## Thursday, August 21, 2008

### Sometimes We Do Agree...

Not often enough do opposing views come together in agreement but I did experience such a phenomenon w/ some strangers from Drexel while on vacation - in between reading Mark Garcia's book and tooling around Martha's Vineyard with the family.

For those who enjoy probability and like to see resolution to apparent disagreements, read on.

Date: 07/24/2008 at 09:28:28
From:  Ron DiGiacomo
To: dr.math@mathforum.org
Subject: Boy Girl probability

[Question]
I have a problem w/ this formulation (link below) in which you conclude that if a boy is selected from a family of two children that the probability of the remaining child being a girl is 2/3 http://mathforum.org/library/drmath/view/52186.html

[Difficulty]
In your solution you note four possible combinations of siblings and eliminate one of the possibilities (gg) because a boy was selected. That much is correct. The conclusion, however, would seem to ignore the fact that two of three remaining combinations are mutually exclusive, making one of them as impossible as set (gg), which was eliminated. In other words, given two mutually exclusive remaining sets of children, one is false and is also to be eliminated. See my solution below:

[Thoughts]
You correctly note that the set of (gg) is to be eliminated upon a boy being selected. That leaves three possible groups from which the boy was selected:

(1)bb
(2)gb
(3)bg

Although we don't know the age of the selected boy relative to his sibling (i.e. whether he's older or younger), we do know that the selected boy is *either* older or younger. If older, then (2) is eliminated. If younger, then (3) is eliminated. Consequently, from the entire sample of three possible combinations of siblings (even without knowing whether the selected boy is an older or younger sibling), one of the boy-girl combinations is to be eliminated because it *must* be false. That would make the probability ½ of the other sibling being a girl given a selected boy.

Somewhere in the thread it is correctly noted that more information can change the probability. What is missed, however, is that one need not know whether the the sibling is older or younger in order to know that it he must be older or younger. That he must be older or younger is information-enough to change the sample.

Ron

________________________________________
Date: 07/24/2008 at 13:15:11
From: Doctor Anthony
To: xxxx@comcast.net (Ron DiGiacomo)
Subject: Re: Boy Girl probability

remaining MUST be wrong. That doesn't mean that we know which two. Instead of reasoning in the manner you described from the archive consider the following argument:

We have a population of 2-child families and one such family is questioned. The information is that AT LEAST one child is a boy.

The sample space = 1 - Prob(2 girls)

= 1 - (1/2 x 1/2)

= 3/4

1/2 x 1/2 1/4
Prob(2 boys1 child is a boy) = ---------- = -----
3/4 3/4

= 1/3

and the probability of anything else is 2/3

- Doctor Anthony, The Math Forum

The Math Forum @ Drexel is a research and educational enterprise
of the Drexel School of Education:

________________________________________
Date: 07/24/2008 at 18:42:48
From: @comcast.net (Ron DiGiacomo)
To: dr.math@mathforum.org
Subject: Boy Girl probability

We started with four combinations of siblings (let’s assume all
sitting in a room). Those combinations can be written as follows:

(1)bb
(2)gb
(3)bg
(4)gg

Then one child was randomly selected and it turned out to be a boy.

Consequently, we know that selected sibling did not come from sibling - set #4.

We now know that the three possible combinations of siblings, given the impossibility of ‘gg’ due to a boy having been already randomly selected, are as follows:

(1)bb
(2)gb
(3)bg

The randomly selected boy is one of the boys found within the three sets listed above.

To simplify matters we may rewrite the three possible sets as follows:

(1) Bob, Bill
(2) Gayle, Brian
(3) Bart, Gabriella

NOTE: With the same probably of 25%, the randomly selected boy can either be: Bob, Bill, Brian, or Bart. In other words, given the already randomly selected boy (who must be either: Bob, Bill, Brian, Bart), there are four equally possible sibling outcomes that can obtain:

(A) Given Bob, then 100% chance that sibling Boy-Bill obtains
(B) Given Bill, then 100% chance that sibling Boy-Bob obtains
(C) Given Brian, then 100% chance that sibling Girl-Gayle obtains
(D) Given Bart, then 100% chance that sibling Girl-Gabriella obtains

There is a 25% chance that any one of those four outcomes obtains.

