The Monty Hall paradox/conundrum

[sysprog]

I see this very simply.

The contestant chooses 1 of 3 doors. The probability of selecting the car is 1/3 and thus the probability of not selecting the car is 2/3. This can be restated as the probability that the car is behind the other doors is 2/3. Obviously there are 2 other doors. The host then reveals that there is no car behind one of the two remaining doors, thus eliminating that door as a selection.

Even at this point nothing has really changed in terms of probability. 1/3 probability that the car is behind the selected door and 2/3 that it is not. What has changed is that you are given the opportunity to select another door, whereas in the beginning there were 2 doors now there is only one and thus the probability is 2/3 that the car is behind the remaining door.

Simple.

Note: I didn't read through the 7 pages of posts, just the first and last, so I apologize if this explanation has already been given by others, I imagine it must have.

The joke is on you, because despite knowing how to "calculate" the probability, in our society nobody cares if you made the "rational" selection or not. People only care that you are the winner, the contestant that choses the door with the car gets all the praise, even if the car was in the first door selected. In fact the contestant likely gets more praise because he/she "knew" the car was there from the beginning.

Your post is apparently in agreement with mine one post earlier.

 

[FactChecker]

I see this very simply.

The contestant chooses 1 of 3 doors. The probability of selecting the car is 1/3 and thus the probability of not selecting the car is 2/3. This can be restated as the probability that the car is behind the other doors is 2/3. Obviously there are 2 other doors. The host then reveals that there is no car behind one of the two remaining doors, thus eliminating that door as a selection.

Even at this point nothing has really changed in terms of probability. 1/3 probability that the car is behind the selected door and 2/3 that it is not.

This is where I can never resist a comment because I think you have skipped an essential point. Your statement here is only true because it is guaranteed ahead of time that Monty will not open a door with the prize. If he accidentally bumped a door and it opened, showing a goat, then this is wrong. In that case, the probabilities of both your door and the remaining closed door each increase to 1/2.
In the standard puzzle, it is the intentional avoidance of opening a door with the prize that increases the remaining closed door probability to 2/3 while your door remains at 1/3.

 

[hutchphd]

.
I bid 8 quatloos on door number 3....

 

[sysprog]

I bid 8 quatloos on door number 3...

That reminds me of the Steve Goodman song (further popularized by Jimmy Buffet) that includes the line "my whole world lies waiting behind door number three".

 

[WWGD]

My take: Since doors are opened/chosen randomly, 2/3 of the time , a door with a goat will have been chosen, in which case switching wins. 1/3 of the time a car has been chosen in which case switching loses.

 

[FactChecker]

My take: Since doors are opened/chosen randomly, 2/3 of the time , a door with a goat will have been chosen, in which case switching wins. 1/3 of the time a car has been chosen in which case switching loses.

When something is true, there are many ways to intuit that fact, even if the intuition is incomplete. There never was a side in this argument that thought it was bad to switch. When you say that switching wins, the argument has been whether switching is actually better versus switching just breaks even. IMO, the correct answer to that question must specifically take into account that Monty is forced to select a door without the prize. That is the essential difference between switching being better versus switching just breaking even.

 

[Merlin3189]

When something is true, there are many ways to intuit that fact, even if the intuition is incomplete.

My experience is that, in probability there are even more ways to intuit falacies!

 

[WWGD]

When something is true, there are many ways to intuit that fact, even if the intuition is incomplete. There never was a side in this argument that thought it was bad to switch. When you say that switching wins, the argument has been whether switching is actually better versus switching just breaks even. IMO, the correct answer to that question must specifically take into account that Monty is forced to select a door without the prize. That is the essential difference between switching being better versus switching just breaking even.

Edit:True, I am assuming that in that if a goat is chosen first then a change implies a win ( edit) because Hall will present a goat after the first door has been chosen.

 

[PeroK]

My experience is that, in probability there are even more ways to intuit falacies!

The fundamental issue in this case is to target your thinking towards the right problem. Whether by calculations or "intuition" it's always possible to solve the wrong problem.

The wrong solution to this problem is often the right solution to the different "fall" problem.

 

[FactChecker]

Edit:True, I am assuming that in that if a goat is chosen first then a change implies a win ( edit) because Hall will present a goat after the first door has been chosen.

