In my “money always flows towards the writer” post, I said that if your odds of winning a contest was 1% you had a good chance of winning over 100 tries. A few people took issue with that in the comments so I thought I’d explain the logic here with a fun example:
Your neighbour has two kids. One of them shows up at your door and it’s a girl. What are the odds that the other one is a boy? 50/50 right? R-O-N-G, wrong. It’s 66%. Wanna see why?
These are the possible combination of the sexes of your neighbours kids:
Boy Boy
Boy Girl
Girl Boy
Girl Girl
In our example, one of them is a girl, we know that, so the first one can’t be true. That leaves the other three. In two of them, the other one is a boy. So the odds are 66%.
Now, people say that the sexes of the two kids aren’t related to each other, and here is where we make an important distinction. If I’d said: The oldest one show up at your door and it’s a girl, then the odds would in fact be 50% that the other one is a boy, because at that point, these are the possible combinations.
Girl Girl
Girl Boy
See how that works? 50% chance.
Don’t believe me? You can test this quite easily at home with a couple of coin flip. Flip a coin three times. These are the possible combinations of flips.
H H H
H H T
H T H
T H H
H T T
T H T
T T H
T T T
In only one of those do you NOT get tails, right? So what are your odds of getting all heads? 1/8.
However, if you ask what are the chances of you NEXT flip being heads, it’ll be 50%. Every time.
113 Comments(+Add)
I agree… 50% but I guess it never turns out that way does it?
Iim trying to win the lotto, and chances of winning are 1 in 14 million!!
but hey there’s a chance!
I play the lotto, not with an realistic chance of winning, but because it allows me to dream of what I’d do with the winnings!
I didn’t realize my nickname was “a few people”
You might want to refresh your statistics.
The odds of the sex of one child are not dependent of the sex of the other child, so odds are 50%. If you include the other child in the equations, you are doing it wrong. (Actually I think a random child is slightly more likely to be a girl, but that’s a demographic thing)
And, actually, both your examples are wrong. In the coin flip example, the odds are 1 in 8, not 1 in 9.
Each flip is independent, and the chance of heads on each is 1/2. So to calculate the chance of 3 out of 3 being heads is (1/2)*(1/2)*(1/2), or 1/8.
This isn’t really any different than your method, the reason you came up with the wrong answer (or the “rong” one) is that you listed HHT twice.
And, for further explanation of why the boy-girl solution is wrong, if you already know the kid who visited your door is a girl, then the:
Boy Girl
and
Girl Boy
solutions are equivalent.
Either way, it’s the OTHER kid that is is the boy, there are only two possibilities:
Girl Girl
and
Girl Boy
so it is a 1/2 chance of either sex.
And, when you’re calculating the cost-benefit analysis of the likelihood of profiting from a submission, there are many unknowns, including the number of other submitters, the tastes of the editors, etc… And the quality of a story is not quantifiably measurable, so any two people will come up with different opinions of the story on their own.
Anyway, have a good day.
Signed,
David “a few people” Steffen
Dave. You’re wrong. You can’t just magically make two possible combinations into one by saying “They are equivalent”.
Here, what if I asked you what are the odds that your neighbor has two boys? No sweat you say! It’s 1/2 * 1/2 = 1/4. Recognize that one quarter from anywhere???? Right. It’s the 1/4 situation I outline above. It’s not 1/3 because two of the scenarios are magically equivalent.
Yeah, sorry about the HHT thing, but my reasoning still stands.
Dave, It’s always easier to say “a few people” so that it doesn’t look like I’m picking a personal fight with someone on my blog.
Sorry, answering you comments one-by-one.
Re: the OTHER child.
As I mention above, if I’d say “the oldest, or the first” child, you’d be right. But since I don’t tell you which one it is, the four combination solution stands.
This is a well known statistics problem. I think I can even find the textbook that outlines it.
Re: The cost-benefit analysis. Yeah, I mention in the post that the weak part of it is quantifying your odds. However, you can do some research here.
First, yeah, read the magazine. That’s tell you the editors’ tastes. Second, try and find out how many submissions they get a month? Asimov’s is like 600. GlimmerTrain, if it’s proportional to their payment rate (and it >usually< is) would probably get… 800?
Now, since this is a pay contest (which eliminates guys like Shaun from the slush pile), and it’s for new authors only (eliminating the pros), odds are probably substantially better than for Asimov’s.
I would wager that you are misquoting the wording of the textbook. The way you worded it, the relative ages of the children doesn’t matter.
It could be reworded so that it does matter, such as “The odds that the older child is a girl”. Since you know that this child is a girl, but you don’t know if she’s the oldest, then this makes the boy-girl and the girl-boy into separate cases, and the odds would be 2 out of 3, as you suggest.
By saying “older”, the cases aren’t independent, because you don’t know which is older.
However, by saying “other” instead of “older”, the ordering doesn’t matter in the slightest. There is one other child, and that child’s sex does not depend on this one.
So, you’re still wrong. If you find the textbook, I’d be interested in seeing the question typed out exactly as they worded it.
signed,
a few people
You can tell the editor’s tastes by reading the mag, but I insist that you can’t quantify them, especially in regards to a particular slush story.
Payment rate is not necessarily proportional to submission rate. A magazine with more widespread demographic of viewers will have more submissions, and the reputation and publishing history of the mag count into it as well. Let’s say The New Yorker pays 5 cents a word, and let’s say Asimov’s does too. I guarantee The New Yorker will get significantly more submissions because they are so well established and appeal to a larger portion of the population.
>>”You can’t just magically make two possible combinations into one by saying “They are equivalent”.”
There’s no magic about it. They are two equivalent ways of looking at the SAME solution.
Just as 4*x is the same thing as x+x+x+x, so are these solutions the same. One is more concise, one is quicker to draw, but they are the same thing.
For instance, if you wanted to expand your three flips into 6 flips, I could use my method very quickly: 1/2 to the 6th power equals 1/64, which is the odds of coming up with 6 tails in 6 flips. To do it your way you’d have to list out 64 different possibilities, which would both take a long time and be prone to error. If done right, both would always come up with the same solution, so they are equivalent.
