Online executable Python solution to the Tuesday boy problem(ideone.com)
ideone.com
Online executable Python solution to the Tuesday boy problem
http://ideone.com/qmcjVy
1 comments
This answer is incorrect. There is a 1/4 chance that both children are boys, and each child has 1/2 chance of being borne a boy, regardless of what day his sibling was or will be borne on. The program arrives at the probability of 13/27 by excluding the possibility that both children might be borne on tuesday and both might be boys.
No, it does not exclude that possibility (print the final list of pairs if you want to check). The calculation is correct. The result is counter-intuitive to almost everyone, which is why I made the program, so you can convince yourself.
Nope, sorry, it's wrong. Here are the possibilities for someone with 2 children were at least one of them is a boy borne on tuesday:
01) 1BT 2BM
02) 1BT 2BT
03) 1BT 2BW
04) 1BT 2BR
05) 1BT 2BF
06) 1BT 2BS
07) 1BT 2BY
08) 1BT 2GM
09) 1BT 2GT
10) 1BT 2GW
11) 1BT 2GR
12) 1BT 2GF
13) 1BT 2GS
14) 1BT 2GY
15) 1BM 2BT
16) 1BT 2BT
17) 1BW 2BT
18) 1BR 2BT
19) 1BF 2BT
20) 1BS 2BT
21) 1BY 2BT
22) 1GM 2BT
23) 1GT 2BT
24) 1GW 2BT
25) 1GR 2BT
26) 1GF 2BT
27) 1GS 2BT
28) 1GY 2BT
Actually you are still wrong.
Try something easier, for example two dice. The probability to roll 1 1 is less than to roll 1 2, because 1 2 can be rolled either by first rolling a 1 or a 2, while 1 1 can be rolled only by first rolling a 1. That information is used often in dice games like backgammon.
Same with children: BT BT can only be "rolled" if your first child is BT, whil BT BW can be "rolled" if your first child is either BT or BW.
Do you see?
Try something easier, for example two dice. The probability to roll 1 1 is less than to roll 1 2, because 1 2 can be rolled either by first rolling a 1 or a 2, while 1 1 can be rolled only by first rolling a 1. That information is used often in dice games like backgammon.
Same with children: BT BT can only be "rolled" if your first child is BT, whil BT BW can be "rolled" if your first child is either BT or BW.
Do you see?
The probability that you will roll 1 1 is indeed less then rolling 1 2 or 2 1, but that's only because you are allowing for twice as many acceptable outcomes.
The Wiki article you reference in your own code explains it well. The problem is in the assumptions made by the person answering a riddle. Just because you have a child, and it's a boy, and he is born on Tuesday, it does not mean that your next child is more likely to be a girl. If you agree with that statement, then the original answer must be false.
The Wiki article you reference in your own code explains it well. The problem is in the assumptions made by the person answering a riddle. Just because you have a child, and it's a boy, and he is born on Tuesday, it does not mean that your next child is more likely to be a girl. If you agree with that statement, then the original answer must be false.
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.
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.
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;I whipped together something here, before I saw that someone else did it too: 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 ;)
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 ;)
Here's my attempt in python: http://pastebin.com/tSDK1uRv
the range(0,10000) may not be correct, it might need to be 1,10000
the range(0,10000) may not be correct, it might need to be 1,10000
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!
This is really counter intuitive, the open vs. closed probability set through me off. Hey, thanks for putting up with my stubbornness!