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Originally posted by mtthwWrong mtthw. Again there are only 3 possibilities, again you are regurgitating a worthless and in this case wrong explanation.
He's right, you know.
Writing the eldest first, there are four possibilities with equal probability. BB, BG, GB, GG. They are equal in probability because we assume that the sex of the first child is independent of the second child.
The additional information rules out one possibility. So we have three possibilities..but still with equal probability.
A 1/3 chance that there are two boys, given that we know there's at least one. Yes.[/b]
BB
BG
GG
the third possibility can be excluded.
So that leaves only 2 possibilities.
BB
BG
You're thinking permutations, that is not implicity implied in the problem statement. Your whizz bang solution is in error, your reasoning and explanation doesn't make any sense.
"A 1/3 chance that there are two boys, given that we know there's at least one. Yes."
So that means that there is a 2/3 chance that there is a girl and a boy, given that there is one boy.
This doesn't even make sense.
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Originally posted by eldragonflyEquivalent problem..
GB and BG are the same, one girl and one boy. And GG is impossible.
I toss a fair coin two times in succession. What are the odds at least one of the two tosses comes up tails if...
1) The FIRST toss is heads?
2) At least one of the tosses are heads?
The first toss and the second toss are distinct events and we can differentiate between the two. Head-tails is therefore a distinctly different possibility than tails-heads, and given each toss is an independent event, they are equally likely.
The upshot is that given just the setup, you have a 50% that both heads and tails come up, and a 25% chance for two heads, and a 25% chance for two tails..
In the first question, we are given the first toss results, and no information on the second toss, explicit or implied. Since the second toss is independent of the first and the coin is fair, it's fairly obvious that either side has a 50% chance of occurring.
In the second question, we aren't told WHICH toss was heads (and it could be both). But we are told heads came up at least once, and therefore it is the case both coin tosses aren't tails.
That leaves us with the following possibilities..
1) Heads came up twice,
2) Heads was thrown first, followed by tails.
3) Tails was thrown first, followed by heads.
Do note that the third case was eliminated in the first question, but isn't eliminated here. At any rate, we know all three of these started as equally likely, and two of these feature tails being thrown once.
The child problem is very much the same.
The second question includes those cases where the girl was born before the boy, whereas the first question does not include that possibility. The eldest IS a boy.
Two girls, of course, is eliminated as a possibility in both questions.
Originally posted by deriver69That's one's not particularly hard either.
Everytime I come back here there are another 50 or so comments. Have we done the one where if the four aces (2 red and 2 black) are face down on the table, what is the probability that two cards turned over will be of the same colour?
Actually I think it might be similar the boy/girl problem but it is getting late
There's only three cards left after you make your initial draw, and the color of the card (black/red) has symmetrical properties, since there are two of each, so in the end, in really doesn't matter which ace is drawn.
To eldragonfly in regards to the Child problem..
Let me pose three more questions to you with the same family with two non-twin children, without being told any additional information.
1) What are the odds both are boys?
2) What are the odds both are girls?
3) What are the odds they have one of each?
The answers to these questions are highly relevant when you start throwing in additional conditions, especially in the second part of the question, where no reference to made to elder/younger.
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Originally posted by geepamoogleDo the math geepamoogle. You can't have a 2/3 chance that the next child is a girl and only a 1/3 chance that the next child is a boy, when the true odds are 50-50% by definition.
Equivalent problem..
I toss a fair coin two times in succession. What are the odds at least one of the two tosses comes up tails if...
1) The FIRST toss is heads?
2) At least one of the tosses are heads?
The first toss and the second toss are distinct events and we can differentiate between the two. Head-tails is therefore a distinctly diff [b]IS a boy.
Two girls, of course, is eliminated as a possibility in both questions.[/b]
Your coin toss analogy is wrong. Each coin toss in an independent event.
The upshot is that given just the setup, you have a 50% that both heads and tails come up, and a 25% chance for two heads, and a 25% chance for two tails..
The binomial expansion is wrongly referenced here, has no place in this problem. Once again you are addressing this as a permutations problem instead of a combinations problem.
Originally posted by geepamoogleWrong. Again.
1) What are the odds both are boys?
2) What are the odds both are girls?
3) What are the odds they have one of each?
The answers to these questions are highly relevant when you start throwing in additional conditions, especially in the second part of the question, where no reference to made to elder/younger.
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Originally posted by eldragonflyPART 1
Do the math geepamoogle. You can't have a 2/3 chance that the next child is a girl and only a 1/3 chance that the next child is a boy, when the true odds are 50-50% by definition.
Your coin toss analogy is wrong. Each coin toss in an independent event.