Dave,
sheesh you’re combative today. Re glimmertrain, see my weasel word “usually”
re: girl boy. Let’s label those columns “a” and “B”, shall we? That way it’s clear that we have two distinct human beings here, child A and child B. Since I don’t tell you if you’re looking at child A or child B, the child could be EITHER. Therefore both possibilites 2 & 3 are distinct and possible, and NOT equivalent, because the columns represent actual real people who can’t be swapped around.
Yup, I’m combative because you’re wrong.
Labelling columns doesn’t change the facts. The sex of the other child is independent of this one. You cited a textbook as the source of this question. If you want to prove your point, I suggest typing out the question verbatim. I would be willing to wager that it is not saying the same thing you are saying.
Instead of sister, let’s say you pick two names off an author list to meet at a convention. You don’t know their gender, and they both use initials for first names, like LP Hopland, and HM Hogenstern. Let’s say for the sake of this problem that half of writers are female and half are male. You meet LP and she turns out to be a woman.
At this point you want to figure out the odds of HM being a woman also. This problem would produce the SAME chart with the SAME columns as you suggested. What you are saying is that there is a 2/3 chance that HM is a woman.
Numbers don’t justify themselves. For the gender of one person to affect the gender of the other, there has to be a REASON for it. If there’s no reason, then the numbers are wrong.
Heh, David, hate to break it to you but you’re actually wrong on this. It’s a variant of the famous game show problem:
http://socialmode.com/2008/05/29/game-show-problem-a-common-problem/
The Boy-Girl and Girl-Boy situations are not equivalent. One way to say this is that any couple with 2 kids, assuming 0.5 probabilities at each birth, is more likely to have a boy and a girl than two girls. (You can only get two girls by first having a girl, then having a girl – 0.25 chance. You can get a boy and a girl by first having a girl, then a boy or first a boy, then a girl – each 0.25 chance, for a combined 0.5 chance). And once you find out one is a girl, without knowing if she came first or second, the likely answer is that the other one’s a boy.
Some very smart people first come to the same conclusion as you, though. It really takes stepping back and thinking about it to figure out why it works the way it works.
(By the way, though, Jordan’s not right about the contest probabilities. . .the probability curve there isn’t smooth. If you lack a certain level of skill that marks basic competence, your chance of winning the contest is 0. If you’re above a certain level of skill, it’s more like 1/20 or 1/30 or whatever).
Dave,
saying that A B is equivalent to B A is so fundamentally wrong I just don’t know where to begin. That’s like saying, in the coin example, that hht = thh=hth. I’m not going to dig around in boxes to prove something I’ve already proven.
If you think that thh=hht then I don’t know how to proceed.
(though now that i look at what Jordan said in the previous post, he’s not calculating directly from the number of entries so it’s a somewhat better argument. still, i’d say it’s a lot more about skill than about probabilities. . .)
Another way to say it – one person’s gender doesn’t affect the other person’s gender. But if you know that, upon picking out any random 2 people from the crowd, you’re more likely to get a male-female combo than a single gender combo, and then you isolate two people and learn the gender of one of them, you know what gender the other one is more likely to be.
(so in your writers of example, you’d actually expect the other writer to more likely be a man)
Yes, if you pick two random people, they are more likely to be male-female. But in this problem, you already know the gender of one, it is no longer unknown. The only unknown is the single person’s gender. That changes everything.
Right, but the point is, you picked the two people BEFORE you learned their gender. So if you know the pair you’ve picked is more likely to be male-female, and you learn one of them is male, well of course the other one is more likely to be female.
Allow a math teacher — one with a masters in mathematics — to step in. You’re both right.
Dave is correct, the probability of any child being male or female is 50%, regardless of the possible combinations.
I also see Jordan’s point. Unfortunately it is based off of experimental probability and not theoretical probability. ASSUMING that 25% of two-child families have two girls, two boys, older girl/ younger boy, and older boy/younger girl, then his 66% is reasonable because we don’t know if the girl is older or younger. But there is no evidence of this. For instance, families with two girls may be more likely to have a third child, eliminating them from the group.
If we knew the older child was agirl, even this experimental model would dictate a 50/50 chance of the younger being a boy.
Still, these are independent events and one does not dictate anything about the other.
A much more fun version of this is the Let’s Make a Deal paradox (http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html) where the events are technically dependent.
The gameshow problem, while interesting, is not at all the same problem. Knowing the contents of ONE door does affect the contents of ANOTHER door. But knowing the gender of ONE person does not tell you the gender of the other.
By the way, fascinating debate.
I may have to write a flash piece regarding the Let’s Make a Deal paradox. Hadn’t thought of that…
Oh, didn’t see the previously-existing link. Oh well, love being redundant.
Tom,
just putting the phrase “of course” into a sentence doesn’t make it true, I’m afraid.
Oso, thanks for chipping in.
I’M disagreeing with you her, Tom.
The columns in dave’s example are lik this:
lp. Hm
girl. Girl
girl. Boy
boy. Girl
boy. Boy
dave said that lp is a girl, eliminating choices 3 & 4, therefore odds that HD is a girl is 50/50. The difference is that we know which column we’re looking.
But thanks for bringing up the Monty hall problem. That illustrates the solution nicely!
Jordan said: “That’s like saying, in the coin example, that hht = thh=hth.”
Nope. In the coin example, it’s like saying that if the first coin comes up heads, then that fact does not change the probably of the second one coming up heads, which is still 50%. And the fact that the first one came up heads does not change the probability of the third one coming up heads, which is also still 50%.
Jordan said: “I’M disagreeing with you her, Tom.”
Explain how the author example is fundamentally different than the child example. According to your original explanation, you don’t KNOW which “column” LP is in, so it would be a 2/3 chance.
Sorry, I’ve benn writing this all on my iPhone. Dave, lp is in the lp column because it’s called the $&@! “lp” column. She is the $&@: column.
You labeled the columns, not me. The columns are merely a representation of the problem.
