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#1
08-10-2006, 02:08 PM
 onetimepad Guest Join Date: Aug 2006 Posts: 2
Boy/Girl probability in Monty Hall

Dear Cecil,

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

There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)

The possible gender combinations for two children are:

(1) Child A is female and Child B is male.

(2) Child A is female and Child B is female.

(3) Child A is male and Child B is female.

(4) Child A is male and Child B is male.

We know one child is female, eliminating choice #4. In 2 of the remaining 3 cases, the female child's sibling is male. QED.
This is a fallacy. It remainds me of an old coin toss problem.

Let's throw a coin three times. If it comes up all the same side you win, otherwise I win. The odds are fifty/fifty as you can see:

2. All tails
=YOU WIN

3. Two heads and a tail
4. Two tails and a head
=I WIN

Out of four options we each have two winners. But intuitively you've probably guessed there's something wrong here. It's not that easy to get three heads/tails. Our list of options should look like this (with letters for heads/tails):

1.H H H
2.T T T
=YOU WIN

3. H H T
4. H T H
5. T H H
6. T T H
7. T H T
8. H T T
=I WIN

So the odds are actually 1/4 for you and 3/4 for me. The fallacy is assuming that T T H is the same as H T T, and T H T.

In the gender problem you list the options as:

1. F M
2. F F
3. M F
4. M M

While this is accurate as far as it goes, your final statement immediately shows up the fallacy. "In 2 of the remaining 3 cases, the female child's sibling is male"

You seem to forget that in the real world the females in option 2. are in fact two different people, who BOTH have a sister. So let's call the named child Debbie, in which case your options are:

1. Debbie + M
2. M + Debbie
3 Debbie + F
4. M M

But we're leaving out the important fact that another option exists:

5. F & Debbie

[and also in theory, 6. M M reversed]

Taking this into account, Debbie has potential for an older/younger brother OR an older/younger sister--making the odds 50/50, as nature intended it.

Thanks,
David, Ireland
#2
08-10-2006, 03:32 PM
 borschevsky Guest Join Date: Sep 2001 Location: Canada Posts: 1,864
Nope, Cecil is right. Of families that have two children, 1/4 will have two daugters, 1/4 will have two sons, and half will have a son and a daughter. So, of the families that have a daughter, 2/3 also have a son.
#3
08-10-2006, 03:58 PM
 John W. Kennedy Charter Member Join Date: Apr 1999 Location: Chatham, NJ, USA Posts: 4,990
As soon as you choose one child (e.g., Debbie), the situation changes. But as long as all you know is "One child is a daughter," the odds continue to be 2/3 that the other child is a son.
__________________
John W. Kennedy
"The blind rulers of Logres
Nourished the land on a fallacy of rational virtue."
-- Charles Williams. Taliessin through Logres: Prelude
#4
08-10-2006, 04:24 PM
 Giles Charter Member Join Date: Apr 2004 Location: Newcastle NSW Posts: 12,834
My way of looking at it:

(1) If you pick children at random, and find you have picked a girl witb one sibling, then it's a 1/2 chance that her sibling is a boy.

(2) If you pick families at random, and find you have picked a family with two children, at least one of whom is a girl, then it's a 2/3 chance that the family contains a boy as well as a girl.

So it depends on what your random selection process is.
#5
08-10-2006, 06:03 PM
 Find Friends Charter Member Join Date: Aug 2002 Location: Image City on North Coast Posts: 2,355
Quote:
 Originally Posted by John W. Kennedy As soon as you choose one child (e.g., Debbie), the situation changes. But as long as all you know is "One child is a daughter," the odds continue to be 2/3 that the other child is a son.
I agree, BUT it's important that phrasing the puzzle be clear. "One child" may mean there is AT LEAST one child (or at least one of whatever you may be talking about -- For example: "This solid figure has one right-angle.") That is, I believe, the way mathematicians speak.

But in common parlance it is often unclear. Rather than "one child" or "a child" perhaps it should be spelled out as "at least one."

Even saying, "They are not both boys" wouold be clear enough, albeit an unusual way to say it.

But you are correct and onetimepad is not. My point here is one about semantics, not mathematics.

I found his example of all-the-same-way flipped coins interesting, though.

* * * * * * * * * * *

True Blue Jack
#6
08-10-2006, 06:35 PM
 caernavon Guest Join Date: Aug 2006 Posts: 4
Quote:
 Originally Posted by onetimepad There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.) The possible gender combinations for two children are: (1) Child A is female and Child B is male. (2) Child A is female and Child B is female. (3) Child A is male and Child B is female. (4) Child A is male and Child B is male. We know one child is female, eliminating choice #4. In 2 of the remaining 3 cases, the female child's sibling is male. QED.
Isn't it true that 1 and 3 are in fact the same scenario? 1 boy, 1 girl. If that's the case, then the odds are 1/2, not 2/3.

If they're not the same, and if child A is the first child mentioned (the girl), and B is the other child, then 3 is invalid because child A must be a girl.

In either case, the answer is 1/2, not 2/3.
#7
08-10-2006, 06:44 PM
 gazpacho Guest Join Date: Oct 1999 Posts: 5,628
Quote:
 Originally Posted by caernavon Isn't it true that 1 and 3 are in fact the same scenario? 1 boy, 1 girl. If that's the case, then the odds are 1/2, not 2/3. If they're not the same, and if child A is the first child mentioned (the girl), and B is the other child, then 3 is invalid because child A must be a girl. In either case, the answer is 1/2, not 2/3.
1 and 3 are not the same. Lets list them as follows

1) first born child girl second born child boy

2) first born child girl second born child girl

3) first born child male second born child female

4) first born child male second born child male

case 1 the girl has a younger brother
case 3 the girl has an older brother

So they are clearly different families.
#8
08-10-2006, 07:00 PM
 caernavon Guest Join Date: Aug 2006 Posts: 4
Quote:
 Originally Posted by gazpacho 1 and 3 are not the same. Lets list them as follows 1) first born child girl second born child boy 2) first born child girl second born child girl 3) first born child male second born child female 4) first born child male second born child male case 1 the girl has a younger brother case 3 the girl has an older brother So they are clearly different families.
You're adding an explanation that isn't relevant to the puzzle. It doesn't matter if child B is older or younger, because the puzzle only mentions the other child.

There only three ways of having two children:

1. two boys
2. two girls
3. one of each

1 is invalid, and so the answer to the puzzle is 1/2.

