Monty Hall Problem... For Newbies

[This is The End]

Yes. From before you make your initial pick, we know that the car will wind up either behind the door you choose, or the one Monty doesn't open. 50/50

Wrong. That is the marbles fallacy as I explained to you earlier.

Say you have a bag with 99 black marbles and 1 white marble.

Just because there are only 2 colors does not mean that the odds of picking a white one is 50/50. Even if you play a monty hall game with the bag.

The odds of picking the white marble with the first pick is always 1/100, and the odds of it being the marble you switch to is 99/100.

It is never 50/50.

 

[This is The End]

Once there are only two doors left, the odds of the car being behind either door are in fact 50/50.

Nope. Not in the problem being discussed.

 

[This is The End]

No. the remaining two doors are 50/50 to anybody.

If it was a completely different game with just 2 doors and 1 prize then yes.

But in the game being discussed it is not ever 50/50.

 

[This is The End]

I was thinking about the 50/50 last night and I came up with another way to look at it.

The 50/50 does actually exist but it is not the odds of the doors.

The 50/50 is the choice you have between the two strategies if you are ignorant of them.

So you have a 50/50 chance to go with "always change strategy" and "never change strategy".

Yes, but there are a multitude of statistics that surround everything. Bringing up unrelated ones is silly.

You might as well say: "50/50 does actually exist because one door is on the right and the other is on the left!!" Or that one is slightly taller. Or a different color. Or that one is closer to Kansas.

 

[HumptyDumpty]

Your scenario C is irrelevant, because in that case the positioning of the car is not random. The car can be behind any door, otherwise the original probability is not 1/3.

If you want to change the rules then that's fine, but don't pretend that the result you get applies to the original problem.

How is scenario C irrelevant? The car can obviously be behind any door, and 2/3rd of the time it won't be behind Door2, and in those instances Monty opens Door2. In the 1/3rd of the time that the car is behind Door2 Monty has to open Door3, and if he does that you know with 100% certainty the car is behind Door2.
I'm not changing the rules , I'm assuming a bias in how Monty opens a goat door given he has a choice of 2 goat doors, and the results certainly do apply to the problem.

 

[ynot]

Your scenario C is irrelevant, because in that case the positioning of the car is not random. The car can be behind any door, otherwise the original probability is not 1/3.

If you want to change the rules then that's fine, but don't pretend that the result you get applies to the original problem.

Scenario B is also irrelevant. The rules say Monty has to reveal a goat after the first guess and he can only guarantee to do this if he knows where the goats and car are

 

[GlennB]

Thanks – That makes the scenario much clearer.

If Monty always opens the rightmost door when it hides a goat then when he doesn’t open that door you know with 100% certainty that it hides the car. The odds of the first guess being correct would therefore change to 0/0 not 1/2 and the remaining door would have odds of 1/1.

Not sure it’s a credible scenario in practice however and I think the “spirit” of the Monty game is to analyse the odds without them being effected by a Monty bias.

Yeah. Not credible, just thread drift But my digression into bridge could have been a more generalised one relating to many games - "Why did my opponent do that instead of something different/better? Ah ... they were unable to do anything different/better".

 

[Roboramma]

How is scenario C irrelevant? The car can obviously be behind any door, and 2/3rd of the time it won't be behind Door2, and in those instances Monty opens Door2. In the 1/3rd of the time that the car is behind Door2 Monty has to open Door3, and if he does that you know with 100% certainty the car is behind Door2.
I'm not changing the rules , I'm assuming a bias in how Monty opens a goat door given he has a choice of 2 goat doors, and the results certainly do apply to the problem.

Yeah, makes sense. Thanks for the explanations.