In the original boy-girl problem, I could ask the girl her name, she could say “Leslie” and I could label the first column the “Leslie” column. It doesn’t change a thing. The other child is still 50% chance.
The difference, Dave, is that in the lp problem, you know which column to eliminate because you can SEE lp. In the girl A/girl B problem you don’t know which of them is in front of you.
As long as we’re throwing credentials around, I’m a recipient of a national scholarship (canadian millenium excellence award), won the dean’s award and was nominated valedictorian. Plus I do this shit for a living as Manager of Development at my job.
Who was throwing credentials around? I haven’t mentioned my education or occupation, only facts. Congratulations on your academic success, and that you do “this shit” for a living. I’m not impressed by your credentials, nor am I impressed by cursing. I’m impressed by facts.
You put too much importance on the columns. The columns are your manner of representation and they are misleading you.
Do you still have the textbook for this handy? Can you type out the problem?
Jordan, not sure what part you’re disagreeing with.
David, sorry for the of course. This is simply such a familiar problem I was surprised to see it generating so much argument. And yes, the underlying principle _is_ essentially the same as the Monty Hall/gameshow example.
And Oso, yes, boys-girls isn’t ideal because there’s biology etc involved. The coin toss is a much better example:
Let’s say you toss two coins at once but don’t look at either. We both agree, I think, that the most likely combination is tails-heads. We would both agree that if you look at one of them first, it’s going to be EITHER tails or heads. So are you saying that simply looking at one of the coins first alters the fact that tails-heads is more likely? You always look at one of the two first. Even after you’ve looked at one, tails-heads is still more likely. Looking doesn’t change that (haha, except in quantum physics).
P.s. heh, as long as we’re comparing certs, I’ve a degree in physics, did research on black holes, and wrote my thesis on special relativity. . .but that doesn’t help my argument much! We’ve already said all that needs to be said here. At some point further argument doesn’t change anything. One has to be willing to consider that one might be wrong (which I did several times during the writing of this post) before one can be convinced. And it’s hard to do that in the middle of a heated argument. . .
A couple has two children. In the order they had them, there are four possibilities:
B B
B G
G B
G G
The question was, if the odds of having a boy or girl is 50-50, then if a girl shows up at the front door, what are the odds the other sibling is a boy?
By saying a girl showed up at the door, it does NOT make the LH column above = “Leslie = Girl”, because two of the entries in the LH column say B. You are NO LONGER looking at the full set of possibilities. You are looking at the subsets which have at least one girl:
B G
G B
G G
Since you don’t know whether the girl at the door is the first child or the second child, there are 2 possibilities the the other child is a boy, and only 1 that the other child is a girl.
Note that other child is not the same as second child.
The probability that the other child is a boy is 2/3.
Once you made a choice, as in a girl showed up at your door, then you are dealing with a subset of the complete set.
A DIFFERENT PROBLEM: Consider a Schroedinger’s Cat Box inside a Schroedinger’s Cat Box. Each box has a 50-50 chance that when you open the box the cat is dead or alive.
If you open the outer box and the cat is alive, what are the odds the cat in the inner box is alive? Answer: 50-50. Because in this case, the living cat in the inside box cannot be the first cat you run into.
But a girl in the family who comes to the door could be either the first or second born. So these are not the same problem. Sorry.
NOTE: We are not considering outside effects such as the family choosing to have a second or third child based on the gender of the first, or any chemical or familial traits which make the preference not 50-50 in the first place.
Dr. Phil
Tom, no need to apologize for the “of course”, I just didn’t have much to say in response to it. .
It is not the same as the Monty Hall show, because in the monty Hall show, the first choice changes the probability of the other.
Yes, looking at one of the coins, does change the overall probably of heads-tails, because you have more information. Having more information changes the probability changes the problem.
to apply that same question to the probability of getting three heads in a row, as Jordan pointed out that is 1/8. But if the first coin comes up heads, that changes the remaining odds of getting 3 heads in a row, because more things are known, and less are unknown. Instead of eight possibilities, you have 4:
HHH
HHT
HTH
HTT
Note that the first value is always heads, because that is a known quantity. That means that the overall probability of getting 3 heads is now 1/4 because you know more.
BUT it still doesn’t change the probability of the second coin alone coming up heads, which is still 1/2.
Hi Dr. Phil.
I’m afraid you’re also wrong, for reasons I’ve already explained. You’ve said pretty much the same things as Jordan has said and I still disagree for the same reasons.
In any case, this has turned out to not be quitesuch a “basic” lesson after all, since a bunch of educated folks haven’t come to a consensus.
Oh, and as for collapsed waveforms in quantum mechanics, yes you can get two degenerate waveforms, but there are still two of them, not one. So in the classic helium atom with two electrons, the spin possibilities are UU, UD, DU, DD. But the UD and DU solutions end up being reformatted as two mixed states of UD +/- DU. There are still four states, however.
Simplified, “of course”, and I’ll stop this here. (grin)
Dr. Phil
Saying that I am wrong and me being wrong doesn’t make me wrong.
John Jill
John Bob
Joan Bob
Joan Jill
Joan or Jill comes to the door.
If Joan comes, it could be Bob or Jill.
If Jill comes, it could only be John.
The statistics are different here because Joan is not equal to Jill, these are discernible individuals and not identical particles.
And yes, it really does make a difference.
Dr. Phil
Saying that I am wrong and me being wrong doesn’t make me wrong.
Gah! Tangled syntax.
Saying that I am wrong and me actually being wrong are not equivalent.
But one does have to be careful about how one states the problem. I learned this in Statistical Mechanics, Fall 1977.
Dr. Phil
“Saying that I am wrong and me actually being wrong are not equivalent.”
True enough. But if you use the same arguments as Jordan, then I will just be using the same counter-arguments, so there’s not much point in me posting the same things all over again.
Thanks for all the awesome comments, guys, but. Tom’s right. I think we’re going in circles here. I sent a private email to Dave because he’s awesome and I wanted to make sure he wasn’t taking this personally.
Let’s drop the statistics comments for now and confine ourselves to awesome physics problems!