There are four ways to order two children, or four different ways they can come to your house:

1. boy then boy
2. boy then girl
3. girl then boy
4. girl then girl

1 is invalid. 2 is also invalid, because the first child in the puzzle was a girl. Only 3 and 4 remain as possibilities, and so the answer to the puzzle is 1/2.
#9
08-10-2006, 07:44 PM
 gazpacho Guest Join Date: Oct 1999 Posts: 5,628
Quote:
 There only three ways of having two children: 1. two boys 2. two girls 3. one of each
True enough. If we assume that girls and boys are equally likely then the probabilities for these families are

1/4 two boys
1/4 two girls
1/2 one of each

In a population of 1000 families

250 have two boys
250 have two girls
500 have one of each

There are 750 families that have at least one chiled a girl
250 have two gilrs
500 have one of each

500 out of these 750 have a boy or 2/3 of them.
#10
08-10-2006, 07:54 PM
 gazpacho Guest Join Date: Oct 1999 Posts: 5,628
I should expand on the probabilities a little bit.
The probabilities can be shown as follows again the same 1000 families

500 families have the first child a boy of those families 250 have the second child a boy and 250 of them have the second child a girl.
250 are two boys and 250 one of each

500 families have the first child is a girl of those families 250 have the second child a boy and 250 of them have the second child a girl.
250 are two girls and 250 one of each

This makes
250 families have two boys
250 families have two girls
500 families have one of each
#11
08-10-2006, 08:02 PM
 JR Brown Guest Join Date: Mar 2006 Location: Boston, MA Posts: 1,032
Quote:
 Originally Posted by caernavon There only three ways of having two children: 1. two boys 2. two girls 3. one of each 1 is invalid, and so the answer to the puzzle is 1/2.
The point you are missing is that case #3 (one boy, one girl), is TWICE as common as the other two cases. 25% of two-child families will have two girls, 25% will have 2 boys, and 50% will have one of each. So in the first puzzle (the probability that a pair of siblings, of which at least one is a girl, also includes a boy) is, as stated, 2/3.

The second problem (a girl with one sibling visits, what is the probability that she has a brother) is a little trickier. What many people have forgotten to take into acount is that a family with two girls has two girls who each have a sister, whereas a familiy with one of each only has one girl who has a brother.

To make following this easier, let us consider these families:

1. Aaron and his brother Adam (2 boys)
2. Betty and her brother Bill (1 girl, 1 boy)
3. Cedric and his sister Cecilia (1 boy, 1 girl)
4. Daisy and her sister Dot (2 girls)

You are expecting one set of siblings to visit, but you don't know which set. You hear a knock at the door. Your roomate answers the door and says there is a girl looking for you. What is the probability that she will have a brother?

There are four possible girls (Betty, Cecilia, Daisy and Dot). Two of them have brothers (Betty and Cecilia) and two of them have sisters (Daisy and Dot). Therefore the probability that the girl at the door has a brother is 2/4, or 1/2. QED.

JRB
#12
08-10-2006, 08:04 PM
 JR Brown Guest Join Date: Mar 2006 Location: Boston, MA Posts: 1,032
Incidentally, onetimepad's explanation is the solution to the second puzzle, not the first.
#13
08-11-2006, 12:18 AM
 wissdok Guest Join Date: Aug 2005 Posts: 107
I wrote a lengthy amount of posts on this puzzle in the several other threads and I won’t bore everyone with rehashing all of stuff again. This puzzle appears to have been around at least 60 years, and it was shot down by both Scientific America and the guru of puzzles, Martin Garner. Oddly enough, Martin Garner was use by Marilyn as the judge of her work in the Monty Haul puzzle.

This are the facts:

The four possible family combinations:
Column A ….Column B
1)…..Girl……….. Girl
2)…..Girl……….. Boy
3)…..Boy……….. Girl
4)…..Boy……….. Boy

Statements of Equality

*If the families are equal, then the children are not. If we are given a girl and the families are equal, then each families (#1, #2, #3) are each likely 33.3% of the time… making each girl in family #1 worth half the value of the girls in families #2 and #3.

*If the girls are equal, then the families are not. If we are given a girl and the girls are equal, then each girl has an equal chance… making family #1 50% likely to be the home of the girl we are talking about.

Now if we are given a girl, then that girl is either the girl from column A or the girl from column B. From this we can conclude that there isn't two boys in the family because... AGAIN we know either column A is a girl or Column B is a girl but we don't know that both are girls. If there are no boys in column A then neither families #3 and #4 are possible; If there are no boys in column B then neither families #2 and #4 are possible. The idea that we can just get rid of family #4 without throwing out either of the corresponding families(#2 or #3) with it , is poor logic.

Before we meet a child, the families are equal, but the childrem are not. Once we meet a child, the children become equal and the families become unequal.
We are twice as likely to meet a girl from family #1 than we are in either family #2 or #3. In fact, half the time with #2 and #3 we will meet a boy. The reverse of this is turn as well, if we meet a boy then that boy is twice as likely to be from family #1 than he is from families #2 or #3. These events are true because the children are equal not the families.

Having meet a girl then the possible combinations are:
Known girl and another girl
Known girl and another boy
A girl and the known girl
A boy and the known girl

In the four combinations, the other child is a boy 50% of the time and a girl 50% of the time.
#14
08-11-2006, 12:41 AM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
Quote:
 Originally Posted by wissdok ... the guru of puzzles, Martin Garner.
It's actually Martin Gardner.
#15
08-11-2006, 01:15 AM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
Quote:
 Originally Posted by wissdok Before we meet a child, the families are equal, but the childrem are not. Once we meet a child, the children become equal and the families become unequal.
This may not be relevant, as the original problem did not mention meeting a child. But it can point toward the following line of thinking.

There are two possible problems here; they are related and can sound as if they're the same, but they are different. Here's one way to state them:

From a population of two-child families:
1. A family is chosen at random and found to include at least one daughter. What is the probability that the family also includes a son?

2. A child is chosen at random and found to be female. What is the probability she has a brother?
You are giving the correct answer to Problem #2: 1/2. But the problem posed in the OP is #1, and Cecil has given the correct answer there: 2/3
#16
08-11-2006, 01:22 AM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
And note that J.R. Brown is examining the same two (distinct) problems.
#17
08-11-2006, 03:15 AM
 tim314 Charter Member Join Date: Mar 2004 Posts: 4,460
Look, whether you count four possibilities or three doesn't matter. Whether or not you consider birth order doesn't matter. Either way, Cecil is right.

If the odds of having one girl are 50%, then the odds of having two girls must be 25%. (This assumes the sex of the two kids is independent.) This is because 50% times 50% is 25%.

Similarly, the odds of having two boys must be 25%.

