In Boy1Tu/Boy2Tuesday, the boy referred to as BTu in the original statement is boy 1, in Boy2Tu/Boy1Tuesday the boy referred to in the original statement is boy2.
That’s why the “born on tuesday” is a red herring, and doesn’t add any information. How could it?
This sounds like you are trying to divide “two boys born on Tuesday” into “two boys born on Tuesday and the person is talking about the first boy” and “two boys born on Tuesday and the person is talking about the second boy”.
That doesn’t work because you are now no longer dealing with cases of equal probability. “Boy 1 Monday/Boy 2 Tuesday”, “Boy 1 Tuesday/Boy 2 Tuesday”, and “Boy 1 Tuesday/Boy 1 Monday” all have equal probability. If you’re creating separate cases depending on which of the boys is being referred to, the first and third of those don’t divide into separate cases but the second one does divide into separate cases, each with half the probability of the first and third.
doesn’t add any information. How could it?
As I pointed out above, whether it adds information (and whether the analysis is correct) depends on exactly what you mean by “one is a boy born on Tuesday”. If you picked “boy” and “Tuesday” at random first, and then noticed that one child met that description, that rules out cases where no child happened to meet the description. If you picked a child first and then noticed he was a boy born on a Tuesday, but if it was a girl born on a Monday you would have said “one is a girl born on a Monday”, you are correct that no information is provided.
The only relevant information is that one of the children is a boy. There is still a 50% chance the second child is a boy and a 50% chance that the second child is a girl. Since you already know that one of the children is a boy, the posterior probability that they are both boys is 50%.
Rephrase it this way:
I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?
Now, to see why Tuesday is irrelevant, I’ll re-state it thusly:
I have flipped two coins. One I flipped on a Tuesday and it came up heads. What is the probability that both are heads?
The sex of one child has no influence on the sex of the other child, nor does the day on which either child was born influence the day any other child was born. There is a 1⁄7 chance that child 1 was born on each day of the week, and there is a 1⁄7 chance that child 2 was born on each day of the week. There is a 1⁄49 chance that both children will be born on any given day (1/7*1/7), for a 7⁄49 or 1⁄7 chance that both children will be born on the same day. That’s your missing 1⁄7 chance that gets removed inappropriately from the Tuesday/Tuesday scenario.
I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?
1⁄3 (you either got hh, heads/tails,or tails/heads). You didn’t tell me THE FIRST came up heads. Thats where you are going wrong. At least one is heads is different information then a specific coin is heads.
This is a pretty well known stats problem, a variant of Gardern’s boy/girl paradox. You’ll probably find it an intro book, and Jiro is correct. You are still overcounting. Boy-boy is a different case then boy-girl (well, depending on what the data collection process is).
If you have two boys (probability 1⁄4), then the probability at least one is born on Tuesday (1-(6/7)^2). ( 6/7^2 being the probability neither is born on Tuesday). The probability of a boy-girl family is (2*1/4) then (1/7) (the 1⁄7 for the boy hitting on Tuesday).
No one is suggesting one flip informs the other, rather that when you say “one coin came up heads” you are giving some information about both coins.
I have two coins. I flip the first one, and it comes up heads. Now I flip the second coin. What are the odds it will come up heads?
This is 1⁄2, because there are two scenarios, hh, ht. But its different information then the other question.
If you say “one coin is heads,” you have hh,ht,th, because it could be that the first flip was tails/the second heads (a possibility you have excluded in the above).
No, it’s the exact same question, only the labels are different.
The probability that any one child is boy is 50%. We have been told that one child is a boy, which only leaves two options—HH and HT. If TH were still available, then so would TT be available because the next flip could be revealed to be tails.
It’s exactly the same as the coin flip, because the probability is 50% - the same as a coin flip. This isn’t the monty hall problem. Knowing half the problem (that there’s at least one boy) doesn’t change the probability of the other boy, it just changes what our possibilities are.
No, it’s the exact same question, only the labels are different.
No, it isn’t. You should consider that you are disagreeing with a pretty standard stats question, so odds are high you are wrong. With that in mind, you should reread what people are telling you here.
Now, consider “I flip two coins” the possible outcomes are
hh,ht,th,tt
I hope we can agree on that much.
