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Topic: Faites vos jeux messieurs, dames (Read 3 times) previous topic - next topic

radman

I was surprised that nobody quizzed me when I wrote that a single individual could have two different sets of DNA. Anyway, for our purposes here we can assume that babies come in two flavours, male and female, and that statistically these appear in equal numbers.

Now lets get the easy, boring bit out of the way;

PART 1 - I have two children. One of them is a boy. What are the odds that the other is also a boy?

Lay your cards on the table.

Coding Badly

...that statistically these appear in equal numbers.


But they don't.

Quote
PART 1 - I have two children. One of them is a boy. What are the odds that the other is also a boy?


If your wife typically eats a big breakfast about 55%.

radman


...that statistically these appear in equal numbers.


But they don't.

Quote
PART 1 - I have two children. One of them is a boy. What are the odds that the other is also a boy?


If your wife typically eats a big breakfast about 55%.


@CodingBadly I agree, but here we have rules. There are boy babies and girl babies born statistically at the same rate. So I have two children one is a boy what are the odds that the other is also a boy?

Come on this is the boring part, boy/girl lets move on.

Nick Gammon

I don't think it matters. If I throw a coin and it comes up heads, what are the odds that the next throw comes up heads?
Please post technical questions on the forum, not by personal message. Thanks!

More info:
http://www.gammon.com.au/electronics

radman


I don't think it matters. If I throw a coin and it comes up heads, what are the odds that the next throw comes up heads?


Nick, Nick you are a "Brattain Member", come on, you are up there with Dr Spock, we want logical, correct, computer programmer thinking not wooly, nonsense. This is the easy part.

PART 1 - I have two children. One of them is a boy. What are the odds that the other is also a boy?

Are you really saying the odds are 50/50 ?

Nick - I would add, good on you for playing the game and taking a position.
P.S. would you like to take a bet on this  $)


Nick Gammon

Oh I see. You are asking what are the odds, given you already have the children? Not, what would the next one be?

Quote
Nick you are a "Brattain Member" ...


That just means I make a lot of posts, not that I know what I am talking about. ;)
Please post technical questions on the forum, not by personal message. Thanks!

More info:
http://www.gammon.com.au/electronics

robtillaart

50% as the birth of children is independent.

The other way of reasoning can be
there are 4 possible combinations
BG
BB
GG
GB

one is a boy so GG is removed, the list of possible pairs is
BG
BB
GB
one is a boy so the other is
G
B
G
that would imply 33%


Rob Tillaart

Nederlandse sectie - http://arduino.cc/forum/index.php/board,77.0.html -
(Please do not PM for private consultancy)

radman

PART 1 - I have two children. One of them is a boy. What are the odds that the other is also a boy?

@robtillaart is correct the odds of my other child being a boy are 1:3 - give the man a coconut

robtillaart

Note that this question would have a different answer if you said the eldest one is a boy...
Then the chance of the youngest being a boy is ...
Rob Tillaart

Nederlandse sectie - http://arduino.cc/forum/index.php/board,77.0.html -
(Please do not PM for private consultancy)

mmcp42

nope
you seek to imply that the birth of one child has an influence on the birth of the next
leaving aside genetic tendencies (our family only ever has boys, twins, kittens)
the odds for the 2nd child are the same as the odds for the first
there are only 10 types of people
them that understands binary
and them that doesn't

robtillaart

on contrary, in the original question there was no order information, in the variation there is.
Due to the order information the reasoning would be:

There are 4 possible combinations
BG
BB
GG
GB

The oldest is a boy so GG AND GB is removed, the list of possible pairs is
BG
BB

This leaves a 50% chance ...
Rob Tillaart

Nederlandse sectie - http://arduino.cc/forum/index.php/board,77.0.html -
(Please do not PM for private consultancy)

Nick Gammon

Quote

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
Please post technical questions on the forum, not by personal message. Thanks!

More info:
http://www.gammon.com.au/electronics

mmcp42

ok so after 4 children what are the odds
by your logic you can't have more than 4
and even after three (if by then you have BBG)
there is still no guarantee that the fourth will complete the set
there are only 10 types of people
them that understands binary
and them that doesn't

Nick Gammon

Please post technical questions on the forum, not by personal message. Thanks!

More info:
http://www.gammon.com.au/electronics

Simpson_Jr


P.S. would you like to take a bet on this  $)


You would need to test that with loads of mothers/kids to get an accurate result, If... you have a load of nice good looking moms to choose from and you're willing to raise the kids, I may be interested.... :P

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