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For your entertainment, inspired by a protracted debate with our SM1 lecturer. The basic idea of these questions isn't original to me, but I've modified them somewhat to make the phrasing less ambiguous.*

1. A friend mentions that they have two kids. With no other information, what are the odds that they have (a) two boys, (b) one boy and one girl, or (c) two girls? (For the sake of these problems, assume that half of children are boys, half are girls, and people don't have a natural tendency to children of one sex or the other.)

2. Looking at one of their many bookshelves, you spot a Saddle Club book, which (applying gender stereotypes) you may take as indication that at least one of their children is a girl. Based on this information, what are the odds that they have (a) one boy and one girl, or (b) two girls?

3. You mention the Saddle Club book to your friend. He replies "Yeah, that's Mary's." What are the odds that both their kids are girls?

4. As above, but the reply is "Yeah, that's Mary's, she's my eldest."

5. Different friend, same dilemma - two kids, you don't recall their sexes. Being sneaky, you ask "Would your oldest like to come to my kid's birthday party?" and the response is "Yes, she'd love to." What are the odds that both their kids are girls?

6. As above, but the response is "Mary? Yes, she'd love to."

7. You find a Saddle Club book lying on the ground, inscribed 'To Mary'. You ask around the local schools and in a flagrant breach of privacy they give you the addresses of several dozen Saddle-Club-age 'Mary's in the neighbourhood. You go to the first house on the list and, by examining shoes again, deduce that the family has two children. What are the odds that both are girls?

*Which was surprisingly difficult to do - it's actually very hard to convey no more and no less information than one means to when writing these things.

Date: 2008-04-09 11:58 am (UTC)
From: [identity profile] nefaria.livejournal.com
Ugh, I'm gonna embarrass myself if I mess these up, considering that I used to be a math major long, long ago, in a galaxy far, far away.

1. a. 1/4, b. 1/2, c. 1/4
2. a. 2/3, b. 1/3
3. 1/3
4. 1/2
5. 1/2
6. 1/2
7. 1/3

Date: 2008-04-09 01:54 pm (UTC)
From: [identity profile] lederhosen.livejournal.com
Mostly right, but not all... will post answers in a couple of days.

How do you keep an idiot in suspense?

Date: 2008-04-09 02:45 pm (UTC)
From: [identity profile] nefaria.livejournal.com
Days?!?! You're crueller than I imagined!

Re: How do you keep an idiot in suspense?

Date: 2008-04-09 07:33 pm (UTC)
From: [identity profile] notasquirrel.livejournal.com
You're crueller than I imagined!

*happychitter*

Re: How do you keep an idiot in suspense?

Date: 2008-04-12 12:23 pm (UTC)
From: [identity profile] nefaria.livejournal.com
OK Mr. Smarty-Squirrel, your homework is to prove the Riemann zeta hypothesis. Do it and win a million dollars!

http://en.wikipedia.org/wiki/Riemann_hypothesis
Edited Date: 2008-04-12 12:23 pm (UTC)

Re: How do you keep an idiot in suspense?

Date: 2008-04-13 02:59 pm (UTC)
From: [identity profile] nefaria.livejournal.com
** takes the $1 off the $1,000,000 prize and leaves the ,000,000 for Buzzy **

Date: 2008-04-09 11:36 pm (UTC)
From: [identity profile] chaos-crafter.livejournal.com
I think #3 has to be 1/2 not 1/3. All you've now done is establish the gender of one child, The other is now 50/50.

Date: 2008-04-09 12:35 pm (UTC)
From: [identity profile] tenner.livejournal.com
Gonna give this a shot (without reading the previous commenter) so here goes, though I could be wrong, I hate conditional probability:

1. 25% 2B, 50% 1B1G, 25% 2G

2. At least one child is a girl, so the 25% BB case goes away. Thus, now it's 2/3 likely there's one boy and one girl, and 1/3 likely there are two girls.

3. 1/3.

4. Now you're making the kids distinguishable. The only possibilities are 50% GB and 50% GG, so it's 50% likely both kids are girls.

5. Same as 4. 50%.

6. I can't see how this is different from 5. You're still getting the same information, that the eldest is female, and that the gender of the youngest is still unknown (and equally likely to be boy or girl). So, 50% again.

7. I'd say this is the same case as 2. All you know is that there's one girl in the house, and the three remaining cases (GB, BG, GG) are equally probable. The odds that both children are girls is 1/3.

Date: 2008-04-09 12:35 pm (UTC)
From: [identity profile] tenner.livejournal.com
Wow... can't believe my entries exactly matched the previous commenter. I didn't look until afterwards, I swear.

Date: 2008-04-09 01:55 pm (UTC)
From: [identity profile] lederhosen.livejournal.com
As above: most of these are right, but not all.

Date: 2008-04-09 01:30 pm (UTC)
From: [identity profile] silmaril.livejournal.com
Before I post the real answers, obSmartMouth: It's a given thing that Mary is a girl's name, right? *grin*

Date: 2008-04-09 01:55 pm (UTC)

Date: 2008-04-09 02:43 pm (UTC)
From: [identity profile] nefaria.livejournal.com
This might be a good place to bring up the classic Monty Haul problem.

You have a choice of three doors, two of them hide old goats and one hides a brand new car. You want the car, not the goats.

You pick one of the doors. Monty Haul opens one of the other doors to reveal a goat. Now he offers to let you switch your choice to the remaining door before he opens it. Should you switch or not, and why?

Date: 2008-04-09 10:22 pm (UTC)
From: [identity profile] lederhosen.livejournal.com
The stock answer to this one is 'switch' (1/3 vs 2/3) but it depends on assumptions.

If you know that Monty always shows you a goat after you make the initial choice, the answer is as above. If you don't know that he had to show you a goat... well, the worst case is that Monty has decided he'll only show you a goat if you picked the right door first time, and he's trying to dissuade you.

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