B The Monty Hall paradox/conundrum

[DaveC426913]

Ah, I just realized you have the bogey games in there. Games 5, 7, 9 and 11 cannot happen. You have the host opening door #1 to reveal the car! He never does that. Which in a way is the source of the paradox.

That's actually the Monty fall problem, where one game in three is spoiled by the host.

Good point. Although eliminating those rows that does not change the outcomes.

I think you're on to something though. I can't just eliminate those rows, since I've already stipulated that they're all equally likely.

Updated:

Game #​	Setup​	I picked​	Host opened​	I chose to​	to Door #​	Outcome​	Stick​	Switch​
1​	Car, Goat, Goat​	1​	2​	Stick​	1​	Win​	1​	​
2​	Car, Goat, Goat​	1​	3​	Stick​	1​	Win​	1​	​
3​	Car, Goat, Goat​	1​	2​	Switch​	3​	Lose​	​	​
4​	Car, Goat, Goat​	1​	3​	Switch​	2​	Lose​	​	​
5​	Car, Goat, Goat​	2​	3​	Stick​	2​	Lose​	​	​
6​	Car, Goat, Goat​	2​	3​	Stick​	2​	Lose​	​	​
7​	Car, Goat, Goat​	2​	3​	Switch​	3​	Lose​	​	​
8​	Car, Goat, Goat​	2​	3​	Switch​	1​	Win​	​	1​
9​	Car, Goat, Goat​	3​	2​	Stick​	3​	Lose​	​	​
10​	Car, Goat, Goat​	3​	2​	Stick​	3​	Lose​	​	​
11​	Car, Goat, Goat​	3​	2​	Switch​	2​	Lose​	​	​
12​	Car, Goat, Goat​	3​	2​	Switch​	1​	Win​	​	1​
​	​	​	​	​	​	​	2​	2​
​	​	​	​	​	​	​	​	​
13-24​	Goat, Car, Goat​	​	​	​	​	​	2​	2​
​	​	​	​	​	​	​	​	​
25-36​	Goat, Goat, Car​	​	​	​	​	​	2​	2​

 

[PeroK]

Good point. Although eliminating those rows that does not change the outcomes.

I think you're on to something though. I can't just eliminate those rows, since I've already stipulated that they're all equally likely.

Updated:

Game #​	Setup​	I picked​	Host opened​	I chose to​	to Door #​	Outcome​	Stick​	Switch​
1​	Car, Goat, Goat​	1​	2​	Stick​	1​	Win​	1​	​
2​	Car, Goat, Goat​	1​	3​	Stick​	1​	Win​	1​	​
3​	Car, Goat, Goat​	1​	2​	Switch​	3​	Lose​	​	​
4​	Car, Goat, Goat​	1​	3​	Switch​	2​	Lose​	​	​
5​	Car, Goat, Goat​	2​	3​	Stick​	2​	Lose​	​	​
6​	Car, Goat, Goat​	2​	3​	Stick​	2​	Lose​	​	​
7​	Car, Goat, Goat​	2​	3​	Switch​	3​	Lose​	​	​
8​	Car, Goat, Goat​	2​	3​	Switch​	1​	Win​	​	1​
9​	Car, Goat, Goat​	3​	2​	Stick​	3​	Lose​	​	​
10​	Car, Goat, Goat​	3​	2​	Stick​	3​	Lose​	​	​
11​	Car, Goat, Goat​	3​	2​	Switch​	2​	Lose​	​	​
12​	Car, Goat, Goat​	3​	2​	Switch​	1​	Win​	​	1​
​	​	​	​	​	​	​	2​	2​
​	​	​	​	​	​	​	​	​
13-24​	Goat, Car, Goat​	​	​	​	​	​	2​	2​
​	​	​	​	​	​	​	​	​
25-36​	Goat, Goat, Car​	​	​	​	​	​	2​	2​

For game #7, you have picked door #2, the host opens door #3 to reveal a goat and your switch strategy involves switching to door #3, where you can already see the goat!

Are you sure you understand this game? If you see a goat, you switch to the unopened door. In both games #7 and #8, therefore, the switch is to door #1, which wins both times.

 

[Ibix]

I can't just eliminate those rows, since I've already stipulated that they're all equally likely.

They're not equally likely, though. As far as your initial choice they're equally likely. But if the host has a choice which door to open, those games are half as likely as those with no choice for the host.

The easiest thing to do is point out that (given the rules the host must follow) the stick/switch choice could be known before the host opens a door.

