Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

Login

Username

Password

Not registered yet?

#1 2013-03-24 23:28:49

anna_gg
Full Member

Offline

Shared birthdays

What are the chances that 6 people celebrate their Birthday in the same 2 months? Assume all months are equal.

Last edited by anna_gg (2013-03-25 00:27:25)

 

#2 2013-03-25 00:09:04

bobbym
Administrator

Offline

Re: Shared birthdays

Hi;

Last edited by bobbym (2013-03-25 00:13:40)


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#3 2013-03-25 00:28:11

anna_gg
Full Member

Offline

Re: Shared birthdays

bobbym wrote:

Hi;

See my corrected description.

 

#4 2013-03-25 00:48:05

bobbym
Administrator

Offline

Re: Shared birthdays

Hi;

That is a little different. I hope I am understanding what you want. It seems you want all six people to have their birthdays in only two months.

For that I am getting.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#5 2013-03-29 00:42:08

anna_gg
Full Member

Offline

Re: Shared birthdays

Right smile

 

#6 2013-03-29 10:24:04

bobbym
Administrator

Offline

Re: Shared birthdays

Hi;

Whamo! Wunderbar! Doron Zeilberg eat your heart out. Am I the king of comby/proby or what?


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#7 2013-03-29 10:50:27

anonimnystefy
Real Member

Offline

Re: Shared birthdays

How'd you get that answer?

And, I am pretty sure his last name is Zielberger.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#8 2013-03-29 10:51:56

bobbym
Administrator

Offline

Re: Shared birthdays

Hi;

No, it is Zeilberger. The way I get all the answers I get.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#9 2013-03-29 10:56:02

anonimnystefy
Real Member

Offline

Re: Shared birthdays

Yes. I permuted the i and the e. It seems to be common with me. I have a tough time remembering whether it's Liebniz or Leibniz.

Direct count?


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#10 2013-03-29 11:05:58

bobbym
Administrator

Offline

Re: Shared birthdays

No, all combinatoric problems fall into categories. Surely you have read the 12 fold way or even better the 30 fold way. I have my own set in addition to those. These templates help solve many types of problems.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#11 2013-03-29 11:12:58

anonimnystefy
Real Member

Offline

Re: Shared birthdays

Would you mind sharing?


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#12 2013-03-29 11:20:55

bobbym
Administrator

Offline

Re: Shared birthdays

Of course you were half right I already had the answer before I even began.

Of course, I will put down the formula-template here.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#13 2013-03-29 11:23:54

anonimnystefy
Real Member

Offline

Re: Shared birthdays

Great, thank you!


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#14 2013-03-29 11:33:13

bobbym
Administrator

Offline

Re: Shared birthdays

This looks like some of Feller's work or maybe Rose, I am not sure.



Surely you recognize that?!


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#15 2013-03-29 21:54:19

anonimnystefy
Real Member

Offline

Re: Shared birthdays

What the hell is that?


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#16 2013-03-29 22:18:03

bobbym
Administrator

Offline

Re: Shared birthdays

Looks like a formula! So you do not recognize it?


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#17 2013-03-29 22:32:15

anonimnystefy
Real Member

Offline

Re: Shared birthdays

It looks like some stuff I've seen on Wiki's distributions pages.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#18 2013-03-29 22:33:24

bobbym
Administrator

Offline

Re: Shared birthdays

Better than that. With that you compute the answer to anna's problem quickly.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#19 2013-03-31 03:58:00

anna_gg
Full Member

Offline

Re: Shared birthdays

I will try to share my solution in a more explanatory way (Sorry, Bobby, I don't imply that yours is not easy to understand - it is just that I am a novice and not very familiar with complicated solutions!!):
We first must calculate all the possible ways to get 2 months out of 12, that is



Then we must calculate all the different ways by which we can arrange the birthdays of 6 people in these 2 months: Either 5 people have their birthday in the first month and 1 in the second, or 4 in the first and 2 in the second etc. Obviously we do not consider the case of 0/6 or 6/0. For the first case, we first get 1 out of 6 (for the first person’s birthday) and then for the second person’s it will be 5 out of 5, and so on.
Here is the calculation:



The total probability is the product of the first two (66 x 62) divided by the total number of all different ways by which 6 people can have their birthdays in 12 different months, that is, 12^6.

So we have

Last edited by anna_gg (2013-03-31 03:58:57)

 

#20 2013-03-31 04:11:42

bobbym
Administrator

Offline

Re: Shared birthdays

Hi anna_gg;

Sorry, Bobby, I don't imply that yours is not easy to understand

No problem. I am glad to see your solution. Also, anyone who can do that problem I do not characterize as a novice.

Have you tried Codecogs?


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#21 2013-03-31 04:57:52

anna_gg
Full Member

Offline

Re: Shared birthdays

No I haven't; actually I wrote the solution in a Word doc, but when I copied it here, the formatting was screwed up sad(

 

#22 2013-03-31 05:35:29

bobbym
Administrator

Offline

Re: Shared birthdays

Hi anna_gg;

I latexed it up for you.

When you have time try this site

http://latex.codecogs.com/editor.php

perfect math every time!


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#23 2013-03-31 06:35:15

anna_gg
Full Member

Offline

Re: Shared birthdays

Thank you! Very useful!

 

#24 2013-03-31 07:04:32

bobbym
Administrator

Offline

Re: Shared birthdays

Hi;

You are welcome and happy latexing.


In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
 

#25 2013-03-31 09:35:21

anonimnystefy
Real Member

Offline

Re: Shared birthdays

Hm, then the solution for n people seems to be 66*(2^n-2)/(12^6).

And thank you both for showing your methods.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

Board footer

Powered by FluxBB