Now, I give you more information and I say “one of the coins is heads,” so we Bayesian update by crossing out any scenario where one coin isn’t heads. There is only 1 (tt)
hh,ht,th
So it should be pretty clear the probability I flipped two heads is 1⁄3.
Now, your scenario, flipped two coins (hh,ht,th,tt), and I give you the information “the first coin is heads,” so we cross out everything where the first coin is tails, leaving (hh,ht). Now the probability you flipped two heads is 1⁄2.
I know it’s not the be all end all, but it’s generally reliable on these types of questions, and it gives P = 1⁄2, so I’m not the one disagreeing with the standard result here.
Do the math yourself, it’s pretty clear.
Edit: Reading closer, I should say that both answers are right, and the probability can be either 1⁄2 or 1⁄3 depending on your assumptions. However, the problem as stated falls best to me in the 1⁄2 set of assumptions. You are told one child is a boy and given no other information, so the only probability left for the second child is a 50% chance for boy.
Did you actually read it? It does not agree with you. Look under the heading “second question.”
Do the math yourself, it’s pretty clear.
I did the math in the post above, enumerating the possibilities for you to try to help you find your mistake.
Edit, in response to the edit:
I should say that both answers are right, and the probability can be either 1⁄2 or 1⁄3 depending on your assumptions.
Which is exactly analogous to what Jiro was saying about the Tuesday question. So we all agree now? Tuesday can raise your probability slightly above 50%, as was said all along.
However, the problem as stated falls best to me in the 1⁄2 set of assumptions. You are told one child is a boy and given no other information, so the only probability left for the second child is a 50% chance for boy.
And you are immediately making the exact same mistake again. You are told ONE child is a boy, you are NOT told the FIRST child is a boy. You do understand that these are different?
The paradox arises because the second assumption is somewhat artificial, and when describing the problem in an actual setting things get a bit sticky. Just how do we know that “at least” one is a boy? One description of the problem states that we look into a window, see only one child and it is a boy. This sounds like the same assumption. However, this one is equivalent to “sampling” the distribution (i.e. removing one child from the urn, ascertaining that it is a boy, then replacing). Let’s call the statement “the sample is a boy” proposition “b”. Now we have:
P(BB|b) = P(b|BB) P(BB) / P(b) = 1 1⁄4 / 1⁄2 = 1⁄2.
The difference here is the P(b), which is just the probability of drawing a boy from all possible cases (i.e. without the “at least”), which is clearly 0.5.
The Bayesian analysis generalizes easily to the case in which we relax the 50⁄50 population assumption. If we have no information about the populations then we assume a “flat prior”, i.e. P(GG) = P(BB) = P(G.B) = 1⁄3. In this case the “at least” assumption produces the result P(BB|B) = 1⁄2, and the sampling assumption produces P(BB|b) = 2⁄3, a result also derivable from the Rule of Succession.
We have no general population information here. We have one man with at least one boy.
I’m not at all sure you understand that quote. Lets stick with the coin flips:
Do you understand why these two questions are different:
I tell you- “I flipped two coins, at least one of them came out heads, what is the probability that I flipped two heads?” A:1/3
AND
“I flipped two coins, you choose one at random and look at it, its heads.What is the probability I flipped two heads” A: 1⁄2
For the record, I’m sure this is frustrating as all getout for you, but this whole argument has really clarified things for me, even though I still think I’m right about which question we are answering.
Many of my arguments in previous posts are wrong (or at least incomplete and a bit naive), and it didn’t click until the last post or two.
Like I said, I still think I’m right, but not because my prior analysis was any good. The 1⁄3 case was a major hole in my reasoning. I’m happily waiting to see if you’re going to destroy my latest analysis, but I think it is pretty solid.
Yes, and we are dealing with the second question here.
Is that not what I said before?
We don’t have 1000 families with two children, from which we’ve selected all families that have at least one boy (which gives 1⁄3 probability). We have one family with two children. Then we are told one of the children is a boy, and given zero other information. The probability that the second is a boy is 1⁄2, so the probability that both are boys is 1⁄2.
The possible options for the “Boy born on Tuesday” are not Boy/Girl, Girl/Boy, Boy/Boy. That would be the case in the selection of 1000 families above.
The possible options are Boy (Tu) / Girl, Girl / Boy (Tu), Boy (Tu) / Boy, Boy / Boy (Tu).