So we have:
1) Odds of two girls: 25%
2) Odds of two boys: 25%

The only remaining possibility is that the couple had one boy and one girl (in some order). Since the total odds of all possibilities must add up to 100%, then the odds of this possibility must be 50%.

3) Odds of one boy and one girl: 50%

Now, we've all agreed that the sex of one kid won't effect the sex of the other -- no one is saying it does. But if a couple waits until they have two kids, and then tells you "One of our kids is a girl," then they're effectively telling you "We're not in group 2 (two boys)"

That means they must be in group 1 (two girls), or group 3 (one of each). Since having one of each was twice as likely as having two girls, there's a two thirds chance that that's the group they're in.

I said nothing about birth order, I treated "one of each" as a single group, and the answer is still the same.
#18
08-11-2006, 03:35 AM
 tim314 Charter Member Join Date: Mar 2004 Posts: 4,460
Yet another way to look at it: there's a 50% chance of both children being the same sex. That's because whatever sex you get for the first kid, you have a 50% chance of the second one matching it.

That means 50% of the parents have one kid of each sex. The other 50% are split evenly between having two boys and having two girls.

So there are more parents with a boy and a girl than with two girls. Thus, of the parents with at least one girl, most have a boy and a girl.

This all assumes that the sex of each kid is random, and one doesn't affect the other. No one is saying having one girl causes the other child to be a boy. It's just that there are more parents with one of each than with two girls, so if the parents have one girl they're more likely to fall into the "one of each" group. Correlation does not imply causation.

Or, two put it in a less cliche way: The odds don't change because having a girl directly effects the other kid's sex -- the odds change because the parents gave you more information to help guess which group they're in.
#19
08-11-2006, 05:25 AM
 The Riddler Guest Join Date: Aug 2006 Posts: 1
The correct answer is 50%. We've been conditioned with brain teasers that the less obvious answer is always correct, but in this case it really is the simpliest answer.

Both sides are overcomplicating the problem. One child's gender has already been determined. Only one child's gender remains to be "randomly generated" . The fact that there are two children total is irrelevant. The order they were born in is irrelevant. The only fact that matters is there is ONE child left up to the 50/50 gender assignment. The odds that the child is female (making both children daughters) is 50%.

To go by the combination charts people have been posting (overcomplicating the simple answer):

1. M,M
2. M,F
3. F,M
4. F,F

The first (first as in first to have their gender "generated", not first born) child is locked into being female, so options 1&2 are impossible. Only the second child can be either gender. Since only one child's gender remains subject to statistics, there are only 2 possibilities, or 50/50.

Other examples

Coin toss-

Question: A coin is tossed. What are the odds that it will be tails?

Q: You flip a coin. Another coin is already lying tails side up on the table. What are the odds that both coins will end up tails?

A: 1/2. The fact that there is a second coin is irrelevant to determining the outcome of one single toss. The first toss' outcome has already been locked and only the second is still up to chance.

Die roll-

Q: What are the odds of rolling a 12 with two six sided dice?
A: 1/36 (1/6 for a '6' on the first toss multiplied by 1/6 for the second)

Q: One die is on a table showing a '6'. You roll a second die. What are the odds that the dice will add up to 12?
A: 1/6. Only one die is subject to chance.

I'm really surprised Cecil debunked the game show prize door question and got the correct 50/50 answer instead of the incorrect 2/3, but then went on to get this similar question wrong with an answer of 2/3 instead of the correct 50/50.

Of course, the wording of the question is a bit unclear. "they have a daughter" could mean they have AT LEAST one daughter or that they have ONLY one daughter. In the latter case, the answer would be a 100% chance that they have a son. In either case, the answer still would never be 2/3.
#20
08-11-2006, 09:46 AM
 bubbajoe Guest Join Date: Aug 2006 Posts: 1
I believe some folks are confusing independent events in statsitics from this word problem.

In a truly independent event world, the answer is indeed 1 in 2. No matter how many kids walk through a door, the next one has a 50/50 shot of being a boy.

We're not in an independent world here. There are four cases, each with a 25% probability:

1- FF
2 - FM
3 - MF
4 - MM

Someone TELLS you that one is a girl (ruining total independence). I would assume they know exactly what the family has. They have essentially eliminated case 4. Assuming this person picked the family at random (A big assumption), 2 of the 3 remaining cases has a boy. Therefore, if some person TELLS you a family has one girl, the odds of a boy for the other is 2 in 3.
#21
08-11-2006, 10:04 AM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
Quote:
 Originally Posted by The Riddler One child's gender has already been determined. Only one child's gender remains to be "randomly generated" .
You have changed the problem to: "A family is selected at random, and then one of the children in the family is selecetd at random and found to be female. What is the probabilty that the other child is male?"

The answer to this is indeed 50%, but this is not the same as the original problem.
#22
08-11-2006, 10:11 AM
 wissdok Guest Join Date: Aug 2005 Posts: 107
Xema, I disagree, there is but one problem here, not two. The families are random when the problem starts, but once a child is given (or meet) the child becomes the focus and the families are no longer random. This given child is equally able to be any child that is of that same sex. If we are given a girl (like Cecil’s puzzle) then the girl can be either of the girls in the family with two (25% each) or either of the girls in the two families that have a son also(25% each). If we are given a boy(like Marilyn’s puzzle) then that boy can be either of the boys from the family with two(25% each) or either of the boys from the two families that have a girl also(25% each).

As an example of proof that 2/3 is wrong, this problem cannot be repeated and guarantee a 2/3 result. As I explained above Cecil chose girls and Marilyn chose boys, these opposite events cannot be right. Above, Gazpacho spoke of 1000 families. If we repeat this puzzle to all 1000 families using the 2/3 “rule” of choosing the opposite, we should be right 66.67% of the time.
But you can easily see that it wouldn’t be 66.67% because:
On all GG families we would be wrong. (25% of the families)
On all GB families we would be right. (25% of the families)
On all BG families we would be right. (25% of the families)
On all BB families we would be wrong. (25% of the families)

We would be right 50% of the time and wrong 50%.

As Martin Gardner (correct spelling) explained in his opinion of this puzzle, this can only be 2/3 if we assume that that we are given a child and all families with the sex of child always tell (or are given) that sex of child. That assumption is not listed in the puzzle.
#23
08-11-2006, 12:06 PM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
Quote:
 Originally Posted by wissdok As I explained above Cecil chose girls and Marilyn chose boys, these opposite events cannot be right.
They're just different ways of phrasing the same problem - rather like the difference between b = a + 2 and y = x + 2.

