The Monty Hall paradox/conundrum

Ibix · Jan 3, 2020

DaveC426913 said:

that's not going to help a layperson (like me) see why those probabilities are what they are. (eg. Think of showing this to my family over the dinner table.)

A suggestion - get a piece of lined paper and draw a table eighteen lines high, plus a header row. Divide it into four columns headed "Car behind door...", "You choose door...", "Host opens door...", and "You win if you...".

Divide the first column into three equal sized boxes labelled 1, 2, 3. The heights of the boxes represent the equal 1/3 probabilities of the three outcomes.

Copy the divisions in the first column into the second, then subdivide each one for the three options you have at this stage. You should end up with nine two-line high boxes, representing the nine 1/9th probabilities of each choice.

Copy the divisions in the second column into the third and subdivide according to the options the host has. Sometimes this will be one two-line high box and sometimes two one-line high boxes.

Finally, copy the divisions from the third column into the fourth and label each box stick or switch.

This is just the tree mfb drew, but with probability represented by height. I think it's probably the same as one of Dale's tables, but with some cells merged.

If you count the stick and switch entries, you'll find they are 50:50. But you would be making the same mistake cmb made - forgetting to weight by the probability of the outcome. If you count stick and switch entries multiplied by height (i.e. by the probability of the outcome) you'll get the expected 33.3:66.6 ratio.

Note also that your choice is completely constrained by the first and second column. The third one is actually irrelevant, which is to say that "sticking is only right if you guessed right first time" - trivially 1/3.

I'm sorry I can't draw this at the moment - my LaTeX table-fu is not strong enough on a tiny phone screen.

sysprog · Jan 4, 2020

Ibix said:

I'm sorry I can't draw this at the moment - my LaTeX table-fu is not strong enough on a tiny phone screen.

Not to worry, @Ibix; @stevendaryl presented a full matrix in this (now closed) thread in his 02-17-2019 post #79 therein:

stevendaryl said:

The correct analysis is that there are 9 possibilities for the situation (x,y) where x is the location of the car (behind doors 1, 2, or 3) and y is the door you pick. All 9 possibilities are equally likely (if everybody plays randomly). They are:

1. (1,1)
2. (1,2)
3. (1,3)
4. (2,1)
5. (2,2)
6. (2,3)
7. (3,1)
8. (3,2)
9. (3,3)

The strategy of always switching wins in cases 2, 3, 4, 6, 7, 8. It loses in cases 1, 5, 9. That's the correct analysis.

Now, all of those possibilities have probability 1/9. Now, let's take possibility 1. It has two subcases:

1A: Monty opens door 2.
1B: Monty opens door 3.

The probabilities for those two subcases have to add up to the probability for case 1. They are equally likely. So we have:

P(1A) = P(1B) = 1/2 P(1)

Similarly, there are two subcases for 5 and 9.

So the complete probability matrix looks like this:

##\left( \begin{array} \\ choice & car & open & probability \\ 1 & 1 & 2 & \frac{1}{18} \\ 1 & 1 & 3 & \frac{1}{18} \\ 1 & 2 & 3 & \frac{1}{9} \\ 1 & 3 & 2 & \frac{1}{9}
\\ 2 & 1 & 3 & \frac{1}{9} \\ 2 & 2 & 1 & \frac{1}{18} \\ 2 & 2 & 3 & \frac{1}{18} \\ 2 & 3 & 1 & \frac{1}{9} \\ 3 & 1 & 2 & \frac{1}{9} \\ 3 & 2 & 1 & \frac{1}{9}
\\ 3 & 3 & 1 & \frac{1}{18} \\ 3 & 3 & 2 & \frac{1}{18} \end{array} \right)##

I think that all the probabilities add up to 1.

And indeed they do sum to 1, and furthermore (just as LaPlace would say, when omitting a deriviation, "il est aisé à voia que"), it is easy to see that:

the probability sum of the 50% of the rows on which the first 2 row-column cells are equal is 1/3 and represents the probability that sticking wins,

and that

the probability sum of the 50% of the rows on which the first 2 row-column cells are unequal is 2/3 and represents the probability that switching wins.

