Good luck.

Quote:FinsRuleSince I'm in HR, I track a lot of employee demographic data. One thing that stood out was that out of our 60 full time employees, there is a span of 24 days that no one has a birthday. My question is, in a group of 60 people what is the average longest span between birthdays one would expect. (Assuming birthdays are randomly distributed throughout the year)

Good luck.

6.083 days

Quote:Ayecarumba6.083 days

That's the minimum. Definitely not the average.

I'm not sure how to figure out the maximum. Brute forcing is not really a good option (365 choose 60 is big, to say the least). It's too late at night for me to see a shortcut if one exists. I'm not sure if this is a well-known problem or not (there may be some function that approximates it)

Quote:AxiomOfChoice365 choose 60 is big, to say the least

It's 365 choose 59, since you must wrap the year. For an average estimate (since no-one cares for the exact solution), simulation should be fast enough.

Quote:MangoJIt's 365 choose 59, since you must wrap the year. For an average estimate (since no-one cares for the exact solution), simulation should be fast enough.

Sure, you can fix one of the birthdays. Actually I'm not sure that the question specified that all birthdays must be different so it's actually 365^59.

Simulation is pointless because the answer to the question doesn't really matter, and it won't give any insight into the actual question.

Anyways, for what it’s worth, a simulation has popped out 27.9 days as the average longest span between birthdays.

Warning: I may be mistaken.Quote:FinsRuleMy question is, in a group of 60 people what is the average longest span between birthdays one would expect. (Assuming birthdays are randomly distributed throughout the year)

This is the continuous equivalent of the problem. The discrete (birthday) problem should be harder to tackle.

Consider a length of 1 (or 360° or 365 days)

Take 59 random draws from a uniform distribution f(x)=1: Xi, i = 1 to 59

Take 0 as the position of the 60th person (or 0th for the computer-minded): x0 = 0

Y=minX is the position of the closest neighbour (in positive direction). The distribution of a minimum is well-known.

f

_{Y}(y) = 59 (1-y)^58

Now, taking each person j in turn, their Yj (next neighbour position) are NOT independent: they are bound by SUM(Y

_{j}) = 1

This creates a 60-variate vector distribution.

We are looking for the distribution of the maximum value among these 60.

We want the distribution of y, given that it is the maximum span: p(y|max)

Bayes gives

p(y|max) = p(max|y) p(y) / p(max)

p(y) = fY(y) is given above.

p(max) is simply 1/60, from symmetry.

This leaves us to find p(max|y): knowing that your span is of length y, what is the probability that it is the maximum span?

It is the probability that all subsequent spans are shorter or equal: P(Yj<=y for all j)

OR... the probability that, on a stretch of size (1-y), the maximum span among 59 is <= y.

We have a recursive formula, here. :-/

Quote:FinsRuleSince I'm in HR, I track a lot of employee demographic data. One thing that stood out was that out of our 60 full time employees, there is a span of 24 days that no one has a birthday. My question is, in a group of 60 people what is the average longest span between birthdays one would expect. (Assuming birthdays are randomly distributed throughout the year)

This is like asking how the losses or wins bunch up. How often you can expect to win, and how often you can expect to lose.

Might be a way to help avoid the long spaces between wins, on average. To build a system around that average wait for another win?