Quote:
 As an example of proof that 2/3 is wrong, this problem cannot be repeated and guarantee a 2/3 result.
I believe this can easily be done.

Quote:
 If we repeat this puzzle to all 1000 families using the 2/3 “rule” of choosing the opposite, we should be right 66.67% of the time. But you can easily see that it wouldn’t be 66.67% because: On all GG families we would be wrong. (25% of the families) On all GB families we would be right. (25% of the families) On all BG families we would be right. (25% of the families) On all BB families we would be wrong. (25% of the families)
This would be correct if BB families were included in the problem; in fact, they are eliminated by the condition that the selected family includes a daughter.

We thus are left with the first three of the four cases you list - 750 of the original 1000 families. In 500 of these (66.67%), the daughter has a brother.

Quote:
 ... this can only be 2/3 if we assume that that we are given a child and all families with the sex of child always tell (or are given) that sex of child.
We are not given a child - we are given a family and some information about the family: that it includes at least one daughter (which is exactly equivalent to saying that this is not a BB family). The family does not tell us anything, nor do we meet anybody. And yet the answer is 2/3.
#24
08-11-2006, 12:30 PM
 Larry Borgia Charter Member Join Date: Mar 2000 Location: Washington DC Posts: 9,888
Once you start specifying the order of the children, you change the problem. In the problem you are given a family with one daughter and one unknown child. That's it. One quarter of all possible families have two boys, one quarter have two girls, and one half have a boy and a girl. Eliminate the 2B families and you have 2/3 BG or GB families. All the problem does is eliminate the 2B families. It doesn't do anything else.

The problem could have been phrased "Given the set of all two child families with at least one daughter, what is the probability that a family picked at random will have a son?" answer: 2/3.

wissdok, you're right that if I pick a daughter at random the probability that she belongs to a GG family as opposed to a {GB, BG} family is 2/3, cancelling the effect of eliminating the BBs and giving a probability of 1/2 that she has a brother. But that's not the problem we were given. We're picking families, not daughters.
#25
08-11-2006, 01:04 PM
 CJJ* Guest Join Date: Nov 2004 Location: Chicago Posts: 2,268
The key point here is that the initial question "Is one of your children a boy/girl", weeds out any two-child family where the answer is "no". It is therefore not surprising (assuming two-child families are split evenly between BB, BG, GB, and GG--written in birth order) that once your initial question eliminates either BB or GG, the odds are in favor of guessing the opposite sex for the other child since 2 of the 3 remaining possibilities are mixed-gender families. By this time, I'd think this logic would be obvious.

What interests me is how this "weeding out" requirement can be short-circuited without careful consideration. Suppose instead of my asking specifically "Is one of your children a boy?" I instead ask "What is the gender of one of your children?" It seems to be a similar process, but it's not. By assumption, half of all two-child families are BB or GG. In either case, assuming the opposite gender for the other child is a loser, but you win if you happen to catch a parent with BG or GB, i.e. half the time. In essence, the answer to the gender question didn't give you any information to weed out possibilities.

As Larry Borgia points out, another way to short-circuit the value of this weeding-out is to specify the gender of the oldest child only. Suppose you ask "Is your oldest child a boy?" An answer of "yes" here weeds out both GG and GB. If you follow the "guess the opposite gender" strategy, this weeding eliminated half your winners and half your losers. Unlike the previous case--where the strategy failed because no weeding out was actually performed--this question does cause some possibilities to be weeded out, but does so ineffectively for the strategy you adopted.

The correct approach relies on effectively weeding out only GG families by specifically asking "Is one of your children a boy". This then eliminates 1/4 of guesses which would have resulted in a loss by guessing the opposite gender. All my winning chances remain, half my losers are weeded out, so I have a 2:1 incentive to switch.

I appreciate the problem for the attention it brings to subtlety in language (balanced, of course, by the number of posts required to elucidate that subtlety)
#26
08-11-2006, 01:16 PM
 tim314 Charter Member Join Date: Mar 2004 Posts: 4,460
Quote:
The Riddler, I'm sorry, but you're wrong. If you think the problem is equivalent to flipping one coin, observing the outcome, and then flipping another, then you're misunderstanding the problem Cecil is trying to state.

Have you read my explanations above? If so, and you disagree with them, please quote the part you think is wrong and explain why.

Anyway, here's why it's not equivalent to flipping one coin and flipping another, explained as well as I can.

In your coin flip example: You flip one coin and get heads. Then you flip another coin, and there's a 50-50 chance it's heads.

In the children example: The couple has two kids. Then, after both kids are born, they tell you that one of them is a girl. That's not the same as telling you the first one was a girl.

To be equivalent to the coin flip, they couple would tell you they had had a girl after having the first child, and then would ask you to guess the odds of the second child being a girl. In that case, it would be 50-50. But do you see how that's different than the stated problem?

Here's a description of a coin flip problem that's equivalent: Flip a pair of coins and write down the outcome of both. 25% of the time, they will both be heads. Do you see why? Likewise, 25% of the time, they will both be tails. So that means 50% of the time, you get a heads and a tails. Please tell me what you disagree with in this description, if anything.

Assuming you agree with the above, then consider this: You get a pair that contains heads and tails twice as often as you get a pair where both are tails. This follows from assuming the outcomes of the two flips are independent. So that means, if you tell me that your pair of coins had at least one come up tails, then I can consider it more likely that you got heads-tails (in some order) than tails-tails, since heads-tails happens more often.

Please explain what if anything you think is wrong with this argument. If it still doesn't make sense, I urge you to get a pair of coins and flip them 100 times. I promise you it will come up with one heads and one tails more often than it comes up two tails. If you don't believe it, see for yourself.
#27
08-11-2006, 03:57 PM
 wissdok Guest Join Date: Aug 2005 Posts: 107
Quote:
 Xema post: We are not given a child - we are given a family and some information about the family: that it includes at least one daughter (which is exactly equivalent to saying that this is not a BB family). The family does not tell us anything, nor do we meet anybody. And yet the answer is 2/3.
I stand by my earlier post, Cecil’s puzzle included a girl, Marilyn's a boy. Their argument is that it doesn’t matter which child we are given, it is the same for either sex. As we looked at the earlier 1000 families, the 2/3 “rule” can’t be correct for both boys and girls at the same time. While the families with two boys or two girls will always give you the same “clue”, the families with both a boy and a girl can’t give an answer that makes both Cecil’s and Marilyn’s puzzle true.