The upshot is that given just the setup, you have a 50% that both heads and tails come up, ...[text shortened]... Once again you are addressing this as a permutations problem instead of a combinations problem.
Let me get straight to the point. With the children, each birth is a distinct and independent event. I made this abundantly clear by letting you know they weren't twins (and hence that the gender of the first birth does not affect the odds of the second birth at all.)
Both births HAVE occurred in the problem, although we are not told the exact result of the two births. We aren't predicting the birth of the "next" child.
Now it would seem we are in agreement on the first question, at least on the answer, so I shant do that math again.
However, I will note that in the second question, we don't know that the ELDEST is a boy, like we did in the first. This allows for the possibility that a girl was born, followed by a boy, a case which does not meet the pre-condition for the first question. So in addition to all of the cases fall into the first question, you have additional cases which are added to the second question, ALL OF WHICH INCLUDE A GIRL.
You know what though, I see now my chances of getting any point across are very slim. But let me explain to you my reasoning behind the relevancy of the three additional questions.
PART 2
Just as with two tosses of a fair coin, you're most likely result will be 1 heads and 1 tails (50% applying binomial formulas), with two independent births the odds are 50% that a family with two children will have 1 boy and 1 girl. (Twins are excluded from this calculation as they are linked together.)
The chances of two boys is 1/2 * 1/2 or 25%. Same with two girls.
Press 1 if you disagree with these numbers.
If you agree, let me continue then...
Next step is to take into account information on the actual results. The additional information will eliminate some possibilities that were possible given the set-up alone. I take this step independently afterwards so that I don't let information on results confuse my calculation on the odds of the set-up. At any rate, since in the second question, I am told there is at least one boy in the family, I can now take out the 25% of the time the family has two girls, leaving the other 75% of the time in my sample space.
Press 2 if you disagree with this reasoning and why.
Now that we've limited ourselves to the sample space which meets the pre-condition (where the results fit with the information provided on what happened), we can then examine what percentage of the sample space meets the criteria we are looking for in our question.
In this case, we want to find the odds of the family having at least one girl. This occurs 50% out of the 75% of the sample space left.
50% / 75% = 2-in-3... (Yes, this is Bayes formula at work again..)
Press 3 if you diagree with this and explain why you think the 50% of the time a family has 1 boy/1 girl should be considered equal in likelihood to the 25% of the time the family has 2 boys and no girls...
Here once again I will state the logical fallacy I see you falling into..
Just because there are only 2 possibilities does not necessarily mean that they have an equal chance of occurring.
For instance, if I enter the lottery, I will either win or I will lose. That does not mean I have a 50% chance at the jackpot.
mtthw, geepamoogle, LemonJello, Aethereal, deriver69...I love all you guys. Even if eventually, one of you think I'm stupid, or I think you're stupid, I'll respect your opinions and be glad that I've known you.
eldragonfly, you're the biggest laughing stock since ACME Novelties went public. To be honest, I'm surprised you're kicking up such a fuss, because you did answer the dice problem correctly. One of these days you're going to have to tell me how you grew so thick on a diet gruel and bologna.
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Originally posted by PBE6lmao PBE6. love you too 😉 i'm a pretty patient person, and consider myself a pretty clear and reasonable teacher... but eldragonfly seems to have given up on trying to give the nuance of the question its due consideration.
mtthw, geepamoogle, LemonJello, Aethereal, deriver69...I love all you guys. Even if eventually, one of you think I'm stupid, or I think you're stupid, I'll respect your opinions and be glad that I've known you.
eldragonfly, you're the biggest laughing stock since ACME Novelties went public. To be honest, I'm surprised you're kicking up such a fuss, because ...[text shortened]... days you're going to have to tell me how you grew so thick on a diet gruel and bologna.
that being said, i will try once more with the boy/girl question. i'll start with a direct quote from eldragonfly, who in turn used a quote:
"A 1/3 chance that there are two boys, given that we know there's at least one. Yes."
So that means that there is a 2/3 chance that there is a girl and a boy, given that there is one boy.
This doesn't even make sense.
this indeed makes perfect sense: if you think of the children as independent, equally probable events (i.e. that the chance of getting a boy or a girl, for a given birth, is 50/50), then clearly you could get either two boys, a boy/girl combo, or two girls. But, the boy/girl combo is twice as likely as each of the other two cases. Not to bring in the coin toss analogy and open up a can of worms, but it's similar - your odds of getting HH is 1/4, TT is 1/4, and a combination of the two is 1/2... twice as likely as each of the other two events.
Now the question says that we know the family has at least one boy. This means it's either two boys, or a boy/girl combo. But the boy/girl combo is twice as likely as having two boys. This means that IF THE FAMILY HAS AT LEAST ONE BOY (the known, required information behind the problem), that it's a 2/3 chance it's a combination of girl and boy, and a 1/3 chance that it's two boys.