Yes. This is a good example of why I am a big fan of methodically applying Bayes' Rule on problems like this. It forces one to think systematically of the probability of the condition (probability=1 that Monty picks a door with a goat) and the probabilistic consequences of the condition. I think that helps to point out where some people's intuition goes wrong.

 

[supernova1054]

First rebuttal is experiment. This has been tested and the standard analysis is correct. The switching strategy does in fact measurably increase your odds in the amount predicted.

The second rebuttal is that your analysis is incomplete. You need to include the cases where the goat is behind 2 and where the goat is behind 3. And look at the probabilities for each case.

Think of this. Suppose you could pick 2 doors. Your odds are 2/3. That is the same as the host necessarily picking an empty door if you switch doors.

 

[gmax137]

EDIT
Sorry, I misunderstood your post. Never Mind.

 

[Frodo]

There are many ways to convince people unfamiliar with probability calculations.

The best is undoubtedly to draw up the table of every possibility, where the car is placed behind each door in turn, and where the player chooses each door in turn. There are nine possible outcomes as in the table below. The player who keeps his choice wins three times (1/3) and loses six times (2/3). If the player swaps his choice, he wins six times (2/3) and he loses 3 times (1/3).

A simple way to convince people (it may have been said before bit I haven't waded through all the posts) is to re-order the game slightly. Remember the game is "You choose a door. I will open one of the remaining two doors which I know does not have the car. Do you now want to swap?" So play it as:
1. Choose your door.
2. I ask "Do you want to keep your choice? Or do you want to swap and have both the other doors?"
3. I will now open one of the two doors which I know does not have the car.

Everyone chooses to swap at Step 2.

 

[sysprog]

Another honesty version:

 

[BrassOctopus]

Is this a situation where the probability is being adjusted in the middle of an operation? Can we do that in mathematics?

(General question - I know very little about math)

 

[etotheipi]

Is this a situation where the probability is being adjusted in the middle of an operation? Can we do that in mathematics?

You could analyse it using Bayes' theorem, yes. That theorem will tell you how the probability of a certain event updates, given some new evidence (the new evidence being, that you now know a door behind which there is definitely no car).

 

[sysprog]

You could analyse it using Bayes' theorem, yes. That theorem will tell you how the probability of a certain event updates, given some new evidence (the new evidence being, that you now know a door behind which there is definitely no car).

I think that you don't need to do such analysis if you simply recognize that given the rules Monty always offers the 2 other door chances in exchange for your original 1 door chance.

 

[etotheipi]

I think that you don't need to do such analysis if you simply recognize that given the rules Monty always offers the 2 other door chances in exchange for your original 1 door chance.

Yes. I don't want to re-start this thread since the topic has been discussed to death, but by far the easiest way to think about it is just to notice that if your strategy is to switch then if you initially pick the goat you get the car, and if you initially pick the car you get the goat. The $2/3$ probability falls right out.

But I think @BrassOctopus was asking about a Bayesian analysis, and the answer is that yes, you can do such an analysis if you like.

 

[BrassOctopus]

But I think @BrassOctopus was asking about a Bayesian analysis, and the answer is that yes, you can do such an analysis if you like.

Yeah, I looked this up and, well, it's a bit out of my intellectual reach. But I was just curious if it could be done.

 

[sysprog]

Yeah, I looked this up and, well, it's a bit out of my intellectual reach. But I was just curious if it could be done.

I agree with @etotheipi that the Bayesian approach would work in this instance; however, it's a more subtly principled approach than is actually required in this case: if you want the easy answer it's in Monty's speech bubble in post #189.

 

[sysprog]

Oh and @BrassOctopus and anyone else who might be reading, you might be pleasantly surprised to learn how much more of mathematics than you had previously supposed would be, turns out to indeed be, within your intellectual reach, if not yet within your firm grasp, upon visiting https://www.khanacademy.org/ $-$ it might take some time and effort $\cdots$

 

[JeffJo]

In case anybody, like me, doesn't want to read through all of this? But still wants to see it resolved correctly (which very seldom happens)? I decided to add a response at the end of the thread. Hopefully it will be closed soon.

Let's say that just at the moment you are about to decide whether to swap doors or not, you go sick and your quiz partner steps in. They are now presented with two doors, which, for sure, one has the car and one has the goat. There is simply no possible way that the probability for them is not 50:50! Yet, somehow the probability for you was different?