OK, so most of the arguments in the other directions have been based on column labeling, so I’ll address that.
In your representations, you’ve been labeling the columns by age of the child. That’s fine, but nothing in the problem suggests that age is significant, so let’s say I choose a different labeling system.
Instead of:
Youngest child, Oldest Child
I will be labelling them as:
First child I meet, second child I meet.
The first child I meet (column 1) is a girl, so the remaining possibilities are:
GB
GG
The columns are just the way to represent the problem to make it easier to solve, so changing the representation to an equally correct form should not change the solution. Yet, using Jordan (and Phil’s) method with this column labelling system creates a 50% chance.
And really, labelling the columns in order of the event of meeting the child is more consistent than labelling them by age. In the coin flip example, you are ordering the events in the order they happen, not by some arbitrary secondary characteristic such as the date printed on the penny.
Dave but that’s the whole point. Numbering by age matters. And in the coin example, it’s NOT a coin flip example. The coins are flipped _before you see them_. You don’t know the order. When you look at one of them, you don’t know if that one was flipped first or second. It makes all the difference in the world.
I apologize. I cross-posted the last comment with Jordan’s request to stop the topic.
I’m enjoying myself in this intellectual discussion, and I do think the argument added something new to the discussion and is interesting, but I don’t want to irritate people so I am willing to say I think I am wrong and Jordan thinks he is right and leave it at that.
haha yeah and I couldn’t resist, sorry.
This brings me right back to college.
Tom, I’m still happy to keep discussing this, but since Jordan’s asked for the discussion to stop, I think I’d better stop. If you’re interested in continuing, we can figure out another venue.
I thought we were getting offended. Obviously, I underestimated us. >:) please let’s continue!
One last post, but I SWEAR it’s not an argument.
I just think this statistics problem is PARTICULARLY interesting, because here we have two groups of intelligent, educated people who have differing answers and both sides think their answer is the “obvious” one.
Actually, David, that’s the thing. It’s not obvious. I’ve been where you are (I guessed the same). My dad, who’s a university professor, gave the same answer at first. It takes almost everybody a while to really figure it out.
It’s just that I’ve seen iterations of this problem so many times over the years that I can’t help being surprised when it flares up again
Jordan, well if you say it’s okay.
“Numbering by age matters.”
It doesn’t matter in the problem as stated. You could just as easily number them by height, weight, or grade point average. None of these factors were mentioned in the problem so presumably they’re not relevant. If you ordered by one of these other things, and used your method, the answer could be different every time. To me that says the method is wrong.
I will concede that the age of the coin was perhaps not the best example.
Now if the problem had mentioned age, then that would be different.
What if you can’t determine the ages at a glance? What if the kids are, say, 20 and 21. Even if you saw them both standing together, you might not be able to tell the age order. But that still doesn’t change the odds of gender.
Dave, Dr Phil is right. How you ask the question is important. That why you can use statistics to prove anything.
This question relies on the fact that you don’t know the order of the columns. In our example you could put column A before column B and nothing would change, because you don’t know who belongs in which column even after you meet them. In your example you do know because you assign them a column when you meet them.
Age per se is not the key point, but rather sequence – which comes first. There is only one point at which things are _decided_ – when the coin flips, when the baby is conceived, etc. Which one is decided first matters. It doesn’t have to be mentioned in the problem. Its inherent in it, because that’s what gives you the 4 different scenarios:
1) girl pops out first, then boy
2) boy then girl
3) boy then boy
4) girl then girl
(at each conception/toss there’s a 50/50 chance that gives you this set of outcomes. And if you don’t know if the girl you meet was born first or second, 1 and 2 are distinct items for probability purposes, which is the crux of the problem).
Anyway, I’ve about 20000 words to read and critique for a workshop tomorrow so I’m turning off my comments subscription and signing off for the night. Have fun!
(haha, I’m lying again)
although Jordan has a good point. It doesn’t _have_ to be sequence. It could be names too, or anything else that distinguishes one from the other – as long as _you_ don’t know which is which. So let’s say we have Jo and Mo
1) Jo is girl, Mo is boy
2) Jo is boy, Mo is girl
3) Jo is boy, Mo is boy
4) Jo is girl, Mo is girl
As long as you see a girl but don’t know if it’s Mo or Jo, 1 and 2 are distinct cases. If you know the name, that reduces the possibilities.
Basically, Mo and Jo have a distinguishing characteristic that you don’t know the value of. (If they were identical the statistics would be different, which does indeed happen in quantum with identical particles, as Dr. Phil mentions).
“There is only one point at which things are _decided_ – when the coin flips, when the baby is conceived, etc.”
Okay, sure the genders were decided back then, but the whole point is determining what is unknown from the perspective of the problem. To an omniscient viewer, both of the genders are known by the time they reach this age, therefore there is no probability involved.
However, since the viewer is not omniscient, the gender of the children is unknown until he/she meets the kids. From his point of view, the moment of determination of gender is when he MEETS the kid, not when the kid is conceived.
The point of view does matter. If you asked the first kid, let’s say her name is Leslie, “Leslie, what is the probability that your sibling is a girl.” If she knows what probability means, her answer would be either 100% or 0%, because she KNOWS the answer.
Because the observer does not know everything, the whole point is to calculate chances based on what is known and unknown.
If you wanted the birth order to be the ordering factor, an equivalent problem could be:
“A couple has a baby girl. 2 years later, the mother is pregnant again. What is the probability that the second child is a girl.”
Again, it’s still 50/50, because the gender of the first child is irrelevant to the determination of the second child’s gender. They are independent variables, so knowing the actual value of one does not affect the probability of the other.
“although Jordan has a good point. It doesn’t _have_ to be sequence. It could be names too, or anything else that distinguishes one from the other – as long as _you_ don’t know which is which.”
If I were setting up this problem, I would set the columns up in the order that I determined their age. If you are right, then there must be a logical reason why I’m not allowed to order my columns this way. What is that reason?
The flaw in this is the way the question is asked. The girl is a given before we are asked to predict the second child.
Consider the Game Show problem. If one donkey is revealed BEFORE a door is picked, the odds of winning from the other two doors are 50/50.
