Monty Hall Problem... For Newbies

[Meed]

Tried it, the odds are 2/3 1/3 switch stay. Which has been proven many times in many ways above.

How did you try it? I just ran an R simulation and got 1/2 and 1/2.

Nope. Not true. No matter how Monty opens the door, your initial choice is still 1/3.

Here is where we disagree. After Monty's opened a door, then the odds of your initial pick being right may or may not go up, depending on what his door-opening rules are.

The odds at that point are 1/2 for staying only because you are throwing out the 1 out of 3 times when Monty reveals the car. If you just select a door (no switching), you have a 1/3 chance of winning the car no matter what Monty does with the doors.

Correct. I'm talking about the odds of switching or sticking once he has revealed a goat, not at the start of the game. The odds of the initial pick being right will always be 1/3 at the time it is made, however, it may go up or down as we gain new information. Or, as in the standard Monty Hall problem, it may remain the same, even after a goat is revealed.

 

[ynot]

Yep. And the rules were laid out clearly way back in the thread.

Once again - 'revealing a goat at random' is only available to Monty in 1/3 trials, the one where the contestant guessed the car at the outset. In 2/3 trials he's obliged to open the non-car door. That isn't random, it's forced. In those same 2/3 trials the car must be behind Monty's unopened door.

Exactly.

The odds of the Monty Game are that many people will get it wrong to begin with, but after a basic explanation of how the odds work in practice, they will then get it right. Some will never get it right no matter how many times and ways the odds are explained. Then there are those (regardless of whether they get it right or not) that want to change the clearly described and defined rules of the game (for some mysterious reason) and debate a new and different game as if they were debating the original.

 

[Red Baron Farms]

How did you try it? I just ran an R simulation and got 1/2 and 1/2.

Run it again then. Somehow you messed up. My guess is you only gave the simulation 2 options. Which is effectively "reshuffling" AFTER Monty picks.

Different game, different odds.

 

[Meed]

Run it again then. Somehow you messed up. My guess is you only gave the simulation 2 options. Which is effectively "reshuffling" AFTER Monty picks.

No, it starts with the initial pick arbitrarily set to door 1 and runs from there. First the car is randomly placed. Then Monty makes a random pick between the doors that haven't been picked. Then of those trials for which Monty opens a goat door (which is the situation we're talking about), the number of wins for sticking vs switching are recorded. There is no reshuffling. If you like, check my code in the spoilers below.

Input

nrep <- 10000
opencar <- rep(0,nrep)
stickwin <- rep(0,nrep)
switchwin <- rep(0,nrep)
for (irep in 1:nrep) {
pick <- 1
car <- sample(1:3, 1)
open <- sample(2:3, 1)
if (car==open) {
opencar[irep] <- 1
}
else if(car==1) {
stickwin[irep] <- 1
}
else if (car==2) {
switchwin[irep] <- 1
}
else if (car==3) {
switchwin[irep] <- 1
}
}
stickwins <- sum(stickwin)
switchwins <- sum(switchwin)
opencars <- sum(opencar)
total <- nrep - opencars
prop.stickwin <- (stickwins/total)
prop.switchwin <- (switchwins/total)
total;stickwins;switchwins;prop.stickwin; prop.switchwin

Output

total: 6735
stickwins: 3343
twitchwins: 3392
prop.stickwin: 0.4963623
prop.switchwin: 0.5036377

 

[malicus]

Run it again then. Somehow you messed up. My guess is you only gave the simulation 2 options. Which is effectively "reshuffling" AFTER Monty picks.

Different game, different odds.

It's 50/50.

Monty, by randomly choosing, has the same odds of choosing a car as you do. Rather, look at the odds of picking a goat.

You pick a door, you have 2/3 chance of picking a goat.
Monty stumbles, and picks a door at random, he has 2/3 chance of picking a goat.

The unopened door also has a 2/3 chance of being a goat, even when Monty opens his random door and reveals a goat (because Monty's random choice had a 1/3 chance of being the car, as opposed to the original MH problem when Monty's door had a 100% chance of being a goat).

 

[dlorde]

If you discard trials where Monty has randomly happened to pick the car, you're discarding that many trials where you picked a goat, so of course in the remaining trials, the probability that the door you picked the conceals the car will be higher.