There are two Boy/Boy combinations, not one. You don’t have enough information to throw one of them out.
As long as you realize there is a difference between those two questions, fine. We can disagree about what assumptions the wording should lead us to, thats irrelevant to the actual statistics and can be an agree-to-disagree situation. Its just important to realize that what the question means/how you get the information is important.
We don’t have 1000 families with two children, from which we’ve selected all families that have at least one boy (which gives 1⁄3 probability). We have one family with two children. Then we are told one of the children is a boy, and given zero other information.
If we have one family with two children, of which one is a boy, they are (by definition) a member of the set “all families that have at least one boy.” So it matters how we got the information.
If we got that information by grabbing a kid at random and looking at it (so we have information about one specific child), that is sampling, and it leads to the 1⁄2 probability.
If we got that information by having someone check both kids, and tell us “at least one is a boy” we have different information (its information about the set of kids the parents have, not information about one specific kid).
This is NOT a case of sampling.
If it IS sampling (if I grab a kid at random and say “whats your Birthday?” and it happens to be Tuesday), then the probability is 1⁄2. (we have information about the specific kid’s birthday).
If instead, I ask the parents to tell me the birthday of one of their children, and the parent says ‘I have at least one boy born on Tuesday’, then we get, instead, information about their set of kids, and the probability is the larger number.
Sampling is what leads to the answer you are supporting.
The answer I’m supporting is based on flat priors, not sampling. I’m saying there are two possible Boy/Boy combinations, not one, and therefore it takes up half the probability space, not 1⁄3.
Sampling to the “Boy on Tuesday” problem gives roughly 48% (as per the original article), not 50%.
We are simply told that the man has a boy who was born on tuesday. We aren’t told how he chose that boy, whether he’s older or younger, etc. Therefore we have four possibilites, like I outlined above.
Is my analysis that the possibilities are Boy (Tu) /Girl, Girl / Boy (Tu), Boy (Tu)/Boy, Boy/Boy (Tu) correct?
If so, is not the probability for some combination of Boy/Boy 1/2? If not, why not? I don’t see it.
BTW, contrary to my previous posts, having the information about the boy born on Tuesday is critical because it allows us (and in fact requires us) to distinguish between the two boys.
That was in fact the point of the original article, which I now disagree with significantly less. In fact, I agree with the major premise that the tuesday information pushes the odds of Boy/Boy closer 50%, I just disagree that you can’t reason that it pushes it to exactly 50%.
Is my analysis that the possibilities are Boy (Tu) /Girl, Girl / Boy (Tu), Boy (Tu)/Boy, Boy/Boy (Tu) correct?
No. For any day of the week EXCEPT Tuesday, boy and girl are equivalent. For the case of both children born on Tuesday you have for girls: Boy(tu)/Girl(tu),Girl(tu)/Boy(tu), and for boys: boy(tu)/boy(tu).
That was in fact the point of the original article, which I now disagree with significantly less. In fact, I agree with the major premise that the tuesday information pushes the odds of Boy/Boy closer 50%, I just disagree that you can’t reason that it pushes it to exactly 50%.
This statement leads me to believe you are still confused. Do you agree that if I know a family has two kids, I knock on the door and a boy answers and says “I was born on a Tuesday,” that the probability of the second kid being a girl is 1/2? And in this case, Tuesday is irrelevant? (This the wikipedia called “sampling”)
Do you agree that if, instead, the parents give you the information “one of my two kids is a boy born on a Tuesday”, that this is a different sort of information, information about the set of their children, and not about a specific child?
This statement leads me to believe you are still confused. Do you agree that if I know a family has two kids, I knock on the door and a boy answers and says “I was born on a Tuesday,” that the probability of the second kid being a girl is 1/2? And in this case, Tuesday is irrelevant? (This the wikipedia called “sampling”)
I agree with this.
Do you agree that if, instead, the parents give you the information “one of my two kids is a boy born on a Tuesday”, that this is a different sort of information, information about the set of their children, and not about a specific child?
I agree with this if they said something along the lines of “One and only one of them was born on Tuesday”. If not, I don’t see how the Boy(tu)/Boy(tu) configuration has the same probability as the others, because it’s twice as likely as the other two configurations that that is the configuration they are talking about when they say “One was born on Tuesday”.