Experimental probability and theoretical probability have a complex relationship.
I am going to perform an experiment on Monday. I’ll ask my female students with ONLY one sibling what kind of sibling they have. Consider it a field test. I’ll even toss out the twin girls I have so they don’t skew the results.
Would this work on the roulette wheel? How many people come over and bet on the color that didn’t just come up? How well does it work?
Full confession: I hate prob & stats, mostly because they have no effect on individual events, only on large numbers.
Oso,
keep in mind that a real life sample probably won’t provide the same results, because our example is so simplified.
It doesn’t take into account the fact that females are slightly more abundant than males, all other things equal. It also doesn’t take into account that a particular set of parents may be more likely to produce girls or boys because of genetics. And it doesn’t take into account social situations–if I remember correctly, China has a larger male:female ratio because the limit on number of children and the emphasis on patriarchy have applied unique social pressures.
Also, your class is probably not large enough to be statistically significant, nor is it particularly “random”.
I’m not saying it won’t be interesting, but regardless of what it shows I wouldn’t call it conclusive.
Oso,
By the way, I agree with your point “The flaw in this is the way the question is asked. The girl is a given before we are asked to predict the second child.”
Concisely put.
BTW – note that I used different boy/girl names for the two columns in my example. That’s because, if you break it down to each conception being a battle between two sperm, then the second set of two sperm doesn’t yet exist when the first two sperms fight it out — so because these are people they ARE distinguishable particles, so it does matter.
In other words, the actual combinations can only be:
John Jill
John Bob
Joan Bob
Joan Jill
You can’t have Joan John or Bob Jill. In other words, distinguishable particles.
And while you say it doesn’t matter or you can’t tell whether the girl at the door is the older or the younger child, it doesn’t matter — the statistics are based on the actual two events of conception, not on your perception of who is at the door.
As far as coin flips go, there are two cases:
(1) just have two coins
H H
H T
T H
T T
If you know one coin is Tails, then the other is either Heads or Tails — 50-50.
(2) The coins are individuals by date or type:
1972H 1998H
1972H 1998T
1972T 1998H
1972T 1998T
If you see that one coin is Tails, then you get:
1972T 1998H
1972T 1998T
or
1998T 1972H
1998T 1972T
But 1972T 1998T = 1998T 1972T so you get 3 possibilities (2H 1T):
1972T 1998H
1972T 1998T
1998T 1972H
OR you only get the 1972T coin and:
1972T 1998H
1972T 1998T
you are back to 50-50. But again, that’s a different problem.
As I tell my students, “Clear as mud, right?”
Dr. Phil
Phil,
You are distinguishing between what is unknown and what is unknowable, which does not apply here.
In statistics there is the known and the unknown, but no distinguishing between the unknown and unknowable.
So you’re saying the difference here is that the children’s gender is already determined? But is still unknown.
The outcome of a coin toss is unknowable until the event occurs, so to illustrated my point, let’s say that someone flips two coins in the other room, then brings the results of one of the tosses out for you to observe. Assume this person is absolutely trustworthy.
They show you that it is heads.
The value of the other toss is still unknown to you in this case, though it is KNOWABLE. Your friend knows the outcome. But in the end it is still unknown to you and that is all that matters.
So, according to you guys, the fact that the outcome of the other coin toss is ALREADY determined affects the probability? What if you don’t KNOW if the other coin has been flipped or not. Should that determine it’s probable outcome? No. Either way, it’s outcome is unknown so is still up in the air. And the two events are still independent, so one still does not affect the other.
Two births are a classic example of a statistically independent variable, just like two coin tosses.
Fo say that two events are independent intuitively means that the occurrence of one event makes it neither more nor less probable that the other occurs.
http://en.wikipedia.org/wiki/Independence_(probability_theory)
Great post by Dr. Phil. It illustrates why the coin example I was using wasn’t quite right – which is what I was slowly arriving at through my ‘distinguishing characteristic’ argument, but never quite explicitly stated. You’d have to be able to say ‘1972 vs 1998 coin’ or ‘coin always flipped first’ vs ‘coin always flipped second’ (otherwise they’re like identical particles), without the observe knowing which is which. So I apologize for using a confusing example in many of the posts above.
Haha, but if anything, this reinforces the point that, no, this isn’t particularly obvious.
If the outcome of one event can affect the probability of outcomes of another event, then those events are by definition, statistically dependent. With me so far?
Statistical dependence ALWAYS has a reason.
Examples of statistical dependence:
1. Draw a card from a deck, look at it, and set it aside. Draw another card. The probability of the value of the second card is dependent on the first draw, because there are now less cards in the deck.
2. Have a 3rd party measure a single person’s height and weight. Have them tell you the person’s height. The weight is related to the person’s height to some degree, so they are dependent.
Examples of independent variables:
1. Flipping a coin. How would the coin know what the previous result was? There’s no way one could affect the other.
2. Draw a card from the deck, then replace and SHUFFLE before drawing again. Odds are 1:52 each time for particular card.
3. This child birth example.
Unless there’s a REASON for dependence, then there IS no dependence.
And by the way, saying they are dependent because the chart sez so does not count. The chart is a tool to help you understand. If it doesn’t reflect the facts, then it has failed in its purpose.
Scott,
What you missed is that I didn’t say which child shows up at your door. If I’d said the oldest child shows up, what is the sex of the other one, then yes it would be 50/50.
Dave, as soon as you unique identify the child at your down, then the cases become unrelated at the chances go to 50/50. But by NOT KNOWING, the two cases are linking, so the chances are 60/30
OK, you’re right, the problem should say ‘one of the children is a girl’. Then the chances of the other being a boy would be 2/3 (I’m hoping David agrees with this one at least
). As stated, the chances are 50/50.
The reason: as stated, the assumption would have to be that the other family sent one of the two kids over at random. This is what we’ve been ignoring.
50% of families have a boy and a girl. Chance of sending a girl – 50% of 50%, or 25% of all possibilities.
25% of families have two girls. Chance of sending a girl – 100% of 25%, or 25% of all possibilities.
