Please Explain (actually explain) The Monty Hall Problem

[Ibix]

The host's knowledge does not change the problem after you have been shown a door with a goat, it just prevents the awkward situation of him revealing the car.

Note that your modified game is only the same as the normal one if this "awkward situation" is dealt with somehow - perhaps re-playing the game. So it still depends on the host's knowledge of where the prize is - it just changes when the host learns that fact.

 

[BWV]

Note that your modified game is only the same as the normal one if this "awkward situation" is dealt with somehow - perhaps re-playing the game.

Agree with this

So it still depends on the host's knowledge of where the prize is - it just changes when the host learns that fact.

If the host and the player have the same info at the same time, then the game cant depend on the host's knowledge

 

[PeroK]

I remember someone posted the following analysis for the Monty Hall game with 100 doors.

Let's say the contestant picks door 16 and the car is behind door 59. Monty goes down the doors in order. Opens doors 1-15, misses out door 16. Then opens doors 17-58, coughs discreetly while missing out door 59, before opening doors 60-100.

Now, @Timbre might be happy to stick with door 16, but the rest of us would have the strong suspicion that the car is behind door 59.

 

[jedishrfu]

Your friend or internet source is using flawed logic. It's the same logic people use and they get it wrong. It's why probability is often counter-intuitive to people.

Suppose we modify the game and bring a second contestant up who is free to choose one of the two closed doors. Then that second contestant has a 1/2 chance of getting it right. Now if he knows which door the first contestant chose, thought and said "hey I'm picking the other door." Then he would be using prior knowledge to inform his choice.

As has been mentioned earlier if the first contestant does nothing then chance of winning is 1/3 of the time.

This has been beat to death, but thought I would chime in that it actually does not matter whether the host knows what door the car is behind, if you pick door 1 and the host opens 2 or 3 without knowing where the goat is, there is a 1/3 chance that the host reveals the car and a 2/3 chance he reveals the goat.

As an aside, if the host periodically picked the car then I think this show would be very short-lived.

 

[jedishrfu]

Thinking further, we could always just send you a goat. Though I don't know that I could do that since I have neither a goat nor a mailman willing to deliver it.

---

In Texas, they have an agg exemption if you raise a tribe of 3-5 goats on five acres or more. And you get property tax break. You do however have make effort make money off of the goats by selling them, or selling their milk for goat cheese.

 

[PeroK]

As an aside, if the host periodically picked the car then I think this show would be very short-lived.

He would soon run out of space in his garage.

 

[Ibix]

If the host and the player have the same info at the same time, then the game cant depend on the host's knowledge

Well, your game is longer than the standard version because it sometimes involves replays, and whether you have a replay depends on revealing where the prize is. So in that case it depends on somebody knowing the location of the prize and that it is not known to the contestant when they actually choose.

A game that is statistically identical to Monty Hall is: contestant chooses a door, then may choose to open that door or open both the others. In that case switching is the obvious strategy, but there's no counter-intuitive behaviour.

 

[Ibix]

As an aside, if the host periodically picked the car then I think this show would be very short-lived.

I believe that in the actual game show Monty had options other than "open a door and offer a switch", and was a skilled cold reader who would typically take the choice that he reckoned would maneuver the contestant away from the prize. So he offered you a swap if he thought you'd take it and lose, or if he thought you'd stick and lose, and he didn't think one of his other options was better.

The first article I read about commented that the experience of watching the game show may have messed with people's intuition about the puzzle when it was first stated.

 

[Timbre]

This has been beat to death, but thought I would chime in that it actually does not matter whether the host knows what door the car is behind, if you pick door 1 and the host opens 2 or 3 without knowing where the goat is, there is a 1/3 chance that the host reveals the car and a 2/3 chance he reveals the goat. From your standpoint, if he does not show you the car, the problem is the same (if he does open the door with the car, then you definitely should switch ;) - there is a 2/3 chance you win by switching. The host's knowledge does not change the problem after you have been shown a door with a goat, it just prevents the awkward situation of him revealing the car.

The problem can be reduced to one decision you either bet that the car is behind one door or either of the remaining doors - so of course it makes sense to bet that the car is in the larger subset

To my knowledge this is not how the game is played. On the first round of questions, the host will never reveal the car. That's why the first round is not an actual question/answer that lends any new information to the game.

 

[Ibix]

That's why the first round is not an actual question/answer that lends any new information to the game

Why do you keep ignoring us when we point out this isn't true?

 

[berkeman]

[Note -- the OP is on a 10-day vacation from PF.]

 

[Dale]

That's why the first round is not an actual question/answer that lends any new information to the game.

Another way to show that this is a false claim is to directly calculate the amount of information entropy before and after opening the door. Information entropy (in bits) is given by $$-\Sigma \ p \ \log_2 p$$ where the sum is taken over all the events in the probability space. This is easily calculable for the Monty Hall problem.

Before opening the door the probabilities for the prize being behind doors 1, 2, and 3 respectively are $\{ \frac{1}{3},\frac{1}{3},\frac{1}{3} \}$. This gives an information entropy of 1.58 bits.

Suppose that the contestant chooses door 1 and Monty opens door 2. Then, after opening the door, the probabilities are $\{ \frac{1}{3},0,\frac{2}{3} \}$. This gives an information entropy of 0.92 bits. This reduction of entropy from 1.58 to 0.92 bits means that opening the door gave 0.67 bits of information.

However, suppose that you disagree about the probabilities in the previous paragraph and insist that the post-opening probabilities are equal, then you would (incorrectly) claim that the probabilities are $\{ \frac{1}{2},0,\frac{1}{2} \}$. This gives an information entropy of exactly 1 bit. So there was still a decrease in entropy from 1.58 to 1 bits. This means that even under the incorrect claim, opening the door gave 0.58 bits of information.

In fact, there is no value of $p$ for which $\{ p,0,1-p \}$ can give an entropy larger than 1 bit. Therefore by direct calculation you can show that opening the door provides some information. The only question is how much information it provides. But an assertion that it provides none is wrong, as is any proof or chain of reasoning based on such an assertion.

 

[PeroK]

I believe that in the actual game show Monty had options other than "open a door and offer a switch", and was a skilled cold reader who would typically take the choice that he reckoned would maneuver the contestant away from the prize. So he offered you a swap if he thought you'd take it and lose, or if he thought you'd stick and lose, and he didn't think one of his other options was better.

The first article I read about commented that the experience of watching the game show may have messed with people's intuition about the puzzle when it was first stated.

I imagine that takes us out of the realm of elementary probability theory into game theory.

 

[FactChecker]

A mathematical proof (there might be others) is based on Bayes' Rule. You can use that to directly calculate the probabilities. If you understand and accept the proof of Bayes' Rule, then you will understand the proof of this problem solution.

 

[DaveC426913]

Another way to show that this is a false claim is to directly calculate the amount of information entropy before and after opening the door. Information entropy (in bits) is given by $$-\Sigma \ p \ \log_2 p$$ where the sum is taken over all the events in the probability space. This is easily calculable for the Monty Hall problem.

Before opening the door the probabilities for the prize being behind doors 1, 2, and 3 respectively are $\{ \frac{1}{3},\frac{1}{3},\frac{1}{3} \}$. This gives an information entropy of 1.58 bits.

Suppose that the contestant chooses door 1 and Monty opens door 2. Then, after opening the door, the probabilities are $\{ \frac{1}{3},0,\frac{2}{3} \}$. This gives an information entropy of 0.92 bits. This reduction of entropy from 1.58 to 0.92 bits means that opening the door gave 0.67 bits of information.

However, suppose that you disagree about the probabilities in the previous paragraph and insist that the post-opening probabilities are equal, then you would (incorrectly) claim that the probabilities are $\{ \frac{1}{2},0,\frac{1}{2} \}$. This gives an information entropy of exactly 1 bit. So there was still a decrease in entropy from 1.58 to 1 bits. This means that even under the incorrect claim, opening the door gave 0.58 bits of information.

In fact, there is no value of $p$ for which $\{ p,0,1-p \}$ can give an entropy larger than 1 bit. Therefore by direct calculation you can show that opening the door provides some information. The only question is how much information it provides. But an assertion that it provides none is wrong, as is any proof or chain of reasoning based on such an assertion.

Damn that's well put together.

 

[BWV]

Before opening the door the probabilities for the prize being behind doors 1, 2, and 3 respectively are $\{ \frac{1}{3},\frac{1}{3},\frac{1}{3} \}$. This gives an information entropy of 1.58 bits.

