First a
message from David Johnson for proposals on locations for SODA 2012 both in and outside the US.
Here's an interesting approach to the birthday paradox using variances.
Suppose we have m people who have birthdays spread uniformly over n
days. We want to bound m such that the probability that there are are
at least two people with the same birthay is about one-half.
For 1 ≤ i < j ≤ m, let Ai,j be a random variable taking
value 1 if the ith and jth person share the same birthday and zero
otherwise. Let A be the sum of the Ai,j. At least two
people have the same birthday if A ≥ 1.
E(Ai,j) = 1/n so by linearity of expectations,
E(A) = m(m-1)/2n. By Markov's
inequality, Prob(A ≥ 1) ≤ E(A) so if
m(m-1)/2n ≤ 1/2 (approximately m ≤ n1/2), the
probability that two people have the same birthday is less than 1/2.
How about the other direction? For that we need to compute the
variance. Var(Ai,j) = E(Ai,j2)-E2(Ai,j)
= 1/n-1/n2 = (n-1)/n2.
Ai,j and Au,v are independent random
variables, obvious if {i,j}∩{u,v} = ∅ but still true
even if they share an index:
Prob(Ai,jAi,v = 1) = Prob(The ith, jith and vth person all share the same birthday) = 1/n2 =
Prob(Ai,j=1)Prob(Ai,v=1).
The variance of a sum of pairwise independent random variables is the sum of
the variances so we have Var(A) = m(m-1)(n-1)/2n2.
Since A is integral we have
Prob(A < 1) = Prob(A = 0) ≤ Prob(|A-E(A)| ≥ E(A)) ≤ Var(A)/E2(A)
by Chebyshev's
inequality. After simplifying we get
Prob(A < 1) ≤ 2(n-1)/(m(m-1)) or approximately
2n/m2. Setting that equal to 1/2 says that if m ≥
2n1/2 the probability that everyone has different birthdays is at most 1/2.
If m is the value that gives probability one-half that we have at
least two people with the same birthday, we get
n1/2 ≤ m ≤ 2n1/2, a factor of 2
difference. Not bad for a simple variance calculation.
Plugging in n = 365 into the exact formulas we get 19.612 ≤ m ≤ 38.661
where the real answer is about m = 23.
Enjoy the Thanksgiving holiday. We'll be back on Monday.