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I Love Solving Math Questions

 
 
harpazo
 
Reply Mon 30 Dec, 2013 01:02 pm
I am 48 years old. I have two college degrees in areas other than math. I got an A minus in precalculus back in 1993. I am divorced.

I am single. I am lonely. My circle of friends is very small. I play hymns and praise songs on my yamaha classical guitar. In the midst of the valley (my life is like a dry valley), I find time for math. I find time to do what so many people hate---Math.

I am criticized by friends and family for practicing math on my days off at the local library. They do not understand that math is my passion. I just love solving for x.

I am a good guy. No police record. A very clean member of society. My only "crime" is loving mathematics. What is your reply?
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Type: Question • Score: 12 • Views: 2,882 • Replies: 50

 
dalehileman
 
  0  
Reply Mon 30 Dec, 2013 03:34 pm
@harpazo,
Harp, this is purely mathematical q

In an infinite universe where all the "rules"are the same throughout and which exists forever, isn't it true that anything that can happen, will happen

If so, won't the very same thing happen in different places at once (though mostly quite far apart); and in fact isn't there an infinite number of galaxies simultaneously identical to ours in ever respect, a Harp chatting with a Dale by means of an A2k Forum

Carrying the idea of infinity a step further, wouldn't there also be an infinite number differing from this one only in one quantum--that is absolutely identical at this very moment except for instance one hair on Harp's head is 0.00000000….01324 in. longer

If all the above strikes you intuitively as absurd is your reaction not a sort of evidence that the Universe probably isn't infinite after all

Thanks, Harp, for entertaining my ruminations
jespah
 
  3  
Reply Mon 30 Dec, 2013 03:36 pm
@harpazo,
Can I hire you to do my taxes?
Romeo Fabulini
 
  0  
Reply Mon 30 Dec, 2013 03:45 pm
There was an article about luck and chance in the paper recently, i meant to save it but forgot.
It said something like if there are a certain number of people in a room, the chance is almost 100% that two of them shared the same birthday.
Can anybody explain that further?
Frank Apisa
 
  1  
Reply Mon 30 Dec, 2013 04:16 pm
@Romeo Fabulini,
Romeo Fabulini wrote:

There was an article about luck and chance in the paper recently, i meant to save it but forgot.
It said something like if there are a certain number of people in a room, the chance is almost 100% that two of them shared the same birthday.
Can anybody explain that further?


If there are 366 people in the room...there is a 100% chance. Obviously.

But if there are as few as 25 people in the room...the chances are probably 50/50 that there are two people with the same birthday.

Harp could probably give you the math formula.

timur
 
  2  
Reply Mon 30 Dec, 2013 04:29 pm
Obviously not 100%!

What's the chance having two people, born February the 29th, in a crowd of 366?

Birthday problem
Frank Apisa
 
  1  
Reply Mon 30 Dec, 2013 05:06 pm
@timur,
timur wrote:

Obviously not 100%!

What's the chance having two people, born February the 29th, in a crowd of 366?

Birthday problem


Two share the same birthday...is not that two share 2/29...or any other date.

If there are 366 people in a room...how could there not be at least one match? It is 100% for 366. It is about 50% for 25 people.
Frank Apisa
 
  1  
Reply Mon 30 Dec, 2013 05:07 pm
@Frank Apisa,
By the way...the probability is not as strange as it may seem. If there is a probability person here...he/she will explain it.
0 Replies
 
chai2
 
  3  
Reply Mon 30 Dec, 2013 05:23 pm
@jespah,
jespah wrote:

Can I hire you to do my taxes?


Just don't wear a short skirt when you have an appointment with him.
jespah
 
  1  
Reply Mon 30 Dec, 2013 05:25 pm
@chai2,
I'll keep that in mind.

Can I wear a short skirt if you do my taxes, chai?
0 Replies
 
raprap
 
  1  
Reply Mon 30 Dec, 2013 05:26 pm
@Romeo Fabulini,
30. It is a near certainty that if you have 30 people in a room, two of them will share a birthday.

