## Game Show Problem

**Moderator:** Marilyn

### An easy way to look at the problem...

Perhaps here is another way to explain why Marilyn's answer is correct that may be more obvious to many people. Compare the game as described where the host will open one losing door to a game where the player is given the opportunity to open a second door if his first door is a dud (and the host does not select a door). The only difference between these two scenarios is that the host by definition will open a losing door. When the player gets to open two doors we can see clearly that the odds are 2/3. The original problem is just a variation of this problem.

Mike

- Mike12061
- Thinker
**Posts:**1**Joined:**Sat Aug 05, 2006 7:22 pm**Location:**East Greenbuh, NY

### Re: An easy way to look at the problem...

Mike12061 wrote:Perhaps here is another way to explain why Marilyn's answer is correct that may be more obvious to many people. Compare the game as described where the host will open one losing door to a game where the player is given the opportunity to open a second door if his first door is a dud (and the host does not select a door). The only difference between these two scenarios is that the host by definition will open a losing door. When the player gets to open two doors we can see clearly that the odds are 2/3. The original problem is just a variation of this problem.

Right! By being allowed to switch and the host always choosing a losing door, the player is effectively being the choice of two doors: the two other than his original selection. He takes the better of the two.

Kyle

- kyle
- Intellectual
**Posts:**124**Joined:**Wed May 10, 2006 4:09 pm**Location:**Louisiana

I first read about this problem at straightdope.com, and it makes perfect sense. I've tried explaining it to others with similar disbelief. I considered illustrating it with the computer and finally decided to.

After starting the script it became clear just how simple the problem really is. (Clear to me anyway. I could be babbling.)

--

The problem used is the classic. The host knows where the prize is and will never open that door. You get the chance to switch and (obviously) will never pick the hosts door.

Thus leaving only the two possibilities:

If you pick the prize on the first try you win by staying. Otherwise you will always win by switching.

Marilyn's diagrams demonstrate those two scenarios.

--

Start out both stayWins and switchWins to 0.

Pick a large number of random pairs of numbers from 1 to 3, pick and prize.

if pick = prize

stayWins := stayWins + 1

else

switchWins := switchWins + 1

You don't have to pick the door for the host because that will always be a booby prize. The outcome is always a switchWin if you guess wrong on the first try.

It could be further reduced to pick always being 1. The odds of picking the prize on the first try are still 1 in 3.

--

Thinking about how to write the script seems even more clear than thinking of statistics and odds. For me at least.

It isn't very surprising, but the result was:

Staying won 3331 times

Switching won 6669 times

After starting the script it became clear just how simple the problem really is. (Clear to me anyway. I could be babbling.)

--

The problem used is the classic. The host knows where the prize is and will never open that door. You get the chance to switch and (obviously) will never pick the hosts door.

Thus leaving only the two possibilities:

If you pick the prize on the first try you win by staying. Otherwise you will always win by switching.

Marilyn's diagrams demonstrate those two scenarios.

--

Start out both stayWins and switchWins to 0.

Pick a large number of random pairs of numbers from 1 to 3, pick and prize.

if pick = prize

stayWins := stayWins + 1

else

switchWins := switchWins + 1

You don't have to pick the door for the host because that will always be a booby prize. The outcome is always a switchWin if you guess wrong on the first try.

It could be further reduced to pick always being 1. The odds of picking the prize on the first try are still 1 in 3.

--

Thinking about how to write the script seems even more clear than thinking of statistics and odds. For me at least.

It isn't very surprising, but the result was:

Staying won 3331 times

Switching won 6669 times

- klhuillier
- Thinker
**Posts:**2**Joined:**Wed Aug 09, 2006 6:45 pm

Imagine my amusement to see today's straightdope.com article is...

«

Today's Question: On "Let's Make a Deal," you pick Door #1. Monty opens Door #2--no prize. Do you stay with Door #1 or switch to #3? (A Straight Dope Classic by Cecil Adams)

»

Cece made excellent points. His initial answer, that the odds are 1/2 to switch if the host opens a door, were based on the show format where the host does not always open a door. Adding the constraint that the host always opens a door, the odds work out to 2/3.

http://www.straightdope.com/classics/a3_189.html

I modified my script to see how Cece's rules change the odds. I don't know how the show ran, but i assumed you could still switch doors if he doesn't open one. With these rules it comes close to even.

Keeping the 2/3 odds in mind, the host would tend to open a door more often when the contestant picks the prize. Switching is a 2/3 chance, right? What contestant could resist?

