Fraternal Twins

[rstrats]

A woman is about to give birth to fraternal twins. What is the chance that one of the twins will be a girl?

 

[Mageth]

One in three, the way the problem's stated.

(Three outcomes with equal probability, assuming all else being equal: one of the twins is a girl, both are girls, or neither is a girl)

 

[Tom Sawyer]

It depends on how weak and cowardlly the father is because that's what determines whether he has pathetic sperm which give him female offspring or awesome sperm which give him some nice, strong sons.

 

[Keith&Co.]

Well, i'm weak and cowardly, but all three of my boys are boys. Including the twins.

Oh, damn! That means i've been cheating on my wife...

 

[rstrats]

Megeth,

re: "Three outcomes with equal probability..."


Actually, there are four possible outcomes.

 

[Tom Sawyer]

Well, i'm weak and cowardly, but all three of my boys are boys. Including the twins.

Oh, damn! That means i've been cheating on my wife...

Well, apparently your opinion of yourself is faulty because the evidence suggests that you're actually strong and brave.

But that's OK. You're a man, so if you want to inaccurately view yourself as weak and cowardly, then you damn well go ahead and do that and anyone who has a problem with it can go and fuck themselves.

 

[Mageth]

Megeth,

re: "Three outcomes with equal probability..."


Actually, there are four possible outcomes.

Not the way you stated the problem, there's not. Consider the "truth table":

GB
BG
GG
BB

Now, that looks like four, but is only three, as GB/BG are the same as order is ignored/not specified in the problem, and are thus considered as one possible outcome (one is a girl).

So only three possible and equally probable outcomes...one is a girl, both are girls, or neither is a girl, and the probability that one is a girl is one in three.

 

[Keith&Co.]

To be four, the question would have to be "What is the chance that the first twin born will be a girl?"

Then the difference between GB and BG matters.

- - - Updated - - -

You're a man, so if you want to inaccurately view yourself as weak and cowardly, then you damn well go ahead and do that and anyone who has a problem with it can go and fuck themselves.

Well, i'm certainly not going to argue with you...

 

[Mageth]

I'll note that the way the problem is stated:

"What is the chance that one of the twins will be a girl?"

...I'm reasonably interpreting as "[only] one of the twins will be a girl".

Alternatively, if interpreted as "at least one", then the probability that at least one twin is a girl would be two in three.

 

[Alcoholic Actuary]

Megeth,

re: "Three outcomes with equal probability..."


Actually, there are four possible outcomes.

Not the way you stated the problem, there's not. Consider the "truth table":

GB
BG
GG
BB

Now, that looks like four, but is only three, as GB/BG are the same as order is ignored/not specified in the problem, and are thus considered as one possible outcome (one is a girl).

So only three possible and equally probable outcomes...one is a girl, both are girls, or neither is a girl, and the probability that one is a girl is one in three.

I think I disagree.

The problem specifically mentions fraternal twins - which is when two separate eggs are fertilized by two separate sperm. Thus there are two separate chances for a girl. No different than flipping two coins simultaneously (order still doesn't matter). The probability of exactly one girl is .5 and the probability of at least 1 girl is .75.

(Or to say it differently, there are only 3 outcomes, but they are not all equally likely)

aa

 

[Mageth]

Megeth,

re: "Three outcomes with equal probability..."


Actually, there are four possible outcomes.

Not the way you stated the problem, there's not. Consider the "truth table":

GB
BG
GG
BB

Now, that looks like four, but is only three, as GB/BG are the same as order is ignored/not specified in the problem, and are thus considered as one possible outcome (one is a girl).

So only three possible and equally probable outcomes...one is a girl, both are girls, or neither is a girl, and the probability that one is a girl is one in three.

I think I disagree.

The problem specifically mentions fraternal twins - which is when two separate eggs are fertilized by two separate sperm. Thus there are two separate chances for a girl. No different than flipping two coins simultaneously (order still doesn't matter). The probability of exactly one girl is .5 and the probability of at least 1 girl is .75.

(Or to say it differently, there are only 3 outcomes, but they are not all equally likely)

aa

If you flip two coins simultaneously, there are three possible and equally likely outcomes if order is not considered - one is heads, both are heads, or neither are heads.

The only way to get the .5/.75 result/truth table is if you specify order is significant.

I will gladly play poker with anyone who disagrees.

