

If a company gives holidays on people's birthdays, how many people will it hire? - strategy
http://mindyourdecisions.com/blog/2011/06/07/birthday-laws-probability-puzzle/

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patio11
If you don't immediately knock out a 13 line Ruby script to solve this with a
Monte Carlo situation, and the calculus gets heady, try this hypothetical:

a) A special act of Congress has made the year only one day long. How many
employees would you hire?

b) Congress extends the year by one day. Think quick: hire, fire, or no
action?

c) Congress extends the year by one day. Think quick: hire, fire, or no
action?

d) Congress extends the year by one day. Think quick: hire, fire, or no
action?

...

zzzz) Congress extends the year by one day. Think quick: hire, fire, or no
action?

~~~
jules
I don't see how that line of thought helps you solve this problem. Can you
elaborate?

~~~
patio11
It's the start of a very, very informal argument that the optimum number of
randomly assigned objects for any N containers to maximize the number of
individually packed containers is N. This problem asks you for what happens if
N = 365.

~~~
jules
It's just that I don't see the pattern:

a) A special act of Congress has made the year only one day long. How many
employees would you hire?

It doesn't matter, because the people you hire will have their birthday on
this day.

b) Congress extends the year by one day. Think quick: hire, fire, or no
action?

Hiring one person gives you 1 workday. Hiring two people gives you 2 workdays
with probability 1/2 and 0 workdays with probability 1/2, so the expected
value is 1 workday. Hiring 3 people gives you 3 workdays with probability 1/4
and 0 workdays with probability 3/4. So that gives you expected 3/4 workdays.

c) Congress extends the year by one day. Think quick: hire, fire, or no
action?

    
    
        1 person: 2 workdays
        2 people: 1/3*4 + 2/3*2 = 8/3
        3 people: 1/9*6 + 6/9*3 = 8/3
        4 people: 1/27*... = ...
    

d) Congress extends the year by one day. Think quick: hire, fire, or no
action?

That is getting complicated. Obviously I'm missing something here. Can you
reveal the insight?

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jules
With n employees the probability that any given day is a working day is
(364/365)^n. So in a year you get 365n(364/365)^n man working days. Maximizing
this continuously gives uniquely n=364.5, so either n=364 or n=365 is the best
we can do. Computing both it turns out it doesn't matter which.

~~~
euroclydon
I would think the probability that any given day is a work day is
((364/365)(5/7))^n

~~~
jules
Well, if you're thinking about weekends it is (364/365)^n*5/7, so it doesn't
make a difference for the maximum.

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esrauch
When I read the naive comments that didn't consider the birthday problem
properly I scoffed, but I actually think they may have accidentally had some
insight that the "correct" answers didn't.

The value with the highest expected value of man hours worked would be to hire
365 people, but that might not actually be the best strategy since the
variance is going to be pretty high. It's possible that you would be better
off with a lower expected value that is also more of a "sure thing" venture

~~~
delinka
And not one of them considered salaries. For example, the answers that said
182 or 183 would get more work for their money hiring 182 employees. But that
also assumes everyone makes the same salary. Differing salaries also
complicate matters.

Adding complexity is generally not the purview of the theoretical
mathematician, e.g. "OK, imagine every horse is identical and spherical..."

~~~
jules
The story is made up to serve the mathematical problem, not the other way
around.

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ohyes
Don't hire any employees, just use contractors.

You get the added benefit of not having to pay for their
healthcare/dental/other benefits.