Here were my rules: Roll a die. If the contestant chose incorrectly, open a door if you rolled 1. If the contestant chose correctly, open a door if you rolled 1, 2, or 3.

That reduced the winning moves of 10,000 games to ...

Stay and win 3331 times (no surprise)

Switch and win 3865 times

I ran several games and the mean for switching seems to be around 3875.

Not _quite_ 50/50 for the contestant. It still favors switching, but only slightly.

It makes me wonder how the real show worked out.

«

Today's Question: On "Let's Make a Deal," you pick Door #1. Monty opens Door #2--no prize. Do you stay with Door #1 or switch to #3? (A Straight Dope Classic by Cecil Adams)

»

Cece made excellent points. His initial answer, that the odds are 1/2 to switch if the host opens a door, were based on the show format where the host does not always open a door. Adding the constraint that the host always opens a door, the odds work out to 2/3.

http://www.straightdope.com/classics/a3_189.html

I modified my script to see how Cece's rules change the odds. I don't know how the show ran, but i assumed you could still switch doors if he doesn't open one. With these rules it comes close to even.

Keeping the 2/3 odds in mind, the host would tend to open a door more often when the contestant picks the prize. Switching is a 2/3 chance, right? What contestant could resist?

Here were my rules: Roll a die. If the contestant chose incorrectly, open a door if you rolled 1. If the contestant chose correctly, open a door if you rolled 1, 2, or 3.

That reduced the winning moves of 10,000 games to ...

Stay and win 3331 times (no surprise)

Switch and win 3865 times

I ran several games and the mean for switching seems to be around 3875.

Not _quite_ 50/50 for the contestant. It still favors switching, but only slightly.

It makes me wonder how the real show worked out.

- klhuillier
- Thinker
**Posts:**2**Joined:**Wed Aug 09, 2006 6:45 pm

### Boys and Girls

niles wrote:Vosh,

I'm not quite sure how to respond to this. My purpose was to present my analysis of why so many people in the field of probability were at odds with Marilyn. My analysis, which is merely my opinion, is that they misinterpreted the problem and solved a similar but different problem. Clearly you believe that my explanation is a "stretch" and if so I am sure you have the ability to present a well-reasoned argument which is supported by more than an "everyone can see" assertion.

I don't agree with your assessment of my response as "non-conciliatory" since I stated quite clearly that Marilyn had correctly analyzed the first problem. I think that it is clear without my saying so that although the problem was ill-defined, Marilyn immediately clarified her assumptions and proceeded correctly based upon them. On the other hand, some people who work in the field of probability have a predisposition to view problems as conditional probability problems, and would tend to formulate and analyze the alternate problem (where the door is opened at random, and the player has conditional knowledge of that outcome), stopping only to make sure that the original statement of the problem does not exclude that interpretation.

As for ego, certainly there is no shortage of ego in the mean-spirited responses which Marilyn chose to highlight in her posting. Marilyn exhibits extraordinary grace in her eloquent and straightforward response to those criticisms, with never a word of condescension or anger. Although I cannot hope to attain Marilyn's high genius, I can certainly try to learn from her ego-free demeanor. I would never expect to read the "r" word in any of Marilyn's responses, and I question why you would use it, negated or not, to describe me or my comments on those who misinterpreted the problem.

The only reason I decided to comment was that I read the thread, and it seemed to me that people were happily jumping to the conclusion that all of those who wrote in to disagree with Marilyn in 1990-91 were incompetent in spite of whatever impressive titles or degrees they might have. The comments were overwhelmingly of the "Marilyn was right, PhD's were wrong, ha, ha" genre. I thought, perhaps incorrectly, that a little bit of reasoned analysis might be welcome.

Niles,

Marilyn was right about the goats, but wrong about the probability for two boys. She answered 1/3, she did not mention ambiguity.

Your paraphrase gave a scenario. When you give a scenario, it alters the question, sometime that alters the answer. Marilyn's question had no stated scenario, "The woman has two children and at least one is a boy." That's all, just a statement about a woman.

In your scenario, you meet a woman with a son in a stroller. Then you say that, because we don't know the probability that a woman with one of each would bring a son, the problem is ambiguous.

The probability that a woman with one of each would bring a son is a number between zero and one. Us not knowing that number, does not make the problem ambiguous.

Eldon:)

Challenged Marilyn in 1997 on the sex of the other sibling question. Told her I would put $1000 into her favorite charity if my argument was wrong. She said, "You're wrong, send the money to the American Heart Association."

- Elmo
- Intellectual
**Posts:**722**Joined:**Thu Aug 10, 2006 4:42 pm**Location:**Abilene Texas

### Ambiguity

Don't even try to answer that without further clarification. It is AMBIGUOUS! The answer depends on how she chose which of her two children to take in the stroller! There are at least two valid interpretations of this question, each of which has a different answer. I'll do my best to quickly analyze each.

Niles,

The probability that a woman with one of each would select a son is a number between zero and one. Not knowing that number does not make the question ambiguous. In Marilyn's question, she did not plead ambiguity, she opted for 1/3. She was wrong.

Elmo

Niles,

The probability that a woman with one of each would select a son is a number between zero and one. Not knowing that number does not make the question ambiguous. In Marilyn's question, she did not plead ambiguity, she opted for 1/3. She was wrong.

Elmo

Challenged Marilyn in 1997 on the sex of the other sibling question. Told her I would put $1000 into her favorite charity if my argument was wrong. She said, "You're wrong, send the money to the American Heart Association."

- Elmo
- Intellectual
**Posts:**722**Joined:**Thu Aug 10, 2006 4:42 pm**Location:**Abilene Texas

Hi Eldon, nice to have you here, and I mean that.

When I wrote my original post several weeks back, I had no idea who you, Eldon, were. I had no idea of the original wording of the "two child" problem. My exposure to the controversy in which you were involved was indirect, and I was relying on my memory going back almost a decade. In the late nineties, I had a subscription to a financial newsletter, and one of the columnists decided to use the "two child" problem to illustrate some point about how our intuition is often wrong. However, when he paraphrased it, he reformulated it. He told us that despite our intuition, if a woman known to have two children is observed strolling her son in the park, the probability that her other child is a son is only 1/3. Personally I think it is obvious that the scenario of observing a woman strolling a son is NOT equivalent to the scenario where "a woman is selected at random from the population of women who have two children, at least one of whom is a boy." I wasn't alone, since he received quite a bit of mail, to which he replied only that "reasonable minds can disagree."

Personally, I don't think reasonable minds can disagree on the problem the way he posed it. I originally stated that the problem was ambiguous, and you quoted me. I would like to point out that after I wrote what you quoted, I discussed it with a somewhat better educated colleague who pointed out the "principle of indifference" in probability which states something like in the absence of information to the contrary, the choice should be random. So when you say, "us not knowing the number, doesn't make the problem ambiguous," I agree, and furthermore, by the principle of indifference, the probability must be 1/2, from which you can quickly deduce that the financial columnist has posed a problem for which the answer is 1/2.

It is similar to the problem where I roll a pair of fair dice, where one drops in front of me showing a two, and the other drops out of my sight. What is the probability that I have rolled a seven? Would you say 2/11 because that is the conditional probability of rolling a seven given that at least one die shows a two? Of course not, the odds are 1/6, and few (I hope) will argue.

After I wrote that original post which you quoted, I learned about your $1000 offer (notice that I carefully did not call it a bet) that Marilyn summarily dealt with. I read your logic carefully. I also read the original article and problem statement, and found out that it had nothing to do with observing parents strolling children. In short, I am better informed now than when I wrote in early July.

Obviously, we tread on thin ice to discuss this well worn issue in Marilyn's own forum. If that is what you are here to do, and you want someone to bounce your ideas off of, I will happily provide a dialogue for you within the rules of the forum. After all, this is a discussion forum, and nowhere are we required to be mind-numbed robots. The issue between you and Marilyn is a tough one, and I'd certainly like to be part of a polite discussion on the topic. If so, maybe you should start a new thread for it.

Elmo wrote:In your scenario, you meet a woman with a son in a stroller. Then you say that, because we don't know the probability that a woman with one of each would bring a son, the problem is ambiguous.

The probability that a woman with one of each would bring a son is a number between zero and one. Us not knowing that number, does not make the problem ambiguous.

When I wrote my original post several weeks back, I had no idea who you, Eldon, were. I had no idea of the original wording of the "two child" problem. My exposure to the controversy in which you were involved was indirect, and I was relying on my memory going back almost a decade. In the late nineties, I had a subscription to a financial newsletter, and one of the columnists decided to use the "two child" problem to illustrate some point about how our intuition is often wrong. However, when he paraphrased it, he reformulated it. He told us that despite our intuition, if a woman known to have two children is observed strolling her son in the park, the probability that her other child is a son is only 1/3. Personally I think it is obvious that the scenario of observing a woman strolling a son is NOT equivalent to the scenario where "a woman is selected at random from the population of women who have two children, at least one of whom is a boy." I wasn't alone, since he received quite a bit of mail, to which he replied only that "reasonable minds can disagree."

Personally, I don't think reasonable minds can disagree on the problem the way he posed it. I originally stated that the problem was ambiguous, and you quoted me. I would like to point out that after I wrote what you quoted, I discussed it with a somewhat better educated colleague who pointed out the "principle of indifference" in probability which states something like in the absence of information to the contrary, the choice should be random. So when you say, "us not knowing the number, doesn't make the problem ambiguous," I agree, and furthermore, by the principle of indifference, the probability must be 1/2, from which you can quickly deduce that the financial columnist has posed a problem for which the answer is 1/2.

It is similar to the problem where I roll a pair of fair dice, where one drops in front of me showing a two, and the other drops out of my sight. What is the probability that I have rolled a seven? Would you say 2/11 because that is the conditional probability of rolling a seven given that at least one die shows a two? Of course not, the odds are 1/6, and few (I hope) will argue.

After I wrote that original post which you quoted, I learned about your $1000 offer (notice that I carefully did not call it a bet) that Marilyn summarily dealt with. I read your logic carefully. I also read the original article and problem statement, and found out that it had nothing to do with observing parents strolling children. In short, I am better informed now than when I wrote in early July.

Obviously, we tread on thin ice to discuss this well worn issue in Marilyn's own forum. If that is what you are here to do, and you want someone to bounce your ideas off of, I will happily provide a dialogue for you within the rules of the forum. After all, this is a discussion forum, and nowhere are we required to be mind-numbed robots. The issue between you and Marilyn is a tough one, and I'd certainly like to be part of a polite discussion on the topic. If so, maybe you should start a new thread for it.

- niles
- Scholar
**Posts:**45**Joined:**Thu Jul 06, 2006 2:32 am

This is interesting, but I have no idea what you guys are on about. Is this stroller problem posted somewhere? Many thanks.

"A new scientific idea does not triumph by convincing its opponents and making them see the light, but rather because its opponents eventually die and a new generation grows up that is familiar with it." -Max Planck

- Vosh
- Intellectual
**Posts:**389**Joined:**Sun May 28, 2006 1:49 pm**Location:**Earth, just visiting on my way somewhere...

### Two child problem

Vosh,

The woman with a child in a stroller was introduced earlier. I don't know where it went, or how to find it.

Niles,

I need special access to start a new topic. I don't have it. I agree that when a woman is selected from a group whereas the statement would be true, then her chances are 1/3.

What Marilyn did, she asked one question, then answered another. "A woman has two children and at least one is a boy." There is nothing in that statement about selection of the woman. All we know is that the statement was made about the woman. That is different information than a woman who was chosen from a select group.

The key to the goat in the gameshow question is understanding what the host did. He knew where the car was, he opened a goat door on purpose, the odds are to switch.

The key to the two child question is when was the boy/girl decision made. Boys girls start out equally likely. They stay equally likely until one or the other is chosen. Our woman has bb and the statement was made, or bg and the statement was made, or gb and the statement was made. When the choosing of boy was after inspection, then the answer is 1/2.

Marilyn had a similar question one time about two puppy dogs, she got it right. She had a similar question pitching pancakes from a hat. She got it right. The one about the woman with two boys, she got wrong.

I'm willing to operate within the rules of the forum. There should be someway to get this question in it's own thread. Maybe Vosh can figure it out.

Elmo

The woman with a child in a stroller was introduced earlier. I don't know where it went, or how to find it.

Niles,

I need special access to start a new topic. I don't have it. I agree that when a woman is selected from a group whereas the statement would be true, then her chances are 1/3.

What Marilyn did, she asked one question, then answered another. "A woman has two children and at least one is a boy." There is nothing in that statement about selection of the woman. All we know is that the statement was made about the woman. That is different information than a woman who was chosen from a select group.

The key to the goat in the gameshow question is understanding what the host did. He knew where the car was, he opened a goat door on purpose, the odds are to switch.

The key to the two child question is when was the boy/girl decision made. Boys girls start out equally likely. They stay equally likely until one or the other is chosen. Our woman has bb and the statement was made, or bg and the statement was made, or gb and the statement was made. When the choosing of boy was after inspection, then the answer is 1/2.

Marilyn had a similar question one time about two puppy dogs, she got it right. She had a similar question pitching pancakes from a hat. She got it right. The one about the woman with two boys, she got wrong.