Suppose that the contestant chooses door 1 and Monty opens door 2. Then, after opening the door, the probabilities are $\{ \frac{1}{3},0,\frac{2}{3} \}$. This gives an information entropy of 0.92 bits. This reduction of entropy from 1.58 to 0.92 bits means that opening the door gave 0.67 bits of information.

To put it another way, the contestant is presented with $\{ \frac{1}{3},\frac{1}{3},\frac{1}{3} \}$ but before a choice is made, the host adds information that makes it $\{ \frac{1}{3},0,\frac{2}{3} \}$ - the interesting bit is why is this a paradox that so many struggle with (myself included when first learned it)? It does not require any mathematical training or knowledge of Bayes rule.

There is a whole field of behavioral finance which analyzes biases and flaws in decision making under uncertainty and there are a few documented effects in play here -

A) the endowment effect - people anchor to their original choice and refuse to change if given new information
B) Confirmation bias - when the host reveals a goat, it reinforces the belief that the original choice was correct
C) an equiprobability heuristic - this is correct at first, but then it gets incorrectly applied to the updated information set

 

[DaveC426913]

the contestant is presented with $\{ \frac{1}{3},\frac{1}{3},\frac{1}{3} \}$ but before a choice is made, the host adds information that makes it $\{ \frac{1}{3},0,\frac{2}{3} \}$

The naive analyst might wonder why you have it changing to $\{ \frac{1}{3},0,\frac{2}{3} \}$ and not $\{ \frac{1}{2},0,\frac{1}{2} \}$

Although, it still highlights the import of @Dale's comment:

... by direct calculation you can show that opening the door provides some information*. The only question is how much information it provides. But an assertion that it provides none is wrong, as is any proof or chain of reasoning based on such an assertion.

* even $\frac{1}{3}$ to $\frac{1}{2}$ proves some information was emitted.

 

[PeroK]

To put it another way, the contestant is presented with $\{ \frac{1}{3},\frac{1}{3},\frac{1}{3} \}$ but before a choice is made, the host adds information that makes it $\{ \frac{1}{3},0,\frac{2}{3} \}$ - the interesting bit is why is this a paradox that so many struggle with (myself included when first learned it)? It does not require any mathematical training or knowledge of Bayes rule.

There is a whole field of behavioral finance which analyzes biases and flaws in decision making under uncertainty and there are a few documented effects in play here -

A) the endowment effect - people anchor to their original choice and refuse to change if given new information
B) Confirmation bias - when the host reveals a goat, it reinforces the belief that the original choice was correct
C) an equiprobability heuristic - this is correct at first, but then it gets incorrectly applied to the updated information set

I think the answer is quite subtle. As Dave says:

The naive analyst might wonder why you have it changing to $\{ \frac{1}{3},0,\frac{2}{3} \}$ and not $\{ \frac{1}{2},0,\frac{1}{2} \}$

Then you have to sit down and calculate. I.e. analyse the possible outcomes and their likelihood. A friend of mine at work was unconvinced until he started writing a computer program. He didn't even have to run it. Simply designing the code highlighted what was happening. And, highlighted the distinction between Monty Hall and Monty "Fall".

 

[DaveC426913]

... unconvinced until he started writing a computer program. He didn't even have to run it. Simply designing the code highlighted what was happening.

Yup. That's exactly what happened to me.

The process of writing the program forces you to untangle and define the logic. By the time I was only half way through, there was no point in finishing, because it was plainly obvious what it was going to produce.

I only finished it for the benefit of others.

 

[FactChecker]

The naive analyst might wonder why you have it changing to $\{ \frac{1}{3},0,\frac{2}{3} \}$ and not $\{ \frac{1}{2},0,\frac{1}{2} \}$

Although, it still highlights the import of @Dale's comment:


* even $\frac{1}{3}$ to $\frac{1}{2}$ proves some information was emitted.

It's important for an explanation to clearly distinguish between the case where the host performs a "filtering selection" by intentionally opening losing doors, versus the case where the host just opens a losing door by luck. A direct application of Bayes' Rule involves assigning a value to P(doorB opened by host | doorB has the prize). That is the step where the difference is clear.

 

[Ibix]

A friend of mine at work was unconvinced until he started writing a computer program.

