(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:

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 - The Fusion of Science and Community**

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

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: (Probability) The birthday problem P(at least 2) DIRECT APPROACH

Loading...

**Physics Forums - The Fusion of Science and Community**