Why are there 4 combinations and not 3… why does the order of birth impact anything in a situation where birth order was not introduced as a defining characteristic?
So, if we ignore birth order and limiting to B/G, there are only 3 combinations - BB, BG (or GB), and GG. But they’re not all equally likely. BG (or GB) is twice as likely to occur as BB/GG. So if you ignore order, you need to account for the difference in likelihoods - you can’t assume GB is equally likely to occur as BB.
If you take birth order into account, you count BG and GB separately. In this case, BB, BG, GB, and GG are all equally likely.
Oversimplifying, it’s like saying playing the lottery is a 50/50 shot - either you win or you lose. Clearly that’s not correct.
What would those three combinations be? If you propose them to be “both boys”, “both girls” and “one boy and one girl”, what do you suppose the probabilities of those three sets to be?
The standard boy/girl paradox is pretty easy to understand if you just consider that there is a set of 2-children families out there with that have 25% with 2 boys, 25% with 2 girls, and 50% with one of each. If you eliminate the 2-girl families, you are left with 75% of the total, 2/3 of which has one girl in it. So any family that has at least one boy has a 66% chance that the other sibling is a girl.
The only reason for using birth order is that it is a definitive way of differentiating between two children in a set that you know nothing else about. It is completely independent from the gender and allows you to differentiate between the items in a set, which is really the important part of the equation.
One of the key aspects of these type of math problems is that you need to think of the children as a set, not as two individuals. All the questions are about the set, not about the other child. That’s why translating the math into English is so difficult; conversational English isn’t set up to discuss sets of objects.
The best way to state the problem clearly and unambiguously is to go into detail which tends to derail the question and keys the listener that the problem isn’t as simple as you might think.
- Assume boys and girls are equally likely
- Assume all weekdays are equally likely
- Assume gender/weekday of any child is independent of other children
- Start with all the families with exactly 2 children
- Reserve all of the families with at least one boy born on a Tuesday
- What percentage of the these families have a girl?
Ah. That makes sense. I wasn’t considering that breaking out BG and GB (in @YamatoTwinkie ‘s symbology) accounted for the probability of each set. I was ascribing equal weight to BB, GG, and [GB/BG].
And thanks @Telemark , though I think my question was more why the order in the set made a difference, why {G,B} is different than {B,G}. In this case, I was missing the concept that there are going to be twice as many pairs of boys and girls (in whatever order) than there are of pairs of boys or pairs of girls.
“Tuesday” itself isn’t special, Birthday-of-the-week is just a convenient way to categorize a set of individuals into smaller groups with intuitive assumed probabilities. Birthday-of-the-week is also completely independent to the other category chosen (Boy/Girl).
“Human” and “breathes oxygen” don’t further categorize a set of 2 children into smaller sets
“Blue eyes” is statistically problematic because parents with one child with blue eyes are vastly more likely to have a second child with blue eyes than the general world statistic.
But hypothetically, you could get a large room of moms of two children together, have all of them flip a coin associated for each of their children, ask “how many of you have at least one boy that you flipped tails for?”, and then from the parents that raised their hands, you could assume that the probability that their other child was a girl was 57% (4/7)
A few thoughts so far:
- It would be absolutely brilliant if there were a SDMB gender reveal party and we could all sit around a table arguing the probabilities and all of the permutations of judging the probabilities, while pissing off the expectant mother by saying, “don’t pop that balloon yet, we’re still debating the phrasing!”
- Several mentions of sex not being quite 50/50 for human live births, but nobody has yet mentioned that births aren’t distributed evenly by days. I believe studies in the past have shown increases in induced or caesarean births on Fridays, so the doctors don’t have to work over the weekend.
