Skin'Em84 Posted June 16, 2011 Share Posted June 16, 2011 9..4 married couples (8 people) and Jack. Crap, didn't read the whole thing. Link to comment Share on other sites More sharing options...
DeaconTheVillain Posted June 16, 2011 Author Share Posted June 16, 2011 9..4 married couples (8 people) and Jack. She did not shake hands with Jack Link to comment Share on other sites More sharing options...
December90 Posted June 16, 2011 Share Posted June 16, 2011 Jack and his wife went to a party where four other married couples were present. Every person shook hands with everyone he or she was not acquainted with. When the handshaking was over, Jack asked everyone, including his own wife, how many hands they shook. To his surprise, Jack got nine different answers. How many hands did Jack's wife shake ? She did not shake any hands as she was already acquainted with everyone (or else they would not have been invited in the first place) Link to comment Share on other sites More sharing options...
DeaconTheVillain Posted June 16, 2011 Author Share Posted June 16, 2011 She did not shake any hands as she was already acquainted with everyone (or else they would not have been invited in the first place) 5 total couples. Nether couple shook hands with significant other Link to comment Share on other sites More sharing options...
renaissance Posted June 16, 2011 Share Posted June 16, 2011 That thing in the OP isn't a riddle. Seems like more of a math problem to me. Link to comment Share on other sites More sharing options...
Califan007 The Constipated Posted June 16, 2011 Share Posted June 16, 2011 Jack and his wife went to a party where four other married couples were present. Every person shook hands with everyone he or she was not acquainted with. When the handshaking was over, Jack asked everyone, including his own wife, how many hands they shook. To his surprise, Jack got nine different answers. How many hands did Jack's wife shake ? Ok, I just looked up the answer for this riddle lol...and it makes ZERO sense. *EDIT: Ok, I get it now lol :doh:... Link to comment Share on other sites More sharing options...
Gibbs Hog Heaven Posted June 16, 2011 Share Posted June 16, 2011 8, but I'm not as confident as this one being correct. Hail. Link to comment Share on other sites More sharing options...
artmonkforHOF Posted June 16, 2011 Share Posted June 16, 2011 I can see the logic, since the host said "my other two kids.." after you meet the first boy. If he had 2 girls and 1 boy, he would have been more likely to have said "my two girls..", which means that the 2 boys & 1 girl would be the more likely than 1 boy & 2 girls. Or you can just inquire "boys? girls?" and the most likely thing the host will do is give you a truthful response and not have to guess at all. I would be more concerned if I started a new job and the first day someone wanted to take me home to meet his/her family, creepy. Link to comment Share on other sites More sharing options...
daveakl Posted June 16, 2011 Share Posted June 16, 2011 All three actually are girls. You thought the first one was a boy but it really was a girl. Link to comment Share on other sites More sharing options...
sideshow24 Posted June 16, 2011 Share Posted June 16, 2011 The other two are girls. They weren't home yet because they were at an understaffed soccer practice. Link to comment Share on other sites More sharing options...
Darth Tater Posted June 16, 2011 Share Posted June 16, 2011 its been a while since i did probability, but i am pretty sure gibbs is correct. The remain kids are either boy/girl, girl/boy, or girl/girl because we know he has at least 1 girl. So with that its more likely he has 2 boys and 1 gir Yep, that's the way I remember doing this problem. ---------- Post added June-16th-2011 at 03:49 PM ---------- Ok someone post another one. I like these Do you know the Monte Hall problem? Also, look up Knights and Knaves. Link to comment Share on other sites More sharing options...
PokerPacker Posted June 17, 2011 Share Posted June 17, 2011 I defy someone to read through my post and show me where my 50/50 proof is incorrect. Link to comment Share on other sites More sharing options...
DeaconTheVillain Posted June 17, 2011 Author Share Posted June 17, 2011 I defy someone to read through my post and show me where my 50/50 proof is incorrect. We know the first child is a boy, so we are really only concerned with the two unknown children. Assuming girls and boys are 50/50 the options are: BB GG BG GB However, when we find out that one of the two is a girl that eliminates BB leaving us with: GG BG GB and a 2/3 chance that one of the remaining two is a boy. ---------- Post added June-17th-2011 at 09:33 AM ---------- Don't get me wrong Poker when I saw this problem it was simple. He has 3 kids. One is a boy One is a girl 3rd child is ?? 50/50 chance of boy or girl right? Simple. Link to comment Share on other sites More sharing options...
PokerPacker Posted June 18, 2011 Share Posted June 18, 2011 We know the first child is a boy, so we are really only concerned with the two unknown children. Assuming girls and boys are 50/50 the options are:BB GG BG GB However, when we find out that one of the two is a girl that eliminates BB leaving us with: GG BG GB and a 2/3 chance that one of the remaining two is a boy. ---------- Post added June-17th-2011 at 09:33 AM ---------- Don't get me wrong Poker when I saw this problem it was simple. He has 3 kids. One is a boy One is a girl 3rd child is ?? 50/50 chance of boy or girl right? Simple. problem is that you've only shown that there are three scenarios with the way you are doing it. You are making the assumption they follow uniform distribution and each of those scenarios is equally likely. If you look at my post, I analyze each scenario and can give you the exact odds of each scenario. I'd like someone to show me how my proof is flawed. Link to comment Share on other sites More sharing options...
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