Here’s my breakdown with 1000 families, to try to make it clear what I mean:
1000 Families with two children, 750 have boys.
Of the 750, 500 have one boy and one girl. Of these 500, 1⁄7, or roughly 71 have a boy born on Tuesday.
Of the 750, 250 have two boys. Of these 250, 2⁄7, or roughly 71 have a boy born on Tuesday.
71 = 71, so it’s equally likely that there are two boys as there are a boy and a girl.
Having two boys doubles the probability that one boy was born on Tuesday compared to having just one boy.
And I don’t think I’m confused about the sampling, because I didn’t use the sampling reasoning to get my result*, but I’m not super confident about that so if I am just keep giving me numbers and hopefully it will click.
*I mean in the previous post, not specifically this post.
Of these 250, 2⁄7, or roughly 71 have a boy born on Tuesday.
This is wrong. With two boys each with a probability of 1⁄7 to be born on Tuesday, the probability of at least one on a Tuesday isn’t 2⁄7, its 1-(6/7)^2
How can that be? There is a 1⁄7 chance that one of the two is born on Tuesday, and there is a 1⁄7 chance that the other is born on Tuesday. 1⁄7 + 1⁄7 is 2⁄7.
There is also a 1⁄49 chance that both are born on tuesday, but how does that subtract from the other two numbers? It doesn’t change the probability that either of them are born on Tuesday, and both of those probabilities add.
You overcount, the both on Tuesday is overcounted there. Think of it this way- if I have 8 kids do I have a better than 100% probability of having a kid born on Tuesday?
There is a 1/7x6/7 chance the first is born on Tuesday and the second is born on another day. There is a 1/7x6/7 chance the second is born on Tuesday and the first is born on another day. And there is a 1⁄49 chance that both are born on Tuesday.
All together thats 13⁄49. Alternatively, there is a (6/7)^2 chance that both are born not-on-Tuesday, so 1-(6/7)^2 tells you the complementary probability.
I’ve seen that same explanation at least five times and it didn’t click until just now. You can’t distinguish between the two on tuesday, so you can only count it once for the pair.
Which means the article I said was wrong was absolutely right, and if you were told that, say one boy was born on January 17th, the chances of both being born on the same day are 1-(364/365)^2 (ignoring leap years), which gives a final probability of roughly 49.46% that both are boys.
Thanks for your patience!
ETA: I also think I see where I’m going wrong with the terminology—sampling vs not sampling, but I’m not 100% there yet.
“The first coin comes up heads” (in this version) is not the same thing as “one of the coins comes up heads” (as in the original version). This version is 50%, the other is not.
How is it different? In both cases I have two independent coin flips that have absolutely no relation to each other. How does knowing which of the two came up heads make any difference at all for the probability of the other coin?
If it was the first coin that came up heads, TT and TH are off the table and only HH and HT are possible. If the second coin came up heads then HT and TT would be off the table and only TH and HH are possible.
The total probability mass of some combination of T and H (either HT or TH) starts at 50% for both flips combined. Once you know one of them is heads, that probability mass for the whole problem is cut in half, because one of your flips is now 100% heads and 0% tails. It doesn’t matter that you don’t know which is which, one flip doesn’t have any influence on the probability of the other. Since you already have one heads at 100%, the entire probability of the remainder of the problem rests on the second coin, which is a 50⁄50 split between heads and tails. If heads, HH is true. If tails, HT is true (or TH, but you don’t get both of them!).
Tell me how knowing one of the coins is heads changes the probability of the second flip from 50% to 33%. It’s a fair coin, it stays 50%.
Flip two coins 1000 times, then count how many of those trials have at least one head (~750). Count how many of those trials have two heads (~250).
Flip two coins 1000 times, then count how many of those trials have the first flip be a head (~500). Count how many of those trials have two heads (~250).
By the way, these sorts of puzzles should really be expressed as a question-and-answer dialogue. Simply volunteering information leaves it ambiguous as to what you’ve actually learned (“would this person have equally likely said ‘one of my children is a girl’ if they had both a boy and girl?”).
Yeah, probably the biggest thing I don’t like about this particular question is that the answer depends entirely upon unstated assumptions, but at the same time it clearly illustrates how important it is to be specific.