I'm willing to operate within the rules of the forum. There should be someway to get this question in it's own thread. Maybe Vosh can figure it out.

Elmo

Challenged Marilyn in 1997 on the sex of the other sibling question. Told her I would put $1000 into her favorite charity if my argument was wrong. She said, "You're wrong, send the money to the American Heart Association."

- Elmo
- Intellectual
**Posts:**722**Joined:**Thu Aug 10, 2006 4:42 pm**Location:**Abilene Texas

### Re: Two child problem

To the best of my knowledge, the "stroller" version of this problem never appeared in Marilyn's column. You can check out Elmo's "www" link which paraphrases the question as it appeared in Marilyn's column.Elmo wrote:Vosh,

The woman with a child in a stroller was introduced earlier. I don't know where it went, or how to find it.

There are non-controversial ways to state the problem, such as "You meet a man. You ask him if he has any boys. He answers 'Yes.' You ask him how many children he has. He answers, 'Two.' What is the probability that both of his children are boys? Answer: 1/3." I think even Elmo will agree with this, because we decided on boys prior to inspection, i.e. we decided to ask him about boys, not knowing anything about his children.

As I mentioned above, I remember reading a financial columnist who, after reading the problem in Marilyn's column, went on to incorrectly deduce that if you observe a woman known to have two children strolling her son in the park, then the probability that her other child is also a son is 1/3. This is obviously false. It would be true, for instance, if women with both a son and a daughter always took their sons out in the stroller, and never their daughters. In that case, you would see many more sons than daughters overall.

I could not find the original article on the Internet. It was a paid subscription, and although it may be in their archives, those archived articles definitely do not show up in search engines. Anyway, it is not relevant to the discussion, except to show how valid results can sometimes be misinterpreted by otherwise intelligent people.

Elmo wrote:Niles,

I need special access to start a new topic. I don't have it.

I believe ordinary members need to go to the General Discussion area to start threads.

Elmo wrote:I agree that when a woman is selected from a group whereas the statement would be true, then her chances are 1/3.

What Marilyn did, she asked one question, then answered another. "A woman has two children and at least one is a boy." There is nothing in that statement about selection of the woman. All we know is that the statement was made about the woman. That is different information than a woman who was chosen from a select group.

The key to the goat in the gameshow question is understanding what the host did. He knew where the car was, he opened a goat door on purpose, the odds are to switch.

The key to the two child question is when was the boy/girl decision made. Boys girls start out equally likely. They stay equally likely until one or the other is chosen. Our woman has bb and the statement was made, or bg and the statement was made, or gb and the statement was made. When the choosing of boy was after inspection, then the answer is 1/2.

As I have said before, the mathematics behind the problem are not difficult. The controversy is in choosing the proper interpretation of the problem. We have the "before inspection" version of this problem, which leads to 1/3, as illustrated in my "non-controversial" formulation at the beginning of this post. We have the "after inspection" version of the problem, whereby an observer decides to make either the statement "at least one boy" or "at least one girl" after observing one or both of the children. This version leads to the answer 1/2.

So what we're going to discuss here, essentially, are the merits of whether the "before inspection" or "after inspection" interpretation of the problem is better. Elmo has stated that the only valid interpretation of this problem is after inspection. Marilyn apparently believes (as inferred from her answer in the column and her published declaration that Elmo owes $1000 to a charity) that the only valid interpretation of the problem is before inspection.

Personally, I'm on the fence. I could argue either way. There have been times when I have leaned one way or the other. What bothers me most is that people pose these problems in imprecise form.

Elmo wrote:Marilyn had a similar question one time about two puppy dogs, she got it right. She had a similar question pitching pancakes from a hat. She got it right. The one about the woman with two boys, she got wrong.

I'm willing to operate within the rules of the forum. There should be someway to get this question in it's own thread. Maybe Vosh can figure it out.

Elmo

I know the pancakes one. I'm curious about the puppy dogs, though.

- niles
- Scholar
**Posts:**45**Joined:**Thu Jul 06, 2006 2:32 am

### Stating the question

Niles said:

There are non-controversial ways to state the problem, such as "You meet a man. You ask him if he has any boys. He answers 'Yes.' You ask him how many children he has. He answers, 'Two.' What is the probability that both of his children are boys? Answer: 1/3." I think even Elmo will agree with this, because we decided on boys prior to inspection, i.e. we decided to ask him about boys, not knowing anything about his children.

Niles,

The problem is stated in the problem statement. The problem statement states the entire problem. When you say, "You meet a man", you are stating a different question.