And if you have Monty open both remaining doors and discard any trials where he doesn't find two goats, you have 100% chance of winning by sticking.

If you start discarding trials on various criteria, you can make the odds pretty much what you like. So what?

 

[Red Baron Farms]

No, it starts with the initial pick arbitrarily set to door 1 and runs from there. First the car is randomly placed. Then Monty makes a random pick between the doors that haven't been picked. Then of those trials for which Monty opens a goat door (which is the situation we're talking about), the number of wins for sticking vs switching are recorded. There is no reshuffling. If you like, check my code in the spoilers below.

As I said, different game, different odds. 1/2 the time you don't pick the car, ie 2/3 X 1/2 = 1/3 is subtracted from the odds of winning. So instead of 2/3 : 1/3 switch stay.....you get
(2/3-1/3) : 1/3
1/3 : 1/3
50:50
And never even get a chance to switch when your program causes Monty to pick a car because you lost before your second pick. Which explains the missing 1/3

To fix the program simply make an open car result by Monty automatically pick the alternate door instead. Then that 1/3 will be added to the players chances and you'll be back to the 2/3rds 1/3rd ratio

 

[HumptyDumpty]

As I said, different game, different odds. 1/2 the time you don't pick the car, ie 2/3 X 1/2 = 1/3 is subtracted from the odds of winning. So instead of 2/3 : 1/3 switch stay.....you get
(2/3-1/3) : 1/3
1/3 : 1/3
50:50
And never even get a chance to switch when your program causes Monty to pick a car because you lost before your second pick. Which explains the missing 1/3

To fix the program simply make an open car result by Monty automatically pick the alternate door instead. Then that 1/3 will be added to the players chances and you'll be back to the 2/3rds 1/3rd ratio

That's not what everyone else understands by the Random Monty variation. If Monty doesn't know where the prize is, opens a door at random (from the 2 doors you didn't choose) and it happens to be a goat, then it's 50/50 for your door and the door Monty didn't open.
I explained why it's 50/50 earlier - in "Variation (B) Ignorant Monty"

 

[HumptyDumpty]

Here is where we disagree. After Monty's opened a door, then the odds of your initial pick being right may or may not go up, depending on what his door-opening rules are.

Exactly. And since his door-opening rules aren't specified in the OP it's relevant to discuss how different rules affect the probabilities.

In the 2 coin problem the rules for how I announce "at least one is heads" aren't specified either; if flip HH I have no choice, if I flip TT I can't obviously, but if I flip HT or TH then I have a choice, and the most reasonable assumption to make is that I am unbiased and equally likely to announce "at least one is tails" as "at least one is heads" - which makes the answer 1/2.

 

[ynot]

It’s wrong to claim that Monty’s door opening rules aren’t specified. The rules of the game require that Monty reveals a goat every time to give the contestant a chance of winning the car. The only way he can guarantee to reveal a goat every time is by knowing before it's opened that the door he opens hides a goat. The rules of the game require that Monty doesn’t open a door at random and half the time reveal the car.

Besides, the OP clearly specifies that Monty knows where the car is and reveals a goat. . .

You pick one of the three doors at random and the presenter (who knows which door the car is behind) opens a door that you didn't pick, revealing a goat. (This is standard procedure for the show.)

 

[Meed]

To fix the program simply make an open car result by Monty automatically pick the alternate door instead.

The situation we're talking about is "if Monty had opened one of the two remaining doors at random and had happened to reveal a goat" (from post #512). The change you're proposing would take us back to the original Monty Hall problem in which Monty must always reveal a goat and only makes his choice randomly 1/3 of the time. Which is not what we want... My code accurately simulates the problem at hand.

If you discard trials where Monty has randomly happened to pick the car, you're discarding that many trials where you picked a goat, so of course in the remaining trials, the probability that the door you picked the conceals the car will be higher.

Correct. And of course those trials where Monty reveals a car have to be discarded if we're talking about what our odds are when Monty has revealed a goat.

If you start discarding trials on various criteria, you can make the odds pretty much what you like. So what?

There is a dispute over what the odds in a given situation would be, so people are discussing it.

 

[HumptyDumpty]

It’s wrong to claim that Monty’s door opening rules aren’t specified. The rules of the game require that Monty reveals a goat every time to give the contestant a chance of winning the car. The only way he can guarantee to reveal a goat every time is by knowing before it's opened that the door he opens hides a goat. The rules of the game require that Monty doesn’t open a door at random and half the time reveal the car.

