FinsRule
FinsRule
  • Threads: 128
  • Posts: 3914
Joined: Dec 23, 2009
July 19th, 2014 at 12:36:57 PM permalink
Since 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.
OnceDear
OnceDear
  • Threads: 63
  • Posts: 7477
Joined: Jun 1, 2014
July 19th, 2014 at 12:58:52 PM permalink
You'd better bias your recruitment towards candidates with the correct birthdays, or you might get sued for some kind of discrimination.
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
Ayecarumba
Ayecarumba
  • Threads: 236
  • Posts: 6763
Joined: Nov 17, 2009
July 19th, 2014 at 10:23:16 PM permalink
Quote: FinsRule

Since 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
Simplicity is the ultimate sophistication - Leonardo da Vinci
AxiomOfChoice
AxiomOfChoice
  • Threads: 32
  • Posts: 5761
Joined: Sep 12, 2012
July 20th, 2014 at 3:15:58 AM permalink
Quote: Ayecarumba

6.083 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)
MangoJ
MangoJ
  • Threads: 10
  • Posts: 905
Joined: Mar 12, 2011
July 20th, 2014 at 3:22:36 PM permalink
Quote: AxiomOfChoice

365 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.
AxiomOfChoice
AxiomOfChoice
  • Threads: 32
  • Posts: 5761
Joined: Sep 12, 2012
July 20th, 2014 at 3:28:51 PM permalink
Quote: MangoJ

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.



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.
PeeMcGee
PeeMcGee
  • Threads: 1
  • Posts: 115
Joined: Jul 23, 2014
July 23rd, 2014 at 11:07:31 AM permalink
Hey everyone, my first post here…I’ve been enjoying the site and these forums.

Anyways, for what it’s worth, a simulation has popped out 27.9 days as the average longest span between birthdays.
beachbumbabs
beachbumbabs
  • Threads: 100
  • Posts: 14265
Joined: May 21, 2013
July 23rd, 2014 at 11:14:45 AM permalink
Hi ya, PMG! Thanks for the sim.
If the House lost every hand, they wouldn't deal the game.
kubikulann
kubikulann
  • Threads: 27
  • Posts: 905
Joined: Jun 28, 2011
August 1st, 2014 at 1:04:57 PM permalink
Quote: FinsRule

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)

Warning: I may be mistaken.

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.
fY(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(Yj) = 1

This creates a 60-variate vector distribution.
We are looking for the distribution of the maximum value among these 60.
Reperiet qui quaesiverit
kubikulann
kubikulann
  • Threads: 27
  • Posts: 905
Joined: Jun 28, 2011
August 1st, 2014 at 1:22:29 PM permalink
I see a Bayesian approach, along these lines.

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. :-/
Reperiet qui quaesiverit
Actuarial
Actuarial
  • Threads: 12
  • Posts: 105
Joined: Apr 23, 2014
August 1st, 2014 at 7:14:13 PM permalink
I think you can use the exponential function as an approximation for the maximum inter-arrival time of birthdays. Based on a uniform average distribution of birthdays, theta = 365/60 = 6.083. Since you are approximating a discrete distribution, I used the percentile = 59.5/60. If my actuarial exam skills of yesteryear are correct, you're looking at approximately 28 days as an average.
Kerkebet
Kerkebet
  • Threads: 1
  • Posts: 362
Joined: Oct 2, 2014
October 2nd, 2014 at 3:37:43 PM permalink
Quote: FinsRule

Since 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?
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
  • Jump to: