# Thread: probability two dates will be greater than or equal to a given number of dates apart

1. ## probability two dates will be greater than or equal to a given number of dates apart

So...I feel like an idiot.

I'm doing a multi part problem where it appearsI have the second and third part correct, but the first part incorrect, which the next two parts hinge on.

I'm not replicating the problem here, but the kind of problem it is. Can any one school me on how to do this, because I am ready to smack my head with a hammer.

Example: What is the probability that two graduation dates, independently and uniformly distributed are greater than or equal to 40 days apart. Assume 365 days in a year N = infinity

Example 2: What is the probability that two random holidays, independently and uniformly distributed, are greater than or equal to 10 days apart. Assume 365 days in a year

Example 3: What is the probability that two alien people will have egg days (the day they burst from an egg) that are greater than or equal to 30 days apart. Assume a human year of 365 days.

These should all basically be the same problem and if they aren't then its my mistake because I made them up based on the problem that is slowly killing me.

Assume the first date is X, the second is Y so for each problem
absolute value of [X-Y] > selected amount (ie 40, or 10, or 30, etc.)

PS. I have been all over google, and the results being returned are the birthday paradox which is not what I am asking. I'm not asking what is the probability that two people will share the same birthday. Also I ahve been to Khan academy.com to watch videos, Wolframalpha.com, several other statistics forum, looked in my textbook, looked in statistics for dummies, looked in a public health statistics text book, looked in the statistics problem solver, tried to brain think it out.

I came up with an answer ((selected amount - 1) x 2)/365, but I don't think this is right.

2. ## Re: probability two dates will be greater than or equal to a given number of dates ap

So, after thinking about it last night...dreaming about it...and then when I got up, I decided the reason I don't feel comfortable with this is because I'm leaving out the fact that the first birthday and the second birthday could be on the same day.

Also, I made a mistake in the upper post. I mean to say 1-the equation I selected.
So I'm going to put what I decided is the final answer here, so if other's google it (I didn't come up with anything) they have some sort of guide, with a caveat, that I could be completely wrong, and this could be a more complex problem and I just don't know how to do it.
So I would guess that the probability actually equals 1- (((selected amount x 2) -1) /365).
i.e. the first problem would be 1-(((40 x 2) -1)/365) = 78.4%
the second would be 1- (((10 x 2) -1)/365) = 94.79%
the third would be 1-(((70 x 2) -1)/365) = 61.92%

If you are worried about doing my home work for me (like maybe these are my home work problems - even though I guarantee they aren't), the problem set is due today, so if you jump in next week or next month, I think it would still do people good to know if this is the right way or wrong way to do it. For myself, my teacher doesn't tell me where I went wrong and how to fix it, just marks it right or wrong, and I can't learn to correct it that way, so I'm hoping eventually someone will remark if this is how you do this type of problem.

This feels like the right way to do it, but perhaps it is more complicated and I just don't know enough to see the extra steps.

Now I'm on to figuring out what would the probability be of any three dates being greater or equal to x days apart. Yippee. I'm thinking since the days are independent this is an addition problem, not a multiplication problem.