Chance the girl came from a girl-girl family=chance the girl came from a girl-boy family. QED.
Good fun.
At the risk of sounding antagonistic (but I’m not), the dependence is created in the construction of the problem. Take my John/Joan Jill/Bob case. If you see one child arrive at your door and then ask what the probability that the other child is a girl — it’s 50-50:
John Jill
John Bob
Joan Bob
Joan Jill
As soon as it is specified that the child at the door is a girl, you have thrown out the John/Bob pair. And now you have three cases, which cannot be divided 50-50, no matter how you argue it, because John/Jill can never be equal to Joan/Bob:
John Jill
Joan Bob
Joan Jill
or
Girl John
Girl Bob
Girl Girl
Now, where I think it looks like it derails, is that if you try to look at the child at the door, you get with a girl at the door:
Jill John
Jill Joan
Joan Bob
Joan Jill
Then it looks like it’s back to 50-50.
Gurk? What’s the difference? Well, the difference is that there are four possible children, but we didn’t specify which girl shows up.
If you ASK the girl at the door what her name is, then it’s a different problem.
And that’s what drives people mad about statistics. Information changes things.
A cop comes to your door. It would be one of say 120 on the force. If you say it’s a female cop, it might be one of 45 on the force. If you say it’s a detective, it might be one of 15. Any question you ask after you know more information at the beginning changes the probabilities.
I’m still having fun — but I really got to (a) write an exam this weekend and (b) finish my WOTF Q4 story… (grin)
Dr. Phil
Dr. Phil, as per my previous post, I believe we were ignoring a random event (the selection of the child to send over) that changes the odds. Those two cases (BG, GB) get reduced to one because it’s 1/2 as likely that a girl would be sent over then in the GG case. In other words, knowing that at least one of the children is a girl is different from knowing that a child picked at random was a girl. I even ran an excel simulation to test it
.
Of course, this all depends on the assumption that a child was sent over at random. But that’s the assumption you’d make in that situation, with zero other information.
Let’s dissect this a different way. I will make statements that should be agreeable to all and make Jordan’s point. (I think.)
Of a hundred two-child families, there should be half with same-sex children and half with different-sex children. Yes?
Half of the same-sex pairs should be boys. Yes?
Knowing a family has a girl eliminates only families with two boys. Yes?
Thus, the families with two girls are the other half of a half of a half. Yes?
So there’s a half of a half girl-girl while still a whole-half boy-girl (or girl-boy). Yes?
So yes, there is a 2/3 chance that the FAMILY has one of each. But it is based on the fact that any CHILD has a 50/50 chance of being male or female. So the question is actually about the family. If the child had a better than 50% chance of being female, the numbers would change again.
In other words, my next child is no more likely to be male because my first is a daughter. But my family will fit in with one of the groups mentioned above. It’s a trick of what the question is really asking.
So have I changed positions? Not really, just perspectives.
I like that description, Tom.
Bottom line, you can’t use the 50% chance of a child being female to argue that there’s a 66% chance.
I’m getting dizzy.
“OK, you’re right, the problem should say ‘one of the children is a girl’. Then the chances of the other being a boy would be 2/3 (I’m hoping David agrees with this one at least
).”
Nope! According to what you are saying, if this child had been a boy it would alter the chances of the other child being born a boy. That would mean they’re dependent variables, which no one has explained.
“Take my John/Joan Jill/Bob case. ”
That case doesn’t make sense. If you don’t know the genders, you certainly don’t know their names. When a girl arrives at the door she is just a girl, not specified one way or the other.
The age is an irrelevant detail. No one has adequately explained to me why I can’t order the columns by the order I meet the kids, which is an important aspect. If I order the columns that way, then even your system agrees with my solution.
Phil–good luck with your story by the way. I submitted very early so that is not on my agenda.
And, the most important point: You are saying that the genders are DEPENDENT events, by definition, but I have not heard a real life explanation for that.
The purpose of statistics is to describe real life events. If an oddity is present in the statistics, then it better represent something logical in the real world or the statistics are wrong. Hand waving at a chart is not an explanation, unless that chart corresponds with reality.
So the core question is: HOW would the gender of one child affect the gender of the other?
If there is no logical real-world answer to that, then the events cannot be dependent. If the events are independent, then the outcome of one can NOT affect the other.
All right, I’m going to shut my computer off and head for bed. I’ve been posting to this thread for like 9 hours now, so I think it’s time for a break. Rest assured I’ll be back tomorrow.
Cheers!
Haha thanks for giving me something I can argue with, Dave – makes me feel better!
Here’s a simple example (let me go back to the coin toss). I toss 2 coins. Let’s say they’re at my left and right hand. They could be, a priori:
T H
H T
T T
H H
If I do a hundred tosses, each will occur 25 times. (True/False?)
I’m told one coin is tails (but not which one!!). I know this happens in 75 of my tosses. Now I know that in 25 of my tosses both are tails, in 50 only one is, so the chances are 2/3.
CONTRAST: I know which coin is tails. Any given coin is tails in only 50 of my tosses. In only 25 is the other heads. 50/50.
To add something more confusing:
This is not a matter of dependent variables. It’s a matter of probability distributions – if you did this enough times, what would happen? That’s a bit similar to how a digital binomial distribution is formed. Each step in a random walk is random (they’re not dependent), but the sum in aggregate always looks like a bell curve.
Haha, Dave, there’s nothing more real life than this. If you looked at 1,000,000 families with 2 children, 500,000 of them (roughly, adjusting for local propensities for male/female children) would have children of opposite genders. (TRUE/FALSE?)
If you know ONE of the children in a family is a girl, that will occur in 500,000 mixed-gender-kids families and only 250,000 same-gender-kids families. So in those cases the other is more likely to be a boy.
Okay, decided to boot the computer back up, not ready for sleeping yet.
It IS a matter of dependent variables. Given any two events they are either independent or dependent. I am saying these are independent. You are saying they are dependent. You haven’t used that exact word, but if the outcome of one event affects the other then it is dependent. Unless there can be an explanation of HOW the dependency takes effect, then they are not dependent.