. There is still a 50% chance the second child is a boy and a 50% chance that the second child is a girl.
No there’s not. The cases where the second child is a boy and the second child is a girl are not equal probability.
I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?
If you picked “heads” before flipping the coins, then the probability is 1⁄3. There are three possibilities: HT, TH, and HH, and all of these possibilities are equally likely.
I have flipped two coins. One I flipped on a Tuesday and it came up heads. What is the probability that both are heads?
If you picked “heads” and “Tuesday” before knowing when you would be flipping the coins, and then flipped each coin on a randomly-selected day, and you just stopped if there weren’t any heads on Tuesday, then the answer is the same as the answer for boys on Tuesday. If you flipped the coin and then realized it was Tuesday, the Tuesday doesn’t affect the result.
The sex of one child has no influence on the sex of the other child, nor does the day on which either child was born influence the day any other child was born.
If you picked the sex first before looking at the children, the sex of one child does influence the sex of the other child because it affects whether you would continue or say “there aren’t any of the sex I picked” and the sexes in the cases where you would continue are not equally distributed.
In Boy1Tu/Boy2Tuesday, the boy referred to as BTu in the original statement is boy 1, in Boy2Tu/Boy1Tuesday the boy referred to in the original statement is boy2.
That’s why the “born on tuesday” is a red herring, and doesn’t add any information. How could it?
This sounds like you are trying to divide “two boys born on Tuesday” into “two boys born on Tuesday and the person is talking about the first boy” and “two boys born on Tuesday and the person is talking about the second boy”.
That doesn’t work because you are now no longer dealing with cases of equal probability. “Boy 1 Monday/Boy 2 Tuesday”, “Boy 1 Tuesday/Boy 2 Tuesday”, and “Boy 1 Tuesday/Boy 1 Monday” all have equal probability. If you’re creating separate cases depending on which of the boys is being referred to, the first and third of those don’t divide into separate cases but the second one does divide into separate cases, each with half the probability of the first and third.
As I pointed out above, whether it adds information (and whether the analysis is correct) depends on exactly what you mean by “one is a boy born on Tuesday”. If you picked “boy” and “Tuesday” at random first, and then noticed that one child met that description, that rules out cases where no child happened to meet the description. If you picked a child first and then noticed he was a boy born on a Tuesday, but if it was a girl born on a Monday you would have said “one is a girl born on a Monday”, you are correct that no information is provided.
The only relevant information is that one of the children is a boy. There is still a 50% chance the second child is a boy and a 50% chance that the second child is a girl. Since you already know that one of the children is a boy, the posterior probability that they are both boys is 50%.
Rephrase it this way:
I have flipped two coins. One of the coins came up heads. What is the probability that both are heads?
Now, to see why Tuesday is irrelevant, I’ll re-state it thusly:
I have flipped two coins. One I flipped on a Tuesday and it came up heads. What is the probability that both are heads?
The sex of one child has no influence on the sex of the other child, nor does the day on which either child was born influence the day any other child was born. There is a 1⁄7 chance that child 1 was born on each day of the week, and there is a 1⁄7 chance that child 2 was born on each day of the week. There is a 1⁄49 chance that both children will be born on any given day (1/7*1/7), for a 7⁄49 or 1⁄7 chance that both children will be born on the same day. That’s your missing 1⁄7 chance that gets removed inappropriately from the Tuesday/Tuesday scenario.
1⁄3 (you either got hh, heads/tails,or tails/heads). You didn’t tell me THE FIRST came up heads. Thats where you are going wrong. At least one is heads is different information then a specific coin is heads.
This is a pretty well known stats problem, a variant of Gardern’s boy/girl paradox. You’ll probably find it an intro book, and Jiro is correct. You are still overcounting. Boy-boy is a different case then boy-girl (well, depending on what the data collection process is).
If you have two boys (probability 1⁄4), then the probability at least one is born on Tuesday (1-(6/7)^2). ( 6/7^2 being the probability neither is born on Tuesday). The probability of a boy-girl family is (2*1/4) then (1/7) (the 1⁄7 for the boy hitting on Tuesday).
Lets add a time delay to hopefully finally illustrate the point that one coin toss does not inform the other coin toss.
I have two coins. I flip the first one, and it comes up heads. Now I flip the second coin. What are the odds it will come up heads?