Yes, this was the leap that really helped me to understand what is going on here. In my little toy script the set idea is (to me) became obvious, because
flips <- data.frame(
flip1 = sample(1:2, size, replace=TRUE),
day1 = sample(c("SUN", "MON", "TUE", "WED", "THU",
"FRI", "SAT"), size, replace=TRUE),
flip2 = sample(1:2, size, replace=TRUE),
day2 = sample(c("SUN", "MON", "TUE", "WED", "THU",
"FRI", "SAT"), size, replace=TRUE)
)
creates a dataframe called flips which contains the paired results. Even though each individual birth/flip is independent of the others, two independent events are linked together to make a new event—two separate births become a family.
If the problem were setup in such a way that the births/flips/cards were completely independent, then the simulation could do anything it wanted to the first set of births, such as filtering by boys on Tuesday or Thursday. Then the second set of births would be totally unrelated to whatever filtering was happening. So the quantifying lines would then be something like
girls <- flips[(flips1 == 2 | flips2 == 2),]
nrow(girls)/(nrow(flips$flips1)+nrow(flips$flip2))
Which just says, count up the number of girls and divide by the total number of births. Absolutely nothing about the other paired birth is known or cared about.
My lightbulb moment was “Monty’s FIRST door will NEVER contain the car,” but that wasn’t made explicit in the original problem, you as the reader had to intuit that to answer it correctly.
Same with the treadmill thing, since I assumed people were insisting a STATIONARY plane was trying to takeoff (which I found absurd, natch), and nobody bothered for some time in the relevant thread to explicitly state that the plane can still roll forward, wheels going double speed. Someone, Cecil himself IIRC, then came up with a scenario which most closely matched my original intuition, that a superduper treadmill with a non-superduper plane could roll at a ferocious speed and the wind generated would keep the plane stationary (until its wheels if not engines etc. disintegrated).
Same song third verse with the dress (tho that one is admittedly a different set of issues): I never imagined that people actually perceived it as dark blue, and assumed everyone who said so saw it as light blue/white (like I did), but were intuiting that it was actually dark blue. Eventually I realized that they actually SAW it as dark blue. [always “light” blue with “gold” trim to me, never managed to flip it]
And if you go the other way and use day of the year i.e. 365 possibilities instead of 7 you get 50.03427%.
Think I’ve found the problem with the original riddle, which includes “at least”
“Mary has 2 children and at least one is a boy born on a Tuesday. What’s the probability the other child is a girl?”
There are three possible outcomes, not two.
A boy born Tuesday, and a boy not born Tuesday
A boy born Tuesday, and a girl
Two boys, both born Tuesday, in which case there is no “other child”.
So the riddle becomes clearer if we change it to
“Mary has 2 children and at least one is a boy born in an even-numbered year. What’s the probability the other child is a girl?”
The original poster left out the “at least” in his version of the riddle, so the answer to that version is 50%.
No good, because you’re still saying “the other child”. If both of her children happen to be full-of-woe boys, which one is the “other” one?
Hermitian can relax – the explanation in his link is wrong.
The way he quoted the riddle (with no “at least”) the probability is clearly 50%. The only confusion is if you include the “at least”, like the riddle in the link. But Presh Talwalkar is the one who made the mistake, when he said “We have already listed b2b2, so there are only 6 additional cases for the second child’s day of the week.” He thinks there are only 27 total possibilities, and 14 of them are girls; actually there are still 28 possibilities, including 14 girls.
Try this riddle. “Mary has two children. She tells you that at least one is a boy born in an even-numbered year. What’s the probability the other child is a girl?”
Using his reasoning, there are just three possibilities and two of them include a girl.
Guess the source of the problem is talking about “the other child” – when the riddle includes “at least”, the “other child” may not exist.
Impossible to answer, because Mary’s rationale for making the statement has not been given.
Mary always makes ambiguous statements about her children. What is the probability that you correctly interpret her?
I absolutely expect better from a surgeon operating on one of her kids.
There is a probability problem on the internet that I really don’t understand. Please explain it to me in a way that makes rational sense to me.
….
Please someone help me make rational sense of the answer!
I hope to. I haven’t read the other responses, because I know what they will say and how those who wrote them will respond if they don’t like what I say. But I do know all the correct reasons.
