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#1 2015-12-23 07:28:59

realgains
Member
Registered: 2015-12-23
Posts: 2

Probability question

I want to know what the chances are that someone in a room of 30 has the exact same birthday as ME......

The way I understand the below  calculation, which was sent to me in answer to my question,  is that the chances of ANY PAIR in a   room of 30 that share the exact birthday is 76.6%.
That is comparing ereryone in  room to everyone else.

This was and is my question.....
What I want to know is what are the chances of anyone in the room of 30 having the exact birthday as ME...one specific person in the group?  (not the chances of any PAIR in the group having the same b-day.)



There are 30 people in a room ... what is the chance THAT ANY TWO of them celebrate their birthday on the same day? Assume 365 days in a year.


The chance of not matching:

364/365 × 363/365 × 362/365 × ... × 336/365 = 0.294...


And the probability of matching is 1- 0.294... :

The probability of sharing a birthday = 1 - 0.294... = 0.706...

Or a 70.6% chance, which is likely!

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#2 2015-12-23 12:55:45

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Probability question

Hi;

that the chances of ANY PAIR in a   room of 30 that share the exact birthday is 76.6%.

I think that you mean 70.6% as you stated later.

What I want to know is what are the chances of anyone in the room of 30 having the exact birthday as ME...one specific person in the group?  (not the chances of any PAIR in the group having the same b-day.)

For your question, this probability is given by the formula:

where n is the number of other people. Since you are in a room of 30 people, there are 29 other people.

So we have a 7.6% chance of someone in a room of 29 other people will share your birthday. Welcome to the forum.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2015-12-23 20:10:01

realgains
Member
Registered: 2015-12-23
Posts: 2

Re: Probability question

Yes that is right...thanks!


What about this twist....

What is the probability that one person in that 29  has their birthday on the day of my wedding June 29,1986(month and day, not the year) ...when we view my wedding as ONE date in history and not repeating like a birthday or anniversary?

Not sure it makes a difference ...hmmm

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#4 2015-12-23 20:22:22

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Probability question

The answer should be the same as your birthday.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#5 2015-12-24 00:00:48

Relentless
Member
Registered: 2015-12-15
Posts: 631

Re: Probability question

It will not make a difference, since the relevant date still does repeat yearly. If you wanted to work it out with the year included, you would have to know (or have some idea of) the variance in age of the 29 people, which would require estimates of the average age and the spread of ages.

I'd be more interested in including February 29th into this problem. (:

Last edited by Relentless (2015-12-24 00:04:50)

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#6 2015-12-24 03:42:03

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Probability question

Only a slight adjustment for that.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#7 2015-12-24 03:44:13

Relentless
Member
Registered: 2015-12-15
Posts: 631

Re: Probability question

365.25 instead of 365?

Or 365.2425 if we didn't know which century we were in lol


About 7.642803971% if we know we are talking about 1901 to 2096.
Edit: Actually about 7.627735001% with your equation.

About 7.642954961% if we are dropped into a random time.

Last edited by Relentless (2015-12-24 04:00:34)

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#8 2015-12-24 03:57:41

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Probability question

One of the great things in combinatorics is that we never have to deal with such numbers as 365.25. Positive integers are all we do, a mathematicians dream...


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#9 2015-12-24 03:58:31

Relentless
Member
Registered: 2015-12-15
Posts: 631

Re: Probability question

Edit: I don't understand how you know to just add 1. What would you do with 365.2425?

Okay, while we're at it, how about this:

I'd be interested to see how knowledge of the standard deviation of the age of the group can affect the probability, because of the likelihood of leap year birthdays.

I have an odd fascination for factors that have only a slight effect.

Last edited by Relentless (2015-12-24 04:02:34)

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#10 2015-12-24 04:06:35

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Probability question

The sd is a measure of the spread of the data as you know. What meaning does that have when we are talking about birthdays, for instance:

Jan 10, August 1, June 3, May 29... In what way will you get the sd from that sort of data?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#11 2015-12-24 04:12:19

Relentless
Member
Registered: 2015-12-15
Posts: 631

Re: Probability question

Well, I suppose you could get the sd from the mean date and see things like how the birthdays are distributed in the year (more in Summer or Winter, for instance). You just model the dates as numbers, e.g. January 1 is 1, February 1 is 32, etc.

But I actually meant the sd of the ages in years (maybe the age that everyone turned this year).

Last edited by Relentless (2015-12-24 04:14:55)

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#12 2015-12-24 04:20:22

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Probability question

Hi;

That would be a different type problem than what the OP is asking. Open up another thread and ask the question but please be sure to exactly describe what you want solved for.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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