 

[mfb]

You never switch to an opened door with a goat (unless you really want to get a goat). Fixed it:

Game #​	Setup​	I picked​	Host opened​	I chose to​	to Door #​	Outcome​	Stick​	Switch​
1​	Car, Goat, Goat​	1​	2​	Stick​	1​	Win​	1​	​
2​	Car, Goat, Goat​	1​	3​	Stick​	1​	Win​	1​	​
3​	Car, Goat, Goat​	1​	2​	Switch​	3​	Lose​	​	​
4​	Car, Goat, Goat​	1​	3​	Switch​	2​	Lose​	​	​
5​	Car, Goat, Goat​	2​	3​	Stick​	2​	Lose​	​	​
6​	Car, Goat, Goat​	2​	3​	Stick​	2​	Lose​	​	​
7​	Car, Goat, Goat​	2​	3​	Switch​	1​	Win​	​	1​
8​	Car, Goat, Goat​	2​	3​	Switch​	1​	Win​	​	1​
9​	Car, Goat, Goat​	3​	2​	Stick​	3​	Lose​	​	​
10​	Car, Goat, Goat​	3​	2​	Stick​	3​	Lose​	​	​
11​	Car, Goat, Goat​	3​	2​	Switch​	1​	Win​	​	1​
12​	Car, Goat, Goat​	3​	2​	Switch​	1​	Win​	​	1​
​	​	​	​	​	​	​	2​	4​
​	​	​	​	​	​	​	​	​
13-24​	Goat, Car, Goat​	​	​	​	​	​	2​	4​
​	​	​	​	​	​	​	​	​
25-36​	Goat, Goat, Car​	​	​	​	​	​	2​	4​

 

[sysprog]

At the time of the choice to stick or switch being offered, the only 2 doors that can conceal the car are the one initially chosen, and the other one that isn't opened. The one initially chosen still has its original 1/3 chance of concealing the car, so the other unopened one has 2/3 chance of concealing the car. The significant facts are summarized in the following location and outcome table, showing the stick and switch results for each location and each initial choice:
\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline & \mathtt 1 & \mathtt 2 & \mathtt 3 & \mathtt {st\ 1} & \mathtt {sw\ 1} & \mathtt {st\ 2} & \mathtt {sw\ 2} & \mathtt {st\ 3} & \mathtt {sw\ 3}\\
\hline \mathtt 1 & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathbf {\underline{car}}\\
\hline \mathtt 2 & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}}\\
\hline \mathtt 3 & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathbf {\underline{car}} & \mathbf {\underline{car}} & \mathtt {goat} \\
\hline
\end{array}
No matter which door conceals the car, and no matter which door the contestant first chooses, the chance for the car for stickers begins and ends at 1/3, while the chance for the car for switchers begins at 1/3 and ends at 2/3. This can be verified by counting the relevant cells; each stick column has one car cell and 2 goat cells, while each switch column has 2 car cells and one goat cell.

The same numbers are used for the columns over the door contents as are used for the rows

The more complicated tables that have been provided make reference to the ruse of opening of one of the doors, which is not important. The host could simply offer the choice of (1) sticking or (2) switching to both of the other two doors, and that would make no difference for the chance of getting the car: 1/3 for sticking; 2/3 for switching.

 

[DaveC426913]

You never switch to an opened door with a goat (unless you really want to get a goat). Fixed it:

Ah!

(These tables are harder than I thought!)

Thank you mfb, for correcting my table.

And thanks everyone else for your contributions. I know there are lots of ways of visualizing this in terms of what shall happen (probabilities), but I wanted to see it as factual - in terms of what did happen (experiment).

The code samples proved what did happen, but they're inscrutable - they don't show how/why.

..

mfb showed how switching is the better strategy.

You can see that some rows are identical. That's where they double their probability. The reason they're there is because, the host is constrained to pick the same door twice (the only door I haven't picked and also not the car) See games 7,8 and again games 11,12.

Sticking wins 2 games, switching wins 4 games. Switching is twice as likely to win as sticking.

Game #​	Setup​	I pick​	Host opens​	I choose to​	to Door #​	Outcome​	Stick​	Switch​
1​	Car, Goat, Goat​	1​	2​	Stick​	1​	Win​	1​	​
2​	Car, Goat, Goat​	1​	3​	Stick​	1​	Win​	1​	​
3​	Car, Goat, Goat​	1​	2​	Switch​	3​	Lose​	​	​
4​	Car, Goat, Goat​	1​	3​	Switch​	2​	Lose​	​	​
5​	Car, Goat, Goat​	2​	3​	Stick​	2​	Lose​	​	​
6​	Car, Goat, Goat​	2​	3​	Stick​	2​	Lose​	​	​
7​	Car, Goat, Goat​	2​	3​	Switch​	1​	Win​	​	1​
8​	Car, Goat, Goat​	2​	3​	Switch​	1​	Win​	​	1​
9​	Car, Goat, Goat​	3​	2​	Stick​	3​	Lose​	​	​
10​	Car, Goat, Goat​	3​	2​	Stick​	3​	Lose​	​	​
11​	Car, Goat, Goat​	3​	2​	Switch​	1​	Win​	​	1​
12​	Car, Goat, Goat​	3​	2​	Switch​	1​	Win​	​	1​
​	​	​	​	​	​	​	2​	4​
​	​	​	​	​	​	​	​	​
13-24​	Goat, Car, Goat​	​	​	​	​	​	2​	4​
​	​	​	​	​	​	​	​	​
25-36​	Goat, Goat, Car​	​	​	​	​	​	2​	4​

 

[cmb]

Well **that** OP sure released a flood of responses. I see that cmb is long gone, however...

