Quick Logic Problem

KingCondanomation said:
Do 1000 coin tosses of 2 coins at a time. Eliminate all those tosses where BOTH coins show as heads, leaving you with only those tosses where there are either (your goal here is to see the chances of getting both as tails):
1. heads on first coin, tails on second coin
2. tails on first coin, heads on second coin
3. tails on first coin, tails on second coin

Damo is saying that if I name those coins, I would somehow be able to reverse the last case (to get a 1 in 2 shot of having both as tails, but somehow I am NOT allowed to reverse the first 2 cases).

It's a common mistake and is talked about in the link that I gave. You would find that each case would occur about 1 out of 3 times, including the case with both coins showing as tails.
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King, it's not a binomial, so don't even entertain the argument.
 
Damocles said:
It doesn't matter if "both sexes are known by the parent" we are figuring the probability of the unknown quantity.

One is a given. I tried to help you out by "naming" that Given. Well, now his name is given.

Given - Boy
Given - Girl
Boy - Given
Girl - Given

Are the only possible scenarios. The only ones. BECAUSE you have Given, you have limited the variable to only one child and only two possibilities for outcome.
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No man, given is a boy, there is only one way a mother can have 2 boys, her first child must be a boy and her 2nd child must be a boy. That's it, you agree?

Your example above doesn't make any sense because you have written 2 different ways of having 2 boys:
Given - Boy
Boy - Given

when in reality there is only ONE possible way.
 
KingCondanomation said:
No man, given is a boy, there is only one way a mother can have 2 boys, her first child must be a boy and her 2nd child must be a boy. That's it, you agree?

Your example above doesn't make any sense because you have written 2 different ways of having 2 boys:
Given - Boy
Boy - Given

when in reality there is only ONE possible way.
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It makes sense because of "birth order"...

If Given was born first, there are two possible scenarios. Girl or boy.

If given was born second there are two possible scenarios. Boy or Girl.

The second set is superfluous and redundant, but since it seemed so important to Dano to understand it.

One of the children is not a variable. Charts notwithstanding. This is not a binomial equation.
 
Thorn said:
Each toss is independent of all others, and whatever has occurred previously or subsequently has no influence at all on the outcome of a single toss. Each toss has the probability of 0.5 for each side.

Similarly, in this question, one boy in each family has been established and has no bearing on the gender of the other child, whether that child is older or younger. The probability for each family is 0.5 that the other child is one or the other gender. EOS.
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NOT EOS.

The conditional statement disallows one of the possible outcomes - it is NOT equally likely, there fore the application of a binomial, which applies to variable with a constant probability of occurring, is not valid.
 
Troglodyte27 said:
NOT EOS.

The conditional statement disallows one of the possible outcomes - it is NOT equally likely, there fore the application of a binomial, which applies to variable with a constant probability of occurring, is not valid.
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What conditional statement? Each family had at least one boy. That one is older than his sibling is irrelevant. The gender of the other child is what is at question, and the probability of that is just like the coin toss, M or F, Two possibilities, so 50% probability of each. No other possibilities unless we're being silly.

Mathematically, the probability of both gender-unidentified children being male is .5 x .5 = 0.25, or 25%.
 
Thorn said:
What conditional statement? Each family had at least one boy. That one is older than his sibling is irrelevant. The gender of the other child is what is at question, and the probability of that is just like the coin toss, M or F, Two possibilities, so 50% probability of each. No other possibilities unless we're being silly.

Mathematically, the probability of both gender-unidentified children being male is .5 x .5 = 0.25, or 25%.
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Did you read the problem in Wikipedia? Please do so, you'll see where you went wrong and probably easier than me trying to explain it.
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
 
Damocles said:
The chances for "each" would be the same since they'd have to be figured separately.

It is a 50% chance that each would have two boys.
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This answer is correct. Since each of the women has a child whose sex is already determined, we are only concerned with the possibility of the 2nd child being a male. Therefore 1 out of 2.
 