Rap
Romeo Fabulini
 
  1  
Reply Mon 30 Dec, 2013 09:04 pm
Quote:
Raprap said:@ RF- It is a near certainty that if you have 30 people in a room, two of them will share a birthday

Fascinating, but what does "near certainty" mean?
What number of people must there be to make it an absolute certainty?
And how did you arrive at 30 anyway, it seems very small to me?
I mean, there are 365 days in the year, and 30 is only about one-twelfth of 365, so 30 is only a very small sample.
But what the hell do I know, I was always at or near the bottom of the class in maths.
Brandon9000
 
  1  
Reply Mon 30 Dec, 2013 09:37 pm
Welcome to A2k. Every day on my lunch hour, I go to Barnes & Noble, or Starbucks, or the mall and work through books on math and physics. I find it pleasant and relaxing. I had never met or heard of anyone else doing this until I re-established contact with a long lost cousin I hadn't spoken to in many years and found out that she does it too, although I suspect that she substitutes chemistry for physics, since that was her major. I completely understand your passion for math. It is very, very cool and beautiful.
0 Replies
 
raprap
 
  1  
Reply Mon 30 Dec, 2013 10:06 pm
@Romeo Fabulini,
Near certainty is nearly 1, on the order of 0.99999 ( 99,999 times out of one hundred thousand). One (1) is a certainty, but in probability, like in life, a certainty is a limit, but its a very convergent limit.

Oh! And if you are dealing with 30 people in a room that share my birthday you are right.

But that's not the problem. The problem is that any two individuals in a room share a birthday. To solve that problem you have to consider how many possibilities there are and count them all. Fortunately there is a method of counting (combinations/permutations) that easily used to count.

Granted if you had a maximum improbability drive of "The Hitchhikers Guide' you'd probably hit that exception for any number greater than say 25.

Rap
0 Replies
 
Romeo Fabulini
 
  1  
Reply Mon 30 Dec, 2013 10:18 pm
@Raprap: -Okay i've got that. I think...

Moving on, another puzzle I once heard involved a big bag of different coloured marbles (say red blue and green).
I don't remember exactly, but the question went something like- "if you were blindfolded, what's the minimum number you'd have to take out to be absolutely certain that there was at least one of each colour among them?
I don't remember the answer but I remember it was only a very small number, and I couldn't figger out why it was so low.
Anybody know the exercise I'm talking about?
tomr
 
  1  
Reply Mon 30 Dec, 2013 10:28 pm
@timur,
Quote:
Obviously not 100%!

What's the chance having two people, born February the 29th, in a crowd of 366?



P(n;d) = 1 - [(d - 1)/d]^n

n = 366; d = 4*365 + 1 = 1461

P(366;1461) =1 - (1460/1461)^366
P(366;1461) = .222

I get a 22% chance. Am I right?
markr
 
  1  
Reply Mon 30 Dec, 2013 11:06 pm
@Romeo Fabulini,
Birthdays: 367 guarantees success since there are 366 different days in a leap year.
Marbles: The number would be small if you are to guarantee a small number of the same color. However, if you are to guarantee at least one of each color, then you'd have to pull all but N-1, where N is the number of balls of the least-represented color.
0 Replies
 
Finn dAbuzz
 
  1  
Reply Mon 30 Dec, 2013 11:27 pm
@harpazo,
Absolutely nothing wrong with loving math.

Math is the language of God. I wish I was more fluent in it.

I'm betting that your problems, though, have nothing to do with math.

It's tough to give advice, so far removed, in response to your issues.

Why are you divorced?
0 Replies
 
maxdancona
 
  1  
Reply Mon 30 Dec, 2013 11:54 pm
@Frank Apisa,
Quote:
If there are 366 people in a room...how could there not be at least one match? It is 100% for 366. It is about 50% for 25 people.


There are 366 possible days on which you can be born (including 2/29).

If one person was born on every day (including exactly one on 2/29) then there would be 366 people with no two sharing a birthday.

The 367th of course, would be a repeat.
0 Replies
 
maxdancona
 
  1  
Reply Tue 31 Dec, 2013 12:09 am
@raprap,
The way to solve this (ignoring the problem of leap year for a moment since it won't change the answer much) is this.

It is easier to combine probabilities then to tease them apart, so we change the problem a bit. Instead of caculating the odds that two people will share a birthday, it turns out it is easier to calculate teh odds that some number of people won't share a birth day (and then just subtract from 1).

The first person has a birthday.

The second person can be born on any of the 365 days, 364 of them won't be a match, so the odds that the second person doesn't share a birthday with the first is 364/365.

The third person can be born on any of the 365 days. Of all the times the first two people have unique birthdays, so the odds of the third person having a unique birthday is 363/365. The so the total odds of 3 people having unique birthdays is 364/365 * 363/265


So the get 30 people with unique birthdays you multiply all of these probabilities together.

364/365 * 363/365 * 362/365 ..... * 335/365

This comes out to about 29.4% (which is the odds of 30 people having unique birthdays). This means the odds that two of them will share a birthday is about 70.6%.

I would say this falls short of "near certainty".
 

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