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Monty Hall Problem

CurtC's case A is not the problem presented in the OP ..
No, but it's the problem that the OP intended to present.

BillyKid's solution is valid for the OP .. Whether the problem is usually presented some other way, or regardless of what assumptions someone chooses to make.
No, with the problem as it's presented in the OP, all three of my cases (A, B, and C), as well as many others, are reasonable interpretations of the OP, and they all give different probabilities.

It is very easy to run a computer simulation to show this ..
I completely agree, although a pencil and paper simulation is equally satisfactory and quicker.
 
If the OP was carefully stated, then the correct answer is "I'm sorry, there is not enough information provided."
Then you are free not to play..

If he wanted the "2/3 by switching" answer, he should have more carefully constrained Monty to always revealing a goat.

The answer will not be 2/3 if a car is revealed , because the game will be over ..
Switching is no longer an option..

But if you continue to play long enough, and discard the null iterations.. ( car revealed ) then the success when switching will approach 66% ...

....And prove, --- that when given the opportunity to switch, switching gives you a 66% chance of winning a car ..
 
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Uh, I know I'm repeating myself, but the reason I say switch is that we don't know Monty's motivation.

If we consider the possibility that he only intends to allow the player to switch when he knows the player has already picked the car, this should display quite clearly that his motivation has every bearing on the outcome.

It is up to us to decide whether we think this is what is going on. However, I would call that behaviour cheating, on the host's part. So, I would propose that this possibility may be excluded from the scenarios to be considered, as the very fact of posing the question tends to imply that the game will be fair.

Monty opening either of the two unchosen doors at random, would be considered a "fair game". If that's what he's doing, then switching confers no advantage. You have a 1/3rd chance of being right, so you'd be wrong to switch; a 1/3rd chance of being wrong, so you'd be right to switch; and a 1/3rd chance that Monty would reveal the car when he opened his random door. You know that the third possibility hasn't happened, but the two remaining ones are still of equal probability.

Monty deliberately choosing not to reveal the car (if you've chosen wrongly to begin with), would also be regarded as a "fair game". In this case there is a 1/3rd chance you've chosen correctly and a 2/3rds chance it is behind one of the other doors. If the second of these possibilities is the case, opening the one door of the two remaining to show where the car isn't, transfers the 2/3rds probability on to the remaining door.

Both these scenarios are fair play. It may be more probable that Monty is playing scenario 1, in which case part of the fun will be opening an unchosen door to reveal the car - oh sorry sir, better luck next time, end of game. But even if he is doing that, switching will not make your chance of winning poorer.

It is possible he may be playing scenario 2, in which case switching just doubled your chance of winning.

So switch anyway. What have you got to lose?

Rolfe.
 
CurtC's case A is not the problem presented in the OP ..

It is consistent with it, and it is also the standard Monty Hall problem you can find anywhere on the internet or in print.

BillyKid's solution is valid for the OP .. Whether the problem is usually presented some other way, or regardless of what assumptions someone chooses to make.

It is consistent with the OP, yes.

With three closed doors, the chances of picking the car are 1/3 .. ( out of a sufficiently number of tries, the car will be picked 33% of the time )

Once a door is opened, without the car behind it, the chances go to 2/3 that it is behind the remaining door . ( out of a sufficiently number of tries, the car will be picked 66% of the time )

It is very easy to run a computer simulation to show this ..

Or you can do it by blindly pulling marbles out of boxes .. You just have to do it long enough to see the result converge toward 33 or 66% ..

The computer makes it a lot easier ..

There is no reason to run a simulation - that result is totally obvious once the problem is stated clearly. There are only three cases - why in the world would you need a computer?
 
Then you are free not to play..



The answer will not be 2/3 if a car is revealed , because the game will be over ..
Switching is no longer an option..

But if you continue to play long enough, and discard the null iterations.. ( car revealed ) then the success when switching will approach 66% ...

....And prove, --- that when given the opportunity to switch, switching gives you a 66% chance of winning a car ..

Nope. Lets say you discard the iterations where he reveals a car. Your chances are 50/50, assuming a coin-flipping Monty.

You choose Door 1 (door designations are arbitrary)

Car is behind door 1 (33%)
Door 2 Opened - switching loses
Door 3 Opened - switching loses

Car is behind door 2 (33%)
Door 2 Opened - Start Game over
Door 3 Opened - switching wins

Car is behind door 3 (33%)
Door 2 Opened - start game over
Door 3 Opened - switching wins

The 2/3rds only shows up if we morph the 'start game over' chances into 'switching wins' chances by forcing Monty to open the goat door, at which point it collapses to:

Car is behind door 1 (33%)
Door 2 Opened - switching loses
Door 3 Opened - switching loses

Car is behind door 2 (33%)
Door 3 Opened - switching wins

Car is behind door 3 (33%)
Door 3 Opened - switching wins

That's when the 2/3rds shows up.
 