Our question states, " A woman has two children and at least one is a boy." That's all we know about her. We know that the statement was made about her.

Woman number two was selected from a group of women for whom the statement would be true. That's different information.

Our woman: We know that she has two boys, and the statement was made about her, OR, she has one of each and the statement was made about her.

With boys and girls equally likely, a woman with bg is not as likely to have the "boys" statement said about her as a woman with bb.

To get the 1/3 answer, the probability for a boys statement is one at bb, and bg, and gb.

When all we know is that the woman appeared, she had gb, and the boys statement was made, then the probability for a boys statement was 1/2. When this is true, then the probability for two boys is 1/2.

Some facts:

"Do you have at least one girl?" The asker selected 'girl', prior to inspection. The answerer of that question must inspect both children before answering "no". Could answer "yes" with maybe one, sometime two.

"There are two lights, at least one is red." This statement gives no evidence of, knowledge of, the other light.

"At least one is a boy". The maker of that statement gives no evidence of knowledge of the other boy.

"At least one is" does not communicate prior selection.

Elmo

There are non-controversial ways to state the problem, such as "You meet a man. You ask him if he has any boys. He answers 'Yes.' You ask him how many children he has. He answers, 'Two.' What is the probability that both of his children are boys? Answer: 1/3." I think even Elmo will agree with this, because we decided on boys prior to inspection, i.e. we decided to ask him about boys, not knowing anything about his children.

Niles,

The problem is stated in the problem statement. The problem statement states the entire problem. When you say, "You meet a man", you are stating a different question.

Our question states, " A woman has two children and at least one is a boy." That's all we know about her. We know that the statement was made about her.

Woman number two was selected from a group of women for whom the statement would be true. That's different information.

Our woman: We know that she has two boys, and the statement was made about her, OR, she has one of each and the statement was made about her.

With boys and girls equally likely, a woman with bg is not as likely to have the "boys" statement said about her as a woman with bb.

To get the 1/3 answer, the probability for a boys statement is one at bb, and bg, and gb.

When all we know is that the woman appeared, she had gb, and the boys statement was made, then the probability for a boys statement was 1/2. When this is true, then the probability for two boys is 1/2.

Some facts:

"Do you have at least one girl?" The asker selected 'girl', prior to inspection. The answerer of that question must inspect both children before answering "no". Could answer "yes" with maybe one, sometime two.

"There are two lights, at least one is red." This statement gives no evidence of, knowledge of, the other light.

"At least one is a boy". The maker of that statement gives no evidence of knowledge of the other boy.

"At least one is" does not communicate prior selection.

Elmo

- Elmo
- Intellectual
**Posts:**722**Joined:**Thu Aug 10, 2006 4:42 pm**Location:**Abilene Texas

### Re: Stating the question

Elmo wrote:The problem is stated in the problem statement. The problem statement states the entire problem. When you say, "You meet a man", you are stating a different question.

Our question states, " A woman has two children and at least one is a boy." That's all we know about her. We know that the statement was made about her.

Woman number two was selected from a group of women for whom the statement would be true. That's different information.

I'll agree that the information is different when I say, "I meet a man." That is why I phrased it that way.

When the question states, "A woman has two children and at least one is a boy," then we have to ask, "As opposed to what?"

The way I phrased the question (ask the woman whether she has at least one son), it is clear that the information is given "as opposed to not having at least one son." Based on that question, we've filtered out all of the "GG" cases and only those cases, leaving only "BG" "GB" and "BB", from which the answer of 1/3 follows.

Elmo wrote:Our woman: We know that she has two boys, and the statement was made about her, OR, she has one of each and the statement was made about her.

With boys and girls equally likely, a woman with bg is not as likely to have the "boys" statement said about her as a woman with bb.

To get the 1/3 answer, the probability for a boys statement is one at bb, and bg, and gb.

When all we know is that the woman appeared, she had gb, and the boys statement was made, then the probability for a boys statement was 1/2. When this is true, then the probability for two boys is 1/2.

Continuing with the "as opposed to what" question which I raised above, you now (correctly) illustrate that the question could also be interpreted "she has at least one son as opposed to having at least one daughter." In other words, if the woman has two daugters, it might not make sense to tell you how many sons she has, so the observer would say, "at least one daughter" in that case.

My question to you, is why don't you just state that the question is ambiguous or ill posed? Instead you seem to be insisting that the interpretation which gives 1/3 is not a valid interpretation.