 

[HumptyDumpty]

Scenario B is also irrelevant. The rules say Monty has to reveal a goat after the first guess and he can only guarantee to do this if he knows where the goats and car are

Irrelevant to what?
I included the Ignorant Monty variation only to show how looking at the problem by analysing why Monty opens the door he does leads to the correct answer. The same approach gives the correct answer for the standard problem statement and also variation C.
The Bayes equation I gave only applies to variations A and C though

 

[ynot]

So, those saying that the initial probabilities don't change are assuming (whether they realise it or not) that given a choice of 2 goat doors to open, Monty picks one at random.

In practice game shows don’t normally do anything to help the contestants win the main prize. To assume that Monty wouldn’t have a predictable goat door bias would therefore be a reasonable default assumption if not a given.

 

[ynot]

Irrelevant to what?
I included the Ignorant Monty variation only to show how looking at the problem by analysing why Monty opens the door he does leads to the correct answer. The same approach gives the correct answer for the standard problem statement and also variation C.
The Bayes equation I gave only applies to variations A and C though

Irrelevant to the rules of the actual game.

 

[HumptyDumpty]

In practice game shows don’t normally do anything to help the contestants win the main prize. To assume that Monty wouldn’t have a predictable goat door bias would therefore be a reasonable default assumption if not a given.

I agree, it is a reasonable default assumption by applying the Principle of Indifference (although it's generally not stated in the problem definition).

Whilst that assumption is almost always implicitly made when discussing the MHP, in other 'paradoxes' where it is equally applicable it is often ignored. For instance in this problem:

'If I toss two fair coins, examine both of them and truthfully tell you that at least one of them landed heads. What is the probability both coins landed heads?'

Most people will answer 1/3. But making exactly the same assumption as in the MHP gives the (correct) answer of 1/2.

 

[GlennB]

Count me up for one of the interested, I did play a lot of some bridge about 30 years ago. I found this interesting article, I'm not sure I get it, and what is RHO and "drop"? I think I know "finesse". Maybe you'll elaborate for a rusty danish bridge player?

http://www.acbl-district13.org/artic003.htm

RHO is "right hand opponent". "To play for the drop" means, for example, that where declarer is missing (say) the QJ32 of a suit he just cashes the A and then the K, assuming that the missing cards were divided 2-2 between LHO and RHO all along. Then they "drop" with no complicated fussing required. But if the Q or J appears on the first round then declarer has extra information.

 

[Brian-M]

'If I toss two fair coins, examine both of them and truthfully tell you that at least one of them landed heads. What is the probability both coins landed heads?'

Most people will answer 1/3. But making exactly the same assumption as in the MHP gives the (correct) answer of 1/2.

Why 1/2 and not 1/3?

Maybe I'm not seeing the answer because it's almost midnight here and I need to get some sleep, but it seems to me that the following outcomes of the coin-toss would be equally likely...

Heads + Heads
Heads + Tails
Tails + Heads
Tails + Tails

Since each of the three outcomes containing heads are equally likely and only one of those outcomes has two heads, it seems to me that the chances of both coins being heads is 1/3.

But I'm assuming that the question is only going to be asked if at least one of the coins comes up heads. If you were going to report the outcome of the first coin regardless of whether it comes up heads or tails and then ask us to guess if the other coin came up the same, then the odds of the other coin being the same would be 1/2.

So maybe the people who are getting the "wrong" odds are misunderstanding the question?

 

[JohnnyG]

It has to do with probabilities and the assumptions we (consciously or unconsciously) make in calculating those probabilities.



In the example given (you pick Door1, Monty reveals a goat behind Door2) many people assume the reason the probability the prize is behind Door1 is 1/3 is because that was the initial or 'a priori' probability, and Monty opening a goat door doesn't or cannot change that.
However that reasoning doesn't explain why, if Monty is ignorant of where the prize is and opens Door2 at random, the probability of Door1 hiding the car increases to 1/2, nor in the scenario where Monty has a preference for opening Door2 the probability of Door1 similarly increases from 1/3 to 1/2.

If you approach the problem from the viewpoint of "why did Monty open Door2?", then the answer to all variations becomes clearer.