No one is suggesting one flip informs the other, rather that when you say “one coin came up heads” you are giving some information about both coins.
This is 1⁄2, because there are two scenarios, hh, ht. But its different information then the other question.
If you say “one coin is heads,” you have hh,ht,th, because it could be that the first flip was tails/the second heads (a possibility you have excluded in the above).
No, it’s the exact same question, only the labels are different.
The probability that any one child is boy is 50%. We have been told that one child is a boy, which only leaves two options—HH and HT. If TH were still available, then so would TT be available because the next flip could be revealed to be tails.
Here’s the probability in bayesian:
P(BoyBoy) = 0.25 P(Boy) = 0.5 P(Boy|BoyBoy) = 1
P(BoyBoy|Boy) = P(Boy|BoyBoy)*P(BoyBoy)/P(Boy)
P(BoyBoy|Boy)= (1*0.25) / 0.5 = 0.25 / 0.5 = 0.5
P(BoyBoy|Boy) = 0.5
It’s exactly the same as the coin flip, because the probability is 50% - the same as a coin flip. This isn’t the monty hall problem. Knowing half the problem (that there’s at least one boy) doesn’t change the probability of the other boy, it just changes what our possibilities are.
No, it isn’t. You should consider that you are disagreeing with a pretty standard stats question, so odds are high you are wrong. With that in mind, you should reread what people are telling you here.
Now, consider “I flip two coins” the possible outcomes are hh,ht,th,tt
I hope we can agree on that much.
Now, I give you more information and I say “one of the coins is heads,” so we Bayesian update by crossing out any scenario where one coin isn’t heads. There is only 1 (tt)
hh,ht,th
So it should be pretty clear the probability I flipped two heads is 1⁄3.
Now, your scenario, flipped two coins (hh,ht,th,tt), and I give you the information “the first coin is heads,” so we cross out everything where the first coin is tails, leaving (hh,ht). Now the probability you flipped two heads is 1⁄2.
I don’t know how to make this any more simple.
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
I know it’s not the be all end all, but it’s generally reliable on these types of questions, and it gives P = 1⁄2, so I’m not the one disagreeing with the standard result here.
Do the math yourself, it’s pretty clear.
Edit: Reading closer, I should say that both answers are right, and the probability can be either 1⁄2 or 1⁄3 depending on your assumptions. However, the problem as stated falls best to me in the 1⁄2 set of assumptions. You are told one child is a boy and given no other information, so the only probability left for the second child is a 50% chance for boy.
Did you actually read it? It does not agree with you. Look under the heading “second question.”
I did the math in the post above, enumerating the possibilities for you to try to help you find your mistake.
Edit, in response to the edit:
Which is exactly analogous to what Jiro was saying about the Tuesday question. So we all agree now? Tuesday can raise your probability slightly above 50%, as was said all along.
And you are immediately making the exact same mistake again. You are told ONE child is a boy, you are NOT told the FIRST child is a boy. You do understand that these are different?
Re-read it.
The relevant quote from the Wiki:
We have no general population information here. We have one man with at least one boy.
I’m not at all sure you understand that quote. Lets stick with the coin flips:
Do you understand why these two questions are different: I tell you- “I flipped two coins, at least one of them came out heads, what is the probability that I flipped two heads?” A:1/3 AND “I flipped two coins, you choose one at random and look at it, its heads.What is the probability I flipped two heads” A: 1⁄2
For the record, I’m sure this is frustrating as all getout for you, but this whole argument has really clarified things for me, even though I still think I’m right about which question we are answering.
Many of my arguments in previous posts are wrong (or at least incomplete and a bit naive), and it didn’t click until the last post or two.
Like I said, I still think I’m right, but not because my prior analysis was any good. The 1⁄3 case was a major hole in my reasoning. I’m happily waiting to see if you’re going to destroy my latest analysis, but I think it is pretty solid.
Yes, and we are dealing with the second question here.
Is that not what I said before?
We don’t have 1000 families with two children, from which we’ve selected all families that have at least one boy (which gives 1⁄3 probability). We have one family with two children. Then we are told one of the children is a boy, and given zero other information. The probability that the second is a boy is 1⁄2, so the probability that both are boys is 1⁄2.