I actually initially convinced myself by drawing the probability tree:

Step 1: pick a door to hide the prize (3 equi-probable branches)
Step 2: contestant picks a door (3 equi-probable branches from each of the first three)
Step 3: Monty decides which door to open (2 equi-probable branches from the three branches where the contestant picked a winner, one branch from the other six)

This is, of course, another way of formally writing out the algorithm, but one that doesn't require any knowledge of programming. Laying it out formally leads you to the realisation that Monty sometimes has a choice and sometimes doesn't depending on the contestant's first choice, and that his actions don't change the probability that the contestant guessed right first time. So switching is always the better strategy.

 

[jack action]

Fact:
He has knowledge of his first chosen door, that it's 1/3 the winning door..

And he has knowledge of the other remaining door that it's also 1/3 the winning door.


So he has no useful knowledge to help him in the second choice. Because he has knowledge that his chosen door is 1/3, and he also has the same knowledge of the other optional door has a 1/3 chance from the previous round of choices. So this knowledge, that both doors previously had 1/3 chance, is immaterial, useless, not really information. Now he knows they both have a 1/2 chance, and that's all the probability ever is.

The proponents state the fact he has knowledge of his chosen door after the first round (1/3), but they imply that he has no similar knowledge of the second remaining door. That implication is false. The only relevant knowledge would be if he knew something about his door that he DIDN'T know about the other door. Then he could use that knowledge. But he has no such knowledge of the DIFFERENCE between the two doors.

Look at it the other way around; what are the probabilities it is a losing door instead of a winning door? Facts:

• On the first round, he has knowledge of his first chosen door, that it's 2/3 a losing door.
• On the second round, he has knowledge of the remaining door, that it's 1/2 a losing door.

2/3 > 1/2, therefore his first chosen door has a bigger probability that it is a losing door; he better select the other one.

Repeating the process with 100 doors and Monty opening 98 doors makes it even clearer:

• On the first round, he has knowledge of his first chosen door, that it's 99/100 a losing door.
• On the second round, he has knowledge of the remaining door, that it's 1/2 a losing door.

 

[DaveC426913]

Y'know, we have all been making some pretty bold assumptions about the range of possible outcomes...

 

[WWGD]

Maybe this perspective will help too, a Frequentist approach. Edit:Might need dotting some i's and crossing some t's: The only way the player can avoid winning is if he selects a car as his first choice. Then Monty will open a goat, and switching will lead you to another goat. Else, by the way Monty set up the game, if you select a goat, Monty will open up another goat, and switching will lead you to the car.

 

[sysprog1]

After the random 1 in 3 car placement and the random 1 in 3 contestant door selection the code need check no further than whether the selected door has a car behind it when selected (3/9=1/3 chance), and whether the contestant switched: all car-holders who switch lose; all others who switch win:

if selected-door = car (1/3 chance) then switch (always) loses
(and stick wins) 
else switch (always) wins

or (alternatively)

if selected-door ##\neq## car (2/3 chance) then switch (always) wins 
(and stick loses)
else switch (always) loses

 

[FactChecker]

After the random 1 in 3 car placement and the random 1 in 3 contestant door selection the code need check no further than whether the selected door has a car behind it when selected (3/9=1/3 chance), and whether the contestant switched: all car-holders who switch lose; all others who switch win:

if selected-door = car (1/3 chance) then switch (always) loses
(and stick wins)
else switch (always) wins

or (alternatively)

if selected-door ##\neq## car (2/3 chance) then switch (always) wins
(and stick loses)
else switch (always) loses

If I understand what you are saying, this is wrong. You are gaining no advantage from knowing that Monte Hall (might have) intentionally avoided opening one of the doors. Why not use that knowledge to adjust your guess?

Consider more extreme examples that myself and others have mentioned. Suppose you pick 1 door out of 100 and Monte opens 98 doors, leaving one unpicked door closed. You know that Monte can not open the prize door. Don't you think it looks suspiciously like that one door has something good behind it?
A simple application of Bayes' rule gives the adjusted odds.

PS. It never hurts to switch, unless Monte Hall is trying to tempt you to switch more often when you have the prize door. In that case, it's an example of game theory.

 

[sysprog1]

If I understand what you are saying, this is wrong. You are gaining no advantage from knowing that Monte Hall (might have) intentionally avoided opening one of the doors. Why not use that knowledge to adjust your guess?