With Cecil’s puzzle it is possible for 2/3 to be true IF we assume that all families with girls tell us of them. But at the same time that would mean that Marilyn would be wrong because only the family with two boys would tell us of boys. The reverse is true as well, when Marilyn’s answer is true, Cecil answer must be wrong. This is rather a unique scenario, as more often than not, neither could be true. If the families are random and we are told of a random child of that family, then most likely the families with both a boy and a girl would half the time tell you of either. Because of randomness, half of the 500 families in the earlier group (that have both sexes as children) would tell of girls and the other half, boys.
250 Family Group #1 GG---always gives the clue girls
125 Family Group #2 GB---gives the clue girls
125 Family Group #2 GB---gives the clue boys
125 Family Group #3 BG---gives the clue girls
125 Family Group #3 BG---gives the clue boys
250 Family Group #4 BB---always gives the clue boys

In true randomness, families in group #2 or #3 will be equally likely to tell you of either child. The result of this in either Cecil’s or Marilyn’s puzzle only around 500 of the families will ever give the same clue. Of that 500, half will automatically be from the families with two boys or two girls.

I again repeat: Cecil’s answer (with girls) and Marilyn’s (with boys) cannot ever be correct simultaneously. In fact, either can only be true when both families with mixed child tell of the same sex child AND we just so happen to be looking for that same type child.

Quote:
 Cecil’s original statement: There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)
Diagram of the puzzle:
Step #1
Chose a random family with two children. The family possibilities with two children are:
1) Girl, Girl
2) Girl, Boy
3) Boy, Girl
4) Boy, Boy
The families are random but the children themselves are irrelevant other than to help separate the families.

Step #2
We are given(meet, told, …etc) the sex of a child in a two child family. The focus here is on the child and the families are no longer equal. Like both Cecil and Marilyn pointed out one type of family(GG or BB) can no longer be possible. The randomness is now on the child. The child can be any of the children of that given sex.
Type of families………If we are given a girl then……..If we are given a boy then
…………………………….(% chance the child given)………………………….
Family #1 –Girl, Girl……………25 / 25………………………………...0 / 0……………
Family #2 -Girl, Boy……………25 / 0………………………………….0 / 25…………..
Family #3 -Boy, Girl…………….0 / 25…………………………………25 / 0………….
Family #4 -Boy, Boy……………0 / 0…………………………………..25 / 25…………

Here again the families with two girls or two boys are twice a likely to be the family where the child is from.
#28
08-11-2006, 04:14 PM
It's doomed, but let me try to explain this

Let's talk red and black cards rather than genders.

If you set up, face down, two columns of cards in the following pattern:

R R
B B
R B
B R

You have the four possible distributions.

If you randomly select one card and turn it over, what is the likelihood the paired card is the same color?

THESE COMBINATIONS ARE PAIRED! That's the key statistical concept many people are missing here.

It doesn't matter which side of the pair you choose to reveal, but once you've picked a side, that determines which other pairs are left in the possible solution set. Many of you folk want to keep all "girl" rows once you "turn over" a "girl" card. Doesn't work that way. Once you select from one column, you have to consider only those rows from that column that match what you picked.

If you picked a B card out of the first column, you're left with

B B
and
B R as the possible solution sets; or 50-50. You can't keep the R B row!

If you picked an R card out of the second column, you're left with

R R
and
B R as the possible solution sets; or 50-50. You can't keep the R B row here either!

And yes, I TOOK Statistics at the University of California, but I'm never taught it.
#29
08-11-2006, 04:35 PM
 CJJ* Guest Join Date: Nov 2004 Location: Chicago Posts: 2,268
Quote:
 Originally Posted by adbadqc Let's talk red and black cards rather than genders. If you set up, face down, two columns of cards in the following pattern: R R B B R B B R You have the four possible distributions. If you randomly select one card and turn it over, what is the likelihood the paired card is the same color? THESE COMBINATIONS ARE PAIRED! That's the key statistical concept many people are missing here.
The key point you seem to be missing here is that, as the original problem is stated, you do not know whether the child identified by gender is the older or younger child. In your card scenario, this is equivalent to not knowing whether you've turned over a card from the first column or the second column

In your card problem, I agree that the odds are 50-50 (as long as you can guarantee the specific distribution of cards in column 1 and column 2. However suppose you ask your buddy to pick a card, tell you the color, but NOT tell you whether he picked from the first or second column. Suppose he says he picked B. The only three combinations available then are RB, BR, or BB: 2 to 1 in favor of the other card in the pair being the opposite color. Note that if you further ask him "Which column is the B card in?", and he says "the first column", you can eliminate RB as well and drop the odds back to 50-50.

This is really not that hard, and the subtlety of the language has been explained every which way know to man. The connundrum is in the hidden assumption, and whether or not they are reasonable. The analogy of picking cards in this way with identifying one of two unspecified children's gender does not hold up because there are different assumptions written into the rules of each game.
#30
08-11-2006, 04:35 PM
 CJJ* Guest Join Date: Nov 2004 Location: Chicago Posts: 2,268
Quote:
 Originally Posted by adbadqc Let's talk red and black cards rather than genders. If you set up, face down, two columns of cards in the following pattern: R R B B R B B R You have the four possible distributions. If you randomly select one card and turn it over, what is the likelihood the paired card is the same color? THESE COMBINATIONS ARE PAIRED! That's the key statistical concept many people are missing here.
The key point you seem to be missing here is that, as the original problem is stated, you do not know whether the child identified by gender is the older or younger child. In your card scenario, this is equivalent to not knowing whether you've turned over a card from the first column or the second column

In your card problem, I agree that the odds are 50-50 (as long as you can guarantee the specific distribution of cards in column 1 and column 2). However suppose you ask your buddy to pick a card, tell you the color, but NOT tell you whether he picked from the first or second column. Suppose he says he picked B. The only three combinations available then are RB, BR, or BB: 2 to 1 in favor of the other card in the pair being the opposite color. Note that if you further ask him "Which column is the B card in?", and he says "the first column", you can eliminate RB as well and drop the odds back to 50-50.

This is really not that hard, and the subtlety of the language has been explained every which way know to man. The connundrum is in the hidden assumption, and whether or not they are reasonable. The analogy of picking cards in this way with identifying one of two unspecified children's gender does not hold up because there are different assumptions written into the rules of each game.
#31
08-11-2006, 04:42 PM
 Bytegeist Guest Join Date: Jul 2003 Location: Maryland, US Posts: 2,377
Quote:
 Originally Posted by adbadqc Let's talk red and black cards rather than genders. If you set up, face down, two columns of cards in the following pattern: R R B B R B B R You have the four possible distributions. If you randomly select one card and turn it over, what is the likelihood the paired card is the same color?
Fifty percent, as you say, but you're not solving the same problem as the one stated.