Remember, I'm on the fence here. So now I will shift gears and play devil's advocate for the remainder. (This is of course not to imply that Marilyn is the devil )

But the answerer of the question of course knows what both of her children are, so what is the point?Elmo wrote:Some facts:

"Do you have at least one girl?" The asker selected 'girl', prior to inspection. The answerer of that question must inspect both children before answering "no". Could answer "yes" with maybe one, sometime two.

It simply states at least one is red. There is no concept of "the other light." Why are we not to take the statement at face value? For instance, there are two lights, which can either be red or off. I can't see the lights, but I can see a red glow on a distant wall, from which I correctly deduce that at least one is red. That's all it means. How I got the information is irrelevant, but since you insist that there must be an "inspection," I came up with a sensible way in which the information could have been obtained, and I didn't have to inspect either or both lights. Now, assuming each light is either on or off independently with probability 1/2, what is the probability that both lights are on? I think you'll agre it is 1/3.Elmo wrote:"There are two lights, at least one is red." This statement gives no evidence of, knowledge of, the other light.

Elmo wrote:"At least one is a boy". The maker of that statement gives no evidence of knowledge of the other boy.

Again, where is "the other boy?" There is no concept of "the other boy." By making the "at least one is" statement, we have not distinguished any boy from "the other."

Elmo wrote:"At least one is" does not communicate prior selection.

Elmo

It can. If your very own example involving the red light didn't convince you, I'll leave you with one final example, from Mendelian genetics, analogous to the original problem.

There is a trait in humans called the Widow's peak. It is a V-shaped formation in the hairline directly above the nose. You either have it, or you don't.

This trait is controlled by a pair of genes that each of us has. As is usual in Mendelian genetics, let's call the gene for widow's peak T, and and the gene for absence of the widows peak t.

We each have a pair of these genes; one inherited from our father, the other from our mother. The pairs can be either TT, Tt, tT, or tt. Let's assume that that the first item of the pair is the one inherited from the father, and the second item of the pair is the one inherited from the mother.

The widow's peak is a dominant trait. A person will have the widow's peak if they have at least one T gene. Specifically, TT, Tt, and tT combinations will produce a widow's peak, while tt will not. If T and t genes are randomly and independently present throughout a population, then we would expect each of the four combinations to be equally likely, and consequently 75% of the people would have a widow's peak, and 25% would not have one.

As Mendel discovered, you randomly inherit one of your father's genes, and one of your mother's genes. If a person has TT or tt, he is called pure. If a person has Tt or tT, he is called hybrid. If both of a person's parents are hybrid, it follows that it is equally likely for the person to be TT, Tt, tT, or tt.

Now for the question. A man and a woman (unrelated) each have both parents hybrid for the Widow's peak. The woman has the widow's peak, while the man is known to have inherited the T gene from his father. Do they have equal chances to be pure for the T gene?

Note that the distribution of zero, one, or two T genes is exactly the same as the distribution of zero, one, or two sons in the original problem.

Note that the statement, "the woman has a widow's peak" is exactly equivalent to "the woman has at least one T gene" and is analogous to saying "the woman has at least one boy" in the original problem.

Note that the statement "the man inherited the T gene from his father" is analogous to saying, "the man's oldest son is a boy," in other words, one specific unknown has been revealed.

Why should the answer to the "boys" problem be any different from the answer to the "widows peak" problem?

- niles
- Scholar
**Posts:**45**Joined:**Thu Jul 06, 2006 2:32 am

Of course the examples are analogous and the probability is 1/3,

not 1/2. Let's try it another way.

There are two regular six-sided fair dice, one blue, one purple.

Roll them randomly and don't look. I'LL look.

I TELL YOU (honestly) that the case ODD-ODD did NOT occur.

So WE ALL can exclude that case.

Now I ask you to pick one die randomly (left hand or right hand)

and I TELL you (honestly) whether that die came up EVEN or ODD.

Say, for arguments sake, that I say EVEN.

THEN THE PROBABILITY that EVEN-EVEN occurred is not 1/4,

not 1/2, but 1/3 !!!!!!!!!!

This is called SAMPLE SPACE ANALYSIS.

Since the FOUR cases EE,EO,OE,OO are all equally likely,

when I EXCLUDE one of them, the other three remain

equally likely AS EACH OTHER. But their probabilities

MUST STILL ADD to ONE. So P+P+P == 1 --> P == 1/3.

And that's exactly what Marilyn said.

not 1/2. Let's try it another way.

There are two regular six-sided fair dice, one blue, one purple.

