B The Monty Hall paradox/conundrum

[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.

 

[sysprog]

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.

Yes. So if I didn't initially pick the car, then switching wins the car.

 

[WWGD]

Yes. So if I didn't initially pick the car, then switching wins the car.

Yes, I think we're both saying the same in different ways.

 

[PeterDonis]

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