To translate from the original Boy/Girl Problem, you are not looking at any particular card. You are presented with a pair of cards, randomly chosen from the four you list above, and then told that there's at least one red card in the pair. (So the B-B pair is actually eliminated from consideration.) You're then asked what the probability is that there's a black card in the pair. This probability is 2/3.
#32
08-11-2006, 04:54 PM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
Quote:
 Originally Posted by wissdok As we looked at the earlier 1000 families, the 2/3 “rule” can’t be correct for both boys and girls at the same time.
It can.

The problem is perfectly symmetrical. If we randomly select a family and are told this it includes at least one son, we can say that the chance this boy has a sister is 2/3. If we are instead told that it includes a daughter, the probability of this girl having a brother is again 2/3.

There is no contradiction or difficulty here.
[/QUOTE]
#33
08-11-2006, 05:01 PM
Sorry, wrong.

Sorry: I'm not making my point very well.

Speaking from the standpoint of statistics as a science (and I have to warn the listening audience my UC diploma was signed by a California Governor who was an actor and his first name isn't Arnold... so it's been a few decades), there are two different concepts here; possible outcomes and distributions.

The possible outcomes (sticking with cards) are B B, R B, and R R. These are not order sensitive and do not address how frequently a particular outcome comes up.

The distributions are
25% B B
25% R R
25% R B
and
25% B R

These ARE order sensitive. They have to be, in order to accurately reflect the distribution.

You can choose to label the columns anything you want (oldest, first in the door, what my friend chose, etc.), but if you're attempting to analyze statistical odds you have to deal with the distribution in the manner I suggest. Look at it this way; consider the label of the first column "what was revealed to me".

If you still disagree, I'd like to meet you on a streetcorner for a game of 3-card Monte; bring lots of 20s.

#34
08-11-2006, 05:12 PM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
Quote:
 Originally Posted by wissdok With Cecil’s puzzle it is possible for 2/3 to be true IF we assume that all families with girls tell us of them.
I think you are introducing an unnecessary and unhelpful addition with this concept of the family "telling" something. Is the family truthful? Do all families know the sex of all their children?

Irrelevant! The family is chosen at random, and you are told (assumption: truthfully, by whom it doesn't matter) that at least one of their children is a daughter. This allows you to conclude that they are not among the 25% of families that have two sons, and thus there is a 2/3 probability that their daughter has a brother.
#35
08-11-2006, 05:23 PM
 CJJ* Guest Join Date: Nov 2004 Location: Chicago Posts: 2,268
Quote:
 Originally Posted by adbadqc The possible outcomes (sticking with cards) are B B, R B, and R R. These are not order sensitive and do not address how frequently a particular outcome comes up. The distributions are 25% B B 25% R R 25% R B and 25% B R These ARE order sensitive. They have to be, in order to accurately reflect the distribution.
No; we could just as easily stick with the non-order sensitive grouping B B, R B, and R R and say the distributions are (based on our knowledge of how the groups are created):

25% B B
25% R R
50% R B

You agree these are the distribution percentages, right? From here its obvious that for a group with, say, R in it (R R or R B), the mis-match occurs twice as often, so it's 2:1 in favor of claiming the unseen card is the opposite.

Knowledge of how the cards are distributed helped us determine the percentages, but once these are determined we no longer care in what order the cards were dealt.

Quote:
 You can choose to label the columns anything you want (oldest, first in the door, what my friend chose, etc.), but if you're attempting to analyze statistical odds you have to deal with the distribution in the manner I suggest. Look at it this way; consider the label of the first column "what was revealed to me".
This would be true only if you could guarantee that the "revealor" will always chooses from the same column first--equivalent to always identifying the gender of the older or younger child in the original problem. The whole point of the original problem is that it is set up such that you cannot specify in advance which child (or which card column) is chosen first.

Once again, from your distribution, pick all pairs which have at least one black card. Two of these pairs are matched with a red card, one is not. Thus, if you do not know in advance which column is chosen first, the odds are 2:1 in favor of saying the matched card is red.

Quote:
 If you still disagree, I'd like to meet you on a streetcorner for a game of 3-card Monte; bring lots of 20s.
Sure thing; be prepared to lose your shirt!
#36
08-11-2006, 05:24 PM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
Quote:
 Originally Posted by adbadqc Look at it this way; consider the label of the first column "what was revealed to me".
But what was revealed was exactly this: One, or the other, or both children are girls. Your analysis isn't properly dealing with these three possibilities, each of which is equally probably.
#37
08-11-2006, 05:49 PM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
For Riddler, wissdok and adbadqc (and anyone else who'd like to have a go):

Suppose I take 4 sheets of paper, number them 1 through 4, and write the following on them:
Sheet 1: BB
Sheet 2: GG
Sheet 3: BG
Sheet 4: GB
I proceed to take sheet #1 and burn it, taking care to stomp the ashes into the ground.

I then ask a friend to randomly select one of the remaining sheets of paper by a process that makes each of the three selections equally likely.

1. What is the probability that he selects a sheet of paper with two different letters on it?

2. How does this differ from the original problem?
#38
08-11-2006, 06:10 PM
 Bytegeist Guest Join Date: Jul 2003 Location: Maryland, US Posts: 2,377
Quote:
 Originally Posted by adbadqc The distributions are 25% B B 25% R R 25% R B and 25% B R These ARE order sensitive. They have to be, in order to accurately reflect the distribution.
Agreed with this bit.

Quote:
 You can choose to label the columns anything you want (oldest, first in the door, what my friend chose, etc.), but if you're attempting to analyze statistical odds you have to deal with the distribution in the manner I suggest. Look at it this way; consider the label of the first column "what was revealed to me".
And here I think is the snag. No specific card has been revealed to you. No specific card has been identified as red. You are never given any information about any particular, designated card at all. You are only given aggregate information about a pair of cards: that there's at least one red card in there, somewhere. Could be they both are. And the only thing this information lets you do is eliminate the B-B pair from consideration.

Out of the allowed population of possible card-pairs, exactly 1/3 are R-R (and have no black), and 2/3 are mixed (with one black). You are being asked what the probability is that there's one black card in the pair — or effectively, the probability that the pair is mixed.
#39
08-11-2006, 06:21 PM
 Bytegeist Guest Join Date: Jul 2003 Location: Maryland, US Posts: 2,377
Quote:
 Originally Posted by adbadqc If you still disagree, I'd like to meet you on a streetcorner for a game of 3-card Monte; bring lots of 20s.
Quote:
 Originally Posted by CJJ* Sure thing; be prepared to lose your shirt!
As an aside . . .