I'm definitely here, I just wanted to see what the responses were first, see how far people got down this conundrum.

 

[cmb]

Well, you need to add a probablity entry to the table. Each door pick has a probablility of 1/3, so when you split the pick of door 1, each of the first two entries only have a probability of 1/6. Then it all works out. With "I stick" you win 1/3 of the time, and with "I switch" you win 2/3 of the time.

I agree, this was what I intended to include.

The four options were meant to illustrate scenarios, not probabilities.

 

[cmb]

There have been so many threads on this that I hesitate to respond to this one. But I will tell you what made it clear to me. Suppose, instead of 3 doors, that there are 1,000,000 doors, with 1 car and 999,999 goats. You choose a door, so your odds of choosing the car are clearly 1/1,000,000. Then the host opens 999,998 doors which all contain goats. Do you really believe that the odds that your door has the car has increased to 50:50? The car didn't move, so how could your odds have increased? Think about it from this standpoint.

This does, actually, add considerable clarity in my mind (whether others think so?), and I think might have, in combination with the last respondent's reply, resolved this for me (whether it satisfies anyone else, I can't say!).

"Do you really believe that the odds that your door has the car has increased to 50:50?" Sort of... they were already 50:50 before the game started. I'll explain this, in this way:

We have two questions at play here, and I think this might be where the non-intuitive fallacy comes in. There is;
1) 'a game strategy, planning out what a contestant should do once he is in the studio', and
2) 'what to do next, after Monty opens a door'.

These are two questions which appear to be the same thing, which aren't, and I think there is the dilemma. Some people are answering the first, some the second.

OK, so on point 1), let's say the host explains what is going to happen first [in the 1,000 door contest] before the show starts, he says "You will pick one door of 1,000, I will then open up 998 doors that don't have the prize in. You can then choose to swap if you want." I would rest easy, knowing that the odds were certainly not 1/1000. I'd have intuitively said this is immediately at least 50:50 chance, because whatever door I pick, I will also still end up with two doors, and a car behind one of them.

However, it is better than 50:50 because also I have the original chance of my door choice. I don't just have the chance to swap, I have the original chance that might have been right too! That would be for me to decide.

The chance of me picking 'the car' in the 3 door game, and Monty picking the particular door he does, is 1/6. So the overall chance of me winning the game with a fixed strategy at the outset is 1/6 + 1/2 = 2/3.

However the chance of me winning, once and after Monty has opened the door is 50:50.

I hold this principle as axiomatic, not as intuition. If there are two options and two doors the chance is 50:50. If that is not true then it is equally wrong to claim the chance is 1/3 if there are three options and three doors.

For n choices behind n doors, for which there is one 'win' option, either the probability is 1/n or it is not 1/n. It can't be both, this is axiomatic.

In a way, I think this is more deeply satisfying. On the one hand, we don't need to question the axiom that 1 choice in n is 1/n. That would be really troubling. So on the other, we have two ways to approach the problem both summing up to 2/3. In the one case, we look at the 'overall game strategy' and in the other we consider the situation at the moment where the contestant is given the option to swap. Yes, it is better and a 2/3 chance if he decides on a fixed strategy before entering the game studio, and yet (non-intuitively) the chance is still 50:50 at the moment he has to make a decision whether to swap.

Is this at all a more satisfactory view of the scenario, or do we need to look more closely at the occasions when "1 choice in n is not 1/n"?

 

[cmb]

Yes, their probability is different unless they have been told what occurred.

They know exactly what occurred.

The question is would the probability be the same for the 'stand-in' contestant who's been watching from the curtains or does he get a 50:50 shot at two doors with one win behind one of them?

My answer is that this is different because the stand-in contestant only gets the 50:50 choice of doors, he does not benefit from the additional option of choosing a door in the first round.

... and, no, that doesn't sound that convincing to me either, but I am far less convinced that being told what happened before changes one's probability. I mean, is there something that the contestant wasn't told about which happened in the studio the day before which affects the '1/3' assumption on his first pick? Sorry, I can't accept that, the previous history makes no difference to the choice of 1 in n 'at that particular moment in time' if the n are unbiased and independent. I think this is the logical 'conflict', that there is a difference between a 'game plan' and a decision at a particular moment.

It'd be a bit like walking out into the middle of the road and saying 'only 1 in a million pedestrians a year get hit and killed, so this truck that is coming straight at me has no chance of hitting me', which is obviously nonsense at the time, but the 'game plan' for the probability of you getting killed on the road when walking out of your house at the beginning of the day would be quite different to standing in the road trying to make a decision when to move.

