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

Applejacks01

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:

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:

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:

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 365^5

Code:

For a triplet, there are C(5,3) ways to assign the 
matching day. The # of days are 365*1*1*364*363

Code:

For a quadruplet, there are C(5,4) ways to 
assign the matching day. the # of days are 365*1*1*1*364

Code:

For a quintuplet, there is only 1 way to 
assign the matching day, and there is 365 ways to choose the day.

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

awkward

Have you considered the possibility of one pair and one triplet?

Applejacks01

Quote by awkward

Have you considered the possibility of one pair and one triplet?

Yes! I realized last night I was missing that one! (and now my solution for the 5 person case is correct, thank you)

Okay guys, so I thought all was well, but to really test my combinatoric skills I decided to try the 8 person case. Well...once again there is a discrepancy! I will attach my chart.
Edit: Please ignore "quadrupleruple" lol...I used find/replace to change words around and that happened...oops
Edit 2: I didn't actually sum rows 21-23...oops. Answer is still off by roughly the same margin.

Bacle2

Are the people in your group indistinguishable from each other?

I mean, if you label them 1,2,3,4,5 , are you distinguishing between ,say,

1,2 having the same birthday or 2,4, etc?

Also, in case you're interested in a realistic model --and not just practicing

your counting--there is data that strongly suggests that birthdays are not

distributed uniformly.

Applejacks01

Okay for example:
If 1,2 have the same birthday as 3,4 , then that is considered a quadruple, not 2 pairs.
For a group of 4 people, we can have the following cases:
(0 same): 1 <> 2 <> 3 <> 4
(1 pair) 1=2 and 3<>4 , 1 = 3 and 2<>4, 1=4 and 2<>3. 2=3 and 1<>4, 2=4 and 1<>3, 3=4 and 1<>2
2 pairs 1=2 and 3=4, 1=3 and 2=4, 1=4 and 2=3
(1 triplet) 1=2=3 and 4 is different, 1=2=4 and 3 is different, 1=3=4 and 2 is different, 2=3=4 and 1 is different
(1 quadruplet) 1=2=3=4

And for example, suppose we have 2 birthdays: A and B

_ _ _ _ = 1,2,3,4

There are only 3 ways to have 2 pairs of birthdays:
AABB
ABAB
ABBA

Note that BABA is identical to ABAB (because A can be B and B can be A, but not at the same time).

Does that answer your question?

And yes I am aware, but I am solving this to work on my counting skills really.

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