** STICKY ** League of Reason? PROVE YOURSELF!

[UrbanMasque]

Here's a fun puzzle:

A guy has two children. One of them is a boy, born on a Tuesday. What's the probability that his other child is a boy?

Wow.. that's all we get? At the risk of sounding dumb - I'd argue that the Tuesday bit is IRRELEVANT.
Child 1 = 50/50 boy/girl.
Child 2 = 50/50 boy/girl
It stays at 50/50. Always. You can argue the chance of having a boy/girl after the 1st one is a boy, and I'd say There is a 25% chance the 2nd child is a boy. (that's assuming I know the order of birth)

 

[Case]

Actually the chance of a child to be a boy is slightly less than 50% because more male than female offspring die before birth (there are also chromosomal anomalies which don't really allow for that dichotomous classification). By the way, any specification of a day (read: additionally supplied condition) changes the conditional probability, this doesn't have anything to do with tuesdays per se. In general terms: in sampling without replacement, outcomes of different trials are dependent.

 

[No1Mensan]

This might help people with their reasoning.

 

[Pulsar]

Here's a fun puzzle:

A guy has two children. One of them is a boy, born on a Tuesday. What's the probability that his other child is a boy?

Wow.. that's all we get? At the risk of sounding dumb - I'd argue that the Tuesday bit is IRRELEVANT.
Child 1 = 50/50 boy/girl.
Child 2 = 50/50 boy/girl
It stays at 50/50. Always. You can argue the chance of having a boy/girl after the 1st one is a boy, and I'd say There is a 25% chance the 2nd child is a boy. (that's assuming I know the order of birth)

No, No1Mensan is correct. You can find the answer by writing down all possible combinations (gender and days).

 

[Aught3]

I don't get it. I understand the Monty Hall game but I don't see how it is relevant.

Even when I draw out the possibilities I get 1/2 probability of a second boy.

 

[Case]

I don't get it. I understand the Monty Hall game but I don't see how it is relevant.

You're not allowed to sample [male born on tuesday / male born on tuesday] twice out of all the possible [x born on day soandso / y born on day soandso], so you end up with 13/27 combinations where [male born on tuesday / male born on soandso] is true instead of 14/28.

However, this whole calculation is obviously flawed since the chance of a child being born on a Tuesday (read: any weekday) is significantly higher than the chance of it being born on weekends whereas this calculation assumes P(child born on x)=1/7. I say this has to do with doctors performing c-section and I know I'm right because 30% of all births are by c-section in the US (see my thread about it) and because of this

. q.e.d.

 

[Aught3]

You're not allowed to sample [male born on tuesday / male born on tuesday] twice out of all the possible [x born on day soandso / y born on day soandso], so you end up with 13/27 combinations where [male born on tuesday / male born on soandso] is true instead of 14/28.

I don't understand this.

 

[Case]

You can have:

boy born on monday, boy born on monday
boy born on monday, boy born on tuesday
girl born on monday, boy born on...

and so on.
For "boy born on tuesday, x born on..." you are left with 28 possibilities:

boy born on tuesday, boy born on monday
boy born on tuesday, boy born on tuesday
boy born on tuesday, boy born on wednesday
boy born on tuesday, boy born on thursday
boy born on tuesday, boy born on friday
boy born on tuesday, boy born on saturday
boy born on tuesday, boy born on sunday
*2

boy born on tuesday, girl born on...
...
*2

Since you cannot sample "boy born tuesday, boy born tuesday" twice, you are actually left with 27 possibilities. Of these, 13 involve two boys. Therefore, the conditional probability for the one child to be a boy if the other is a boy born on a tuesday - is 13/27.

 

[Aught3]

Okay I understand what you are saying now. I'll keep going until I get it.

Ah got it, cheers!

 

[No1Mensan]

I'll give you a less complicated puzzle to solve.

What's the next letter in this sequence: q p w o e i r

 

[Case]

u (2 ez)

 

[No1Mensan]

u (2 ez)

I know it works better when people don't have a computer in front of them. That was a warm up one anyway. Here's the real one.

110 Vancouver <-------------> Bali 51
? Corsica<-------------------> Maldives 1556

How far is it to Corsica on this sign?

 

[ArthurWilborn]

No one is going to take my challenge?