Besides, the OP clearly specifies that Monty knows where the car is and reveals a goat. . .

I meant Monty's door-opening rules when he has a choice of 2 goat doors (i.e. when the car is behind the door you picked) aren't specified.

 

[ynot]

I meant Monty's door-opening rules when he has a choice of 2 goat doors (i.e. when the car is behind the door you picked) aren't specified.

That's only relevant to a Monty preference, not a Monty ignorance.

 

[ynot]

Correct. And of course those trials where Monty reveals a car have to be discarded if we're talking about what our odds are when Monty has revealed a goat.

The discarding is what directly changes the odds, not what Monty indirectly does. What Monty does only creates the need to discard.

 

[HumptyDumpty]

That's only relevant to a Monty preference, not a Monty ignorance.

It's relevant to the standard Monty as described in the OP, but you're right it's not relevant to Ignorant Monty. In Ignorant Monty it doesn't matter how he chooses to open a door - if it's a goat door he opens then it's always 50/50 between the 2 remaining closed doors.

 

[ynot]

It's relevant to the standard Monty as described in the OP

Only if the standard Monty was also a preference Monty, and that wouldn’t make him very standard on my opinion. He probably wouldn’t keep his job for very long.

but you're right it's not relevant to Ignorant Monty. In Ignorant Monty it doesn't matter how he chooses to open a door

Ignorant Monty wouldn’t get the job in the first place. The show couldn’t be properly played with an ignorant host.

if it's a goat door he opens then it's always 50/50 between the 2 remaining closed doors.

Only because the games in which he reveals the car are all discarded.

 

[HumptyDumpty]

Only if the standard Monty was also a preference Monty, and that wouldn’t make him very standard on my opinion.

Well since the door-opening rules (when Monty has a choice of 2 goat doors) in the standard Monty problem as per the OP aren't defined, then you don't know whether Monty has a preference or not. It's reasonable to assume Monty has no preference for which door he opens - but it's still an assumption.

 

[ynot]

Well since the door-opening rules (when Monty has a choice of 2 goat doors) in the standard Monty problem as per the OP aren't defined, then you don't know whether Monty has a preference or not. It's reasonable to assume Monty has no preference for which door he opens - but it's still an assumption.

But where do you stop with assumptions? Monty could also inadvertently give the location of the car door away by body language etc. Thing is in the real game in the real world there would be a production crew that would constantly analyse everything Monty did and if he did anything that changed the odds in favour of the contestant he would quickly be told to stop doing it or get another job.

If you don't know why assume, and why assume silly things?

 

[HumptyDumpty]

But where do you stop with assumptions? Monty could also inadvertently give the location of the car door away by body language etc. Thing is in the real game in the real world there would be a production crew that would constantly analyse everything Monty did and if he did anything that changed to odds in favour of the contestant he would quickly be told to stop doing it or get another job.

If you don't know why assume, and why assume silly things?

Well firstly the MHP is a mathematical puzzle and not a real game in the real world (Let's Make A Deal upon which the puzzle is loosely based didn't operate in consistently the same way).
Secondly, in order to calculate the probability the car is behind the door you picked or behind the door Monty didn't open - after Monty has opened a goat door - you have to make an assumption (since it isn't specified) about how he opens a goat door when he has a choice (of 2 goat doors)

(And anyway gameshows ALWAYS favour the contestant - there wouldn't be much point having them if they didn't)

 

[ynot]

Well firstly the MHP is a mathematical puzzle and not a real game in the real world (Let's Make A Deal upon which the puzzle is loosely based didn't operate in consistently the same way).

Perhaps that's why I said "would" rather than "is".

Secondly, in order to calculate the probability the car is behind the door you picked or behind the door Monty didn't open - after Monty has opened a goat door - you have to make an assumption (since it isn't specified) about how he opens a goat door when he has a choice (of 2 goat doors)

(And anyway gameshows ALWAYS favour the contestant - there wouldn't be much point having them if they didn't)

I watch a TV gameshow occasionally in which the host often deliberately helps the contestants. He does this to help ensure there is a reasonable prize offered at the end of the game to help with the ratings. It’s amazing how many contestants still fail even with the help.