Aaaand, I’ve just reached my limit to the amount of stats discussion I can handle. I still believe that I am right but my interest in convincing you guys is waning. It’s been fun, but I’ve just simply had all I can handle. As it is I will probably dream of this tonight!
I’m sure I will see you guys here, there, and everywhere around the Interwebz, but I am now going to remove my subscription to this thread.
Cheers!
Bye, David. Truly sorry if you see it as a spitting contest. I see it as an intellectual discussion over the difference between viewing two random objects (coins) and two discernible individuals (children). The children may be independent, but the family is dependent (there’s a bad IRS joke in there, sorry). Once you decide the sex of one child, you’re dealing with a dependent subset.
Sorry Jordan if our discussions have caused your blog a problem — certainly wasn’t my intent.
Dr. Phil
Certainly hasn’t been a problem, Dr. Phil, though this thread will no longer load on my iPhone (or it takes longer than my patience allows).
Dave’s cool, he’s not offended, but we HAVE been going in circles for an awfully long time.
Tom succumbed to Stockholm Syndrome and went over to the dark side (and we’ll miss him dearly!).
Seriously though, I’m going to find the original wording to that question now and post it up here. Hopefully that settles it once and for all.
Jordan – not entirely the dark side, as you can see! Dave and I still disagree. I just thought he was arguing a much subtler point (I do think the problem as stated is off). I’d be interested to hear your refutation of my post
(#74)
Tom,
I will, but Lord, not tonight! I have a story that I promised I was going to have done by tomorrow. …and don’t have you 20K words to crit by tomorrow.
We’ll cross swords again tomorrow or monday!
Phil–No worries. I’m not offended, nor was I last night.
I’m still not subscribed to the post, but I was discussing this with my good friend Gary Cuba, and he brought up a good point. He’s tried to post to this thread a few times last night and met with failure, so I hope he doesn’t mind if I just repeat it for him.
According to you guys, if it’s a random child answering the door, then the odds of picking the other child are 66%. But what if you ask the first child if they are the oldest or not? Assuming they tell you the truth, that would associate them with these columns you guys are fixated on, and would change the odds of picking the other kid down to 50/50. By learning MORE information, you REDUCE your odds of picking the gender correctly. Doesn’t that strike you as odd?
The reason it is odd is because it is wrong.
And regarding the refuting of post #74, we are back to the age being important. The age is irrelevant. Nobody cares if this is the older kid or the younger kid. The only unknown is the gender of the other kid. The gender of the other kid is not affected by the gender of this kid.
They are statistically INDEPENDENT, which means that the knowledge of one cannot affect the probability of the other. Statistical independence is one of the first things you learn in a stats class, and this is a classic example of statistically independent events.
Again, I will put out the challenge that I’ve put out several times. If you can convince me that the birth gender of one child will AFFECT the birth gender of the other child, then I will concede you are right. That is the only way I can see me agreeing. Otherwise, if one doesn’t affect the other, you are agreeing that they are independent, and independence means that one can’t affect the other.
Anyway, I’m going to try to go back to ignoring this thread again. take care.
And, to expand on #92 a bit: So you’re saying that there are cases where ignorance allows you to make a more accurate decision than knowledge? If that were true, it would undermine the principles of science itself.
“Should we do research?”
“Nah, learning more facts will just make our answers worse. Let’s just wing it!”
#74 doesn’t mention age.
The ‘come to the door example’ is wrong, for the reasons stated in #74. The probabilities there are 50/50.
The “one of the children is a girl” example isn’t wrong. This is combinatorics 101. It isn’t even the Monty Hall problem.
If you know that 75% of all families have at least one girl, and you’re told the family in question has at least one girl, all that you’ve been told is that you’re looking at those 75% of all families. If you know that 2/3rds of those families have one boy, you have your answer.
By your logic, 50% of families with at least one girl have one boy. That means 50% * 75%=37.5% of all families are boy-girl families. But we know that percentage is 50%.
There is no affecting happening. We’re talking about the aggregate result of many different coin tosses, not the result of any single coin toss. All we’re saying is that we know there are more girl-boy families than there are girl-girl families. That’s all.
“We’re talking about the aggregate result of many different coin tosses, not the result of any single coin toss.”
No, we’re talking about a single family.
Tom,
So how do you account for the fact that gaining more knowledge reduces your chances of making a correct decision? I’m presuming you’re not basing your reasoning on a “yay ignorance” plea, so you must have some explanation.
Also, the deciding factor you’re using is the age. Because we don’t know the age, we don’t know the column order, which means there are 3 remaining possibilities.
But if you DO know that this child is the older one, then that leaves two possibilities:
BG
GG
So 50% odds.
And if you DO know that the child is the younger one, there are two possibilites:
GB
GG
No matter which the child tells you, the choice is truly random. Why does the information matter if both possible values of that information suggest the same solution?
AHA!
Ok, trying to address #95 again.
Using your own example, if you have the remaining options:
GB
BG
GG
You are correct in saying that this is 75% of overall families.
You are hinging all of this based on the unknown information that the girl you meet is older or younger.
Let’s say the girl is older, that leaves the possibilities:
GG
BG
But these two possibilities are not equally likely within this remaining 75%. Because, you see, the GG case is also possible if this is the younger girl. So the GG’s probability is split across the two information spaces.
Of that 75% space, if this girl is older:
25% chance that the combo is BG
12.5% chance that the combo is GG (because the GG is also shared in the other space.
Of that same 75% space, if this girl is younger:
25% chance of GB
12.5% chance of GG
Both of these add up to 37.5%, which is within the 75% space.
37.5/75 = 50%
There are two possible facts here, exactly one of which is always true:
“This girl is the older sibling”
or
“This girl is the younger sibling”
Added up, these two must account for the entire space. Your mistake, is that you didn’t account for the fact that GG space is split between these two knowledge spaces.
TESTING.
IS THERE A COMMENT LIMIT?
TESTING2
(TRYING AGAIN)
It’s been fun following this thread. It’s a shame I’m too stupid to figure out how to get my posts to work on this blog, much less solve the more arcane mysteries of probability! Over the last day I’ve been leaning toward both sides of the argument, back and forth. But it strikes me today that there is an assumption behind the original odds table that is not fully validated — namely, that age or birth order or “production sequence” is significant to the problem, and that it imposes a “non-commutative” (is that right?) characteristic to the boy-girl, girl-boy cases. Yes, you could claim that it’s embedded (implied, anyway) in the text of the original problem statement, “Your neighbor has two kids.” The assumption being made is that they are of different ages, and that that feature plays into the odds of correctly guessing the sex of the second child. But there are imaginable situations that invalidate that assumption without changing the problem statement at all. Eg, that they are twins who were delivered simultaneously by c-section (for the medically minded folks), or test tube babies/clones that were decanted at the same time (for the SF folks), or that the “true” age of everyone must be figured based on when their divine souls were created, and that they were _all_ generated at the instant of the universe’s creation in 4004 BCE (for a “religious” example). Or infinitely many other scenarios where the kids’ birth order or relative age “goes away.” That would seem to me to “collapse” the odds table into 3, not 4 combinations, since the concept of unique “ordering” makes no sense in those cases. If you want to introduce “birth sequence” into the solution, I think you must include that in the problem statement. At very least: “Your neighbor has two kids of different ages.” Even then, I ain’t so sure that satisfies me 100%.
Bless you Wikipedia. Dave ur wrong. Tom, you got sucked into the darkside.
As long as you don’t know wcich child you’re looking at, the odds are 2/3
http://en.m.wikipedia.org/wiki/Boy_or_Girl_paradox
Thanks Jordan, that link illustrates my point exactly. I draw your attention to (under Question 2):
“However, if it is assumed that the information was obtained by considering only one child, then the problem is an isomorphism of question one, and the answer is 1/2.”
This is the case when a child comes to the door – you’re just picking one of the children and considering them (a random act).
If you scroll further down, it’s explained better:
“The second question is often posed in a way that leave multiple interpretations open. In response to reader criticism of the question posed in 1959, Gardner agreed that a precise formulation of the question is critical to getting different answers for question 1 and 2[1]. Specifically, Gardner argued that a “failure to specify the randomizing procedure” could lead readers to interpret the question in two distinct ways[1]:
* From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.
* From all families with two children, one child is selected at random, and the gender of that child is specified. This would yield an answer of 1/2, and many experts agree.[3][4]
The ‘comes to the door’ case corresponds to the second case.
Do the odds change if your neighbors are named Schrodinger and you haven’t answered the door yet?
This question has been argued about for decades. It all depends very strongly on the specific wording. If you were to say “A neighbor’s child comes to the door. You know the neighbor has at least one girl child. What are the odds the child at the door is a girl?” (or something along those lines) the answer would indeed be 66%.
However, by randomly choosing one of the children and naming that child as a girl, the answer becomes 50%. You are effectively asking the following question: “The girl at the door has one sibling. What is the probability the girl’s sibling is a boy?” Questions of relative birth rates for boys and girls aside, I assure you the answer is 50%. It doesn’t matter if the girl is older or younger sibling. She has one sibling. There is a 50% chance the sibling is a boy and a 50% chance the sibling is a girl. If the phrasing of the question reduces things to the consideration of one child, the answer is 50%. If the phrasing is such that you need to consider both children, the answer is 66%.
Based on the phrasing of the question posed here, Dave is in fact correct.
A neighbor’s child comes to the door. You know the neighbor has at least one girl child. What are the odds the child at the door is a girl?”
Sorry. Talking about a specific child at the door raises a bunch of different considerations (and I typed girl instead of boy). The question should have been “Your neighbor has two children, one of which is a girl. What are the odds the other child is a boy?”
Wow. Monday morning. That, of course, should have read “at least one of which is a girl”
There exists an interesting aspect to this “paradox” that really messes with your head. Say you learn that the neighbors moving in next door have two kids (thereby representing a randomly selected element from the population of two-child families). You have one son and a large number of daughters and are secretly hoping for a boy for your son to play with. You go through the possible birthing scenarios: BB, BG, GB, and GG. You say to yourself “Good, a 75% chance my new neighbors will have a boy for my son to play with.”
Then someone mentions that at least one of the children is a girl. So, you say to yourself “Too bad, but hey, that still leaves a 66% chance the other child is a boy.”
The neighbors move in and you spot one of their children on the lawn. It is a girl. You tell yourself “That’s okay, I know one of them is a girl, so the probability of a boy is still 66%: she could have an older brother, a younger brother, or a sister.” But oops, that list is incomplete, because she could in fact have an older brother, a younger brother, an older sister, or a younger sister. There are four possibilities. Or, if you remove the now unnecessary question of younger versus older, there are two possibilities: she has a sister or she has a brother. Thus, you now calculate the probability that her sibling is a boy at 50%.
But you already knew one of the children is a girl. So, how does spotting a girl on the neighbor’s lawn reduce the chances of your son getting a boy to play with?
The pat answer would be you are considering two different scenarios, asking two different questions. In the first scenario, you are considering the gender distribution within a randomly selected two-child family (looking at the possible permutations of two variables: gender of child #1 and gender of child #2). In the second scenario, when you spot the girl on the lawn, you are asking “what is the probability her sibling is a boy?” There is only one variable in that question (the gender of the other child). I don’t see a bridge between the two scenarios, and I think that is part of what leads to the enduring controversy from this form of question. And underlines the need for carefully wording the question.
Chris, you’ve a great grasp of the problem. By the way, I think comment #74 gives a good idea of that ‘bridge’ you’re talking about.
In the spot-the-girl-on-the-lawn scenario, or the girl-comes-to-door scenario, there is an extra random event (beyond the original births). One child of the family is selected at random to come to the door, or to go out on the lawn. That changes the odds.
Anyhow, it’s a new week; no more on this
.
regarding items 100 and 101, computers usually work off of powers of 2, so the limit would be set to 63(or 64), 127(or 128), 255(or 256), etc.