No. The experiment is
-1- you pick a door
-2- Monty opens another door with goat behind
-3- you may switch or not.

Revealing a car in step -2- is not compliant to the rules of the experiment. It's totally irrelevant why Monty opens a goat door, he just does per definition.
What you're missing here is that the OP is just a description of what has happened this one time. The host not knowing where the prize is, and just randomly having picked a goat this time, is completely consistent with the problem in the OP. And if that's the case, the answer is that it's a 50/50 proposition.
 
The coin box is weird, conceptually.

You have three boxes, one with two gold coins, one with two silver coins, one with a gold coin and a silver coin.

If you take a random coin out of a random box, and it's gold, what chance is there that the other coin in the box is silver?

Works similar to Monty Hall (yes, I can't spell) only with randomness. Conceptually similar, yet the con man in this case doesn't have to influence the outcome of the event to change the odds of the color of the other coin in the box.


Curse you, you quoted my post before I fixed the typos!

That's an interesting one I haven't come across before. I may not sleep tonight.

It took me quite some time to get my head round the two-goats-and-a-car thing, as I said, back in the 1990s when it was apparently discussed in a newspaper. I was told the problem by someone who had read it in the newspaper, and simply informed that some people thought the motivation of the host was important, but my informant couldn't see how. I didn't have internet access at the time.

At first I couldn't see why opening a door could possibly change the 1/3rd chance. I swore blind that there was no advantage to switching. However, I then imagined the 100-doors scenario, and realised in fact that there must be an advantage. Assuming the host is deliberately avoiding the car at step 2, the odds for both doors are now combined onto the one remaining door.

So then I declared that of course one should switch! You'll double your chances!

Then my informant said, but you don't know whether or not the host has deliberately revealed a goat or not. And how come his motivation can influence the odds?

I got all confused, but realised that it does, again by reference to the 100-doors version. And come to think of it, there are other games I've heard of where a steady profit can be made from a situation where the other player thinks an outcome is 50/50, but the person controlling the choices can skew the odds in his favour by always making the same decision. You don't win every time, but over repeated games, you always come out on top.

Then I thought, the problem has no solution, if we don't know the host's motivation! No advantage to switching if he's not avoiding the car, advantage if he is!

Then, almost finally, I realised that of course that in itself is the answer. Since in neither scenario are your chances reduced by switching, and there is a greater-than-zero chance that he is playing the deliberate avoidance of the car version, then that in itself tips the odds in favour of switching in your favour.

You can't decrease your chances and you may increase them, so go for it!

That took me about a week to work out, almost entirely on my own.

Then the possibility that Monty is cheating was raised in this thread....

There's always a wrinkle you haven't considered.

Hence my final formulation of the answer, if you can be assured that Monty is not cheating, then switch.

So who knows how long it will take me to figure out the coin boxes.

Rolfe.
 
No. The experiment is
-1- you pick a door
-2- Monty opens another door with goat behind
-3- you may switch or not.

Revealing a car in step -2- is not compliant to the rules of the experiment. It's totally irrelevant why Monty opens a goat door, he just does per definition.

No, it is very relevant.

The experiment is not quite as you stated. It is not a forward-looking thing where we will do this, then Monty Hall will do something, then we will have a choice. Instead, we are looking at what did happen. The difference may be subtle, but it is very important.

Assume for a moment Monty is evil incarnate. His strategy is to only show us a goat if we have picked the door with the car. He will not show us what is behind any door otherwise, nor would he give you an option to switch.

So, we are not at the end, and we have picked a door, Monty has shown us a goat from behind another door, and Monty gives us an option to switch. Do you really want to switch your initial choice?
 
Lol, Rolfe, thanks for making me feel better.

The coin-box puzzle is interesting, the answer is 2/3rds (like switching in Monte Hall) despite the fact that at face value it looks like 1/2

(this is because despite your 50/50 chance of having the 2 gold coin/1 silver 1 gold box, you only have a 50/50 chance of grabbing a gold coin from the 1 silver 1 gold box).

It's easier because the boxes don't need motivation.

The final stage to that trick is set it up, and offer to bet $10 that the coin will be the same color - which is why it's the start of a con game too.
 
..............

The 2/3rds only shows up if we morph the 'start game over' chances into 'switching wins' chances by forcing Monty to open the goat door, at which point it collapses to:
.................

That's when the 2/3rds shows up.