Here's the code (from rosettacode.org) for a Python simulation:

Code:

from random import randrange

doors, iterations = 3,100000  # could try 100,1000

def monty_hall(choice, switch=False, doorCount=doors):
  # Set up doors
  door = [False]*doorCount
  # One door with prize
  door[randrange(doorCount)] = True

  chosen = door[choice]

  unpicked = door
  del unpicked[choice]

  # Out of those unpicked, the alternative is either:
  #   the prize door, or
  #   an empty door if the initial choice is actually the prize.
  alternative = True in unpicked

  if switch:
    return alternative
  else:
    return chosen

print "\nMonty Hall problem simulation:"
print doors, "doors,", iterations, "iterations.\n"

print "Not switching allows you to win",
print sum(monty_hall(randrange(3), switch=False)
          for x in range(iterations)),
print "out of", iterations, "times."
print "Switching allows you to win",
print sum(monty_hall(randrange(3), switch=True)
          for x in range(iterations)),
print "out of", iterations, "times.\n"

You can see it run on codepad at http://codepad.org/E7HfuR4s

Dadface · Jan 4, 2020

My problem here is I can't see where there is a problem. I might be overlooking something.

1. If your original choice is the car (1/3 chance) and you switch, you will will win either goat A or goat B.

2. If your original choice is goat A (1/3 chance) and you switch, you will win the car.

3. If your original choice is goat B (1/3 chance) and you switch, you will win the car.

So by revealing a goat the host is giving a 2/3 chance of winning the car.

sysprog · Jan 4, 2020

Dadface said:

My problem here is I can't see where there is a problem. I might be overlooking something.

1. If your original choice is the car (1/3 chance) and you switch, you will will win either goat A or goat B.

2. If your original choice is goat A (1/3 chance) and you switch, you will win the car.

3. If your original choice is goat B (1/3 chance) and you switch, you will win the car.

So by revealing a goat the host is giving a 2/3 chance of winning the car.

The problem that most people who experience a problem experience comes from incomplete reasoning: at the point at which they're offed the option to stick or to switch, they see the facts of there being two doors (chosen and not chosen), two options (stick or switch), and two unknowns (both doors conceal something), as meaning that the two possibilities are equiprobable.

Your not seeing a problem indicates that you didn't disregard the 1/3 original probably of having chosen the car, or the fact that sticking means keeping that probability, or the fact that switching means trading that 1/3 for the other 2/3, and not missing the fact if the car is behind one of the doors that you didn't originally pick, showing you which one it isn't behind doesn't change the fact that it's behind one of them.

It's easy for some people to get distracted by the main non-pivotal fact of the procedure. The act of opening of the only non-car door available to him, or of randomly choosing and opening one of two non-car doors available to him, has no effect on the chance of winning if sticking versus the chance of winning if switching. If the host didn't open a door, but simply asked the contestant whether he'd like to keep the door he's chosen, or switch for the other 2 doors, probably almost no-one would opt to keep the originally chosen door.

Ibix · Jan 4, 2020

sysprog said:

Not to worry, @Ibix; @stevendaryl presented a full matrix in this (now closed) thread in his 02-17-2019 post #79 therein:

Almost - he didn't make the 1/9th row twice the height of the 1/18th rows, nor merge cells.

Ibix · Jan 4, 2020

...and that python code is ugly. I'd use:

Python:

import random

# Specify how many doors and how many times we play
doors = 3
trials = 1000

# Choose where the car is each game... 
car = [random.randint(1,doors) for i in range(trials)]
# ...and the door you choose each game
youChoose = [random.randint(1,doors) for i in range(trials)]

# Define the rules the host follows... 
def hostLeavesClosed(y, c):
    if y == c:
        # You chose the car. Host chooses a random door to leave shut
        hostsChoices = [d for d in range(1, doors + 1) if d != y]
        return random.choice(hostsChoices)
    else:
        # You did not choose the car. Host leaves that door shut
        return c
# ...and work out what he does each game
yourOtherChoice = [hostLeavesClosed(y, c) for (y, c) in zip(youChoose, car)]

