michael

But one door has zero probability and the remaining two are equiprobable.

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.

Michael

But one door has zero probability and the remaining two are equiprobable.

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. — Shamshir

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.

In practice, you might not win one game by switching.

And you might not win one game by not switching. But we're discussing probability theory here.

Shamshir

↪Michael

But you practically can, you can win all 99 by not switching.

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.

Michael

But you practically can, you can win all 99 by not switching. — Shamshir

You can win all 99 by switching as well.

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. — 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.

Shamshir

You can win all 99 by switching as well. — Michael

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.

Michael

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. — Shamshir

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.

Echarmion

I don't think the initial probability of picking a random door is 1/3. Think about it. — Purple Pond

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.

unenlightened

My door. C______G_____G

Monty's G_G___C_G___C_G

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.

Wheatley

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. — Echarmion

Not in this case because the program can loop.

Michael

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. — unenlightened

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).

Michael

↪unenlightened

Actually, you're right. It's 50% for the OP's modified version.

Echarmion

↪Purple Pond

↪Michael

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.

Michael

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. — Echarmion

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.

unenlightened

So there's a game on uk telly called 'Who wants to be a millionaire. A series of questions with 4 possible answers A, B, C, D. On one question, you can ask the computer to remove two wrong answers at random. So you take a guess in your mind - say 'A'. Then the computer takes away - say B and C, leaving A and D. Is there any reason to switch to D?

Michael

↪unenlightened

No, because the computer could have removed your guess (if it’s wrong), and so the logic isn’t comparable to the Monty Hall problem.

Monty Hall Problem - random variation

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