I have zero concern that going into the game show with the 'swap strategy' will result in a 2/3 chance of getting the car. That is well evidenced and proven. The hugely problematic question is about the probability of the moment of making the choice whether to swap. I can't see how that is not 1 independent choice out of 2. Just like standing in the road, to stay there is not a 1 in a million chance of getting hit, it's a lot higher if you don't do something about it! It's the difference between a momentary probability versus the chances of a 'strategy' working out as a whole. Maybe that is what you were trying to say?

 

[sysprog]

Holy X, @cmb, you are clearly evincing befuddlement.

Please trust that no-one here is trying to mislead anyone. If the Mentors/Moderators here were to determine that someone was deliberately doing that, the someone would be not be allowed to persist in doing that on PF.

You start the game with 1/3 chance of choosing the door that conceals the car. We call that door S1, for subset 1.

That leaves the 2 doors, that were not chosen, to contain the other 2 of the 3 possibilities for being the door that conceals the car. We call those 2 doors S2, for subset 2.

We know in advance that no more than 1 member of S2 can conceal the car, because there is only 1 car.
We therefore also know in advance that if the car is behind an S2 door, it's not behind the other S2 door.

There is always either 1 door or 2 doors in S2 that do not conceal a car: 1 if S2 conceals the car, and 2 if S1 conceals the car.

When Monty opens a door, without revealing a car, he thereby reveals nothing that tells you whether there was only 1 door in S2 that he could have opened, or there were 2 doors in S2 that he could have opened.

That means that he has thereby told you exactly nothing that serves to change what you already knew about the car-concealing chances of S1 or S2. You therefore can and should know that S1 remains at 1/3 chance of concealing the car, and that S2 remains at 2/3.

Once S2 has only 1 available member left, i.e., only 1 unopened door, being offered to switch to that door gives you the option to to take the S2 chance instead of the S1 chance. The S2 chance is still 2/3, just as it would have been if Monty had not opened a door from S2, and had given you the option of choosing both doors of S2.

So, after you have chosen, and after a door-opening interlude designed to obfuscate the obvious, you are offered the chance to switch, to the S2 member that at that time contains all of the S2 chance of concealing the car, and that chance is 2/3.

 

[cmb]

Holy X, @cmb, you are clearly evincing befuddlement.

I am, truly, befuddled on this, no question at all!

Not trying to get anyone else to be befuddled, just interested in their answers to the 'stand-in contestant' scenario.

Is their probability different? If not how? Surely 1 choice in n is 1/n, and that is what they would face?

I know this can be dragged back to 'evidencing' the overall game-plan outcome is 2/3 by tables and diagrams. I totally get that, and I don't seek to doubt 'swapping' as the game plan of choice for the contestant walking into the studio. (And, no, I would not take the counter-bet! ;) )

I am interested in whether the 'fresh choice' the stand-in contestant faces with two doors is, at that moment in time, 50:50 or if not how is that possible?

Yes, very befuddled.

In your example, and comparing it with the 1000 door example, presumably that'd mean picking a door, seeing 998 being opened, then the chance of the one other door having the car behind it is 999/1000? ...

hmmm ... yeah, I think I am beginning to see that ... so if the contestant's choice was totally random but all the others weren't, then that 'other' door was 'chosen' by Monty in full knowledge of where it was, thus much more likely ... i.e. not independent. Yeah, OK, really tricky to see with 3 doors, but I think I am comfortable with this now.

Few... befuddlement indeed!

Thanks.

 

[PeroK]

I am, truly, befuddled on this, no question at all!

Not trying to get anyone else to be befuddled, just interested in their answers to the 'stand-in contestant' scenario.

Is their probability different? If not how? Surely 1 choice in n is 1/n, and that is what they would face?

I know this can be dragged back to 'evidencing' the overall game-plan outcome is 2/3 by tables and diagrams. I totally get that, and I don't seek to doubt 'swapping' as the game plan of choice.

I am interested in whether the 'fresh choice' the stand-in contestant faces with two doors is, at that moment in time, 50:50 or if not how is that possible?

Yes, very befuddled.

The stand-in contestant complicates things because, as I pointed out in an earlier post, you now have a variable amount of information available.

1) Monty knows where the car is, so he could win every time.

2) A contestant who knows which door was originally chosen can win 2/3 of the time.

3) A contestant who doesn't know which door was originally chosen, but is simply presented with a choice of two doors, can only win 1/2 of the time.

The mistake is to see this as some paradox on "absolute" probabilities. What are being calculated here are conditional probabilities based on the knowledge available. The amount of information you have determines how often you can win.

 

[Orodruin]

They know exactly what occurred.

The question is would the probability be the same for the 'stand-in' contestant who's been watching from the curtains or does he get a 50:50 shot at two doors with one win behind one of them?

