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I will not change my mind as a result of what others say. You should all know that by now.
Then. Do. The. Math!
Or try using the spreadsheet as I described above.
Or trying tossing two coins 100 times.
Every time you get two heads then cut you left arm.
Every time you get a head and a tail then cut your right arm.
Once you are finished simply measure the blood loss from each arm. You will find the right one has twice as many cuts as the left and will bleed out quicker.
The Mathematicians, Engineers and Scientists are all agreeing, are you a midwife? 😛
I have done the math. I just think you should all be doing different maths.
Smee... it doesn't matter what you say, I know that's what you meant. You lost, and are just too proud to admit it.
I understand probability perfectly well. Many years of doing engineering maths at uni has seen to that.
He is apparently an engineer, but for some reason he uses a different mathematical system from the rest of humanity. I hope you're not involved in anything critical Smee.
You have been given several clear empirical experiments that produce results which fit perfectly with our answer and you have been unable to find fault in them (and instead you've just quietly ignored them).
If your answer (the 50% answer, not the 2:1 answer) is correct then lets hear how we can conduct a simple experiment that backs it up.
Any experiment you like... in your own time...
[i]He is apparently an engineer, but for some reason he uses a different mathematical system from the rest of humanity. I hope you're not involved in anything critical Smee.[/i]
Guess that's why he became a driving instructor.
well i read my way thru that lot but going back to the original rules - how do we know the lady isn't a compulsive liar?
Anything yet Smee? Nope thought not. 😀
The coin toss backs it up.
The coin toss experiment that I outlined a few posts ago? That backs up [u]your[/u] answer?!?
Or have you devised your own perverted version of it?
Please explain in enough detail that we can try it too.
options for two throws
tt
th
ht
hh
when you know that one is a h you have:
one of ht or th - not both.
and
hh left as possible outcomes after the second throw.
The problem is the over-complicated way the question is being asked. As the gender of one child is already known the remaining and simplest unknown and therefore the only question which needs to be asked is what is the gender of the other child? The answer to this question will answer the original question and has a probability of 50%.
Andy.
Erm.. oh dear.
So you're saying "It is one of these four possibilities and if you get information that eliminates one of those possibilities then that leaves you with two."?
one of ht or th - not both
Well actually the result will be one of ht, th or hh - not all three. They are all equally likely though.
No I'm saying that once you know that one is a head then it removes the options that start with a tail.
Go and get two coins.
What are the chances of the first one being a head? 50:50
What are the chances of the second one being a head? 50:50
If you dont know any of the results then all 4 permutations have a 25% chance of happening.
After the first result is known they have a 50% chance.
Look Smee, you have a friend... 😀
By the way, you never did answer me about whether an older girl/younger boy is really the same as an older boy/younger girl...
Yes yes Smee those are independent events. Everyone agrees that, as explained 16 times already.
the first result doesn't influence the second. No one disputes that either.
But you don't know the "first result", you know [u]one[/u] of the results.
That is very different.
Try my 100 coin toss. Explain the results on my spreadsheet.
Or come up with an experiment that matches the question.
How about this then: you agreed above that before you know any results then all four coin combinations have a 25% chance.
So how about I throw the coins and I then tell you "I didn't throw two tails"
What are the odds on the other combinations now?
[i]What are the chances of the first one being a head? 50:50
What are the chances of the second one being a head? 50:50
If you dont know any of the results then all 4 permutations have a 25% chance of happening.
After the first result is known they have a 50% chance. [/i]
I can't believe you don't grasp the coin one.
Come on lads, this isn't over, we can stretch it for another day!
Smee, at some point we need to meet up and do this with coins. I'll bring 50 £1 coins and you bring 50 £1 coins.
If there's two tails, you get to keep the coins. If there's one of each, I get to keep them. If you're right, we'll both come out even.
If you're right, we'll both leave with ~£50. If the rest of the planet is right, I'll leave with ~£67 and you'll leave with ~£33.
right, final answer here
http://mathforum.org/dr.math/faq/faq.boy.girl.html
Surely smee would not dare doubt the wisdom of Dr Math!?
Come on lads, this isn't over, we can stretch it for another day!
250 replies is respectable, but I'm still hoping for my first Triple Century posting. 😀
What happens if we use 20p peices instead of £1 coins?
Would this reduce the odds to 20% or 66%?
From the Dr Math site: "We can also visualize this problem using a probability tree"
Gosh, they look familiar. Mine were prettier though.
[i]right, final answer here
http://mathforum.org/dr.math/faq/faq.boy.girl.html [/i]
But it's wrong the maths wrong, Smee said so.
I know why he's confused I was at first but I never claimed to be right just couldn't see why it wasn't 50/50 1:1 50%. Then it suddenly sunk in that's it's the lack of specify which child that increases the odds, the same way the order the lottery numbers comes out doesn't matter therefore increasing the odds.
Hummm...
I think I'm beginning to understand how the global economy became f*****.
Population too high? Not enough parents with two girls? I think I'm with you here.
I think I'm beginning to understand how the global economy became f*****.
I don't think you can blame Smee for that one.
I think I'm beginning to understand how the global economy became f*****.I don't think you can blame Smee for that one.
I think you can if he's been teaching the banks how to do sums.
when you know that one is a h you have:
one of ht or th - not both.
if you have 1 x 10p and 1 x 2p, then quite clearly 10p heads, 2p tails is not the same as 10p tails, 2p heads. hence two different ways of getting one heads and one tails. therefore twice as likely as getting 10p heads and 2p heads. is this getting through to you?
Lets get it over 300.
It can be 50%, here's how.
What you want to know is whether "she has a girl and a boy?". You can find this out by finding out the gender of both children. As you already know the gender of 1 child you only need to know the gender of the other child. As the only unknown is the gender of 1 child where the two possible outcomes each have a probability of 50% you can determine whether "she has a girl and a boy?" with a probability of 50%.
Andy.
What you want to know is whether "she has a girl and a boy?".
Nope - that's not what the question asked.
The question is about whether her responding that she has at least one girl changes the probability that she is in the group that have one child of each sex. And it does.
If the question said [i]"Here is a woman with a daughter, what is the gender of her other child?"[/i] [u]then[/u] it would have been a straight 50:50 - but that is a very different question.
Also.. Rona is a girl's name. 😛
hi just for the next post
GrahamS is quite correct.
The way the question is asked/worded provides 2 very different answers.
Oooh reading back a few pages, quite a few people are correct apologies.
Yeah, it's pretty much just been Smee that hasn't seen the light so far.
Everyone else has got there - as I'm sure Rona will 🙂
Funny how this is Rona's first ever post.
Drac, been here a very long time.
Graham,
"What you want to know is whether "she has a girl and a boy?".
Nope - that's not what the question asked."
No, indeed but that's why probability is only probability, if you ask a different but relevant question to get the information you need you may get a different probability, i.e. as pointed out before the odds that the remaining child is a boy are 50:50 by simply asking that question the probability of her having a girl and a boy are 50:50.
Andy.
not too many to go now. . .
[i]Drac, been here a very long time.[/i]
Took you a month to post though.
It's been pretty dull.