- #1

- 16

- 0

**Birthday Problem (need help on combinatorics...) URGENT!!**

Question:

There are ''n'' ppl in a room. What's the probability of 2 or more ppl sharing the same birthday?

My Answer: (Looking at 2 ways...)

Let E be Event of 2 or more ppl sharing the same birthday...

then: P(E) = 1 - P(E')

1) Conditional Probability:

n P(E')

1 1

2 364/365

3 364/365 * 363/365

.

.

.

n (364!/365n)/(364-n+1)!

it works for 0<n<366

this is definitely right...

but my combinatorics answer doesn't agree...

2) Combinatorics:

P(E') = 365C(n)/366C(n)

Top part of fraction gives me the no. of ways that n ppl can have completely different birthday.

Bottom part gives me the total no. of ways of possible birthday that n ppl can have.

if my appraoach is not clear for bottom part... try using the analogy of a question: there are 2 balls and 5 baskets, what are the total no. of ways that the balls can be distributed into 5 baskets providing basket can hold 0,1,2 balls:

try putting the basket space as | | | | | |

there are 5 spaces between those lines...

then using ''.'' to represent balls:

so balls can be distributed in these ways:

| . | . | | | |

| . | | . | | |

| . . | | | | |

| | | | . | . |

| | | | . . | |

etc etc...

notice if u ignored the outward most lines:

|( . | . | | | )|

|( . | | . | | )|

|( . . | | | | )|

|( | | | . | . )|

|( | | | . . | )|

etc etc...

u get 6 objects within the brackets and u're actually only trying to fit 2 ''.''s in those 6 objects, hence u get 6C2 for the no. of possible arrangement.

using the same analogy, there are 365 days and ''n'' ppl, the total no. of ways to fit ''n'' ppl in 365 days should be (365-1+n)C(n) = (364+n)C(n)

that's how I get the bottom part. but then... if you try for n = 2

it doesn't fit with the conditional probability...

WHYYYYYYYYY!!!!!!!!!!!!!!!!!?????????????