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The wiki article is not valid, since it considers a subtly different problem than the one I do. If you read the comment at the top of the code, you will see tha

by daivd 13y ago

The wiki article is not valid, since it considers a subtly different problem than the one I do. If you read the comment at the top of the code, you will see that I changed the question to be less ambiguous. In the problem I pose, the person does not randomly come forward and tell me "I have a boy born on a tuesday". In my problem I ask random people who I know have two children if they have "at least on boy born on a tuesday" until someone says yes.

x0054 13y ago
My python is really rusty, but if you convert this sudocode to python, you will see that the answer is 50/50. I'll write this up in Javascript when I have time. If you do this in Python, please post. girl = 0 boy = 0 do 10000 times { child1Sex = randomSex; child1Day = randomDay; child2Sex = randomSex; child2Day = randomDay; if ( (child1Sex == "B" AND child1Day == "Tu") OR (child2Sex == "B" AND child2Day == "Tu") ) { if (child1Sex == child2Sex){ boy++ } else { girl++ } } } print "Ratio of boys and girls is " + boy " : " + girl;
- krapp 13y ago
  Here's my attempt in python: http://pastebin.com/tSDK1uRv http://pastebin.com/tSDK1uRv the range(0,10000) may not be correct, it might need to be 1,10000
  - x0054 13y ago
    Wow, works good. I should learn python :) But yeah, you were absolutely right! It was hard for me to figure out because I was trying to find a mathematical reason for this, and I was wrong. I finally figured it out mathematically. The way you posit the problem the probability that the asked parents have another boy is (1/2)-(1/7/2). This is really counter intuitive, the open vs. closed probability set through me off. Hey, thanks for putting up with my stubbornness!
- daivd 13y ago
  I whipped together something here, before I saw that someone else did it too: http://ideone.com/jyPOhs http://ideone.com/jyPOhs It is the most counter-intuitive statistic problem I know (worse than Monty Hall), it trips everyone up at first. No problem about the stubborness ;)

by daivd 13y ago

x0054 13y ago

My python is really rusty, but if you convert this sudocode to python, you will see that the answer is 50/50. I'll write this up in Javascript when I have time. If you do this in Python, please post. girl = 0 boy = 0 do 10000 times { child1Sex = randomSex; child1Day = randomDay; child2Sex = randomSex; child2Day = randomDay; if ( (child1Sex == "B" AND child1Day == "Tu") OR (child2Sex == "B" AND child2Day == "Tu") ) { if (child1Sex == child2Sex){ boy++ } else { girl++ } } } print "Ratio of boys and girls is " + boy " : " + girl;

krapp 13y ago
Here's my attempt in python: http://pastebin.com/tSDK1uRv http://pastebin.com/tSDK1uRv the range(0,10000) may not be correct, it might need to be 1,10000
- x0054 13y ago
  Wow, works good. I should learn python :) But yeah, you were absolutely right! It was hard for me to figure out because I was trying to find a mathematical reason for this, and I was wrong. I finally figured it out mathematically. The way you posit the problem the probability that the asked parents have another boy is (1/2)-(1/7/2). This is really counter intuitive, the open vs. closed probability set through me off. Hey, thanks for putting up with my stubbornness!
daivd 13y ago
I whipped together something here, before I saw that someone else did it too: http://ideone.com/jyPOhs http://ideone.com/jyPOhs It is the most counter-intuitive statistic problem I know (worse than Monty Hall), it trips everyone up at first. No problem about the stubborness ;)