The Boy or Girl Paradox (riddle) (gtfih)

[HughMyron]

Question: You met a high school friend in a shop, he told you that he had two children and at least one boy. What is the probability that the other child is a boy.

Assumption:
1. Boys to girls ratio is 50:50
2. The sex of each child is independent of the sex of the other.

[Vengeance10]

25%

[BrawnySwoleman]

50%

/thread unless I'm missing something

[Inev]

well, if the ratio is 50:50 and he has 1 boy. Then the odds is 49:50 (boys:girls)

unless you werent trying to be cheeky, then it is just 50% odds.

[gerbilz]

idk, he has three kids the way he stated it

[HughMyron]

25%

Interesting...care to elaborate?

[sFunk]

Boy/Boy
Boy/Girl
Girl/Boy
Girl/Girl

We know at least one is a boy so the Girl/Girl is gone leaving us with 3 possible choices--33.3% chance that the combination is Boy/Boy

[54m0n6]

50%

[MA5Bergey]

50%.

The choices are independent of each other.

The misc makes me wonder why I ever bothered taking a math class above grade 7.

[HughMyron]

Boy/Boy
Boy/Girl
Girl/Boy
Girl/Girl

We know at least one is a boy so the Girl/Girl is gone leaving us with 3 possible choices--33.3% chance that the combination is Boy/Boy

But is Girl/Boy and Boy/Girl not the same thing?

[alphaMotive]

tree fiddy

[asdfqw]

*Will never get a girl's number crew*
*Scared of girls crew*
*Socially awkward crew*
*Forever a virgin crew*
*Have no friends crew*
*Stay at home on weekends crew*
*Beta ******* crew*
*Starvation crew*
*Nail-biter crew*
*Weak jaw crew*
*Cry yourself to sleep crew*
*Never kissed/hugged a girl crew*
*Chemical imbalance crew*
*Lift but still ugly crew*
*Turned down by prostitutes crew*
*Failed community college crew*
*Live with parents crew*
*Idolize serial killers crew*

[LaxinAround]

100%, if it were a girl he would have had an abortion

[Skys-The-Limit]

Lets put this lame ass riddle into real life.

John: Oh hey Jeff, its been years since we've talked. How's life?

Jeff: Hey John, i'm good, I have 2 children, 1 is a boy.

End of real life.
What kunt father ignores the gender of his other child.
But to answer the question without your mathmapotato. If hes going to say one is a boy, than the other is a girl, or else he would have said both are boys.

Rep me

[iluvLP]

0,5 * 0,5 = 0,25 = 25%

[blumnblam]

Misc, I am disapoint.

The options are
-Boy/Girl
-Boy/Boy
-Girl/Boy

Note that Girl/Girl obviously isn't true (though that threesome with me and those 2 ladies would be hot. Like redtube hot. It'd probably be one of them going down on me, the other making out with me. The one who's making out with me then starts riding me and I stick my two fists out along with me tongue. Girl 1 - riding, Girl 2 - sitting on my face, mother - on right hand fist, grandmother - on left hand fist. This is why Girl/girl isn't a statistical option).

2/3 possibilities have the non-"at-least-one's-a-boy" child as a girl. Therefore chances of other child being a boy is 1/3.

[mrkdrt]

2/3 possibilities have the non-"at-least-one's-a-boy" child as a girl. Therefore chances of other child being a boy is 1/3.

wouldnt it be 50/50 as the boy/girl girl/boy are the same

[CutItUp9759]

Boy/Boy
Boy/Girl
Girl/Boy
Girl/Girl

We know at least one is a boy so the Girl/Girl is gone leaving us with 3 possible choices--33.3% chance that the combination is Boy/Boy

order wouldn't matter dawg: boy/girl = girl/boy...

[joblank]

I remember stupid problems like these back in high school math. Could never wrap my head around them even though I got As on the other tests. Farking probabilities.

[marcFE]

Bit of interpretation but it should be 50%.



Same as me saying: If John flips a coin 20 times and if 19 times were heads, whats the probability he get's 20 heads?


If the event had not occurred yet then the problem would be ((P(heads))^19(P(tails))^1 + P(heads)^20) which would be ((.5)^19)(.5)^1)+(.5)^20 but because the event has occurred already it is instead P(heads that have occurred)*(P(getting heads)) or 1^19*.5^1 or simply 50%


In the same way we know that the guy has one kid already and we are trying to find the probability he has two. If we are trying to find the probability of him having at least one male knowing nothing then the problem is very different but we currently know he has one kid with absolute certainty and want to find the probability he will have a second one. That therefore gives him a 50% of having a second kid and not the 33% even if the 3 possible outcomes of male/female, female/male, and male/male. Both 33% and 50% could be correct given the fact that if OP botched the wording even a tiny bit it changes the answer but the way I read it independent probabilities should supersede the implications of at least mathematics.

order wouldn't matter dawg: boy/girl = girl/boy...

Order is actually very important in statistics and that guy is on track since girl/guy and guy/girl are two independent events and therefore have separate probabilities of occurring. In most cases it is relevant, just not under this circumstance.

[iceypain]

1/3 is the answer, there's no argument.

[irollurmom]

1/4 or 0.25 or 25%.

EDIT: Whoa, I read the question wrong. He states that he has at least 1 boy, but he didn't state which child was a boy. Therefore it'll be:
GB
BG
BB
or 1/3. I shall now commit seppuku for my mistake.

[Ka0s]

The answer is 50% you cretins

[Greg14]

1/4 or 0.25 or 25%.

The answer is 1/3

[Ka0s]

If two events A and B are independent, then the conditional probability of A given B is the same as the unconditional (or marginal) probability of A

http://en.wikipedia.org/wiki/Indepen...bility_theory)

[Ufabrah]

100%, since he said the boy to girl ratio is 50/50 and he had 1 boy

its a riddle, you can't think about it logically or mathematically lol

[MA5Bergey]

1/3 is the answer, there's no argument.

He clearly states the second child's gender is INDEPENDENT of the first child's gender. This means the first child being a boy has NO bearing on the probability. Therefore, the second child can simply be a male, or a female. 50-50.

[iceypain]

If two events A and B are independent, then the conditional probability of A given B is the same as the unconditional (or marginal) probability of A

http://en.wikipedia.org/wiki/Indepen...bility_theory)

there is only one event in this case--with 3 possible outcomes; it's not 2 events in sequence.

[Ka0s]

100%, since he said the boy to girl ratio is 50/50 and he had 1 boy

its a riddle, you can't think about it logically or mathematically lol

Dumbest post ITT, and quite frankly that is VERY HARD to do

[Ka0s]

there is only one event in this case.

No there is not

What is the probability of child 2 being boy given that child 1 is boy? It is the same as probability of child 2 being boy (because of independence) = 50%