So the probability the family has a girl is 2/3. QED. If you disagree, that's fine, but actually try to poke a real hole in the logic and don't call me a jerk because i'm really trying here and only being pedantic insofar as is necessary to be absolutely clear of my meaning at each step in the explanation.
cheers 🙂
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OK, one last post for me here..
I just wanted to point out when and where a couple of the formulae from this thread are rightly used.
Bayes The probability of event A given condition B is equal to the the probability of the event A AND condition B occurring over the probability of condition B occuring. Another way to state this would be that it equals the chance of event A occurring within and as a percentage of the sample space that meets condition B.
Bayes formula can be properly applied in a probability problem whenever extra information is revealed which eliminate some of the possible results.
Binomial The explicit formula wasn't stated, but it uses factorials and fractions to powers to derive odds for a specific combination of results for multiple trials.
Binomial equations can be used to find the odds you get a specific number of successes out of a specific number of independent but identically structured events, like coin tosses.
"Successes" in this case could be number of boys, right answers, or anything, so long as there are a number of identically run trials for which the results of one trial do not depend on the results of any other trial.
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Originally posted by eldragonflyOK, I have done the maths (using conditional probability). I have now decided to use Excel to do a monte carlo simulation. Basically 500 families with a random 2 children set up and analysed Results are...
Do the math geepamoogle. You can't have a 2/3 chance that the next child is a girl and only a 1/3 chance that the next child is a boy, when the true odds are 50-50% by definition.
Your coin toss analogy is wrong. Each coin toss in an independent event.
[b]The upshot is that given just the setup, you have a 50% that both heads and tails come up, and ...[text shortened]... Once again you are addressing this as a permutations problem instead of a combinations problem.
First trial : 373 families have at least one boy. 250 or 67% of these have a girl
Second trial :377 families have at least one boy. 252 or 67% of these have a girl
Third trial : 380 families have at least one boy. 228 or 60% of these have a girl
Fourth Trial : 344 families have at least one boy. 242 or 70% of these have a girl
I realise that 4 trials may not be convincing (even if each sample had 500 families). Now I have got excel to repeat it 100(ish) times : the results are as follows (I think it should be 67% not 50% btw)
60% - 1
61% - 3
62% - 2
63% - 6
64% - 11
65% - 10
66% - 15
67% - 14
68% - 12
69% - 11
70% - 4
71% - 7
72% - 2
73%
74% - 1
75% - 1
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Originally posted by eldragonflyI assure you, eldragonfly, I'm not wrong. You can use Bayes Theorem, you can run a simulation, they will all give the same result.
Wrong mtthw. Again there are only 3 possibilities, again you are regurgitating a worthless and in this case wrong explanation.
BB
BG
GG
the third possibility can be excluded.
So that leaves only 2 possibilities.
BB
BG
You're thinking permutations, that is not implicity implied in the problem statement. Your whizz bang solution is in error, your reasoning and explanation doesn't make any sense.
Here's where you are going wrong.
There are two ways of looking at it. In your approach, looking at the combinations, it is true that there are three combinations. BB, BG and GG. The problem is, these are not of equal probability. You are twice as likely two have a boy and a girl as you are to have two boys. You'll accept that the first child born is 50-50? And so is the second? But these are independent events. The chances of getting two boys = 1/2x1/2 = 1/4. The chances of getting two girls are also 1/4. The chances of getting one of each must then be 1/2, as they must add up to 1.
If you accept this, you have to accept that the combinations are not of equal probability. And therefore you can't just count them.
Permutations are also an entirely valid way of looking at the problem, and they have the advantage that each permutation has an equal probability. Both approaches will give you exactly the same answer, though, as long as you do it correctly.
Again, there's a Wikipedia article that explains this. Perhaps you'll believe that where you don't believe me.
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
This really isn't very advanced probability theory. This is school level stuff.
So that means that there is a 2/3 chance that there is a girl and a boy, given that there is one boy.
This doesn't even make sense.
It not only makes sense, it's absolutely true. Go away and simulate it - you can toss a pair of coins - heads it's a boy, tails it's a girl.
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I'm sorry, eldragonfly, but you just don't get it. This really is school stuff as mentioned before. I had to do some basic probability exercises in 9th grade. So my math teacher also is wrong? You are the only one here who thinks you're right. Ok, for a few pages, brobluto appeared but he has disappeared again. Many people have tried to explain these problems to you, but you stick to your own 'truth'. Now meant as a personal attack but if you can't understand, then just drop it.
BTW, mtthw, deriver69, geepamoogle, aetherel, those are some nice explanations. A 4th grader should be able to understand these problems if one read your posts.