Consequently, there is a 25% chance of the second sibling being one of either: Bill, Bob, Gayle or Gabriella – which reduces to a 50% chance that given a selected-boy, the remaining sibling would too be a boy.

If the above reasoning is sound, then your solution must be false because it contradicts the above solution. But rather than leaving you to find the error in your position, please let me try to perform an internal critique of it. You noted that 1 minus the probability of two girls in a family of two children is 75%. Of course that is true. Accordingly, as you noted, the probability of having at least one boy within a family of two children is 75%. That would be relevant if the question were: “Given two children, what is the probability of having
at least one boy?” which is 75%, of course. However, the question that we are to be concerning ourselves with is: “Given a randomly selected boy with a sibling, what is the probability that his sibling is also a boy?” – which is different question entirely. With respect to the problem we are to be solving, the “given” is the randomly selected boy - which can be the first *or* second child of the two children! The sample space that you concerned yourself with, (which gets you off on the wrong track I’m afraid), ignored the relevant statistic that from a two-boy family the randomly selected boy can be either the first or second boy.

From a sample space consideration, the problem is to be viewed thusly:

Girl sibling, randomly selected boy
Randomly selected boy, girl sibling
Randomly selected boy, boy sibling
Boy sibling, randomly selected boy

As I noted in my first post, we know that the randomly selected boy is either older or younger than his sibling. Accordingly, if he has a girl sibling, then there are two, not one, ways in which he might fall within that order. That statistic *is* implied in your reasoning. To be consistent, however, we must also recognize that there are two, not one, ways in which the *already* randomly selected boy with a boy sibling can fall within the order of his family. That statistic is *not* implied by the probability of there being at least one boy in a family of two children!

To simplify this even more, we might consider that there is a
randomly selected boy sitting in a room who has only one sibling:

That boy may have:

(a) An older brother
(b) A younger brother
(c) An older sister
(d) A younger sister

If the randomly selected boy is older than his sibling, then the
probability of (a) and (c) can be eliminated, leaving a 50% chance of his having brother. If the randomly selected boy is younger than his sibling, then the probability of (b) and (d) can be eliminated, leaving also a 50% chance of his having a brother. Added to that, the randomly selected boy has a 50% chance of being older than his sibling, and a 50 chance of being younger than his sibling. Consequently, it the statistically irrelevant whether the randomly selected boy is older or younger than his sibling because the probability of his sibling being a boy is the same 50% no matter whether the randomly selected boy is older or younger, and the
randomly selected boy has the same chance of being older than his sibling as he does being younger than his sibling. Therefore, we need not know the age of the randomly selected boy to know that the probability of his sibling being a boy is 50%.

Finally, we can demonstrate this empirically with playing cards. Let red cards stand for girl and black cards for boys. If a boy is randomly selected, then we can discard the two red 2s.

Red 2, Red 2
Black 3, Red 3
Red 4, Black 4
Black 5, Black 5

The remaining cards are now:

Black 3, Red 3
Red 4, Black 4
Black 5, Black 5

Now place the two red cards, which represent sibling girls, aside and keep the four black cards in your hand. Shuffle the black cards and draw a card at random. If you draw either of the black 5s, then the corresponding sibling would be a boy. If you draw a black 3 or black 4, then the corresponding sibling would be a red card, which represents a sibling-girl.

Probability of drawing a black 3 is ¼ or 25%, corresponding to girl- sibling

Probability of drawing a black 4 is ¼ or 25%, corresponding to a girl- sibling

Probability of drawing a black 5 is 2/4 or 50%, corresponding to a boy-sibling

Ron

________________________________________
Date: 07/25/2008 at 16:28:26
From: Doctor Garramone
To: @comcast.net (Ron DiGiacomo)
Subject: Re: Boy Girl probability

Good day Ron

This is a Bayesian problem (posterior probability given the sample).

