# Anyone up for a funny math problem?

Discussion in 'Community Discussion' started by Jade Cambell, Dec 17, 2007.

1. ### Jade Cambell macrumors 6502

Joined:
Sep 14, 2007
#1
Alright, here's the problem.

My birthday is July 25th. My brothers birthday is October 15th.

I met a girl online about a year ago, and immedietaly noticed that her birthday, oddly enough, was October 15th. When I told her that she had my brothers birthday, she looked at my birthday (on my facebook account), and was stunned to see that my birthday is the same as HER brothers!

The math problem: What are the chances of that happening?

We can call me and my brother a and b, and her and her brother x and y.

The chances of one person having any given birthday is 1 in 365. The number of combinations of any two days in the year, is 365 x 365. That's about as far as I can get. What are the chances and me and my brother share the exact same birthdays with this girl and her brother?

Joined:
Jan 15, 2006
Location:
The Kop
#2
what about all those people born on the 29th February

3. ### runplaysleeprun macrumors 6502a

Joined:
Jul 27, 2004
Location:
Chicago, IL
#3
I think its done by (1/365)X(1/365), then /2, which is 1/266450.

I emphasize, I think

Joined:
Sep 14, 2007
#4
Thanks. Anyone else wanna confirm this? Give it another try? Something?

5. ### CalBoy macrumors 604

Joined:
May 21, 2007
#5
Let's not turn this into a farce!

To answer the question with some seriousness, I think we have to dissect each probability independently.

First probability: Girl online has the same birthday as your brother.

To find this out, we need to realize that the odds of any two random persons having the same birthday is 1/365 (1/1*1/365).

Now, if we assume that the online population is 2 billion persons, we find that there are roughly 5,479,452 with a birthday matching your brother's.

Now, I assume this girl is an American, so let's say that there are 200 million Americans online (or 10% of the world total). This means that there are 547,945 Americans with your brother's birthday who are online. Now, let's assume that girls are 50% of the population (hardly a point of trouble I should think) and we get 273,972 people who are female, American, online, and share your brother's birthday.

Now, if we assume that 10% of these people are in your age demographic (as I assume this girl is) then we can say that 27,397 Americans are female, online, share your brother's birthday, and are in your age group.

Lastly, if we take 27,397 and divide it by the original online population (2 billion) we get: .0000137.

Now I could continue, but I gather you see my logical process (and fudged numbers).

Joined:
Sep 14, 2007
#6
That's not what the problem was... all you did is list an estimated number of american females who are online, and share my brothers birthday.

My question: What are the chances that ONE of those girls who share my brothers birthday, would have a brother who shares MY birthday?

7. ### swiftaw macrumors 603

Joined:
Jan 31, 2005
Location:
Omaha, NE, USA
#7
If you are asking, what is the chance that of four random people a,b,c,d that a and b would have the same birthday and c and d would have the same birthday (but not the same as a and b), then the answer is

365*364/(365^4) = 364/365^3 (pretending leap years don't exist)

If you are asking what is the problem that a girl you meet has the same birthday as your brother and has a brother who has the same birthday as you, that's tricky, because you have to factor in the chance that she actually has a brother.

8. ### runplaysleeprun macrumors 6502a

Joined:
Jul 27, 2004
Location:
Chicago, IL
#8
I think that is what he's asking, and has stated that she does indeed have a brother who has the same birthday as him.

9. ### swiftaw macrumors 603

Joined:
Jan 31, 2005
Location:
Omaha, NE, USA
#9
So then the question becomes, what is the probability that of four people a,b,c,d, a and b have the same birthday, and c and d have the same birthday. I answered that above.

Joined:
Sep 14, 2007
#10
Can you explain your equation? Thanks.

11. ### swiftaw macrumors 603

Joined:
Jan 31, 2005
Location:
Omaha, NE, USA
#11
The total number of ways that 4 people can have birthdays is 365*365*365*365 = 365^4

The number of ways that a and b can have the same birthday is 365, the number of ways that c and d can have the same birthday (but different than a and b) is 364. Thus, the number of ways that a and b can have the same birthday and c and d can have the same birthday but different to a and b is 365*364

Thus, assuming all days are equally likely, the probability of a and b sharing a birthday and c and d sharing a birthday but different to a and b is 365*364/365^4

12. ### CalBoy macrumors 604

Joined:
May 21, 2007
#12
Well I assumed that online had something to do with it since you mentioned it in your intro. However, seeing as how you wanted this:

I'll defer to his post:
Except, I'll factor in leap years to be fancy.

365.25*364.25/365.25^4 = .000007475304

13. ### swiftaw macrumors 603

Joined:
Jan 31, 2005
Location:
Omaha, NE, USA
#13
Unfortunately, probabilistically, you can't do that. It's much harder to incorporate leap years

14. ### CalBoy macrumors 604

Joined:
May 21, 2007
#14
I know, but it felt so satisfying!

15. ### dantherevelator macrumors regular

Joined:
Oct 8, 2007

Joined:
Sep 14, 2007
#16
I was with you almost till the end..

Why did you divide the 365*364 by 365^4?

Aren't the chances 1 in 365*364 and that's it?

17. ### rdowns macrumors Penryn

Joined:
Jul 11, 2003
#17
Wait a minute here, what time did the train leave the station?

18. ### Brianstorm91 macrumors 65816

Joined:
Sep 30, 2007
Location:
Cambridge, UK
#18
That's the chances of the people having the same birthday, then you need to find the probability of your paths crossing (very hard) and multiply them.

19. ### sushi Moderator emeritus

Joined:
Jul 19, 2002
Location:
キャンプスワ&#
#19
And when will it meet the train coming in the opposite direction.

To the OP, that is a very interesting problem statistics wise. Where's Dr. Q? I am sure he can shed some light on this problem. It's right down his alley.

20. ### freeny macrumors 68020

Joined:
Sep 27, 2005
Location:
Location: Location:

Joined:
Aug 6, 2007
#21

*nods*

Nice.

22. ### swiftaw macrumors 603

Joined:
Jan 31, 2005
Location:
Omaha, NE, USA
#22
No

If all possible outcomes are equally likely, then the probability of any event is the number of outcomes that give you that event divided by the total number of possible outcomes.

For your problem, the total number of possible outcomes is 365^4
The number of outcomes that give you the desired result = 365*364

Thus, assuming equally likely outcomes, the probability is 365*364/365^4

23. ### 0007776 Suspended

Joined:
Jul 11, 2006
Location:
Somewhere
#23
I think that the odds are better that she is not actually a girl and is your actually your brother.

Joined:
Sep 14, 2007
#24
So that's about 7.5^-6, which is a tiny decimal number. But what are the chances? 1 in what? I want an answer written out in that format. 1 in what?

25. ### CalBoy macrumors 604

Joined:
May 21, 2007
#25
You don't know how to use the inverse function on your calculator?

And this wouldn't be for school would it?