I don't hang around city street corners much (well not anymore; please don't ask) but is there really any probability analysis to be done for Three Card Monte? It's purely a confidence game, I thought.

Even an "honest" game of Three Card Monte would just come down to a contest of physical skill: the dealer's dexterity versus the player's eye.
#40
08-11-2006, 06:53 PM
sigh

Yeah; you got me on the 3-card-monty(monte?).

Consider this;

Once you know you have one girl (let's use a capital G for the "known girl", and lower case for the unknowns), these are the following possibilities (using the column example):

25% G g
25% g G
25% b G
25% G b

The known girl "card" can be either in the first or the second column, and can be paired with either an unknown boy or an unknown girl in the other column... it's a different distribution than the original 4, but it still results in a 50-50 split.
#41
08-11-2006, 07:01 PM
 tim314 Charter Member Join Date: Mar 2004 Posts: 4,460

Consider the choices:

RR
RB
BR
BB

What you're saying is if one is revealed to be R, then there's a 50-50 chance the other will be revealed to be B. However, this is only true if the one which was revealed was chosen at random. In the following, I will show that this problem (as stated) has the one which is revealed chosen systematically, which changes the odds.

First, consider the case where we reveal a card at random. Without loss of generality, we can rewrite the pairs so that the revealed card is on the left. So the possibilities are:
RR
RB
BR
BB
all equally likely.

Of the cases where the revealed card (the left one) was R, the other card is B half the time and R have the time. 50-50, like you said. But this isn't the same as the given problem, for the reason I'm about to tell you.

In the given problem, the parents tell you whether they have at least one girl child. In other words, if they have even one girl child, they will systematically choose that child as the one to reveal. This is equivalent to systematically chosing to reveal a red card if you have one.

So, the four equally probable possibilities (with the revealed card placed on the left) are:
RR
RB
RB
BB

Note that the BR flipped to a second RB, because they never reveal a B when they have an R.

So now, there are three cases where an R was revealed, and in two of the three the other one was a B. The odds are 2/3.

The point here is that answering the question "Do you have any girl children?" is not the same as answering the request "Tell me the sex of one of your children." In one case, you systematically reveal a girl child (if you have one), in the other you reveal one at random (presumably).

Although on further review, perhaps I'm being unfair. Cecil orignally said "You have been told they have a daughter," which is ambiguous as to whether they systematically revealed a daughter or just happened to reveal that child by chance. His subsequent explanation makes it clear that he meant it to be revealed systematically (as it would be if you had asked them "Do you have any daughters?") but he probably should have been more clear in the original statement.
#42
08-11-2006, 07:09 PM
 tim314 Charter Member Join Date: Mar 2004 Posts: 4,460
So, in summary, it all depends on whether the revealed child is chosen randomly or systematically.

If you always reveal a girl if you've got one, then you'll reveal a girl three times out of four, and two out of those three times the other is a boy. Answer: 2/3

If you reveal a kid at random, then you'll reveal a girl two times out of four, and one out of those two times the other is a boy. Answer: 1/2

So it depends on how you interpret "You have been told this family has a daughter." Cecil's explanation makes it clear he meant it in the way that gives 2/3, but he could have been more clear. If he had said "You ask if they have any daughters, and they say 'yes'" then the answer would be 2/3 unambiguously.

Can we all agree on this?
#43
08-11-2006, 07:21 PM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
Quote:
 Originally Posted by adbadqc Once you know you have one girl (let's use a capital G for the "known girl", and lower case for the unknowns), these are the following possibilities (using the column example): 25% G g 25% g G 25% b G 25% G b
The first two possibilities presented here are in fact one possibility (i.e. one family) with a probability equal to the other two: 1/3.

Focus on the fact that the problem specifices chosing a family, not a child.
#44
08-11-2006, 07:35 PM
Sorry

I really do understand what you're saying, but what I'm trying to say is Statistics doesn't work that way and since it's somewhat counter-intuitive (and I've had a long day) I'm not doing a very good job of explaining it.

You can't flip a BR and make it an RB. The order-sensitive distribution model differentiates between a B showing up before an R and vice versa.

BB
BR
BR
RR

is NOT a statistically accurate distribution model for this problem; RB is not the same thing as BR. So let's try this with the other model (where order doesn't matter) and see where that takes us.

25% BB
50% BR
25% RR

So.... you know you have an R. Therefore since 2/3rds of what's left of the original distribution has a B, then it must be twice as likely as an R, right?

Wrong. Here's why. YOU HAVE TO THROW AWAY HALF OF THE BRs!

???

That's right. Why? Because I said so. No; just kidding.

Because the selection from that subset of the population results in a R only 50% of the time; therefore you can't include the entire subset in the next stage of the problem, you have to take into account that your actual selection (if random) had a 50% chance of selecting if it came out of this cohort*, therefore only 50% can be included in the next sampling, along with 100% of the RR cohort, and 0% of the BB cohort.

That evens up the size between the BR and the RR, and we're back to 50-50.

* cohort: a band or group of people
#45
08-11-2006, 07:55 PM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
Quote:
 Originally Posted by adbadqc BB BR BR RR is NOT a statistically accurate distribution model for this problem; RB is not the same thing as BR.
What in the original statement of the problem makes RB different from BR? Please quote the relevant words.

More specifically, what difference is there between:
1. Choose one of 3 possible families from this list: GG, BG, GB
2. Choose one of 3 possible families from this list: GG, BG, BG
#46
08-11-2006, 09:20 PM
It's a question of the correct statistical model

OK; I think I need to stop trying to explain this and just ask folk a few questions:

1) Based on your (speaking to the audience at large here) understanding of biology and how the world works (put DOWN that statistics textbook and step away from the table), what would you expect the sex ratio of a first child to be? I assume most of you answered 50-50 male/female.

2) Based on your understanding of biology etc. etc. etc, what would you expect the sex ratio of the SECOND child in a family to be? I'm guessing (hoping) most of you said 50-50 male/female.

3) Based on your understanding of biology etc. etc. etc., would you expect the sex of the first child in a family to have an impact on the sex of the second child in a family? (If you answered "yes", please explain the mechanics of this effect).

4) Given the above, and leaving that statistics book lying there; what would you expect the sex ratio of the other sibling in a family to be where there is one daughter? Unless you're cheating, you said 50-50 male/female. Think about it.

So..... if your understanding of biology suggests that if a two child family has one daughter, the likelihood of another daughter is 50% and a son is 50%, and your understanding of statistical manipulation (you can pick up that textbook now) suggests that a brother is twice as likely as a sister.... where's the disconnect?