Monty doesn't open the door with the car, and he doesn't open the contestant's selected door. So he always opens a non-car non-selected door.

The only new thing the contestant learns from Monty opening a door is which door he opened; whether the remaining non-selected non-opened door was eligible for opening, i.e. also didn't have the car, at that juncture is not revealed $-$ if it doesn't have the car, then the selected door does.
Whether the opened door is 1 of 2 non-car non-selected doors, or the only one (1 of 1), is not disclosed. If the former, Monty chooses which door at random, but that in no way affects the chance of the car being behind the selected door; that chance remains 1/3. Switching is the only way to the other 2/3 chance, regardless of which door Monty opens, and regardless of whether he had a choice or not.

Consider more extreme examples that myself and others have mentioned. Suppose you pick 1 door out of 100 and Monte opens 98 doors, leaving one unpicked door closed. You know that Monte can not open the prize door. Don't you think it looks suspiciously like that one door has something good behind it?

Yes, larger n makes (n-1)/n closer to 1

A simple application of Bayes' rule gives the adjusted odds.

In the simulation code, @DaveC426913 checks for whether the contestant's selected door has the car. My point is that once that's known, then if it's so, switching always loses, and if it's not, switching always wins $-$ there is no need for updating probabilities in the Bayesian sense; the originally selected door always has 1/3 chance, and the other 2 doors or the remaining 1 of them always have/has 2/3 chance. That the 2 doors chance is consolidated into 1 remaining door doesn't change their 2/3 chance.

 

[FactChecker]

Monty doesn't open the door with the car, and he doesn't open the contestant's selected door. So he always opens a non-car non-selected door.

The only new thing the contestant learns from Monty opening a door is which door he opened; whether the remaining non-selected non-opened door was eligible for opening, i.e. also didn't have the car, at that juncture is not revealed $-$ if it doesn't have the car, then the selected door does.

What you learn from that is this: Your door was not opened because he is not allowed to open your door. You knew that already. The other door was not opened. WHY? It might be because it has the prize. Now you can calculate the odds of that and adjust probabilities.

Whether the opened door is 1 of 2 non-car non-selected doors, or the only one (1 of 1), is not disclosed. If the former, Monty chooses which door at random, but that in no way affects the chance of the car being behind the selected door; that chance remains 1/3. Switching is the only way to the other 2/3 chance, regardless of which door Monty opens, and regardless of whether he had a choice or not.
Yes, larger n makes (n-1)/n closer to 1
In the simulation code, @DaveC426913 checks for whether the contestant's selected door has the car. My point is that once that's known, then if it's so, switching always loses, and if it's not, switching always wins $-$ there is no need for updating probabilities in the Bayesian sense; the originally selected door always has 1/3 chance, and the other 2 doors or the remaining 1 of them always have/has 2/3 chance. That the 2 doors chance is consolidated into 1 remaining door doesn't change their 2/3 chance.

At this point, don't you mean its 2/3 chance? The remaining door that you didn't chose and he didn't open now has a 2/3 probability of having the prize, all in one door. You should switch to it.

 

[sysprog1]

The remaining door that you didn't chose and he didn't open now has a 2/3 probability of having the prize, all in one door. You should switch to it.

Right $-$ if you always switch, then whenever you first pick the prize door (1/3 of the turns), you lose (by switching from the winner); and whenever you first pick one or the other of the non-prize doors (2/3 of the turns), you win (by switching to the winner (the other loser having been eliminated by the host)); so 'always switch' wins 2/3 of the turns (and loses 1/3), and 'always-stick' wins 1/3 of the turns (and loses 2/3).

 

[sysprog1]

If you didn't pick the prize door, switching wins; if switching loses; you picked the prize door:

~prize-door $\rightarrow$ (switch $\rightarrow$ win)
switch $\rightarrow$ (~win $\rightarrow$ prize-door)

Whether you picked the prize door or not is established when you pick a door.

None of Monty's motions changes the 1/3 chance you have of having picked the prize door. If you don't switch, you keep that chance of winning, and 1/3 of the turns, you win, and on 2/3 of the turns, you lose. You can always stick, and that will never change; you will always have the 1/3 chance of winning.

If you lose when you switch, it's because you picked the prize door and switched; but whenever (2/3 chance) you don't pick the prize door, switching always wins.