If they know exactly what occurred, then the probability is 1/3 if they keep their friend’s door and 2/3 if they switch. Information does matter.

Consider the following: You know that there is a diamond in a piece of rock. The rock is cut into two pieces, one twice as big as the other. If you are allowed a choice between piece A and B without being told which piece is the big one your probability of getting the piece with the diamond is 1/2. If you are told which piece is the big piece, your will pick that one for a 2/3 chance. Nothing changed between the physical setups, but your probability of finding the diamond changed only based on additional information given to you.

Is their probability different? If not how? Surely 1 choice in n is 1/n, and that is what they would face?

Given the extra information, it is not a symmetric choose 1 of n.

 

[sysprog]

I am, truly, befuddled on this, no question at all!

Not trying to get anyone else to be befuddled, just interested in their answers to the 'stand-in contestant' scenario.

Is their probability different? If not how? Surely 1 choice in n is 1/n, and that is what they would face?

I know this can be dragged back to 'evidencing' the overall outcome is 2/3 by tables and diagrams. I totally get that, and I don't seek to doubt 'swapping' as the game plan of choice.

I am interested in whether the 'fresh choice' the stand-in contestant faces with two doors is, at that moment in time, 50:50 or if not how is that possible?

Yes, very befuddled.

From the point of view of the audience, the stand-in post-door-opening contestant who is asked choose 1 of the 2 remaining unopened doors, assuming that he is not told which door was the original contestant's choice, or any other special information, has 1/2 chance of choosing the 1/3 chance door, and 1/2 chance of choosing the 2/3 chance door. The average (arithmetic mean) of 1/3 and 2/3 is 1/2, so the stand-in has a 1/2 chance of choosing the door that conceals the car. That squares with what he should suppose if he is told nothing about the situation other than that there are 2 doors, 1 of which conceals a car, and the other of which conceals a goat, and that he can choose a door, and that he can keep whatever is concealed behind the door he chooses.

[Per @Orodruin pointing out that the stand-in was postulated to have been in on the prior corse of play:]

Assuming that the stand-in contestant knows what the first contestant knows, then he can and should know that: the original door chosen still has 1/3 chance of concealing the car, and the other unopened door has 2/3 chance of concealing the car.

 

[cmb]

Given the extra information, it is not a symmetric choose 1 of n.

Got it.

Thanks everyone, took a while but happy to withdraw the assertions in my OP. Having just the 3 doors does make this fact tricky to see.

... Will still have to go and think about it a bit more!

 

[Orodruin]

assuming tat he is not told which door was the original contestant's choice

OP has stated that the stand-in will be told, see #60.

 

[cmb]

Yeah, I can now see 2/3 is true for the stand-in too, so long as he knows which one was picked by the contestant (who didn't know) and which one was (effectively) picked by Monty who did.

If he didn't know that, then, yes, in that case it'd drop to the 1 in 2. But knowing which one was 'chosen' by each makes the difference, as the 1000 door case shows. In the 1000 door show it becomes 'obvious' that the 999th door was 'chosen' by Monty as the last remaining, thus is 999/1000 likely the car.

... phew ... got there in the end. Thanks for bearing with me. I'd like to think I was befuddled than dumb!

Thanks to gmax for showing me a little sympathy! ;)

Me too. This Monty Hall thing has always bugged me, so I have a lot of sympathy for the previous posters who are harshly judged by others as having "bad intuition."

I don't know if this is a true story, but

https://files.eric.ed.gov/fulltext/EJ1060344.pdf

I wouldn't criticize anyone too harshly for following Erdos' intuition.

 

[sysprog]

... phew ... got there in the end. Thanks for bearing with me. I'd like to think I was befuddled than dumb!

Thanks to you for being good about being contradicted. Many bright people get this one wrong at first. No-one here calls anyone here dumb. Everyone here on PF is here at least partly to learn. That includes those who do a lot of teaching. Anyone here who is not learning something is missing out on a wealth of opportunity. I'm confident that I've learned a great deal here that I wouldn't have picked up elsewhere.

 

[Dale]

I can't just eliminate those rows, since I've already stipulated that they're all equally likely.

This is why, if you are interested in looking at outcomes of games (as opposed to probabilities), the proper tool is a Monte Carlo simulation, not a table. With a table or a decision tree you should calculate the probabilities for each row or path. In my tables I also included the impossible rows and calculated their probabilities which, as expected, was 0.

 

[sysprog]

I can't just eliminate those rows, since I've already stipulated that they're all equally likely.

This is why, if you are interested in looking at outcomes of games (as opposed to probabilities), the proper tool is a Monte Carlo simulation, not a table. With a table or a decision tree you should calculate the probabilities for each row or path. In my tables I also included the impossible rows and calculated their probabilities which, as expected, was 0.