Roll them randomly and don't look. I'LL look.

I TELL YOU (honestly) that the case ODD-ODD did NOT occur.

So WE ALL can exclude that case.

Now I ask you to pick one die randomly (left hand or right hand)

and I TELL you (honestly) whether that die came up EVEN or ODD.

Say, for arguments sake, that I say EVEN.

THEN THE PROBABILITY that EVEN-EVEN occurred is not 1/4,

not 1/2, but 1/3 !!!!!!!!!!

This is called SAMPLE SPACE ANALYSIS.

Since the FOUR cases EE,EO,OE,OO are all equally likely,

when I EXCLUDE one of them, the other three remain

equally likely AS EACH OTHER. But their probabilities

MUST STILL ADD to ONE. So P+P+P == 1 --> P == 1/3.

And that's exactly what Marilyn said.

- davar55
- Intellectual
**Posts:**565**Joined:**Tue Jun 13, 2006 4:24 pm**Location:**New York City

Thanks, davar55, for joining in.

Remember, I'm on the fence, and I'm playing devil's advocate

Tell me, what would you have done if ODD-ODD had occurred?

Some people interpret your scenario as, "ODD-ODD is written in stone. If ODD-ODD occurs, we'll roll again." This makes the answer 1/3.

Some people interpret your scenario as, "He rolled the dice, and now I have to tell him something meaningful about his roll. Since I have told him ODD-ODD on this occasion, it must be that I'll tell him EVEN-EVEN on other occasions, since I have no reason to distinguish ODD from EVEN. If I have a choice, i.e. ODD-EVEN occurred, then I will choose one randomly. This makes the answer 1/2.

Seems to me we have a situation analogous to the Monty Hall problem where the host protocol is undefined. We have seen that if Monty chooses the door to open randomly, the odds of winning by switching are only 1/2, as compared to him always opening a door with a goat, where the probability of winning by switching is 2/3. So we must make it clear that the host protocol includes always opening a door with a goat. (I don't want to discuss MH any more, so please don't interpret me as criticizing Marilyn's answer here.)

In other words, don't we need to firm up the "host protocol" for your actions before we can decide between 1/3 and 1/2?

Remember, I'm on the fence, and I'm playing devil's advocate

Tell me, what would you have done if ODD-ODD had occurred?

Some people interpret your scenario as, "ODD-ODD is written in stone. If ODD-ODD occurs, we'll roll again." This makes the answer 1/3.

Some people interpret your scenario as, "He rolled the dice, and now I have to tell him something meaningful about his roll. Since I have told him ODD-ODD on this occasion, it must be that I'll tell him EVEN-EVEN on other occasions, since I have no reason to distinguish ODD from EVEN. If I have a choice, i.e. ODD-EVEN occurred, then I will choose one randomly. This makes the answer 1/2.

Seems to me we have a situation analogous to the Monty Hall problem where the host protocol is undefined. We have seen that if Monty chooses the door to open randomly, the odds of winning by switching are only 1/2, as compared to him always opening a door with a goat, where the probability of winning by switching is 2/3. So we must make it clear that the host protocol includes always opening a door with a goat. (I don't want to discuss MH any more, so please don't interpret me as criticizing Marilyn's answer here.)

In other words, don't we need to firm up the "host protocol" for your actions before we can decide between 1/3 and 1/2?

- niles
- Scholar
**Posts:**45**Joined:**Thu Jul 06, 2006 2:32 am

I created the problem, I am the host.

In the MHP, Monty Hall created the show,

HE was the host, he had his own protocol.

REMEMBER THE ORIGINAL QUESTION.

Marilyn solved it instantaneously and correctly.

All the rest of the disagreement,

from ordinary readers/people to

eminent mathematicians,

to those who doubt or debate it here,

are due to misreading, misunderstanding,

or ignoring the question

Marilyn Vos Savant was asked.

Controversy be damned.

In the MHP, Monty Hall created the show,

HE was the host, he had his own protocol.

REMEMBER THE ORIGINAL QUESTION.

Marilyn solved it instantaneously and correctly.

All the rest of the disagreement,

from ordinary readers/people to

eminent mathematicians,

to those who doubt or debate it here,

are due to misreading, misunderstanding,

or ignoring the question

Marilyn Vos Savant was asked.

Controversy be damned.

- davar55
- Intellectual
**Posts:**565**Joined:**Tue Jun 13, 2006 4:24 pm**Location:**New York City

Return to Online Articles By Marilyn

### Who is online

Users browsing this forum: No registered users and 1 guest