The 2/3 answer is right, by Bayes' Rule.

Probability of a girl given that you picked a boy = the probability of picking a boy-girl set from the three possible sets (b-b, b-g, g-g) is the probability of picking a b-g set (which is 1/2).

Pr(b-b) = 1/4 this part is not germane to this problem

Pr(b-g) = 1/2 1/2 1/2 2
------------ = --- = ---
Pr(g-g) = 1/4 1/2 + 1/4 3/4 3

Here's a brain teaser:

If there are 3 doors and behind one is a Corvette and you are told to pick one door which you then call out.

A pretty hostess will open a door that is not the door you chose nor will it be the one which has the car.

Now, you are allowed to switch your choice from your original choice to the other closed door or to stay with the first choice.

Which would afford the greatest probability of winning - to stay with your original choice or to switch to the other door?

Why?

- Doctor Garramone, The Math Forum

The Math Forum @ Drexel is a research and educational enterprise
of the Drexel School of Education:

________________________________________
Date: 07/25/2008 at 17:54:47
From: @comcast.net (Ron DiGiacomo)
To: dr.math@mathforum.org
Subject: Boy Girl probability

Hi Doc. Anthony,

The Monty Problem has really nothing to do with this. But for what it’s worth, I’d switch doors given the problem *properly* stated. Also, I’m familiar with Bayes, but we mustn’t apply any formula to a problem that still defies definitive terms. SO, let me state the problem two ways to illustrate the semantic difficulty of this problem. I'm really hopeful we'll come to terms.

Problem 1:

In order to find a family, you randomly select a father of only two children, at least one of which is a boy. Therefore, it is equally probable that any of the three families below is the family indexed to the father.

BG
GB
BB

Consequently, there is a 2/3 probability that the family indexed to the randomly selected father has a girl. That is the way in which you have interpreted the problem.

Problem 2:

In order to find a family, you randomly select a *boy* from a set of families with only two children, at least one of which is a boy.

The boy can live within one of three type families:

BG
GB
BB

NOTE: Since 50% of all boys from families with only two children have a brother, there is a 50% chance that the randomly selected boy has a brother. With that in mind we can more easily find the confusion.

Wherein the confusion lies: Although 2/3 of all 2 children households with at least one boy have a girl, only ½ of randomly selected boys from such households have a sister and ½ don’t. Therefore, when the family in question is indexed to a randomly selected boy, the probability of a two child family with at least one son having a girl is only 50%. I must believe you will agree with that. With that in mind, let’s look at how the problem is usually stated: “If a family has two children and at least one of them is a boy, what is the probability that the family also contains a girl? Given the problem as stated, we need to know how we came across this
family in question before we begin to solve the problem. Is the family in question that has two children, with at least one of whom is a boy, indexed to a randomly selected boy who has a 50% chance of having a brother? Or is the family indexed to a randomly selected father?

Cheers,

Ron

________________________________________
Date: 07/27/2008 at 11:01:19
From: Doctor Garramone
To: @comcast.net (Ron DiGiacomo)
Subject: Re: Boy Girl probability

Hi Ron,

The original supposition is that the families with two children are randomly produced and the distribution of these sets of two would be 1:2:1 according to the binomial distribution. That would lead to the Bayesian calculation of 2/3 which has been discussed in prior threads on this subject.

Looking at boys alone (from two children families) and then looking at the sibling, there is twice much chance of picking up a boy from a 2-boy family than from a boy-girl family EXCEPT that in the random population there are twice as many boy-girl families as there are boy-boy families. So the probability is even or 2:2 or 1:1, however you want to call it. This goes along with the intuition that if you meet a person who has a sib, what is the probability on the gender of the sib? 50-50 of course.

Language is NEVER as precise as mathematical equations. That's one reasons discussions like this one keep occurring.

You are right about switching the choice on the car but it took me a long time thinking about it as to why. The other similar problem is the three chest gold-silver coin problem. This is a classic one.