Tables or decision trees that show all the possible outcomes and how they are arrived at can be used to establish probabilities for each variable, and those probabilities can be verified experimentally by means of a Monte Carlo simulation; however, the rules for the simulation and the rules for constructing the table or decision tree are inter-derivable, and both are derivable directly from the rules of the game.

 

[FactChecker]

The critical issue is the screening process that Monte applies in deciding which door to open. He will intentionally avoid opening the door with the car. That changes the odds. If he had 100,000 doors and opened 99,999 goat-doors, one must ask why he avoided the remaining door.

To emphasize the screening process, consider another problem:
Suppose there is a bucket of sand containing one diamond. You take a random pinch of sand from the bucket. It has a tiny chance of containing the diamond. Now suppose that all the sand in the bucket except a small pinch has been dumped out through a screen that will not allow the diamond to pass. You clearly know that there is an overwhelming chance that the small pinch remaining in the bucket has the diamond, while your pinch of sand still has a tiny chance. You should switch.

 

[Orodruin]

Now suppose that all the sand in the bucket except a small pinch has been dumped out through a screen that will not allow the diamond to pass. You clearly know that there is an overwhelming chance that the small pinch remaining in the bucket has the diamond, while your pinch of sand still has a tiny chance. You should switch.

I do not think that this is true unless you know that somebody deliberately dumped sand that they knew did not contain the diamond. If you just dumped sand randomly from the remainder, the largest probability is that the diamond ends up in the screen. It would be the equivalent of a game show where Monty opens one of the remaining doors at random, ie, the Monty Fall problem - where the probabilities are known to be 50-50.

 

[Dale]

Tables or decision trees that show all the possible outcomes and how they are arrived at can be used to establish probabilities for each variable

The problem with that is that e.g. the first version of this table had four entries two representing wins for “stay” and two representing wins for “switch”. So simply counting entries in the table doesn’t give you probabilities. So the table itself must always be augmented with the probabilities. Even in the case where counting entries in the table gives you the probabilities it is because you know that the probabilities for each row are equal.

 

[FactChecker]

I do not think that this is true unless you know that somebody deliberately dumped sand that they knew did not contain the diamond.

As I said, the sand was dumped through a screen that did not allow the diamond to pass.

 

[PeroK]

As I said, the sand was dumped through a screen that did not allow the diamond to pass.

If you ended up on the Monty Hall show, I'll bet you'd bring a giant magnet and point it at each door and see which one bulges as the car gets pulled towards you. But, that's a different sort of screening process if you do something to find out exactly where something is!

That's nothing to do with probabilities and information. That's to do with physically isolating something.

 

[FactChecker]

If you ended up on the Monty Hall show, I'll bet you'd bring a giant magnet and point it at each door and see which one bulges as the car gets pulled towards you. But, that's a different sort of screening process if you do something to find out exactly where something is!

That's nothing to do with probabilities and information. That's to do with physically isolating something.

So I should have said "the sand was dumped through a screen that would not allow the diamond to pass if it is in the bucket. " Thank you for the English lesson.

 

[Orodruin]

As I said, the sand was dumped through a screen that did not allow the diamond to pass.

This is irrelevant. Somebody would have to pre-screen the sand before the dumping. You would have to explicitly arrange for the diamond to be left in the bucket.

It would be the equivalent of a game show where Monty opens one of the remaining doors at random, ie, the Monty Fall problem - where the probabilities are known to be 50-50.

To put that into mathematics, consider the following:
You pick a door and Monty picks a door not to open at random among N > 2 doors. All the other doors are opened and none of them contains the car. Let A being the event that the car is behind your door, B the event that the car is behind Monty's door and C the event that the remaining N-2 doors do not contain a car. Clearly, P(A∪B∪!C) = 1 and A, B, and !C are exclusive. Both P(A) and P(B) are equal to 1/N because without further information to condition the probabilities on, they are both just choosing one out of N doors. This means that P(C) = 1-P(!C) = 1 - (N-2)/N = 2/N.

We seek P(A|C) and P(B|C).

P(A|C) = P(C|A) P(A)/P(C) = 1 * (1/N)/(2/N) = 1/2

Here, P(C|A) = 1 because if the car is behind your door then the remaining N-2 doors will not contain the car. P(A) = 1/N because there is originally a chance of 1/N of your door containing the car. The same logic can be applied to B.

 

[sysprog]

The problem with that is that e.g. the first version of this table had four entries two representing wins for “stay” and two representing wins for “switch”. So simply counting entries in the table doesn’t give you probabilities. So the table itself must always be augmented with the probabilities. Even in the case where counting entries in the table gives you the probabilities it is because you know that the probabilities for each row are equal.

That criticism seems to me to be apropos of either incorrect table construction or incorrect counting. It is clear from the rules of the game that there are 3 unique possible locations of the car, so 4 lines is immediately apparently 1 too many, and examination of the 3 car locations shows that 1 of them is repeated. The repeated car location line should be merged with its repetition, in the table construction or at least in the count, or the repetition should not be counted.

