07-02-2002 19:52									#61
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quote: Originally posted by Rogan Josh Is it just me or is all this bloody obvious? (Clearly not - since there is so much debate....)

It's bloody obvious, but only if you have some training in it first. We as a species are notoriously bad at probability - possibly because evolutionarily speaking, we had no need to improve. A distinction into "Never", "Unlikely", "Possible", "Likely" and "Certain" is probably all we ever really needed to survive.

As an example, without doing any calculations (and shh if you know the answer already), take a ballpark guess at the following questions.

1) How many people do you need before the probability that at least two share a birthday is greater than 1/2 (ie, when does it become more likely for there to be a coincidence than not)?

2) How many people do you need before the probaility that at least one of them has your birthday is greater than 1/2?

I know the answer to the first by heart, and I know the second one roughly. I'll work it out and post after a few people have had a go. But don't post your reasoning, or do detailed calculations - this is about intuition.

Guesser	First Problem	Second Problem
Victor Gallis	184	730
Boddington's	27	183
Rah	25	183
Richard Burns	19	182
SD	25	250
Paul	20	183
Adam Smith	20	183
Dr. Oogkloot	-	260
Wernazuma	19	183

	07-02-2002 20:08									#62
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quote: Originally posted by Chowlett As an example, without doing any calculations (and shh if you know the answer already), take a ballpark guess at the following questions. 1) How many people do you need before the probability that at least two share a birthday is greater than 1/2 (ie, when does it become more likely for there to be a coincidence than not)? 2) How many people do you need before the probaility that at least one of them has your birthday is greater than 1/2?

Hehe

quote: I know the answer to the first by heart, and I know the second one roughly. I'll work it out and post after a few people have had a go. But don't post your reasoning, or do detailed calculations - this is about intuition.

I know the first one, but not the second.

	07-02-2002 20:24									#63
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quote: Originally posted by Chowlett 1) How many people do you need before the probability that at least two share a birthday is greater than 1/2 (ie, when does it become more likely for there to be a coincidence than not)? 2) How many people do you need before the probaility that at least one of them has your birthday is greater than 1/2?

I know the answer to both of these (don't worry - I will keep quiet ) although I must confess I used my calculator for the first one.

Interestingly though, I tried guessing the first one before working it out (the second one requires no calculation) and was quite a bit out (although surprisingly, I was too low with my guess....).

Edit: I take that back for question 2. I misread it

	07-02-2002 20:28									#64
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quote: Originally posted by Rogan Josh (the second one requires no calculation)

You sure? PM me your answer.

	07-02-2002 20:34									#65
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quote: To those who still don't believe Krazy and Jon, I propose a wager. We'll survey 100 women who have exactly two children, including at least one boy. (Shouldn't be hard to set that up in an online chat room or something.) If at least 51 of them have a girl in addition to the boy, you pay me $1000. If not--if at least 50 of them have two boys--I'll pay you $1500. If Harrison is right and I'm wrong, I'm offering you 3/2 odds on a 50-50 bet. Any takers?

-That's an entirely different problem from that which I suggested that the answer is 1/2. (see my second or third post where I came up with 1/3)

If the question is:

A woman has two children, (the first or the second) of them is a boy, what is the chance that the other is a boy?
The answer is 1/2.

If the question, is we find a population of women with 2 children, at least one boy, then the probability of randomly selecting one that is has another boy is 1/3.

"1) How many people do you need before the probability that at least two share a birthday is greater than 1/2 (ie, when does it become more likely for there to be a coincidence than not)?"

183.125

"2) How many people do you need before the probaility that at least one of them has your birthday is greater than 1/2?"

Hmmm... I'd say 730.5 including yourself. These are rough guesses made between classes.

	07-02-2002 20:41									#66
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I'm now keeping a tally. To clarify things, I'm only considering integer numbers of people, so round up your guesses. I'm also not counting you in the second problem. And just to make things easier, all years are assumed to have 365 days, and birthdays are assumed to be evenly distributed throughout the year (fallacious, but hey)

	07-02-2002 21:05									#67
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rough guesses

1. 20-30 people
2. 183

RAH

	07-02-2002 23:21									#68
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OK, OK, let me get my calculator

	07-02-2002 23:24									#69
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quote: Originally posted by Victor Galis -That's an entirely different problem from that which I suggested that the answer is 1/2. (see my second or third post where I came up with 1/3) If the question is: A woman has two children, (the first or the second) of them is a boy, what is the chance that the other is a boy? The answer is 1/2. If the question, is we find a population of women with 2 children, at least one boy, then the probability of randomly selecting one that is has another boy is 1/3.

I think Victor has a point and this is coming down to an issue of semantics in the question itself. I can certainly see where 1/3 has come from. Let's face it, the wording of the question is ambiguous.

	07-02-2002 23:29									#70
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quote: Originally posted by Chowlett As an example, without doing any calculations (and shh if you know the answer already), take a ballpark guess at the following questions. 1) How many people do you need before the probability that at least two share a birthday is greater than 1/2 (ie, when does it become more likely for there to be a coincidence than not)? 2) How many people do you need before the probaility that at least one of them has your birthday is greater than 1/2? I know the answer to the first by heart, and I know the second one roughly. I'll work it out and post after a few people have had a go. But don't post your reasoning, or do detailed calculations - this is about intuition.

I haven't looked at anything after this post and I only took a basic statistics course, so I'll take a few blind guesses:

1: about 19, or whatever the square root of 365 is.
2: 182

edit: messed up quote tag

	07-02-2002 23:36									#71
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Problem 1: ~25
Problem 2: ~250

	08-02-2002 00:53									#72
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Problem 1: 20
Problem 2: 183

	08-02-2002 02:41									#73
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Edit: I wasn't supposed to put down the answers, eh? Ermm.. my bad..

Last edited by Ramo on 08-02-2002 at 02:53

	08-02-2002 02:44									#74
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Edit that out.

So much for the no calculations rules. You were supposed to estimate on face value!

	08-02-2002 02:50									#75
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Its all well and good being able to calculate the answer, but you won't get far if you don't read the rubric.

	08-02-2002 03:11									#76
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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.

Last edited by Dr.Oogkloot on 08-02-2002 at 03:19

	08-02-2002 03:18									#77
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My second answer is a bit high... but I suppose that was my gut reaction. Probably a lot closer to 184.

	08-02-2002 03:53									#78
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In case you were wondering I PMed my answers to Chowlett

	08-02-2002 04:01									#79
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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.

	08-02-2002 04:41									#80
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First: 19
Second: 183

	08-02-2002 04:51									#81
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I know the gut feeling one, but also realised it is wrong...I would have to calculate the real one...

	08-02-2002 05:12									#82
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EDIT: Nevermind, turns out you aren't still agruing about the original post, afterall.

	08-02-2002 06:52									#83
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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.

	08-02-2002 07:07									#84
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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..

	08-02-2002 07:14									#85
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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.

	08-02-2002 07:19									#86
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quote: 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.

	08-02-2002 07:33									#87
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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

	08-02-2002 09:14									#88
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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

	08-02-2002 09:19									#89
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quote: 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.

	08-02-2002 09:35									#90
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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.