Meiosis in males produces an equal ratio of gametes containing the x or Y chromosome. The female gamete has always got the X chromosome. On ejaculation therefore an equal proportion of male x and Y chromosomes are released. Therefore the chances of a male being conceived is always 1:1.
1:1? That's 100%. I think most people would doubt it's that many.
Let's try this then. The odds on them having two boys were 1/4 before you met them. The odds on them having a girl and a boy were 2/4. Knowing that there's one boy doesn't change this, so the 2/4:1/4 ratio can be simplified as 2:1, which gives odds of 2/3 girl, 1/3 boy. It seems stupid but it's probably what you're looking for. Similar to the Monty Hall problem.
I think he needs to get on and give her one rather than trying to work out the odds, or he'll run out of time (or his missus will fuck off with someone else)
Meiosis in males produces an equal ratio of gametes containing the x or Y chromosome. The female gamete has always got the X chromosome. On ejaculation therefore an equal proportion of male x and Y chromosomes are released. Therefore the chances of a male being conceived is always 1:1.
Not quite. There are factors like the ratio of magnesium to sodium in the mother's diet and acidity of the vagina that affect the chances of the X or Y chromosome sperm surviving to reach the egg. This is why, even though the sperm are produced in equal proportions, males make up just over 50% of births - or at least they used to when we did this in 'O' Level biology in the '70s.
Let's try this then. The odds on them having two boys were 1/4 before you met them. The odds on them having a girl and a boy were 2/4. Knowing that there's one boy doesn't change this, so the 2/4:1/4 ratio can be simplified as 2:1, which gives odds of 2/3 girl, 1/3 boy. It seems stupid but it's probably what you're looking for. Similar to the Monty Hall problem.
No. We were told that the oldest is a boy. That means your 2/4 changes to 1/4, which suggests the odds of girl/boy for the second child return to 50/50. If we'd been told that one child was a boy, but not which one, the odds would be 2:1, so 33% boy, 67% girl.
Think of it this way - There are four couples, each with two kids and these four couples have, respectively, BB, BG, GB, GG. Each of these permutations are equally likely. Now, select one couple at random. If they say their oldest child is a boy, they must be the BB or the BG couple. Therefore, there's an even chance that the second is a boy or girl. However, if they say that they have a boy, but don't say whether it's the older or younger child, then they could be BB, BG or GB. Hence the probability of the other child being male is 33%
Edit: I see the OP has been edited to remove the "oldest" child bit. Cheeky.
I have weighed this up and after careful consideration, and some deliberation, have decided, who the hell gives a flying F**k!. I just want to know how tall this Arsenal lad is, and on what type of train hes coming in on.
I have weighed this up and after careful consideration, and some deliberation, have decided, who the hell gives a flying F**k!. I just want to know how tall this Arsenal lad is, and on which train hes coming in on.
Oh, OK. I was about to give the answer. I won't now.
You meet a couple. They tell you they have two children. One child is a boy. What is the probability that the other child is also a boy?
If we knew which was the first born we could limit it further.
If they were to say 'Our first born was a boy what is the chance the second born is also a boy?' we can then remove girl-boy leaving a 1/2 rating of odds.
You meet a couple. They tell you they have two children. One child is a boy. What is the probability that the other child is also a boy?
If we knew which was the first born we could limit it further.
If they were to say 'Our first born was a boy what is the chance the second born is also a boy?' we can then remove girl-boy leaving a 1/2 rating of odds.
Yes. But that is what was originally said. The OP has been edited to remove the bit that said the first child was a boy. So I refer you back to my answer further up the page.
You're employing the gambler's fallacy, that outcome B is contingent on outcome A (i.e. the roulette wheel was red that spin, so it's more likely to be black this time).
It's not the case, and biologically speaking, is generally 50/50 as to whether a kid is a boy or girl, regardless of gender of previous child.
You're employing the gambler's fallacy, that outcome B is contingent on outcome A (i.e. the roulette wheel was red that spin, so it's more likely to be black this time).
It's not the case, and biologically speaking, is generally 50/50 as to whether a kid is a boy or girl, regardless of gender of previous child.
@Dazzler21 has spelt it out precisely and accurately.
There are four, equally-likely scenarios for people having two children. The fact that one is a boy disqualifies one of those four, leaving three, equally-likely scenarios. It can be 1. Boy-Boy, 2. Boy-Girl or 3. Girl-Boy.
So there are twice as many scenarios in which the "other" child is a girl. Hence, 33%.
You're employing the gambler's fallacy, that outcome B is contingent on outcome A (i.e. the roulette wheel was red that spin, so it's more likely to be black this time).
It's not the case, and biologically speaking, is generally 50/50 as to whether a kid is a boy or girl, regardless of gender of previous child.
@Dazzler21 has spelt it out precisely and accurately.
There are four, equally-likely scenarios for people having two children. The fact that one is a boy disqualifies one of those four, leaving three, equally-likely scenarios. It can be 1. Boy-Boy, 2. Boy-Girl or 3. Girl-Boy.