I've been waiting for months for someone to solve my problem. :lol:

A bit of a mathematical problem:
One fine day you visit your favourite restaurant and find out they're having a contest.
They're offering free dinner for a month to anyone who can give the right answer to a certain problem.
Of course you decide to participate, so here's the fun part:

The restaurant just created the first 4-dimensional cake and they ask you what the highest
possible number of pieces is you can get after cutting this 4d-cake ten times (straight cuts i.e. no figure skating with the dimensional blade ).

What answer wins you the contest (and how did you arrive at the solution)?

Hint 1:

Hint 2: The answer is neither 36, nor 42, nor 144.

I got it.

Sum of xv(n) = xv(n-1) + 2(n-1) from 0 to 10

349

 

[Case]

I got it.

Sum of xv(n) = xv(n-1) + 2(n-1) from 0 to 10

349

Nope. Getting closer now, though.

 

[simonecuttlefish]

DAMNED session timed out and I lost my post!

Anyway ...

NO - you can't determine what colour any of them has. You can only work out a list of possible combinations could have and could not have had,
Like ........
Log. C can't have red if A did or B would know
Log. C can't have red if B did or A would know
etc

You can't determine what they DID have, just a list of combinations individuals could not have. There is no way to determine if they all had green or not.

 

[simonecuttlefish]

Re: League of Reason? PROVE YOURSELF!

We know that C is green.

When A looks around, since he is not able to tell which color he can't be, he either sees that B and C has different colors or that they both have green. B can deduce this from hearing A's statement. So when B looks around, for him not to be able to know which color he can't be, C has to be green, in any other case B would know a color he could not be.

@ iJoe: http://www.youtube.com/watch?v=Cdiz0k0Rudw

WINRAR

I'm so stupid EPIC FAIL! It feels a bit like why I just walked away from learning programming. I never came to terms with people who speak in double, triple, or more negatives and don't understand why end users whine about their interface design or documentation

 

[RichardMNixon]

Since you cannot sample "boy born tuesday, boy born tuesday" twice, you are actually left with 27 possibilities. Of these, 13 involve two boys. Therefore, the conditional probability for the one child to be a boy if the other is a boy born on a tuesday - is 13/27.

Why are you sampling anything twice and why doesn't that apply to double boy tuesdays?

 

[Pulsar]

Why are you sampling anything twice and why doesn't that apply to double boy tuesdays?

Because there are two scenarios: the boy born on a Tuesday is either the youngest child or the oldest child. In the first case, there are 14 possibilities for the other (oldest) child. In the latter case, there are also 14 possibilities for the other (youngest) child. However, combining both scenarios, you've counted the situation 'first child = boy born on a Tuesday, second child = boy born on a Tuesday' twice. So there are in total only 27 combinations, not 28.

The fun thing about this puzzle is that any extra info about the child will alter the probabilities. The basic case would be: 'I have two children, one of which is a boy. What is the probability that my other child is also a boy?' Answer: 1/3, because the combinations are
boy, girl
girl, boy
boy, boy.

Once you add more specific info, like 'the boy is born on a Tuesday', the probabilities change: the more detailed the info, the closer the probability will approach 1/2.

 

[RichardMNixon]

Why are you sampling anything twice and why doesn't that apply to double boy tuesdays?

Because there are two scenarios: the boy born on a Tuesday is either the youngest child or the oldest child. In the first case, there are 14 possibilities for the other (oldest) child. In the latter case, there are also 14 possibilities for the other (youngest) child. However, combining both scenarios, you've counted the situation 'first child = boy born on a Tuesday, second child = boy born on a Tuesday' twice. So there are in total only 27 combinations, not 28.

The fun thing about this puzzle is that any extra info about the child will alter the probabilities. The basic case would be: 'I have two children, one of which is a boy. What is the probability that my other child is also a boy?' Answer: 1/3, because the combinations are
boy, girl
girl, boy
boy, boy.

Once you add more specific info, like 'the boy is born on a Tuesday', the probabilities change: the more detailed the info, the closer the probability will approach 1/2.

Ok thanks, I think I was placing too much importance on the order and trying to figure out why you cared, the important bit is that mixed gender is twice as probable as a specific single gender given no other information.

 

[No1Mensan]

What's the hidden message?

2,1 8,1 4,2 3,2 4,3 7,4 6,1
3,3 6,3 7,3
8,1 4,2 3,2
9,1 4,3 6,2

The answer contains four words.