Instead, let's say I will roll a special six-sided die. The sides are painted either red or green. Two are painted one of those colors, and four are painted the other color. I tell Alice which color is on four sides, but not Bob. I ask both to bet on which color will come up.

If Alice bets on the color that I told her is on four of the sides, she has a 2/3 chance of winning her bet. Bob has no idea which color predominates, and has to guess. So he has a 50:50 chance of guessing right. If he does, he has a 2/3 chance like Alice. If he guesses wrong, he has a 1/3 chance. His overall chances are thus:

(2/3)*(1/2) + (1/3)*(1/2) = 1/2.​

This is exactly analogous to cmb's scenario. According to the usual interpretation, you (like Alice) have information that says one choice has a 2/3 chance. Your quiz partner (like Bob) does not, and must guess. This is how probability works, and it makes perfect sense if you understand probability.

But this doesn't say why your information means that one choice has a 2/3 chance. The usual explanation is that your original chances, 1/3, can't change. That's wrong. They in fact can change, but nothing you know allows this change. To see why, let's examine a slightly modified version of cmb's table:

This isn't the way it should be displayed - we should fix the player's choice and let the location of the car vary - but it works the same as long as we assume the choice is random. I added the probability of reaching each state at two points in time - before a door is opened, and after. Then, under "Stick" and "Swap," I carried over the probability that each choice was a winner. As you can see, the chances that swapping wins total to 2/3.

What cmb did was count the number of rows where Sticking, or Swapping, would win. That only works if there is no difference between the rows. My table shows that there is a difference. But that isn't the only mistake in this table. A better one is this, where I do switch to fixing the player's pick at door #1 while letting the car placement vary:

But I also added a new variable: Q is the probability, when the Host can open either Door #1 or #2, that he chooses #1. The chances for Staying and Swapping still sum to 1/3 and 2/3, respectively, but one piece of information is ignored. The player knows which door was opened. If it was #2, we have to update the probabilities like this:

The point is that the chances when you stay can change, but only if you know that Monty Hall chooses with a bias. If Q=1/2, which means an unbiased choice, the probabilities evaluate to 1/3 and 2/3.

 

[sysprog]

Reiterating:

 

[Moes]

I see this very simply.

The contestant chooses 1 of 3 doors. The probability of selecting the car is 1/3 and thus the probability of not selecting the car is 2/3. This can be restated as the probability that the car is behind the other doors is 2/3. Obviously there are 2 other doors. The host then reveals that there is no car behind one of the two remaining doors, thus eliminating that door as a selection.

Even at this point nothing has really changed in terms of probability. 1/3 probability that the car is behind the selected door and 2/3 that it is not. What has changed is that you are given the opportunity to select another door, whereas in the beginning there were 2 doors now there is only one and thus the probability is 2/3 that the car is behind the remaining door.

Simple.

In other words , the host, by showing you which of the 2 doors not to choose, he showed you which of the 2 to choose. So you are basically choosing 2 doors with the host just showing you which one to point to, since even if you were able to choose 2 doors from the beginning there would be one with a goat behind it. If she was able to choose 2 doors from the beginning the host showing her that one door she chose had a goat behind it obviously wouldn’t change her odds from 2/3.

Yes, it’s simple.

 

[JeffJo]

Reiterating:

View attachment 294328

Reiterating - this ignores information that the contestant has. Plus, it adds information that the contestant does not. The contestant sees what door is opened, which is what it ignores. And the contestant does not know where the car is, so your "If initial guess..." conditional cannot be evaluated by the constestant.

Maybe numbers will help Play the game 30 times. Initially, pick door #1 each time:

• 10 times, the car is behind #3 and the host opens #2.
• 10 times, the car is behind #2 and the host opens #3.
• 10 times, the car is behind #1 and the host has a choice.
  • He opens #3 N times, where 0<=N<=10.
  • He opens #2 the remaining 10-N times.

Now consider three cases:

1. If #3 is opened:
   • Staying wins N out of (10+N) times.
   • Switching wins 10 out of (10+N) times.
2. If #2 is opened:
   • Staying wins (10-N) out of (20-N) times.
   • Switching wins 10 out of (20-N) times.
3. If Monty pushes #2 and #3 together so that they are one door, goes behind it without letting you see, and comes back leading a goat on a leash:
   • Staying wins 10 out of 30 times.
   • Switching wins 20 out of 30 times.