You have three chests, each with 2 drawers. One has a gold coin in each drawer, the next has a gold coin one drawer and a silver coin in the other, and the third has a silver coin in each drawer. If one chooses a draw at random (there are six drawers to choose from) and one finds a silver coin, what is the probability of finding a silver coin in the second drawer of that same chest?

It took me forever to figure that one out, too. Note that here your set distribution {S, S} {S, G} {G, G} is 1:1:1, not the 1:2:1 as with with gender problem.

- Doctor Garramone, The Math Forum

The Math Forum @ Drexel is a research and educational enterprise
of the Drexel School of Education:

________________________________________
Date: 07/27/2008 at 23:12:30
From: Doctor Peterson
To: @comcast.net (Ron DiGiacomo)
Subject: Re: Boy Girl probability

Hi, Ron.

I've been watching this exchange, and want to make sure you have seen that we have dealt with the issue you are concerned about, namely the wording of the problem, both in the page you cite and elsewhere.

Family or Child First?
http://www.mathforum.org/dr.math/faq/faq.boygirl.choose.html
That emphasizes that the wording makes a big difference, specifically because you must be able to determine from the wording that the family is chosen first, not the child, in order to make the answer of 2/3 valid.

Note that in your first message, you presented the problem as
if a boy is selected from a family of two children that
the probability of the remaining child being a girl is 2/3.

This wording pretty clearly says that the family is selected first, though it could be improved. The page you refer to is explicitly focused on the issue of wording, and shows how different ways of stating the problem change the result. But you, in apparently trying to argue that there is no version of the problem in which the answer is 2/3, keep rephrasing the problem to focus on choosing a child, rather than choosing a family:

Then one child was randomly selected and it turned out to be a boy.

and

However, the question that we are to be concerning ourselves with is: “Given a randomly selected boy with a sibling, what is the probability that his sibling is also a boy?” – which is different question entirely.

I think we agree that if the child is selected, the answer is 1/2. Do you agree, however, that the answer to the following version (from the FAQ) is 2/3?

From the set of all families with two children, a family is
selected at random and is found to have a boy. What is the
probability that the other child of the family is a girl?

- Doctor Peterson, The Math Forum

The Math Forum @ Drexel is a research and educational enterprise
of the Drexel School of Education:

________________________________________
Date: 07/28/2008 at 07:05:49
From: @comcast.net (Ron DiGiacomo)
To: dr.math@mathforum.org
Subject: Boy Girl probability

Dr. Peterson,

If we word the problem as "if a boy is selected from a family of two children that the probability of the remaining child being a girl is 2/3" then the conclusion of 2/3 comes from at best an ambiguous question, or the answer is false. It is not true, as you wrote below that "This wording pretty clearly says that the family is selected first". At best the wording is ambiguous. Why is a boy *selected*? Was it because a randomly selected father referred to a son? If so, then the probability is 50%. Also, what is the semantic difference between a boy being selected from a family and a boy being selected from a set of boys all of whom come from such families described in
the problem?

W/ respect to the question: "From the set of all families with two children, a family is selected at random and is found to have a boy. What is theprobability that the other child of the family is a girl?"

How is the family "found to have a boy?" Does it come up in verbal discourse? The way the problem should be worded to conclude a 2/3 probability is: "A family is randomly selected among a set of families all having at least one boy. What is the probability that there is at least one girl?" When we get into the child being "found" or "selected" we are asking another question.

I have not read the link you supplied; I'll do so later.

I'll get back to Dr. Garramone later.

Thanks!

Ron

________________________________________
Date: 07/28/2008 at 07:16:35
From: @comcast.net (Ron DiGiacomo)
To: dr.math@mathforum.org
Subject: Boy Girl probability

Dr. Peterson,

I should clarify that it's not an issue of whether the family was
selected first. The confusion lies within the boy be "selected" (even after). There's an increase probabilty of *no* daughter if the father of such a family refers to one of his children as being a son. This is no different than a random selection of a boy from the set of all families with at least one son.

