In a room full of 30 people, what is the probability that at least 2 of them have the same birthday?
@ganeshie8 @dan815
day as in `monday, tuesday,... ` ?
as in day of the month
30 people where atleast 2 out of the 365 numbers common
what dan said
hold i gota move my laptop first its over heating
so 1/1/1990 and 1/1/2000 are considered same birth days ?
yep
I can tell that it's more than 50% because I only take 23 people for that to happen
if you search birthday paradox you see how it's calculated. The idea is just to do 1 - P(no one share a same birthday)
it looks like a cool question to think about
the goal is to come to the answer without the internet
so you're saying find the prob that none of them share brithdays
yes, it's just 1 - (365/365) * (364/365) * (363/364) * .. all the way to the 30th person
typo 1 - (365/365) * (364/365) * (363/365) * .. all the way to the 30th person
\[\sum \limits_{k=2}^{30}\binom{30}{k} \left(\dfrac{1}{365}\right)^k \left(\dfrac{364}{365}\right)^{30-k}\]
i was thinking it was like that too but is there a closed form for this ganeshie
i feel like theres a nice function or something that gives the sum of that
its a binomial distribution right ? p = 1/365 n = 30 k >= 2
yeah
true okay i get it!
we can integrate the normal distribution curve
yes thats the only simple way i guess ! we can approximate it by converting it to normal curve
ok cool
@geerky42 are you @usukidoll?
@dan815 you could also go back to a permutation by selecting 30 birthdays out of the 365 and assign them to the people which is 365p30 and then divide it by the total possible ways of assigning a birthday to someone?
that of course is the compliment event where everyone has a different birthday but we can subtract it from 1?
if no one share a same birthday, the complement of that is at least two people share a same birthday. So we subtract from 1. If fact, the formula is Pr = 1 - (365Pn) / 365^n in the case of 30, Pr = 70.6%
Yup ^
yes exactly i am curious as to how @ganeshie8 and @dan815 were going to solve it
Forget it, its not a direct binomial distribution - need to tweak some more >.<
No, I am not? @dan815
@dan815 I AM OVER HERE s t h u
@usukidoll :O
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