# Calculate how many times you'd have won by sticking... 
stickWins = sum([1 for (c, y) in zip(car, youChoose) if c == y])
# ...and by switching
switchWins = sum([1 for (c, y) in zip(car, yourOtherChoice) if c == y])

# Display the results
print("With " + str(doors) + " you would have won " + str(stickWins) + " times by sticking and " + str(switchWins) +" times by switching out of " + str(trials) + " trials.")
print("Theory predicts " + str(trials / doors) + " wins by sticking and " + str(trials - trials / doors) + " wins by switching")

sysprog · Jan 4, 2020

Ibix said:

Almost - he didn't make the 1/9th row twice the height of the 1/18th rows, nor merge cells.

In his ##\LaTeX## matrix he had 12 rows, 4 for each door instead of 3, with 2 of each 4 showing 1/18 instead of 1/9, and that's functionally equivalent to, and perhaps more perspicuous than, your device of double height to reflect the bifurcation of the those 9ths into 18ths.

Ibix · Jan 4, 2020

sysprog said:

In his ##\LaTeX## matrix he had 12 rows, 4 for each door instead of 3, with 2 of each 4 showing 1/18 instead of 1/9, and that's functionally equivalent to, and perhaps more perspicuous than, your device of double height to reflect the bifurcation of the those 9ths into 18ths.

But the whole point of my construction is that the height of each row is proportional to the probability of the outcome. I chose to do it that way because Dave wanted - I think - a visual representation of the probabilities. Steven's version doesn't do that - it just lists the probabilities at the end of the table, which Dave already said wasn't what he wanted.

sysprog · Jan 4, 2020

Ibix said:

But the whole point of my construction is that the height of each row is proportional to the probability of the outcome. I chose to do it that way because Dave wanted - I think - a visual representation of the probabilities. Steven's version doesn't do that - it just lists the probabilities at the end of the table, which Dave already said wasn't what he wanted.

To me the difference doesn't matter; however, I do see your point.

hutchphd · Jan 4, 2020

sysprog said:

perspicuous

Excellent new word for my old brain. Thanks.

DaveC426913 · Jan 4, 2020

Ibix said:

Divide the first column into three equal sized boxes labelled 1, 2, 3. The heights of the boxes represent the equal 1/3 probabilities of the three outcomes.

No.

In the case of experimental results, probabilities don't appear.

All you have is the initial conditions, and the results. And the results will literally show two 'switch' wins for every one 'stick' win.

DaveC426913 · Jan 4, 2020

OK, rather than concentrating on probabilities, I'm focusing on results - played games - i.e. the probabilities have already played out - and we are left with the results (after all, that is the only thing that ultimately matters).

To be clear: I am certain of the proof that switching is the winning strategy; I'm now simply trying to (succinctly) demonstrate it as fact rather than theory.

Here is my table showing 36 games. (And yes, I know it is effectively identical to the OP's table.)

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	1	Stick	2	Lose
6	Car, Goat, Goat	2	3	Stick	2	Lose
7	Car, Goat, Goat	2	1	Switch	3	Lose
8	Car, Goat, Goat	2	3	Switch	1	Win		1
9	Car, Goat, Goat	3	1	Stick	3	Lose
10	Car, Goat, Goat	3	2	Stick	3	Lose
11	Car, Goat, Goat	3	1	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

Obviously, the car being behind door 2 or 3 gives identical results, so whatever happens in those scenarios is identical to the first 12.

Question: Are all the games 1-36 equally likely to have happened? i.e. Does every row have the same probability of showing up in the table? If not, what rows am I missing?

PeroK · Jan 4, 2020

DaveC426913 said:

OK, rather than concentrating on probabilities, I'm focusing on results - played games - i.e. the probabilities have already played out - and we are left with the results (after all, that is the only thing that ultimately matters).