 

[doubtingt]

I am guessing this is not a logic problem, but one of those "riddles" that is only hard because the question is worded poorly and ambiguously with the intention of misleading the listener. The phrase "one of them" could mean "either one", "only one", or "one of them in particular". For the first two, the specificity of which one of the twins is female is irrelevant, so there are only 3 equally likely outcomes that differ in relation to the question (with 1 of each incorporating both boy-girl and girl-boy). Common discourse practices would indicate that what was meant is "either one" is a girl unless otherwise specified. That would put the odds at 2 in 3 with only 2 boys not qualifying. IF it is "only one", then the odds are 1 in 3. If what was meant is "one of them in particular" (the most dishonest and/or poorly worded version), then there are actually only 2 equally likely relevant outcomes relevant, because all that matters is whether that specific twin is a boy or girl. So, the odds are 1 in 2.


Of course, technically, either one could be a hermaphrodite, but I'm not sure of the odds of that.

 

[rstrats]

Megeth,

re: "Alternatively, if interpreted as 'at least one..."

Yes, that is the correct interpretation.


re: "...then the probability that at least one twin is a girl would be two in three."

Actually, it's 3 in 4 or 75 percent.

 

[Mageth]

Megeth,

re: "Alternatively, if interpreted as 'at least one..."

Yes, that is the correct interpretation.


re: "...then the probability that at least one twin is a girl would be two in three."

Actually, it's 3 in 4 or 75 percent.

Actually, it's not. It's 2 in 3, unless you're now specifying that order is to be considered as well. See explanations above.

And my name is mAgeth.

 

[Alcoholic Actuary]

Megeth,

re: "Three outcomes with equal probability..."


Actually, there are four possible outcomes.

Not the way you stated the problem, there's not. Consider the "truth table":

GB
BG
GG
BB

Now, that looks like four, but is only three, as GB/BG are the same as order is ignored/not specified in the problem, and are thus considered as one possible outcome (one is a girl).

So only three possible and equally probable outcomes...one is a girl, both are girls, or neither is a girl, and the probability that one is a girl is one in three.

I think I disagree.

The problem specifically mentions fraternal twins - which is when two separate eggs are fertilized by two separate sperm. Thus there are two separate chances for a girl. No different than flipping two coins simultaneously (order still doesn't matter). The probability of exactly one girl is .5 and the probability of at least 1 girl is .75.

(Or to say it differently, there are only 3 outcomes, but they are not all equally likely)

aa

If you flip two coins simultaneously, there are three possible and equally likely outcomes if order is not considered - one is heads, both are heads, or neither are heads.

The only way to get the .5/.75 result/truth table is if you specify order is significant.

I will gladly play poker with anyone who disagrees.

Speaking of gambling, what is the probability that I roll a seven at a craps table? 1 in 11? or 1 in 6? How is the order of the dice considered at craps?

I know what the outcomes are, but why should they be equally likely?

aa

 

[rstrats]

Alcoholic Actuary,

re: "(Or to say it differently, there are only 3 outcomes..."

I assume by that, you mean only 3 outcomes where a girl is involved.



re: "...but they are not all equally likely)"

Why not?

 

[Alcoholic Actuary]

Alcoholic Actuary,

re: "(Or to say it differently, there are only 3 outcomes..."

I assume by that, you mean only 3 outcomes where a girl is involved.


re: "...but they are not all equally likely)"

Why not?

What I mean is that you could say that there are 3 outcomes - 2 boys, 2 girls, or 1 boy 1 girl. However, the probability of 2 girls is (.5)x(.5) = .25 = the probability of 2 boys also. This must mean that the probability of 1 girl and 1 boy = 1 - .25 - .25 = .5

ie, the probability of 1 boy and 1 girl is twice the probability of 2 girls (or 2 boys). So if I define only three outcomes, I don't think all three outcomes are equally likely. Same as rolling dice at a craps table. There are only 11 possible outcomes of 2 dice, but they are not all equally likely.

(You can define 4 outcomes that are all equally likely if you want to - as you have done earlier in the post)

aa

 

[Keith&Co.]

re: "...but they are not all equally likely)"

Why not?

There's only one way to get two girls.
There's only one way to get two boys.
There are two ways to get one girl and one boy (in either order). So the chances of at least one girl are two out of three outcomes, but the one-of-each outcome is twice as likely as either of the other two.

 

[rstrats]

mAgeth,

re: "And my name is mAgeth."

Sorry for the typo.



re: "Actually, it's not. It's 2 in 3..."

What is the chance expressed as a percentage that the woman is going to leave the hospital with at least 1 girl?

 

[rstrats]

Keith&Co.,

re: "There are two ways to get one girl and one boy (in either order)."


And there are 3 combinations out of 4 of having at least 1 girl.