 

[DaveC426913]

...there are 3 unique possible locations of the car, so 4 lines is immediately apparently 1 too many

What? No.

'Possible locations of the car' is not what the number of rows is representing. Your idea that there is one too many rows is in error.

 

[sysprog]

What? No.

'Possible locations of the car' is not what the number of rows is representing. Your idea that there is one too many rows is in error.

In post #55 I included this table:

\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline & \mathtt 1 & \mathtt 2 & \mathtt 3 & \mathtt {st\ 1} & \mathtt {sw\ 1} & \mathtt {st\ 2} & \mathtt {sw\ 2} & \mathtt {st\ 3} & \mathtt {sw\ 3}\\ \hline \mathtt 1 & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathbf {\underline{car}}\\ \hline \mathtt 2 & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} & \mathbf {\underline{car}} & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}}\\ \hline \mathtt 3 & \mathtt {goat} & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathbf {\underline{car}} & \mathtt {goat} & \mathbf {\underline{car}} & \mathbf {\underline{car}} & \mathtt {goat} \\ \hline \end{array}

There are no rows there that repeat a car location and thereby obfuscate the fact that there are exactly 3 different locations for the car. That the 3 locations are equiprobable at 1/3 each is immediately discernible from the stated game conditions. Counting of the 1 car and 2 goats in each of the stick columns, and of the 2 cars and 1 goat in each of the switch columns, shows that the chance for the car is 1/3 if you stick, and 2/3 if you switch.

On the more complicated tables that have more than 1 row per car location, there is still 1/3 probability per car location; not 1/3 probability per reference.

 

[mfb]

This is irrelevant. Somebody would have to pre-screen the sand before the dumping. You would have to explicitly arrange for the diamond to be left in the bucket.

I think the whole bucket is put through the screen which let's everything pass apart from the diamond and a small bit of sand.

 

[DaveC426913]

In post #55 I included this table:

Right. OK. That was your table, not the tables that have been mostly at the centre of the discussion. I might have picked that up if my knee hadn't jerked.

 

[Orodruin]

I think the whole bucket is put through the screen which let's everything pass apart from the diamond and a small bit of sand.

The initial statement makes it clear that not all sand left in the bucket is passed through the filter (my boldface):

Now suppose that all the sand in the bucket except a small pinch has been dumped out through a screen that will not allow the diamond to pass.

 

[sysprog]

Right. OK. That was your table, not the tables that have been mostly at the centre of the discussion. I might have picked that up if my knee hadn't jerked.

Those more complicated tables can be examined for a characteristic repetition: when the car is behind the door initially chosen, they list once for each of the 2 doors that Monty is allowed by the rules to possibly open.

When the car is behind a door not initially chosen, there is only 1 door that Monty can open. There are exactly 3 doors and exactly 1 car, so there are exactly 3 possible outcomes that pivotally depend on where the car is. Listing 4 possibilities is therefore suspect.

If you ensure that you count exactly once per car position, you (presumably) won't wind up with the spurious 1/2 probability assessment that the ceremonious door-opening invites, and you'll (presumably) arrive instead at the correct 1/3 for sticking and 2/3 for switching probability distribution.

 

[Dale]

It is clear from the rules of the game that there are 3 unique possible locations of the car, so 4 lines is immediately apparently 1 too many

The rows of the table indicate more than just the location of the car.

Listing 4 possibilities is therefore suspect.

There are four possibilities. There is nothing suspect about that.

Not all the possibilities are equally probable, but there is no requirement that all entries in a table must be equally probable.

 

[FactChecker]

The initial statement makes it clear that not all sand left in the bucket is passed through the filter (my boldface):

Just to clarify what I meant: All the sand except a pinch is dumped out of the bucket through a filter. The remaining pinch of sand remains in the bucket along with any possible diamond.

 

[mfb]

Just to clarify what I meant: All the sand except a pinch is dumped out of the bucket through a filter. The remaining pinch of sand remains in the bucket along with any possible diamond.

Then you don't have the Monty Hall problem. You have the Monty Fall problem, and both pinches of sand are equivalent.

 

[FactChecker]

Then you don't have the Monty Hall problem. You have the Monty Fall problem, and both pinches of sand are equivalent.

That is not true. If the diamond remained in the bucket after you removed a pinch of sand, it will remain in the bucket after the majority of cand has been filtered out. That gives a very high probability that the remaining bit in the bucket has the diamond.

I have to think that we are not talking about the same thing since what I am saying is trivially true, IMHO, but people are arguing about it.

EDIT: I see what was ambiguous in my statements. I meant that as most of the sand is screened and dumped out, any diamond that hits the screen remains in the bucket. It is not removed.

 

[sysprog]

The rows of the table indicate more than just the location of the car.