In all varaitions the car is either behind Door1 or Door3, and the prior probabilities for both these doors was the same - 1/3.

(A) Classic Monty.
If the prize is behind Door3 there's a 100% chance Monty opens Door2.
(constraint of the game)
If the prize is behind Door1 there's a 50% chance Monty opens Door2 and a 50% chance Monty opens Door3 (applying the Principle of Indifference).
Therefore it is twice as likely Monty opened Door2 because he had to than because he chose to , consequently the prize is twice as likely to behind Door3 as Door1.

(B) Ignorant Monty
If the prize is behind Door3 there's a 50% chance Monty opens Door2 (and a 50% chance he opens Door3)
If the prize is behind Door1 there's a 50% chance Monty opens Door2 (and a 50% chance he opens Door3)
Both scenarios are equally likely, therefore it's 50/50

(C) Preference Door Monty
If the prize is behind Door3 there's a 100% chance Monty opens Door2 (constraint of the game)
If the prize is behind Door1 there's a 100% chance Monty opens Door2 (because it's his preferred door)
Both scenarios are equally likely, therefore it's 50/50

So, those saying that the initial probabilities don't change are assuming (whether they realise it or not) that given a choice of 2 goat doors to open, Monty picks one at random.

The mathematical solution also bears this analysis out. Bayes Theory simplifies to:
Prob(Prize behind Door1) = p/(1+p) and
Prob(Prize behind Door3) = 1(1+p)
where p=probability Monty opens Door2 give the prize is behind Door1.

Since the OP specified Monty knows where the car is, case B is irrelevant. And case C requires the use of information not in the OP, so it is also irrelevant.

Should we also explore the cases where the original car location is not random?

 

[wollery]

Why 1/2 and not 1/3?

Maybe I'm not seeing the answer because it's almost midnight here and I need to get some sleep, but it seems to me that the following outcomes of the coin-toss would be equally likely...

Heads + Heads
Heads + Tails
Tails + Heads
Tails + Tails

Since each of the three outcomes containing heads are equally likely and only one of those outcomes has two heads, it seems to me that the chances of both coins being heads is 1/3.

But I'm assuming that the question is only going to be asked if at least one of the coins comes up heads. If you were going to report the outcome of the first coin regardless of whether it comes up heads or tails and then ask us to guess if the other coin came up the same, then the odds of the other coin being the same would be 1/2.

So maybe the people who are getting the "wrong" odds are misunderstanding the question?

It doesn't matter which coin is heads, as long as one of them is. The probability is then dependent entirely on the other coin. 50/50 heads or tails.

 

[wollery]

Since the OP specified Monty knows where the car is, case B is irrelevant. And case C requires the use of information not in the OP, so it is also irrelevant.

Should we also explore the cases where the original car location is not random?

Maybe we can also explore cases where there's no car, or where MH opens the door with the car and asks you to choose between 2 goats.

Hey, how about a case with only one door?

How far do we change the rules before we decide that it's no longer relevant to the original problem?

 

[Meed]

Brian-M has it right. The probability is 1/3. I challenge anyone who is unconvinced to try a simulation.

If it was "you know the first coin toss is heads, now what is the probability of two heads?" then it would be 50/50.

 

[JohnnyG]

Maybe we can also explore cases where there's no car, or where MH opens the door with the car and asks you to choose between 2 goats.

Hey, how about a case with only one door?

How far do we change the rules before we decide that it's no longer relevant to the original problem?

Exactly.

 

[JohnnyG]

Brian-M has it right. The probability is 1/3. I challenge anyone who is unconvinced to try a simulation.

If it was "you know the first coin toss is heads, now what is the probability of two heads?" then it would be 50/50.

I just ran a simulation of 1000 trials, 756 had one or both coins with heads, 242 had both heads. Ratio of both heads to one or more heads is 0.320106, exacly as Brian-M predicted.

HumptyDumpty: How do you get your probability of 1/2?