Ron
________________________________________
Date: 07/28/2008 at 12:08:36
From: @comcast.net (Ron DiGiacomo)
To: dr.math@mathforum.org
Subject: Boy Girl probability

I looked at the link. The way the two child with at least one boy family is selected is fine. In that case, yes, the probability of the woman having a girl is 2/3. That problem does not entail a child being 'found out' by any means that would entail an increase of probability of any remaining outcome. You are stricly eliminating two- girl families with the inquiry of the mother. It's quite another thing to set the problem up in such a way that entails a discovery of a boy by remembering that one child was a boy. Why would a boy have been remembered? Did the mother refer to a *particular* son? If that were the case, then a particular son would be the *first* child referred to. Accordingly, the prossibilities would be reduced to:

First Son referred to; Second daughter

First Son referred to; Second son

Cheers,

Ron
________________________________________
Date: 07/28/2008 at 23:53:10
From: Doctor Peterson
To: @comcast.net (Ron DiGiacomo)
Subject: Re: Boy Girl probability

Hi, Ron.

Now I think we're on the same page; I wanted to make sure you saw at least some way of stating the problem in which 2/3 is the right answer, because your initial messages gave the impression you were using some nonstandard reasoning that would never yield that answer.

I fully agree that this problem is quite often presented in ways that do not really yield the supposed answer of 2/3 -- perhaps even on our site. Very likely some teachers or authors see the problem as a good one to share students up, but when they present it they either don't quite understand it themselves, or they are trying to oversimplify it, and the result is that they convince a lot of students that math is nonsense.

I just recently answered a question that put it this way, presumably quoted directly from a text:

While shopping in a supermarket you see a woman shopping with her daughter. You are told the woman has two children. What is the probability that the woman has two daughters?

I responded by first pointing out that the question was a little
ambiguous and rephrasing it in a way that was "presumably intended", then solving that; then I showed why the problem as stated was bad:

Are you equally likely to pick any family from the population of all two-child families with at least one daughter (which is the assumption of the solution you were shown)? Or are you equally likely to pick any daughter from such a family (which gives the result you think is more natural)? That is, is the focus on the mother or the daughter? I can't tell!

In reality, in fact, you are simply choosing at random from all
families with two children, at least one of whom is a daughter,
AND in which the mother is shopping with ONE daughter at this moment. That may increase the probability that a one-daughter family is chosen, since a mother with two daughters might often choose to take both shopping. So I think the problem is not stated in its clearest form. We try to state it better in our FAQ.

It may be that some of the versions you find us answering are bad questions that we are trying to make the best of, not always correcting the statement of the problem as we really should. Maybe we just hope that the student misquoted and the book said it better.

I don't want to answer what you've said here, because I'm not sure which version you are referring to. Just to make sure we're talking about the same thing, perhaps you can quote word for word a version or two for which you don't think the answer should be 2/3, but which we or someone else answer in that way, and we can discuss whether it is wrongly worded. We probably won't always agree on the meaning of the
English (English works that way!), but at least we can try to find the best ways to say it that we CAN agree on.

- Doctor Peterson, The Math Forum

The Math Forum @ Drexel is a research and educational enterprise
of the Drexel School of Education:

________________________________________
Date: 07/29/2008 at 08:16:21
From: @comcast.net (Ron DiGiacomo)
To: dr.math@mathforum.org
Subject: Boy Girl probability

Dr. Peterson,

TWE does fine. It's Dr. Anthony, in my estimation that has not been as rigorous as he might have been in answering people. Even when I pointed out some of the ambiguities, he seemed to give no credence to what you and TWE see as a misleading representation of the problem. In turn he would continue to address the problem as (at best) he thought it was intended to be worded. Or (at worst) he didn't recognize the nuance (my issue) that you so obviously picked up on.

For what it's worth, I think that *most* of the time the problem is worded incorrectly *if* the wording is supposed to imply an answer of 2/3. The way the problem is typically worded in my experience demands a 50% answer. I encountered the Monty Hall problem this way too the
first time I heard it (not on your site). My suggestion would be, that we not answer the question according to how it is thought that it was probably intended. Rather, we should take the opportunity to show the problem in two forms and then get into why they yield different answers.