Well really, that's my point ..
Discarding the " car revealed " iterations, is statistically equivalent to " goat is always revealed "... And becomes the implied rule that CurtC is pointing out .. ( fair or not )


CurtC is objecting to the " one round " nature, of the problem as presented ..


The player increasing his chances of winning to 66% by switching when limited to two doors, after starting with 3, is a statistical fact at that point in time, for that iteration of the game, whether the game is ever played again or not ... Or whether he actually wins the car .. In fact he has a 33% of not winning it .. ( :D )

The only way to get a " win 50% of the time " ( 1/2 ) outcome, is to start with 2 doors and pick one..

Computer simulations show this to be true, no matter how you twist it ..
 
Well really, that's my point ..
Discarding the " car revealed " iterations, is statistically equivalent to " goat is always revealed "... And becomes the implied rule that CurtC is pointing out .. ( fair or not )
Not true

CurtC is objecting to the " one round " nature, of the problem as presented ..


The player increasing his chances of winning to 66% by switching when limited to two doors, after starting with 3, is a statistical fact at that point in time, for that iteration of the game, whether the game is ever played again or not ... Or whether he actually wins the car .. In fact he has a 33% of not winning it .. ( :D )

The only way to get a " win 50% of the time " ( 1/2 ) outcome, is to start with 2 doors and pick one..

Computer simulations show this to be true, no matter how you twist it ..
Not true.

Car is behind door 1 (33%)
Door 2 Opened - switching loses
Door 3 Opened - switching loses

Car is behind door 2 (33%)
Door 2 Opened - ?
Door 3 Opened - switching wins

Car is behind door 3 (33%)
Door 2 Opened - ?
Door 3 Opened - switching wins

The scenarios are 33% wins, 33% loses, and 33% ?s. In the forced to pick a goat, the ? becomes wins. If revealing the car becomes an autoloss, then the odds are still 1/3rd to win. If you start the game over, it becomes 50/50. If it's random and you reveal a goat, it becomes 50/50.
 
Lol, Rolfe, thanks for making me feel better.

The coin-box puzzle is interesting, the answer is 2/3rds (like switching in Monte Hall) despite the fact that at face value it looks like 1/2
You mean a third for different color, right?

(this is because despite your 50/50 chance of having the 2 gold coin/1 silver 1 gold box, you only have a 50/50 chance of grabbing a gold coin from the 1 silver 1 gold box).
Hmm. I prefer to calculate. So, I don't have to understand it. :D
 
What you're missing here is that the OP is just a description of what has happened this one time. The host not knowing where the prize is, and just randomly having picked a goat this time, is completely consistent with the problem in the OP. And if that's the case, the answer is that it's a 50/50 proposition.


Curt, you're dictating the rules while appearing not to dictate them. Read your own post again and you'll see what I mean. (Hint: check the word "if".)

What happened that one time cannot, on its own, reveal the entire setup.

Supposing I have a cunningly weighted coin. It will come down heads only 1/3rd of the time. I toss it once, and get heads. I ask you what the probability of that is. You say, 50/50. I say no, the coin is weighted, there was only a 1/3rd chance that was going to happen.

You can't prove me wrong by saying, but it came up heads and that is entirely consistent with what would have happened if you'd used a normal coin!

You can only tell the coin is weighted, and how, by multiple tosses. After you've tossed it 100 times, you'll know whether heads is a 50/50 shot, or just 1/3rd.

So it is with the Monty Hall problem. (I'm so glad it has a name! When I first heard it, it was "the host of a game show in an impoverished South American republic". Calling it "the one about the two goats and the car" can get you some funny looks.)

You can only tell whether or not the coin is weighted by tossing it lots of times. You can only tell which way the Monty Hall choice is set by running the scenario multiple times.

If you do this, and every single time Monty reveals a goat at step 2, then indeed, you double your chances if you switch. If, however, he reveals the car at step 2 1/3rd of the time, then switching confers no advantage.

Please tell me you get it!

Rolfe.
 
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Not true.

Car is behind door 1 (33%)
Door 2 Opened - switching loses
Door 3 Opened - switching loses

Car is behind door 2 (33%)
Door 2 Opened - ?
Door 3 Opened - switching wins

Car is behind door 3 (33%)
Door 2 Opened - ?
Door 3 Opened - switching wins

The scenarios are 33% wins, 33% loses, and 33% ?s. In the forced to pick a goat, the ? becomes wins. If revealing the car becomes an autoloss, then the odds are still 1/3rd to win. If you start the game over, it becomes 50/50. If it's random and you reveal a goat, it becomes 50/50.
What are you talking about? You cannot introduce abritrary '?' scenarios contradicting the experiment. Why don't you add a surprise case where a chimpanzee is revealed from time to time? Or a naked blonde? This is excluded in the very same way a car is.