Either case #1 or case #2 is what happened. So the correct solution has to consider something between 10 and 20 possible games, not all 30. The probability for door #1 does not have to be "1/3 because that is what it was when you initially chose door #1." But unless you have some reason otherwise, you should assume N=5, and the probability for door #1 is 1/3 because that is 5/15.

Your solution is for case #3, and it is an incorrect solution (because case #3 did not happen) that gets the right answer (because you can only assume N=5).

 

[sysprog]

And the contestant does not know where the car is, so your "If initial guess..." conditional cannot be evaluated by the constestant.

It's an 'if then else' statement $-$ you don't need to know where the car is to recognize that the statement is true.

 

[JeffJo]

Say you are on a game show, and offered the choice of four doors. Behind each is a pair of siblings - one door has a pair of brothers, one has a pair of sisters, one has a boy with a younger sister, and one has a girl with a younger brother. You will win a new car if you pick a door with only boys or only girls. You pick a door, say #1, and your chances are 1/2. But the the host, who knows what is behind the doors, tells you that door #4 was a winner, and opens it to reveal a pair of sisters. What are your chances now with door #1?

This is an identical problem to the famous "Boy or Girl Paradox" as originally posed by Martin Gardner in Scientific American. Paraphrased, that one was "You know that door #1 has two children, and that at least one is a boy. What is the probability that both are boys?" But it is now presented in a manner that is identical to the Monty Hall Problem (without the switch, which only complicates it).

• Many people will say your chances, if you stay with door #1, are 1/2. You know that one child there is a boy, so "the other child" has a 1/2 chance to also be a boy. This is an incorrect solution, because no specific child was identified as the boy.
• The solution Martin Gardner's initially gave is that only 1 in 3 of the remaining doors has only one gender, so the answer is 1/3. And that is still though to be correct by many teachers.
  • Yet this is the solution that is called incorrect for the Monty Hall Problem.
  • In fact, Gardner himself withdrew his answer for this very reason. He said the problem was not worded well, because of the "Q" I used originally (or the "N" in the numerical example).
• You could claim that your chances were initially 1/2. That can't change, so they must still be 1/2.
  • This is the solution many give for Monty Hall, and it is just as valid in my game. But I hope you find it to be less satisfying, since one of the winning doors was revealed.
• The correct answer is indeed 1/2. But it is because:
  • The host had to open the door with two sisters if you initially picked the two brothers, This is a 1/4 chance.
  • But he had a choice if you picked a mixed family. That choice was a 1/2 chance, but he could have revealed either the brother or the sisters. Making each choice a final 1/4 chance.
  • In other words, there was a 1/4 chance that you picked the brothers and the Host revealed the sisters, and a 1/4 chance that you picked a mixed family and the Host revealed the sisters.
  • The answer is (1/4)/[(1/4)+(1/4)] = 1/2.

This is why it is important to reject the solution in that picture Sysprog posted.

 

[JeffJo]

It's an 'if then else' statement $-$ you don't need to know where the car is to recognize that the statement is true.

Okay - the condition in the "If...Then...Else" statement cannot be evaluated by the contestant. The statement is indeed true, but it cannot be used by the contestant because it ignores information she possesses that can change the implied answer, and suggests she should decide based on information she does not possess.

Again - the chances for the original choice can change, based on information that is in her possession. The fact that they don't change under the most reasonable application of that information does not mean they can't change.

These two problems - Monty Hall and Two Children - have probably done more to teach incorrect conditional probability methodology to students than any other. The correct methods are not that difficult to comprehend, but they do require attention to details that sometimes seem unnecessary.

1. Create a table of every outcome that is initially possible, even those you later find out did not occur.
2. Assign a probability to each.
3. Remove the outcomes that you later learn did not happen.
4. Add up the remaining probabilities.
5. Divide these remaining probabilities by that sum, so they again add up to 1.

In the Monty Hall Problem, many students who get the wrong answer will do a better job of following these steps than most teachers who get the right answer. They say (assuming the initial choice is 31 and #3 is opened):

1. The car could be behind #1, #2, or #3.
2. Each has a probability of 1/3.
3. But door #3 does not have the car, so remove it.
4. The remaining probabilities sum to 2/3.
5. So each now becomes (1/3)/(2/3)=1/2.