Yours,

Ron

________________________________________
Date: 07/29/2008 at 09:01:27
From: Doctor Peterson
To: @comcast.net (Ron DiGiacomo)
Subject: Re: Boy Girl probability

Hi, Ron.

Exactly.

I concur completely. The main point in any discussion of this
problem, or Monty Hall, should be the translation from English to math, not just the math.

- Doctor Peterson, The Math Forum

The Math Forum @ Drexel is a research and educational enterprise
of the Drexel School of Education:

Counter since: 9/6/2006

Anonymous said...

I love this:

If the above reasoning is sound, then your solution must be false because it contradicts the above solution. But rather than leaving you to find the error in your position, please let me try to perform an internal critique of it. You noted that 1 minus the probability of two girls in a family of two children is 75%. Of course that is true. Accordingly, as you noted, the probability of having at least one boy within a family of two children is 75%. That would be relevant if the question were: “Given two children, what is the probability of having
at least one boy?” which is 75%, of course. However, the question that we are to be concerning ourselves with is: “Given a randomly selected boy with a sibling, what is the probability that his sibling is also a boy?” – which is different question entirely. With respect to the problem we are to be solving, the “given” is the randomly selected boy - which can be the first *or* second child of the two children! The sample space that you concerned yourself with, (which gets you off on the wrong track I’m afraid), ignored the relevant statistic that from a two-boy family the randomly selected boy can be either the first or second boy.

From a sample space consideration, the problem is to be viewed thusly:

Girl sibling, randomly selected boy
Randomly selected boy, girl sibling
Randomly selected boy, boy sibling
Boy sibling, randomly selected boy

As I noted in my first post, we know that the randomly selected boy is either older or younger than his sibling. Accordingly, if he has a girl sibling, then there are two, not one, ways in which he might fall within that order. That statistic *is* implied in your reasoning. To be consistent, however, we must also recognize that there are two, not one, ways in which the *already* randomly selected boy with a boy sibling can fall within the order of his family. That statistic is *not* implied by the probability of there being at least one boy in a family of two children!

Reformed and Reforming said...

Were you a math major?

Ronald W. Di Giacomo said...

No. I majored in time and space.

Reformed Apologist said...

If a man with only two children is asked whether his eldest child is a boy and he answers “yes,” then the probability of him having a girl is ½. But if a man with only two children is asked: "Do you have a boy?" and he answers “yes,” then the probability of him having a girl is not ½ but 2/3.

Regarding the problem at the top of the page - If the children of a man with only two children are sitting in a room and a boy is randomly selected, then there is a ½ chance the other child is a girl. The reason being, the selected boy must be older or younger, larger or smaller, smarter or not smarter etc. If, older (for instance), then not only does the possibility of GG disappear but so does the possibility of GB. (That’s what was missed by the Drexel professors.)The reverse would hold true as well, of course. If the selected boy-child were younger, then GG disappears as well as BG. In either case, GG plus one other set other than BB must disappear. (BB can’t disappear because the set BB reflects a possible outcome whether the boy is older or younger.) So, if the boy is older, then he is from the set of BG or BB; which means his sibling would have ½ chance of being a girl.

Another way of looking at this, yet without eliminating possibilities due to order (e.g. age), is to consider that the randomly selected boy can come from any of the three sets: BB, BG and GB, but not from the set of GG. Without any consideration given to order, the boy would have an equal chance of being any of the four boys. (He would not have an equal chance of belonging to each set(!), which was missed by the Drexel professors. Given that there are four boys that the randomly selected boy can possibly be, he must have 1/2 chance of being from the first set of BB (since he can be one of either two children within that set(!)), and ½ chance from being from one of the other two sets that include a girl (i.e. 1/4 chance of being from either sets with one girl). Accordingly, he’d have ½ chance of having a sister.

Jon said...

Would you be willing to discuss this off line sometime?