True, but using an extra row for 1 of the car locations is misleading to many, who would, as you pointed out, then have to include a column, as you did and as @cmb did not, to show the fact that each of the 2 rows has only half of the probability for the 1 car location that they have in common. With only 1 row per car, you don't need a probability column for this problem, because the rows are simply 1/3 likely each.

There are four possibilities. There is nothing suspect about that.

The fact that there are more than 3 possibilities listed, when all that matters are the 3 possibilities for where the car is, in my view, makes the listing of 4 possibilities suffice as grounds to suspect that something that doesn't matter may be being given undue emphasis. Cursory investigation immediately makes apparent the fact that 1 car position is represented by 2 rows, and to my mind, that reinforces the suspicion. I think that the best correction of the overemphasis is to merge the 2 rows for the same car position into 1 row, but I recognize that your correction of including a probability column was also valid, and was well suited to the explanation that accompanied it, although I liked your earlier terser explanation:

Another way to look at it is as follows: the stick strategy wins if you initially pick the car. The switch strategy wins if you initially pick a goat. The probability of initially picking the car is 1/3 and the probability of initially picking a goat is 2/3.

I was about to post something similar, but I was still thinkig about how to word it, and you beat me to it.

Not all the possibilities are equally probable, but there is no requirement that all entries in a table must be equally probable.

I think that using 4 rows when there are only 3 possible car positions, without noticing that 2 of the rows share the same car location possibility and recognizing the implications of that, maps out exactly what misled @cmb in post #1.

 

[Dale]

With only 1 row per car, you don't need a probability column for this problem

Personally, I disagree with that approach for constructing these tables. I personally think that you should always include one or more probability columns. Even if every row is equal probability it is good to write that down explicitly and make sure that it is reasonable to assume that each is equal probability. Again, I do understand that it is a personal preference, but it never hurts to provide a probability column and in cases like this it is very helpful, so I think you should always do it.

 

[gmax137]

I think you should always do it

Do you mean, like in the next Monty Hall Paradox thread?

Actually that's not fair, this is the first MHP thread where I feel I have really got it straight in my head. So thanks to everyone!

 

[DaveC426913]

Personally, I disagree with that approach for constructing these tables. I personally think that you should always include one or more probability columns. Even if every row is equal probability it is good to write that down explicitly and make sure that it is reasonable to assume that each is equal probability. Again, I do understand that it is a personal preference, but it never hurts to provide a probability column and in cases like this it is very helpful, so I think you should always do it.

If the table shows results - i.e. after an (ideal) set of games are played - then probabilities of what might happen - are moot.

 

[DaveC426913]

Those more complicated tables can be examined for a characteristic repetition: when the car is behind the door initially chosen, they list once for each of the 2 doors that Monty is allowed by the rules to possibly open.

You are still looking at tables of what might happen. Why not look at tables of what did happen (in a set of ideal games)? Then the whole issue of probabilities just goes away.

 

[Dale]

If the table shows results - i.e. after an (ideal) set of games are played - then probabilities of what might happen - are moot.

Only if all results are equally likely. Unless you are talking about Monte Carlo simulations or an actual experiment where you might have thousands of lines in your table. And many will be the same.

 

[DaveC426913]

Only if all results are equally likely. Unless you are talking about Monte Carlo simulations or an actual experiment where you might have thousands of lines in your table. And many will be the same.

The point is to crate a table of ideal results.

For example:
A table of rolling 2 6-sided dice will have exactly 36 rows.
Every row has the same probability.

 

[Dale]

The point is to crate a table of ideal results.

I have no idea what you mean by “ideal” results.

Not all results have equal probability. Does a table of ideal results include duplicates of results with higher probabilities? If not, how does a table of ideal results indicate the results with higher or lower probability?

 

[lavinia]

I have a lot of sympathy for other posters who are sincerely and humbly confused and don't understand the resolution. I have very little sympathy for posters like the OP who come in telling everyone else that the well-known standard resolution is wrong and that everyone else has committed a fallacy to get the wrong answer. The harshness is probably not a reaction to the bad intuition but to the arrogance.

Some people aggressively challenge things as a way of learning - even if they are challenging well understood facts. I do not see this as arrogance.

 

[Dale]

Some people aggressively challenge things as a way of learning

Frankly, that is even worse than arrogance. An aggressive challenge, by nature of being aggressive, provokes defensive replies. It makes the learning environment antagonistic and adversarial instead of cooperative and constructive. Aggression provokes hostility and aggression by a student is stupidly provoking hostility against a superior opponent.

Don’t you think it is better to cultivate strong allies rather than strong enemies? Someone who foolishly chooses this as a learning strategy will surely learn little and will have earned all of the unkind responses from the people who would have otherwise taught them.

 

[mfb]

The point is to crate a table of ideal results.

Here is an extremely simple game: You lose with probability 1/pi, you win with probability 1-1/pi. There is no way to create a table of entries with equal probabilities that accurately reflects your chance to win.