The possible options for the “Boy born on Tuesday” are not Boy/Girl, Girl/Boy, Boy/Boy. That would be the case in the selection of 1000 families above.
The possible options are Boy (Tu) / Girl, Girl / Boy (Tu), Boy (Tu) / Boy, Boy / Boy (Tu).
There are two Boy/Boy combinations, not one. You don’t have enough information to throw one of them out.
This is NOT a case of sampling.
As long as you realize there is a difference between those two questions, fine. We can disagree about what assumptions the wording should lead us to, thats irrelevant to the actual statistics and can be an agree-to-disagree situation. Its just important to realize that what the question means/how you get the information is important.
If we have one family with two children, of which one is a boy, they are (by definition) a member of the set “all families that have at least one boy.” So it matters how we got the information.
If we got that information by grabbing a kid at random and looking at it (so we have information about one specific child), that is sampling, and it leads to the 1⁄2 probability.
If we got that information by having someone check both kids, and tell us “at least one is a boy” we have different information (its information about the set of kids the parents have, not information about one specific kid).
If it IS sampling (if I grab a kid at random and say “whats your Birthday?” and it happens to be Tuesday), then the probability is 1⁄2. (we have information about the specific kid’s birthday).
If instead, I ask the parents to tell me the birthday of one of their children, and the parent says ‘I have at least one boy born on Tuesday’, then we get, instead, information about their set of kids, and the probability is the larger number.
Sampling is what leads to the answer you are supporting.
The answer I’m supporting is based on flat priors, not sampling. I’m saying there are two possible Boy/Boy combinations, not one, and therefore it takes up half the probability space, not 1⁄3.
Sampling to the “Boy on Tuesday” problem gives roughly 48% (as per the original article), not 50%.
We are simply told that the man has a boy who was born on tuesday. We aren’t told how he chose that boy, whether he’s older or younger, etc. Therefore we have four possibilites, like I outlined above.
Is my analysis that the possibilities are Boy (Tu) /Girl, Girl / Boy (Tu), Boy (Tu)/Boy, Boy/Boy (Tu) correct?
If so, is not the probability for some combination of Boy/Boy 1/2? If not, why not? I don’t see it.
BTW, contrary to my previous posts, having the information about the boy born on Tuesday is critical because it allows us (and in fact requires us) to distinguish between the two boys.
That was in fact the point of the original article, which I now disagree with significantly less. In fact, I agree with the major premise that the tuesday information pushes the odds of Boy/Boy closer 50%, I just disagree that you can’t reason that it pushes it to exactly 50%.
No. For any day of the week EXCEPT Tuesday, boy and girl are equivalent. For the case of both children born on Tuesday you have for girls: Boy(tu)/Girl(tu),Girl(tu)/Boy(tu), and for boys: boy(tu)/boy(tu).
This statement leads me to believe you are still confused. Do you agree that if I know a family has two kids, I knock on the door and a boy answers and says “I was born on a Tuesday,” that the probability of the second kid being a girl is 1/2? And in this case, Tuesday is irrelevant? (This the wikipedia called “sampling”)
Do you agree that if, instead, the parents give you the information “one of my two kids is a boy born on a Tuesday”, that this is a different sort of information, information about the set of their children, and not about a specific child?
I agree with this.
I agree with this if they said something along the lines of “One and only one of them was born on Tuesday”. If not, I don’t see how the Boy(tu)/Boy(tu) configuration has the same probability as the others, because it’s twice as likely as the other two configurations that that is the configuration they are talking about when they say “One was born on Tuesday”.
Here’s my breakdown with 1000 families, to try to make it clear what I mean:
1000 Families with two children, 750 have boys.
Of the 750, 500 have one boy and one girl. Of these 500, 1⁄7, or roughly 71 have a boy born on Tuesday.
Of the 750, 250 have two boys. Of these 250, 2⁄7, or roughly 71 have a boy born on Tuesday.
71 = 71, so it’s equally likely that there are two boys as there are a boy and a girl.
Having two boys doubles the probability that one boy was born on Tuesday compared to having just one boy.
And I don’t think I’m confused about the sampling, because I didn’t use the sampling reasoning to get my result*, but I’m not super confident about that so if I am just keep giving me numbers and hopefully it will click.
*I mean in the previous post, not specifically this post.