The error is that there are more possible outcomes. This assumes the Host must open door #3 if it has a goat. The correct solution recognizes that which door was opened is part of the information the contestant possesses.

1. The combinations for (Car,Open) are (1,2), (1,3), (2,3), and (3,2). (Note that #1 can't be opened.)
2. The corresponding probabilities are 1/6, 1/6, 1/3, and 1/3.
3. But, for example, door #3 was opened so two of these are removed. That leaves (1,3) and (2,3).
4. The corresponding probabilities are 1/6 and 1/3, which sum to 1/2.
5. The probability for (1,3) is now (1/6)/(1/3)=1/2, and for (2,3) is now (1/3)/(1/2)=2/3.

Any other solution only serves to confuse students.

In the Two Child Problem:

1. There are six possible outcomes, that include hich gender is the one you learn is included in the family; that is, (Older, Younger; Learn). They are (B,B;B), (B,G;B), (B,G;G), (G,B;B), (G,B;G), and (G,G;G).
2. The probabilities are 1/4, 1/8, 1/8, 1/8, 1/8, and 1/4.
3. You learned about a boy, so only three remain: (B,B;B), (B,G;B), and G,B;B).
4. The corresponding probabilities are 1/4, 1/8, and 1/8 which sum to 1/2.
5. The updated probabilities are now 1/2, 1/4, and 1/4.

Many will disagree with step #2 here, claiming that the probabilities are 1/4,1/4,0,1/4,0, and 1/4. That is the same as when you assume door #3 must be opened in the MHP, and so getting the answer 1/2.

 

[sysprog]

it cannot be used by the contestant

Sure it can.

The chance that I initially picked the car is $\frac 1 3$. If I stick, my chance remains $\frac 1 3$. The chance that I didn't initially pick the car is $\frac 2 3$. If I didn't initially pick the car, and I switch, then I win the car.

 

[WWGD]

Sure it can.

The chance that I initially picked the car is $\frac 1 3$. If I stick, my chance remains $\frac 1 3$. The chance that I didn't initially pick the car is $\frac 2 3$. If I didn't initially pick the car, and I switch, then I win the car.

And (EDIT 2/3 of the time), switching kills your chances. If you unknowingly selected a goat and switch, your chances for the car are 1/2 . So 2/3 of the time, switching makes sense(*Not an argument, just sort of talking out loud).
I guess this can be made more precise with Bayesian theory or just Conditional Probability.

 

[sysprog]

And switching kills your chances. If you unknowingly selected a goat and switch, your chances for the car are 1/2 . So 2/3 of the time, switching makes sense(*Not an argument, just sort of talking out loud).
I guess this can be made more precise with Bayesian theory or just Conditional Probability.

If my initial choice wasn't the car, and I switch, then I win the car.

 

[WWGD]

And switching kills your chances. If you unknowingly selected a goat and switch, your chances for the car are 1/2 . So 2/3 of the time, switching makes sense(*Not an argument, just sort of talking out loud).
I guess this can be made more precise with Bayesian theory or just Conditional Probability.

Choosing at random with 2 Goats and 1 Car, 2/3 of those cases $pm \epsilon$, for a large-enough number of samples, you will choose a Goat G1 or G2 from within a triplet (G1, G2, Car). For each of those, you have (Car, Goat2/Goat1) left.

 

[WWGD]

If my initial choice wasn't the car, and I switch, then I win the car.

True, and if you make that choice randomly, 2/3 of the time, you will choose a goat. Each of those times you will end up with (Goat, Car) left, and will gain by switching doors. So 2/3 of the time you gain by switching. Again, not a rigorous argument, but motivation.

 

[sysprog]

What are you skeptical about? Choosing at random with 2 Goats and 1 Car, 2/3 of those cases $p.m /epsilon$, for a large-enough number of samples, you will choose a Goat G1 or G2 from within a triplet (G1, G2, Car). For each of those, you have (Car, Goat2/Goat1) left.

If I didn't initially pick the car, then I picked a goat. The host always reveals a goat. That accounts for both goats, so if I didn't initally pick the car, switching wins the car.

 

[WWGD]

If I didn't pick the car, then I picked a goat. The host always reveals a goat. That accounts for both goats, so if I didn't initally pick the car, switching wins the car.

I agree. What I mean is in the more general sense, if you're not sure you picked the goat, there is a 2/3 probability you did.