AssHatZombie said:
tails-heads is the SAME as heads-tails, it's not a different outcome.
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That depends: It is a different outcome when the problem is examining permutations (order matters), and the same answer when dealing with combinations (AB = BA).
 
Troglodyte27 said:
That depends: It is a different outcome when the problem is examining permutations (order matters), and the same answer when dealing with combinations (AB = BA).
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If you are only concerned with the number of heads and the number of tails the results are the same regardless.

The actual probability would be 50% would be one of each, and 25% with both heads or both tails.
 
Basically you take a chart and fill it in with the givens and possibilities.

Chart one:
Given1 - Boy
Given1 - Girl


Chart 2:
Given2 - Boy
Given2 - Girl
Boy - Given2
Girl - Given2

(The chart assumes that somebody is concerned with the birth order even though it is irrelevant)...

Each line of the charts have an equal probability to the others on its chart.

Both scenarios have an equal chance of having a boy or a girl because only one of the children are variable.
 
Thorn said:
What conditional statement? Each family had at least one boy. That one is older than his sibling is irrelevant. The gender of the other child is what is at question, and the probability of that is just like the coin toss, M or F, Two possibilities, so 50% probability of each. No other possibilities unless we're being silly.

Mathematically, the probability of both gender-unidentified children being male is .5 x .5 = 0.25, or 25%.
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Even though coin flipping shows girl-girl is a possible outcome and likely 25% of the time (correct), the conditional statement disallows girl-girl as a possible outcome. G-G is no longer equally likely and that negates the underlying assumption.
 
And I thought I could drag a thread out by not letting go.

I am an amateur.
 
a11n said:
well its a 50/50 chance theirs ur answer dumbass.
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You are wrong.

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

"The Boy or Girl problem is a well-known example in probability theory:

A random two-child family whose older child is a boy is chosen. What is the probability that the younger child is a girl? (Or: choose a random two-child family assuring that the older one is a boy. What is the probability that the other one is a girl?)
A random two-child family with at least one boy is chosen. What is the probability that it has a girl? (Or: choose a random two-child family assuring that at least one is a boy. What is the probability that the other one is a girl?)
Investigation of these questions reveals that their answers are very different:

in the first case, there are two equally probable possibilities: the second one is a boy or a girl.
in the second case, there are three equally probable ways in which at least one child can be a boy: only the older one, only the younger one, or both. "
 
KingCondanomation said:
You are wrong.

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

"The Boy or Girl problem is a well-known example in probability theory:

A random two-child family whose older child is a boy is chosen. What is the probability that the younger child is a girl? (Or: choose a random two-child family assuring that the older one is a boy. What is the probability that the other one is a girl?)
A random two-child family with at least one boy is chosen. What is the probability that it has a girl? (Or: choose a random two-child family assuring that at least one is a boy. What is the probability that the other one is a girl?)
Investigation of these questions reveals that their answers are very different:

in the first case, there are two equally probable possibilities: the second one is a boy or a girl.
in the second case, there are three equally probable ways in which at least one child can be a boy: only the older one, only the younger one, or both. "
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wrong. in the second case the older one being a boy or the younger one being a boy are THE SAME OUTCOME.
 
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AssHatZombie said:
wrong. in the second case the older one being a boy or the younger one being a boy are THE SAME OUTCOME.
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Ok, just so you are on record, despite this being a documented WELL-KNOWN example in probability theory with presumably thousands of people with degrees (including the algebra teacher, Dano, MottleyDude, Trog, the smartest woman in the world who does this for a living) , they have all been wrong all this time and you, Damo and Thorn are right, is that right?

Maybe you can't understand it, fine, but don't suggest something so ludicrous as all those who have studied this exact well known case and understand why it is 1 in 3 chance of having 2 boys when one boy is known to exist of 2 kids. Just accept it as something that is true but you can't understand.
For example, I don't understand the theory of relativity, does that mean it's not true?
 
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