Birthday problem using Bernoulli trails

#1
I came up with this problem:

What is the approximate probability that no two people in a group of seven have the same birthday?

So, I set up like this:

P(no 2 ppl have same birthday)
= P(exactly 1 person born on Jan 1st) + P(exactly 1 person born on Jan 2nd) + ...
= 365*P(exactly 1 person born on Jan 1st)
= 365*C(7,1)*(1/365)^1*(364/365)^6

Where C(7,1) is the binomial coefficient of 7 and 1, and a person have 1/365 chance of being born on Jan 1st, and another person have 364/365 chance of not being born on Jan 1st.

But calculating that answer turned out to be about 6.89, which got me completely puzzled because a probability shouldn't be over 1. Am I doing something wrong?
 

Dason

Ambassador to the humans
#2
I don't understand your logic - at all.

The typical way this is done is via conditional probability. Basically you look at each person in order and then find the probability of the event not occuring given the information you already know.

So start with person 1: It doesn't matter when their birthday is so they can have any birthday - Probability: 1
Person 2: They just have to not share the same birthday as person 1 - Probability: 364/365
Person 3: They just have to not share the same birthday as person 1 and 2 - Probability: 363/365
...
Person 7: They just have to not share the same birthday as persons 1, 2, 3, 4, 5, 6: Probability - 359/365

Multiply those all together to get the probability that nobody shares the same birthday
 
#3
I know that, and I was wondering if I can do it just with Bernoulli trails. I just wanted to know what went wrong.

Something just appeared in my mind. The method I just did has double counting because P(person born on Jan 1st) can have two people having same birthday on Jan 2nd.

So I came up with different method:

= P(people born on Jan 1, 2, 3, 4, 5) + P(people born on Jan 1, 2, 3, 4, 6) + ...
= C(365,7)*P(people born on Jan 1, 2, 3, 4, 5)
= C(365,7)*(7!/(1!*1!!*1!!*1!!*1!!*1!!*1!)*(1/365)^7 = 0.9438

Which sounds very reasonable.
 

Dason

Ambassador to the humans
#4
You're only specifying 5 days in your probability though? I'm guessing that is just an oversight?
 
#5
No, what I did is to count all possible 7 different birthdays from 365 days, which is C(365,7). Also, I used multinomial trails instead of Bernoulli trails.
 

Dason

Ambassador to the humans
#6
I'm referring to this:
= P(people born on Jan 1, 2, 3, 4, 5) + P(people born on Jan 1, 2, 3, 4, 6) + ...
= C(365,7)*P(people born on Jan 1, 2, 3, 4, 5)

You only specify 5 days. Like I said I'm guessing it's an oversight...