This is wrong. With two boys each with a probability of 1⁄7 to be born on Tuesday, the probability of at least one on a Tuesday isn’t 2⁄7, its 1-(6/7)^2
How can that be? There is a 1⁄7 chance that one of the two is born on Tuesday, and there is a 1⁄7 chance that the other is born on Tuesday. 1⁄7 + 1⁄7 is 2⁄7.
There is also a 1⁄49 chance that both are born on tuesday, but how does that subtract from the other two numbers? It doesn’t change the probability that either of them are born on Tuesday, and both of those probabilities add.
The problem is that you’re counting that 1/49th chance twice. Once for the first brother and once for the second.
I see that now, it took a LOT for me to get it for some reason.
You overcount, the both on Tuesday is overcounted there. Think of it this way- if I have 8 kids do I have a better than 100% probability of having a kid born on Tuesday?
There is a 1/7x6/7 chance the first is born on Tuesday and the second is born on another day. There is a 1/7x6/7 chance the second is born on Tuesday and the first is born on another day. And there is a 1⁄49 chance that both are born on Tuesday.
All together thats 13⁄49. Alternatively, there is a (6/7)^2 chance that both are born not-on-Tuesday, so 1-(6/7)^2 tells you the complementary probability.
Wow.
I’ve seen that same explanation at least five times and it didn’t click until just now. You can’t distinguish between the two on tuesday, so you can only count it once for the pair.
Which means the article I said was wrong was absolutely right, and if you were told that, say one boy was born on January 17th, the chances of both being born on the same day are 1-(364/365)^2 (ignoring leap years), which gives a final probability of roughly 49.46% that both are boys.
Thanks for your patience!
ETA: I also think I see where I’m going wrong with the terminology—sampling vs not sampling, but I’m not 100% there yet.
“The first coin comes up heads” (in this version) is not the same thing as “one of the coins comes up heads” (as in the original version). This version is 50%, the other is not.
How is it different? In both cases I have two independent coin flips that have absolutely no relation to each other. How does knowing which of the two came up heads make any difference at all for the probability of the other coin?
If it was the first coin that came up heads, TT and TH are off the table and only HH and HT are possible. If the second coin came up heads then HT and TT would be off the table and only TH and HH are possible.
The total probability mass of some combination of T and H (either HT or TH) starts at 50% for both flips combined. Once you know one of them is heads, that probability mass for the whole problem is cut in half, because one of your flips is now 100% heads and 0% tails. It doesn’t matter that you don’t know which is which, one flip doesn’t have any influence on the probability of the other. Since you already have one heads at 100%, the entire probability of the remainder of the problem rests on the second coin, which is a 50⁄50 split between heads and tails. If heads, HH is true. If tails, HT is true (or TH, but you don’t get both of them!).
Tell me how knowing one of the coins is heads changes the probability of the second flip from 50% to 33%. It’s a fair coin, it stays 50%.
Flip two coins 1000 times, then count how many of those trials have at least one head (~750). Count how many of those trials have two heads (~250).
Flip two coins 1000 times, then count how many of those trials have the first flip be a head (~500). Count how many of those trials have two heads (~250).
By the way, these sorts of puzzles should really be expressed as a question-and-answer dialogue. Simply volunteering information leaves it ambiguous as to what you’ve actually learned (“would this person have equally likely said ‘one of my children is a girl’ if they had both a boy and girl?”).
Yeah, probably the biggest thing I don’t like about this particular question is that the answer depends entirely upon unstated assumptions, but at the same time it clearly illustrates how important it is to be specific.
No there’s not. The cases where the second child is a boy and the second child is a girl are not equal probability.
If you picked “heads” before flipping the coins, then the probability is 1⁄3. There are three possibilities: HT, TH, and HH, and all of these possibilities are equally likely.
If you picked “heads” and “Tuesday” before knowing when you would be flipping the coins, and then flipped each coin on a randomly-selected day, and you just stopped if there weren’t any heads on Tuesday, then the answer is the same as the answer for boys on Tuesday. If you flipped the coin and then realized it was Tuesday, the Tuesday doesn’t affect the result.
If you picked the sex first before looking at the children, the sex of one child does influence the sex of the other child because it affects whether you would continue or say “there aren’t any of the sex I picked” and the sexes in the cases where you would continue are not equally distributed.