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


by Applejacks01
Tags: 2direct, birthday, probability, problempat
Applejacks01
Applejacks01 is offline
#1
Feb24-12, 10:40 PM
P: 26
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:
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

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

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

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

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?!?!
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awkward
awkward is offline
#2
Feb25-12, 08:53 AM
P: 325
Have you considered the possibility of one pair and one triplet?
Applejacks01
Applejacks01 is offline
#3
Feb25-12, 05:21 PM
P: 26
Quote Quote by awkward View Post
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.
Attached Thumbnails
Birthdayproblem8people.jpg  

Bacle2
Bacle2 is offline
#4
Feb25-12, 06:51 PM
Sci Advisor
P: 1,168

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


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
Applejacks01 is offline
#5
Feb25-12, 06:56 PM
P: 26
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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