Very high for some reason

>>8077255
Depends on the probability distribution which is not specified here.

>>8077255
Homework goes on >>>/hm/

>>8077255

50/50

Either they do or they don't.

Is being in a room with them relevant? How does it affect the odds?

If your birthday is 9 months after christmas: pretty high

very close to 1

https://en.wikipedia.org/wiki/Birthday_problem

>>8077387
just realized i misread the question, but the distribution i posted is still valid.

Do you have a link to the data of birthday distributions? Without that data I couldn't compute the probability.

>>8077255
1/365 (equally probable for each day)
* 7 (3 days forward, 3 days back, including your birthday)
* 60 (amount of people)
/ 2 (two people needed)
1/365 * 7 * 60 * 0.5 = 57.5%

>>8077255
exactly 2?

>>8077284

It's randomly distributed, dingdong.

[spoiler]Unless we're talking about a city that won a Super Bowl within the last year kek[/spoiler]

>>8077434
Are we ignoring geographic location, culture, religious holidays, secular holidays, historic events, the backgrounds of the people in the room, the reason the people are in the room, time period, whether the location has population control, contraception, age groups, residence s of the people in the room, race and ethnicity, education, income, astrological signs, economic systems, etc?

>>8077255
(60 choose 2) (7/365)^2 (358/365)^58 ~ 0.21

>>8077366
9 months after /02/14

>>8077548
>(7/365)^2 (358/365)^58
what does this mean ?

>>8077455
obviously

>>8077548
>gives a probability when OP asked for odds

>>8077548
oh if its exactly two. question a bit unclear

>>8077434
I'm sure it's not randomly distributed
You could assume so to answer the question, but I definitely believe that there are peaks for certain dates

>>8077660
Anon treated the problem as seeking probability of random variable X=2, where X follows a binom. distribution. The power of 2 item is the probability of two 'successes' and power of 58 is the probability of 58 'failures'.