This is how it is:

Car is behind door 1 (33%) -> switching loses
Car is behind door 2 (33%) -> switching wins
Car is behind door 3 (33%) -> switching wins
 
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The coin box is weird, conceptually.

You have three boxes, one with two gold coins, one with two silver coins, one with a gold coin and a silver coin.

If you take a random coin out of a random box, and it's gold, what chance is there that the other coin in the box is silver?


The coin-box puzzle is interesting, the answer is 2/3rds (like switching in Monte Hall) despite the fact that at face value it looks like 1/2

(this is because despite your 50/50 chance of having the 2 gold coin/1 silver 1 gold box, you only have a 50/50 chance of grabbing a gold coin from the 1 silver 1 gold box).

It's easier because the boxes don't need motivation.

The final stage to that trick is set it up, and offer to bet $10 that the coin will be the same color - which is why it's the start of a con game too.


OK, going to have to think about this one. While making the supper I got as far as realising that intuitively the answer was 50/50. My thinking was that if I've grabbed a gold coin, then I know I have either the gold/gold box, or the gold/silver box. The silver/silver one is excluded. So if I got the gold/gold box, then the other coin is gold, but if I got the gold/silver box, the other coin is silver. 50/50.

Mmmm, light beginning to dawn. It's not that simple.

Yup, sleep no more, GreyICE doth murder sleep.

Thanks a bundle!

Rolfe.
 
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OK, I get it.

Forget the silver/silver box, that's just handwaving.

Two boxes, one with two gold coins and one with one gold and one silver. Pick a coin at random. 75% of the time you get a gold one, 25% of the time you get a silver one.

Break it down.

50% of the time you got a gold coin from the gold/gold box.
25% of the time you got the gold coin from the gold/silver box.
25% of the time you got the silver coin from the gold/silver box.

So, 2/3rds of the times you picked a gold coin, it was from the gold/gold box.

QED.

Phew. MUCH MUCH easier than Monty Bloody Hall.

I will have sweet dreams after all.

Rolfe.
 
OK, going to have to think about this one. While making the supper I got as far as realising that intuitively the answer was 50/50. My thinking was that if I've grabbed a gold coin, then I know I have either the gold/gold box, or the gold/silver box. The silver/silver one is excluded. So if I got the gold/gold box, then the other coin is gold, but if I got the gold/silver box, the other coin is silver. 50/50.

Mmmm, light beginning to dawn. It's not that simple.

Yup, sleep no more, GreyICE doth murder sleep.

Thanks a bundle!

Rolfe.
Your grabbed gold coin is one of the three

GG1 or GG2 (from g-g box) or
GS (from g-s box).

Two out of three .. no, that must suffice.

Edit: Just see you got it quickly and obsoleted my post.
 
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If you say you're playing 200 times, then that means that you started choosing 300 times, and in 100 of them, Monty revealed the prize by his random guess. In the remaining 200 trials, it's in your door 100 of them and in the other door 100 of them: therefore it's a 50/50 choice.
Boah, this weird discussion made my head swirl. I didn't recognize that this is complete nonsense. It's a gambler's fallacy at its best.

How often you skip invalid trials is completely irrelevant. It doesn't say anything about your winning chances in regular trials, each of which is an independent trial not correlated to any of the others.

You argue like somebody repeatedly tossing a coin, and guessing heads must be more likely in the future, because you tossed a tail 5 times in a row.
 
What are you talking about? You cannot introduce abritrary '?' scenarios contradicting the experiment. Why don't you add a surprise case where a chimpanzee is revealed from time to time? Or a naked blonde? This is excluded in the very same way a car is.

This is how it is:

Car is behind door 1 (33%) -> switching loses
Car is behind door 2 (33%) -> switching wins
Car is behind door 3 (33%) -> switching wins

Only if Monty can't reveal a car is this true.

If Monty CAN reveal the car, but DOESN'T because of random chance, this becomes false.

You're actually engaging in the gamblers fallacy here, despite the fact that you think you're not. You're linking two events (the random initial choice of the door, and the random second choice of a door) that are no longer linked.
 
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Only if Monty can't reveal a car is this true.
If Monty reveals anything but a goat (car, gorilla, Elton John, ..) the trial is invalid and discarded. Monty cannot reveal a car in any valid trial per definition.

If Monty CAN reveal the car, but DOESN'T because of random chance, this becomes false.
Discarding invalid trials (car, Madonna, Fidel Castro ..) makes Monty reveal a goat in every valid trial. Each trial is independent from all others, it's a random process of repeated non-correlated trials.

You're actually engaging in the gamblers fallacy here, despite the fact that you think you're not.
How so?
 

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