(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

What is the probability that given a group of 5 people, at least 2 will share the same birthday?

2. Relevant equations

I know that 1-P(0 matches) = answer, but there is a reason I am going about this the head on approach. I am trying to develop my combinatoric skills.

3. The attempt at a solution

EDIT: I think I am missing some cases. What about the case where we have exactly 1 triplet and 1 pair? Hmm..

OKAY so IF I am correct, then here is what happened in my original work:

When I did the case of 1 triplet, I REALLY did the case of 1 triplet and 2 different birthdays. Which means I was missing the case of 1 triplet and 2 same birthdays!

So there are C(5,2) ways to choose the people to be part of the pair = 10 ways. And similar logic yields the final piece of the puzzle. My solutions are equivalent now. Anybody want to verify?

I am calculating the probabilities of the following to solve this:

0 matches

1 pair

2 pairs

1 triplet

1 quadruplet

1 quintuplet

Here is my table of work:

http://img826.imageshack.us/img826/8428/birthday5peopleproblem.jpg [Broken]

The method is as follows:

Code (Text):No matches has 1 way to assign the lack of matches.

The # of days for first person is 365, 2nd is 364,etc..

Multiply everything together and divide by (365^5) to get the probability

Code (Text):1 pair has C(5,2) ways to assign the pairs.

The # of days for first person is 365. 2nd is 1 way.

3rd is 364, 4th 363, 5th 362. Multiply all together and divide by 365^5

Code (Text):2 pairs has 3*5 =15 ways of combinations. this is derived by observing:

A = pair 1

B = pair 2

C = standalone

AABBC

ABABC

ABBAC

There are 5 ways of assigning the standalone(no match) for each group, hence 15 ways.

The way of choosing days is as follows: 365*1*364*1*363

Multiply that with 15, divide by 365^5Code (Text):For a triplet, there are C(5,3) ways to assign the

matching day. The # of days are 365*1*1*364*363

Code (Text):For a quadruplet, there are C(5,4) ways to

assign the matching day. the # of days are 365*1*1*1*364

Please see my chart, and note that the probabilities differ by an extremely small #. I can't figure out where the error is?!?!Code (Text):

For a quintuplet, there is only 1 way to

assign the matching day, and there is 365 ways to choose the day.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: (Probability) The birthday problem P(at least 2) DIRECT APPROACH

**Physics Forums | Science Articles, Homework Help, Discussion**