A woman has two children, at least one of which is a boy...

[Dr.Oogkloot]

1. I know the answer so I won't tell.

2. I'm certain it can't be 183; I'll guess 260.

By the way: the probability in the girl/boy problem is 1/3 and not 1/2, because you were not given the information "the first child is a boy" or "the second child is a boy", but only the information "at least one is a boy". Those of you who think it is 1/2 are simply answering the wrong question.

Switching doors does increase your probability of winning, this is a very infamous problem.

[Victor Galis]

My second answer is a bit high... but I suppose that was my gut reaction. Probably a lot closer to 184.

[Rogan Josh]

In case you were wondering I PMed my answers to Chowlett

[Dauphin]

Its pretty easy to calculate, I knew what had to be done straight away, but its the gut instinct answer that Chowlett wanted and that is what I gave.

I then did a detailed calc and Ramo got the same answer I did .

I neglected leap years though - anyone born on Feb 29th will change the numbers big time.

[Wernazuma III]

First: 19
Second: 183

[Provost Harrison]

I know the gut feeling one, but also realised it is wrong...I would have to calculate the real one...

[General Ludd]

I haven't read much of the thread....

EDIT: Nevermind, turns out you aren't still agruing about the original post, afterall.

[Chowlett]

Ok, I'll give the answers. Firstly, may I say how interesting it is that almost all you answers for the first part were too low. I would have expected much higher guesses.

1. 23: Think of the people coming into the room 1 by 1. Then the first person has a 365/365 chance of being unique. The second has opnly 364 birthdays left, so the chance of no coincindence drops to (365/365)(364/365). Continuing in this way, we see that with n people, the chance for no coincidence occurring is 365!/(365-n)!365ⁿ

Since probabilty of at least one coincidence = 1-probability of no coincedences, the required probality for n people is 1-365!/(365-n)!365ⁿ. Tabulating this, we see that at n=22, this is 0.47, at n=23 it is 0.51

2. 253: Each person has a 364/365 chance of having a birthday different from yours. Therefore, the chances of n people all having birthdays different from yours are (364/365)ⁿ. Therefore, the probability of at least one person out of n having your birthday is 1-(364/365)ⁿ. At n=252, this is 0.4991. At n=253, this is 0.5005

I'm impressed by your first problem guesses, but the second problem shows that the intuition can be waaay off at times. Looks like SD was closest though.

[rah]

Intuition tells me that everyone has a 1/365 chance of having my birthday(assuming a normal distribution of birthdays). If two people enter the room the odds would increase to 2/365, three 3/365, up to 183/365 which is where it passes 1/2. Silly me.
Any answer higher then 183 seems outright wrong.

RAH
At least my 20-30 was a good raw estimate on the first one..

[Rex Little]

Rah, your intuition ignores the fact that there are going to be duplications. 182 people won't cover 182 birthdays, so the odds that they'll cover a particular one are less than 1/2. I must admit I would have guessed the actual number to be quite a bit less than 253, though; more like 190-200.

Anyone care to guess the mathematically expected number of different birthdays which would be covered by a random group of 183 people? I don't know the answer; I might be able to figure it out, or hopefully Chowlett can enlighten us.

[Chowlett]

Originally posted by Rex Little
Rah, your intuition ignores the fact that there are going to be duplications. 182 people won't cover 182 birthdays, so the odds that they'll cover a particular one are less than 1/2. I must admit I would have guessed the actual number to be quite a bit less than 253, though; more like 190-200.

Anyone care to guess the mathematically expected number of different birthdays which would be covered by a random group of 183 people? I don't know the answer; I might be able to figure it out, or hopefully Chowlett can enlighten us.

Not at... good god it's 2 am. Must get sleep.

[rah]

If I have a random number generator that randomly pulls a number between 1 and 365.
And I choose any number, every time I run the number generator will give me a 1 in 365 chance of hitting my number. Every subsequent pull would be additive. I don't see the difference.

RAH

[KrazyHorse]

dm: no insult intended. The was there for a reason. Read upward for my argument.

First question: 20-25?
Second question: 210?

Both rough estimates, sans calculation, but taking into account the "trends" you get a feel for in prob.

Now, those of you still not "getting" the boy-girl problem:

Imagine you have a hundred pairs of children. Given perfect distribution, 25 pairs will consist of 2 girls, 25 will consist of 2 boys, and 50 will consist of 1 boy and 1 girl. By saying that there is at least 1 boy, we eliminate 25 of the pairs of children. This leaves us with 50 bg pairs and 25 bb pairs. Thus, the probability of the "other" child also being a boy is 25/75=1/3

[KrazyHorse]

Originally posted by rah
If I have a random number generator that randomly pulls a number between 1 and 365.
And I choose any number, every time I run the number generator will give me a 1 in 365 chance of hitting my number. Every subsequent pull would be additive. I don't see the difference.

RAH

Not good, rah; your 182 forgets that you "lose" some chances of drawing your number every time because there's a small chance that you stop drawing numbers with your previous turn; taken to the extreme, is the probability of drawing my number exactly 1 after 365 draws? What about the remote possibility that you draw Jan.1 365 times in a row?

The proper prob calculation looks something like 1-(364/365)ⁿ where n is the number of dates drawn.

[Urban Ranger]

KrazyCat,

The difference seems to lie in whether you look at an entire population or an individual.

If you look at an entire population, sure, you are correct. However, for an individual, the answer should be 1/2 since there is indeed degeneracy.