benjer3

Wait, are you admitting that you're a baddie? How else would you know these things?

BrainIAC: Y-san Roster: 1) Hirkala - DEAD [?] Killed by Hawkgirl 2) marquessa - DEAD [?] Killed by Villains 3) curr3nt - DEAD [?] Killed by Villains 4) Molly Mae 5) Yodell 6) TheChad - DEAD [Sinestro] Lynched 7) Panther - voting for Molly Mae 8) Framm 9) MikeD - voting for Molly Mae 10) nana77- voting for Molly Mae 11) TwoaDay - voting for Molly Mae 12) jay 13) Flamebirde DEAD [?] Killed by Villains 14) Aura 15) benjer3 - voting for Molly Mae I guess we're going after Molly.

BrainIAC: Y-san Roster: 1) Hirkala - DEAD [?] Killed by Hawkgirl 2) marquessa - DEAD [?] Killed by Villains 3) curr3nt - DEAD [?] Killed by Villains 4) Molly Mae 5) Yodell 6) TheChad - DEAD [Sinestro] Lynched 7) Panther 8) Framm 9) MikeD 10) nana77 11) TwoaDay 12) jay 13) Flamebirde DEAD [?] Killed by Villains 14) Aura 15) benjer3 - voting for Panther I suppose I'll start the voting....

Though Nana isn't quite cleared yet. She could still be Sinestro. But that doesn't mean I don't trust her. Though that doesn't mean I completely do, either.

BrainIAC: Y-san Roster: 1) Hirkala - DEAD [?] Killed by Hawkgirl 2) marquessa - DEAD [?] Killed by Villains 3) curr3nt 4) Molly Mae 5) Yodell 6) TheChad - voting for Framm 7) Panther - voting for Jay 8) Framm - voting for Molly Mae 9) MikeD - voting for TwoaDay 10) nana77 - voting for TheChad 11) TwoaDay 12) jay 13) Flamebirde DEAD [?] Killed by Villains 14) Aura 15) benjer3 - voting for TheChad Putting needed pressure on, and I trust Nana for now.

BrainIAC: Y-san Roster: 1) Hirkala 2) marquessa 3) curr3nt 4) Molly Mae 5) Yodell 6) TheChad 7) Panther - voting for Benjer 8) Framm - Voting for jay 9) MikeD 10) nana77 - voting for Benjer 11) TwoaDay 12) jay 13) Flamebirde DEAD [?] Killed by Villains 14) Aura - voting for Benjer 15) benjer3 - voting for Hirkala Wow, wasn't expecting that. Well I'm not going to lie, I did target you last night. But you being blocked prevented my action from going through.... As for my vote, it's just random.

Adsum!

BrainIAC: Y-san Roster: 1) Hirkala 2) marquessa 3) curr3nt 4) Molly Mae 5) Yodell 6) 7) Panther 8) Framm 9) MikeD 10) nana77 11) LJ 12) 13) Flamebirde 14) 15) benjer3 Back-ups: 1) Brainiac100 2) Oh alright, I'll join. Though it may mean playing 4 games, hosting a non-mafia, and taking care of the MM wiki all at once.

This yields "Out of 1000 families, there were 239 girl-girls (23.9%), 247 girl-boys (24.7%), 280 boy-girls (28%), and 234 boy-boys (23.4%)." Near-perfect distribution.

In the other 485, all of the "other children" were boys, and the first children could be boys or girls. I could modify the program to show that the first children were equally distributed among boys and girls in this group as well. That leaves 515 "other" girls, 485 "other" boys, (271 + ~0.5 * 485) "first" girls, and (244 + ~0.5 * 485) "first" boys, giving a near perfect distribution. And if you look at the program, I randomly select from the first and second child of the family (though it wouldn't make a difference either way; I just did that to make it more realistic), so I'm not assuming anything about the children I've been introduced to.

On the contrary. Look at this program (and let me know if it's wrong ): class Program { static void Main(string[] args) { Random random = new Random(); int numberOfFamilies = 1000; int numberSecondGirl = 0; int numberGirlGirl = 0; int numberBoyGirl = 0; for (int count = 0; count < numberOfFamilies; ++count) { Family family = new Family(); int childIndex = random.Next(2); bool firstChild = family.Children[childIndex]; bool otherChild = family.Children[(childIndex == 0) ? 1 : 0]; if (otherChild) { ++numberSecondGirl; if (firstChild) { ++numberGirlGirl; } else { ++numberBoyGirl; } } } Console.WriteLine("Out of {0} families, {1} of the \"other children\" were girls. Out of those, {2} of the first children were girls ({3}%), while {4} of them were boys ({5}%)", numberOfFamilies, numberSecondGirl, numberGirlGirl, ((double)numberGirlGirl / (double)numberSecondGirl) * 100D, numberBoyGirl, ((double)numberBoyGirl / (double)numberSecondGirl) * 100D); Console.ReadLine(); } } public class Family { private static readonly Random random = new Random(); public bool[] Children = new bool[2]; public Family() { for (int index = 0; index < Children.Length; ++index) { Children[index] = random.Next(2) == 1; } } } For one run, this resulted in "Out of 1000 families, 515 of the 'other children' were girls. Out of those, 271 of the first children were girls (52.6%), while 244 of them were boys (47.4%)." The results are kinda backwards from what you stated, but I think you get the idea.

Hmm, I think I've had a breakthrough. Let me put it in terms of the OP: A parent comes up to you and says, "I have two children and one of them is a girl." They then proceed to tell you the gender of the other child. Being the analytic people that we are, we wonder what the odds that the second child was a girl were. While we ponder this, another parent happens to come up and says, "I have two children and one of them is a boy." A hundred parents then proceed to do the same, with the gender of the first child varying. (It also just so happens that, if each parent has at least one girl, they will say, "One of them is a girl.") Now I ask, can we just discard the cases where the first child was a boy? I propose not, for in doing so we would be selectively taking members out of a sample group that needs to be complete, not to mention skewing the results by doing so selectively. So what are we to do? What if we changed our question, while keeping the same general idea intact: "Given the gender of one in two children, what are the odds that the other is a girl?" This question encapsulates the original question, as well as allowing for other scenarios. But now the answer is obvious, isn't it. In this question, it is obvious that the gender of the "other child" is completely independent of the gender of the first. And thus its answer can be nothing other than 50%. So think what you will. I did change the question, but as I said, I believe the question is a perfect substitute for the original as it asks the exact same thing but allows for a complete sample group.