So there are twice as many scenarios in which the "other" child is a girl. Hence, 33%.
Yeah. An hour and a half after me. And after you changed the question.
In a week when the BMA have advised their members not to refer to their pregnant patience as 'expectant mothers' and to use instead the more socially acceptable term 'pregnant people', I find this thread highly inappropriate and deeply offensive on so many levels.
Its not really about mutual exclusivity. Its an element of the question, but it's more about an easily misunderstood feature of probability. Basically, it's about how many potential (and in this case, equally likely) outcomes can be represented by the scenario.
As I said earlier, you'd be right if the question hadn't been changed. That is, when we were told that the first child was a boy, the odds of boy or girl for the second child are 50%. Now that we are only told that there is at least one boy, but we don't know whether he is the older or younger sibling, it opens up more possible outcomes. This is because the scenario of both siblings being boys is represented by the set up regardless of whether we have been told the gender of the first or second child. Two birds with one stone, if you like. The scenario of there being one of each gender has two distinct circumstances (older boy, younger girl, or older girl, younger boy). Its counter intuitive, but it does work.
You can actually test this with a shuffled pack of cards. Imagine Black suits are male and red suits are female. Deal two playing cards face up next to each other. If at least one of them is black, make a tally of the colour of the other card. Repeat this process with a full deck four times, shuffling thoroughly each time you start the deck again. When you've done this, you've sampled 104 pairs of cards. You will almost certainly have marked red roughly twice as many times as black. I'd even go so far as to say you'll have made about seventy five to eighty tally marks and over fifty will be for red. If you don't believe me, give it a try.
*Disclaimer* The nature of probability means there is always a chance that this won't happen. In fact, if enough of us do it, it almost certainly won't happen for one of us! But in most cases, it will.
Comments
They didn't give any other clues.
If we'd been told that one child was a boy, but not which one, the odds would be 2:1, so 33% boy, 67% girl.
Think of it this way - There are four couples, each with two kids and these four couples have, respectively, BB, BG, GB, GG. Each of these permutations are equally likely. Now, select one couple at random. If they say their oldest child is a boy, they must be the BB or the BG couple. Therefore, there's an even chance that the second is a boy or girl.
However, if they say that they have a boy, but don't say whether it's the older or younger child, then they could be BB, BG or GB. Hence the probability of the other child being male is 33%
Edit: I see the OP has been edited to remove the "oldest" child bit.
Cheeky.
Statisticians will tell you it is 1/3.
This is because there are four possible combinations:
boy-boy
boy-girl
girl-boy
girl-girl.
Since we know Chizz said one of the children is a boy we can rule out the girl-girl combination, leaving three remaining options.
boy-boy
boy-girl
girl-boy
Only one out of 3 is boy-boy, so we get a 1/3 chance. If we knew which was the first born we could limit it further.
If they were to say 'Our first born was a boy what is the chance the second born is also a boy?' we can then remove girl-boy leaving a 1/2 rating of odds.
Place your bets now, and WSS can tell us what the answer is, if as and when the second comes along.
So I refer you back to my answer further up the page.
*cue nonce comment
You seem to think it's 33%. Why?
You're employing the gambler's fallacy, that outcome B is contingent on outcome A (i.e. the roulette wheel was red that spin, so it's more likely to be black this time).
It's not the case, and biologically speaking, is generally 50/50 as to whether a kid is a boy or girl, regardless of gender of previous child.
There are four, equally-likely scenarios for people having two children. The fact that one is a boy disqualifies one of those four, leaving three, equally-likely scenarios. It can be 1. Boy-Boy, 2. Boy-Girl or 3. Girl-Boy.
So there are twice as many scenarios in which the "other" child is a girl. Hence, 33%.
Disappointed, France.
It is a maths paradox, designed to cause an argument.
Basically, it's about how many potential (and in this case, equally likely) outcomes can be represented by the scenario.
As I said earlier, you'd be right if the question hadn't been changed. That is, when we were told that the first child was a boy, the odds of boy or girl for the second child are 50%.
Now that we are only told that there is at least one boy, but we don't know whether he is the older or younger sibling, it opens up more possible outcomes. This is because the scenario of both siblings being boys is represented by the set up regardless of whether we have been told the gender of the first or second child. Two birds with one stone, if you like. The scenario of there being one of each gender has two distinct circumstances (older boy, younger girl, or older girl, younger boy).
Its counter intuitive, but it does work.
You can actually test this with a shuffled pack of cards. Imagine Black suits are male and red suits are female.
Deal two playing cards face up next to each other. If at least one of them is black, make a tally of the colour of the other card. Repeat this process with a full deck four times, shuffling thoroughly each time you start the deck again. When you've done this, you've sampled 104 pairs of cards. You will almost certainly have marked red roughly twice as many times as black. I'd even go so far as to say you'll have made about seventy five to eighty tally marks and over fifty will be for red. If you don't believe me, give it a try.
*Disclaimer* The nature of probability means there is always a chance that this won't happen. In fact, if enough of us do it, it almost certainly won't happen for one of us! But in most cases, it will.