Road rules2 mins ago
Sex
165 Answers
Following on from the entertaining Dice and Socks threads earlier this week I've now found the problem about the sex of children.
One version goes: "You know that Mr. Smith has two children and that at least one of them is a boy. What is the probability that both children are boys?"
Thoughts please?
One version goes: "You know that Mr. Smith has two children and that at least one of them is a boy. What is the probability that both children are boys?"
Thoughts please?
Answers
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-Talbot- //You are stood with Mr Smith and his son waiting for Mrs Smith to come along with his other child. She will turn up with either a boy or a girl...50/50.
Please tell me how in this actual, physical situation can it be anything other than 50/50. //
Try this, Talbot: you are standing with Mr Smith, Mr Brown, Mr Jones, Mr Grey, Mr White, and Mr Black, each of whom is the father of two children and has one of them - a boy - with him. When the wives come along with the other child, your argument would say that statistically three of them should be boys, and three of them girls.
BUT . . . that would mean that two-child families with two boys were just as likely as two-child families with one of each. That's simply not true, as has been exhaustively explained. When the six wives turn up, we statistically expect them to turn up with four girls and two boys.
Please tell me how in this actual, physical situation can it be anything other than 50/50. //
Try this, Talbot: you are standing with Mr Smith, Mr Brown, Mr Jones, Mr Grey, Mr White, and Mr Black, each of whom is the father of two children and has one of them - a boy - with him. When the wives come along with the other child, your argument would say that statistically three of them should be boys, and three of them girls.
BUT . . . that would mean that two-child families with two boys were just as likely as two-child families with one of each. That's simply not true, as has been exhaustively explained. When the six wives turn up, we statistically expect them to turn up with four girls and two boys.
cheer up, Prudie, I thought it sounded unlikely too, but it does work out to be 1/3.
The "trick" is that we know one is a boy but we don't know which one. That stymies the argument about it being the same as "this one's a boy, the other one has a 50/50 chance of being a boy". We can't say "this one's a boy" because - as per the OP - we know one is but we don't know if it's "this" one.
The "trick" is that we know one is a boy but we don't know which one. That stymies the argument about it being the same as "this one's a boy, the other one has a 50/50 chance of being a boy". We can't say "this one's a boy" because - as per the OP - we know one is but we don't know if it's "this" one.
And to be more serious jno's expansion explains the conundrum perfectly. You are not being asked to assess the probability of what gender the second child will be. You are being asked to assess the probability that the couple will have two children of the same specific gender. It has been adequately explained that there are four possible combinations of genders for their children. One can be ignored (GG) becuase we know that at least one child is a boy. There are two chances that the children are of opposite genders (BG, GB) and only one chance that they are both boys. Simples !!!
If a bookie offered odds on the chances of the other child being a boy he would be delighted to be offering 33.3/1
If someone gave me a million to gamble on the sex of the other child saying I could keep the money on the correct choice, I would put 500,000 on a boy and 500,000 on a girl.
You boffins can divide your million by 3 to cover all possibilities if you like ;)
If someone gave me a million to gamble on the sex of the other child saying I could keep the money on the correct choice, I would put 500,000 on a boy and 500,000 on a girl.
You boffins can divide your million by 3 to cover all possibilities if you like ;)
It can be other than 50/50 because Mrs Smith has twice as much chance of coming along with a daughter as she has with a son.
To understand this one has to leave the 'waiting for Mrs. Smith' scenario, which just serves to confuse, and examine the likelihood of a younger or older sibling being male or female. This has been covered by a number of posts earlier.
To understand this one has to leave the 'waiting for Mrs. Smith' scenario, which just serves to confuse, and examine the likelihood of a younger or older sibling being male or female. This has been covered by a number of posts earlier.
I'm sure it must be fairly easy to write a simple piece of code to show you this is so. A random number generator is the basis of it.
Generate one number, if lower than x call it a boy, if higher than x call it a girl. ensure an equal chance of each, i.e. 50/50.
Generate second number, if they both turn out to be girls, throw the result away as it isn't relevant, and start again.
Otherwise you have a valid combination which we will use.
You bet £500,000 on two boys for each of say, a thousand goes.
Each time you win I'll give you £500,000 and your stake back.
Each time you lose you give me your £500,000 and find another £500,000 to play again.
Generate one number, if lower than x call it a boy, if higher than x call it a girl. ensure an equal chance of each, i.e. 50/50.
Generate second number, if they both turn out to be girls, throw the result away as it isn't relevant, and start again.
Otherwise you have a valid combination which we will use.
You bet £500,000 on two boys for each of say, a thousand goes.
Each time you win I'll give you £500,000 and your stake back.
Each time you lose you give me your £500,000 and find another £500,000 to play again.
Thanks all.
I'm going to struggle to select a Best Answer.
It's good to know that some others found this problem as troublesome as I did, and it's interesting that what seems a simple problem can lead to differences of opinion between people who are accepted as being good mathematicians.
I get the Monty Hall problem, the dice problem and the Sock problem perfectly- but this one just ties my brain in knots.
Now, is it a good time to mention the Two Envelope problem again?
I'm going to struggle to select a Best Answer.
It's good to know that some others found this problem as troublesome as I did, and it's interesting that what seems a simple problem can lead to differences of opinion between people who are accepted as being good mathematicians.
I get the Monty Hall problem, the dice problem and the Sock problem perfectly- but this one just ties my brain in knots.
Now, is it a good time to mention the Two Envelope problem again?
Having seen one has no effect whatsoever on the probability of both having a certain gender. The gender of either is not somehow magically reversed simply by virtue of one having been seen.
Assuming boys and girls are equally likely (and excluding any third alternatives) the probability of both being boys (one in four) remains the same and is not altered in any way simply by virtue of knowing the gender of one of them.
All that aside,
1. They're children
2. They're siblings
They shouldn't be having sex anyway! :o/
Assuming boys and girls are equally likely (and excluding any third alternatives) the probability of both being boys (one in four) remains the same and is not altered in any way simply by virtue of knowing the gender of one of them.
All that aside,
1. They're children
2. They're siblings
They shouldn't be having sex anyway! :o/
Prudie -- I think my plenty of students would beg to differ. If I've lost my patience once on a forum it's hardly a catastrophe, we all do it occasionally.
Nevertheless, I've laid out the solution clearly and in my opinion the problem is only difficult if people make it so by failing to count the outcomes correctly. Once people stop trying to skip essential working and work through the problem systematically, there's no difficulty since in the end it's just working out sums of simple fractions.
Nevertheless, I've laid out the solution clearly and in my opinion the problem is only difficult if people make it so by failing to count the outcomes correctly. Once people stop trying to skip essential working and work through the problem systematically, there's no difficulty since in the end it's just working out sums of simple fractions.
Go on, mention the two envelope problem!
I don't really think it's a difference of opinion. It's one group trying to skip to the end without working through the problem correctly and so getting it wrong, and another group of people who actually work through the solution systematically and so don't slip up.
I don't really think it's a difference of opinion. It's one group trying to skip to the end without working through the problem correctly and so getting it wrong, and another group of people who actually work through the solution systematically and so don't slip up.
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