You have some choices:

1) Your understanding of biology is flawed
2) Your understanding of the appropriate statistical model/manipulation is flawed.
3) Something else (possibly I'm horribly misreading the problem statement).

I've tried to explain the appropriate application of statistical modeling techniques to this problem, and how they adequately explain the expected result; people seem to be having problems with that. So for those of you in that camp, what is it you're suggesting? A daughter in a two child family is twice as likely to have a brother as a sister? That knowing that one child is a daughter increases the likelihood that the other child is a son?

This is taking longer than I thought.
#47
08-11-2006, 10:28 PM
 wissdok Guest Join Date: Aug 2005 Posts: 107
Quote:
 Posted By CJJ In your card problem, I agree that the odds are 50-50 (as long as you can guarantee the specific distribution of cards in column 1 and column 2. However suppose you ask your buddy to pick a card, tell you the color, but NOT tell you whether he picked from the first or second column. Suppose he says he picked B. The only three combinations available then are RB, BR, or BB: 2 to 1 in favor of the other card in the pair being the opposite color. Note that if you further ask him "Which column is the B card in?", and he says "the first column", you can eliminate RB as well and drop the odds back to 50-50.
I once tried to explain this with coins and then cards but I couldn’t get people to agree on how to transpose this into cards. As both children in a family are independent events you should have two piles with each having a red and a black. If your friend draws one card from each pile, then it doesn’t matter which pile the card is from that he reads to you.This is because the other pile had exactly one red and one black. Hence, either card he tells you about the chances of guessing right…50% / 50% In a round about way this works with Adbadqc biology statement, there isn't 3 cards in that other pile, just two...a black and a red.

Quote:
 Posted By Xema The problem is perfectly symmetrical. If we randomly select a family and are told this it includes at least one son, we can say that the chance this boy has a sister is 2/3. If we are instead told that it includes a daughter, the probability of this girl having a brother is again 2/3. There is no contradiction or difficulty here.
These two events are not simultaneously possible. If we take just four families (representing each of the possible types) then if the given is a son and it turns out that we get a 2/3 result…then at the same time if we are given a girl for another of the families then it has to be from the family with only girls and we would automatically get that wrong using the 2/3 “rule”. This happens because to get the a 2/3 result, the two families that have a mixed set of kids, must have given us the clue of “ a boy”…leaving only the family with two girls to answer with the clue of “a girl.” The laws of probability state that the families with both a boy and a girl would randomly give us a clue, half the time boy, half the time girl. If we use a bigger sample group then we will make the 2/3 result even less likely to happen. Either way, this is not a repeatable experiment as even in a group of four families the odds of two families with mixed kids giving matching clue is only 50%. And again I point out that we can’t have a 2/3 result for boys at the same time we have a 2/3 result for girls.

Quote:
 Posted By Xema I think you are introducing an unnecessary and unhelpful addition with this concept of the family "telling" something. Is the family truthful? Do all families know the sex of all their children? Irrelevant! The family is chosen at random, and you are told (assumption: truthfully, by whom it doesn't matter) that at least one of their children is a daughter. This allows you to conclude that they are not among the 25% of families that have two sons, and thus there is a 2/3 probability that their daughter has a brother.
I once thought there was a difference between the puzzle when you used tell, meet, were given, …etc. After I studied this problem enough I came to the conclusion that it doesn’t matter. If we are given a child, told about a child, or even meet the child…we are told all we can base our answer on. The information must be factual; there is no reason to say otherwise. There is an argument that meeting a child allows you to separate it from its sibling, but the puzzle fails long before we have to get that far into the problem.

Quote:
 Posted by Xema Suppose I take 4 sheets of paper, number them 1 through 4, and write the following on them: Sheet 1: BB Sheet 2: GG Sheet 3: BG Sheet 4: GB I proceed to take sheet #1 and burn it, taking care to stomp the ashes into the ground. I then ask a friend to randomly select one of the remaining sheets of paper by a process that makes each of the three selections equally likely. Two questions to answer: 1. What is the probability that he selects a sheet of paper with two different letters on it? 2. How does this differ from the original problem?
Here you have kept the Sheets(families) as a random event, but what is written on them is irrelevant(letters=children). At the being of the boy/girl problem that is true too, but in the second part of the puzzle the children do matter. The children are independent events, not like the letters. While yes this problem is 2/3, there is a big difference between it and the original puzzle because of the lack of a second part.
#48
08-11-2006, 11:05 PM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
Quote:
 Originally Posted by adbadqc So..... if your understanding of biology suggests that if a two child family has one daughter, the likelihood of another daughter is 50% and a son is 50%, and your understanding of statistical manipulation (you can pick up that textbook now) suggests that a brother is twice as likely as a sister.... where's the disconnect?
Where's your acknowledgement that the essential condition of the problem is to exclude one of the four types of families? That nothing posted by those who you disagree with in any way argues aginst the premise that the population of children is half male, half female (indeed, this is stated as one of the conditions of the problem).

It's hard for me to see why, at this stage in this thread, you would devote a bunch of space to a discussion of obvious biology instead of addressing the points that have been made. Instead, why not answer the questions in post #37?

Quote:
 ... what is it you're suggesting? A daughter in a two child family is twice as likely to have a brother as a sister? That knowing that one child is a daughter increases the likelihood that the other child is a son?
How about: that knowing the family isn't BB reduces the possibilities from 4 to 3.
(Has this not already been stated enough?)
#49
08-11-2006, 11:14 PM
 Ivan-Osokin Guest Join Date: Dec 2000 Location: Seattle Posts: 46
I hate you Monty Hall.

Chances that everyone in any thread covering this topic will agree or come to understand that the answer is 2/3?

1 - .9999 repeating.
#50
08-11-2006, 11:21 PM
 Xema Guest Join Date: Mar 2002 Posts: 11,570
Quote:
 Originally Posted by wissdok At the being of the boy/girl problem that is true too, but in the second part of the puzzle the children do matter. The children are independent events, not like the letters. While yes this problem is 2/3, there is a big difference between it and the original puzzle because of the lack of a second part.
Thanks for acknowledging that the probability here is 2/3. What remains is to decide whether or not this problem is congruent to the original one.

You say it isn't because it lacks a second part. I looked back at the original problem and found just one part: we have a family with two children and information that at least one of them is female. I say this is the same as being told that a selected sheet of paper has at least one "G" on it.

I agree if there were a "second part" to the problem - if, for example, we were to select a child at random - then the problem would not be the same as the one with sheets of paper. But I see nothing in the original problem that calls for the selection of a child. If you can find such language in the problem, please let me know.

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