First, a little history. This class of problems dates back to 1889, when Joseph Bertrand tried to warn future mathematicians about a mistake he knew they would try to commit. Few listened. What is now called his “Box Paradox” was not the problem, it was the self-contradictory consequence of that mistake. It is the one you are noticing now. You can read about it here.
It re-surfaced as the Two Child Problem (without “Tuesday”) in the May 1959 issue of Scientific American. Martin Gardner gave the 2/3 answer (well, he asked the complimentary question and answered 1/3). BUT HE WITHDREW IT HALF A YEAR LATER WHEN HE REALIZED IT WAS WRONG. And the incorrect solution resurfaced (with “Tuesday”) in 2010 at a puzzle convention that, ironically, was named for him but forgot his retraction.
In the column where he retracted it, he posed a similar problem with a different number of cases, called the “Three Prisoners Problem” that was identical to the Box Problem. He explained it meticulously. And that problem resurfaced in 1990 as the “Monty Hall Problem” or “Game Show Problem,” which was not explained very well. So lots of people think they know the correct solution, but only spread incorrect answers.
- Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
Most people will, at first, say that initially there were three equally-likely cases (car behind #1, #2, or #3). But by showing a goat behind #3, that case and only that case, is eliminated. The other two remain equally likely, so there is no point in switching. You may recognize this solution at the same one that you “understand [as] the classic answer being 66.6%.” Just with a different number of cases. There were four equally-likely cases (BB, BG, GB, or GG). But by saying there is a boy, the GG case, and only the GG case, is eliminated. The other three remain equally likely, so the probability of BG or GB is 2/3.
This is wrong in both problems.
Going back to the MHP, many solvers who think they have the right idea will say “Your original choice’s probability of 1/3 can’t change.” While true, it can’t just be asserted. It has to be demonstrated. The way to do that is the paradox Joseph Bertrand pointed out. **IF **the probability changes to1/2 because the host opens #3, then it would also change to 1/2 if he would have opened #2. But he can always open one of them. If it changes whichever he chooses, and he can always choose one, then he doesn’t have to open it. Door #1’s chances are already 1/2. This paradox is what proves that it can’t change.
But that isn’t a solution. The correct solution is that the host will open #3 every time the car is behind #2, but only half of the time it is behind #1. So we should not eliminate only the #3 cases, but half of the #1 cases. This is what makes the car twice as likely to be behind #2, than #1.
The same applies to the Two Child Problem. The original chances of a mixed family can’t change. Any potential “Mary” with BB will tell us that she has a boy, but only half of the potential Marys with BG or GB. This makes the two mixed families half as likely as what the incorrect solution claims. The correct answer is 1/2, not 2/3. And the same thing happens if we add Tuesday; the correct answer is again 1/2, eliminating the paradox where irrelevant information changes the answer.
The reason the incorrect answer changes - the issue you were most concerned about - is because that solution is based on assuming that Mary must tell you about a boy, or a boy born on a Tuesday, if she can. That she can’t tell you about a boy born on a Thursday, or a girl born on a Saturday. The probability changes because a two-boy family is almost twice as likely to include a Tuesday Boy as a mixed family. The “almost” is why the answer changes to 51.9% instead of 50%.
If you ask a woman with a boy and a girl if she has a boy, there’s a 50% chance that she’ll say no?
I think, if I understand it, there is a 50% chance the BG/GB mother will tell you about the boy, and a 50% chance she’ll tell you about the girl. I don’t understand how that is making us arrive at the correct answer through the wrong method, though.
Take this to a logical extreme.
Suppose you’re on a game show, and you’re given the choice of a hundred doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens other doors, say Nos. 3 through 100, which have goats. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
The point is, it matters why the woman told us that she has a boy. It’s different if someone asks her “Do you have a boy?” and she answers, than if someone asks her “What gender is your eldest child?”. In the first case, the mother of a mixed pair would say “yes”, while in the second, she might or might not say “boy”.