To be clear: I am certain of the proof that switching is the winning strategy; I'm now simply trying to (succinctly) demonstrate it as fact rather than theory.

Here is my table showing 36 games.

Game #
Setup
I pick
Host opens
I choose to
to Door #
Outcome
Stick
Switch
1
Car, Goat, Goat
1
2
Stick
1
Win
1
0
2
Car, Goat, Goat
1
3
Stick
1
Win
1
0
3
Car, Goat, Goat
1
2
Switch
3
Lose
0
0
4
Car, Goat, Goat
1
3
Switch
2
Lose
0
0
5
Car, Goat, Goat
2
1
Stick
2
Lose
0
0
6
Car, Goat, Goat
2
3
Stick
2
Lose
0
0
7
Car, Goat, Goat
2
1
Switch
3
Lose
0
0
8
Car, Goat, Goat
2
3
Switch
1
Win
0
1
9
Car, Goat, Goat
3
1
Stick
3
Lose
0
0
10
Car, Goat, Goat
3
2
Stick
3
Lose
0
0
11
Car, Goat, Goat
3
1
Switch
2
Lose
0
0
12
Car, Goat, Goat
3
2
Switch
1
Win
0
1

2
2

13-24
Goat, Car, Goat

2
2

25-36
Goat, Goat, Car

2
2

Obviously, the car being behind door 2 or 3 gives identical results, so whatever happens in those scenarios is identical.

(And yes, I know it is effectively identical to the OP's table.)
Question: Are all the games 1-36 equally likely to have happened? i.e. Does every row have the same probability of showing up in the table? If not, what rows am I missing?

I suggest you have confused "equally likely events" with your strategy. Your 12 scenarios are really only 6, in each case you either win by sticking or win by switching.

And, yes, the other 12/24 cases are just the same by with the car in a different position.

DaveC426913 · Jan 4, 2020

PeroK said:

I suggest you have confused "equally likely events" with your strategy. Your 12 scenarios are really only 6, in each case you either win by sticking or win by switching.

No. The above are results. Actual games played. There is no 'either/or'.

'I stuck' is an outcome of a game. 'I switched' is an outcome of another game. The sum total is statistical win for 'Switch'.

If I were to take the results of the actual programmatic experiments, I would see 10,000 results. Many of those rows are identical, but there has to be a minimum number of results that, when sorted for readability, will show the Switch option winning twice as often as the Stick option.

That is what the above table is intended to be. So which rows am I missing?

PeroK · Jan 4, 2020

DaveC426913 said:

No. The above are results. Actual games played. There is no 'either/or'.

'I stuck' is an outcome of a game. 'I switched' is an outcome of another game. The sum total is statistical win for 'Switch'.

If I were to take the results of the actual programmatic experiments, I would see 10,000 results. Many of those rows are identical, but there has to be a minimum number of results that, when sorted for readability, will show the Switch option winning twice as often as the Stick option.

That is what the above table is intended to be. So which rows am I missing?

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.

DaveC426913 · Jan 4, 2020

PeroK said:

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 · Jan 4, 2020

DaveC426913 said:

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 · Jan 4, 2020

DaveC426913 said:

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 · Jan 4, 2020

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 · Jan 4, 2020

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 · Jan 4, 2020

mfb said:

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 · Jan 5, 2020

marcusl said:

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 · Jan 5, 2020

phyzguy said:

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 · Jan 5, 2020

phyzguy said:

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 · Jan 5, 2020

Orodruin said:

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 · Jan 5, 2020

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 · Jan 5, 2020

sysprog said:

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 · Jan 5, 2020

cmb said:

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 · Jan 5, 2020

cmb said:

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.

cmb said:

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 · Jan 5, 2020

cmb said:

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 · Jan 5, 2020

Orodruin said:

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 · Jan 5, 2020

sysprog said:

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 · Jan 5, 2020

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! ;)

gmax137 said:

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 · Jan 5, 2020

... 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 · Jan 5, 2020

DaveC426913 said:

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.

The Monty Hall paradox/conundrum

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