# What is the real answer to the Birthday question?

"How large must a class be to make the probability of finding two people with the same birthday at least 50%?"

I have 360 friends on facebook, and, as expected, the distribution of their birthdays is not uniform at all. I have one day with that has 9 friends with the same birthday. (9 months after big holidays and valentines day seem to be big ones, lol..) So, given that some days are more likely for a birthday, I'm assuming the number of 23 is an upperbound.

Has there been a better estimate to this problem?

• A sample of 360 persons does not make a large sample for the distribution of birthdays over 365 days of the year... You certainly cannot check for uniformity over such a small sample. – Xi'an Jan 31 '12 at 11:00
• A person has a birthday, what are the odds that a second person doesn't share the same birthday? 364/365, what are the odds that a third person doesn't share either birthday? (364/365) * (363/365). Expand on this until you've got a probability < 50%. It would mean the odds that no one has the same birthday, which would in turn mean that the odds for at least two to share a birthday would be > 50%. – zzzzBov Jan 31 '12 at 15:07
• Are we to assume you have random friends? – James Jan 31 '12 at 15:45
• @zzzzBov - you don't understand what the OP is asking for. This is the approach where we assume each birthday is equally likely, each with chance $\frac{1}{365}$ of being yours. The OP is asking for what the estimate would be when say being born on Jan 1 is not as likely as being born on Feb 15. – probabilityislogic Feb 1 '12 at 1:20

Luckily someone has posted some genuine birthday data with a bit of discussion of a related question (is the distribution uniform). We can use this and resampling to show that the answer to your question is apparently 23 - the same as the theoretical answer.

> x <- read.table("bdata.txt", header=T)
> birthday <- data.frame(date=as.factor(x$date), count=x$count)
> summary(birthday)
date         count
101    :  1   Min.   : 325
102    :  1   1st Qu.:1266
103    :  1   Median :1310
104    :  1   Mean   :1314
105    :  1   3rd Qu.:1362
106    :  1   Max.   :1559
(Other):360
> results <- rep(0,50)
> reps <-2000 # big number needed as there is some instability otherwise
> for (i in 1:50)
+ {
+ count <- 0
+ for (j in 1:reps)
+ {
+ samp <- sample(birthday$date, i, replace=T, prob=birthday$count)
+ count <- count + 1*(max(table(samp))>1)
+ }
+ results[i] <- count/reps
+ }
> results
 0.0000 0.0045 0.0095 0.0220 0.0210 0.0395 0.0570 0.0835 0.0890 0.1165
 0.1480 0.1770 0.1955 0.2265 0.2490 0.2735 0.3105 0.3350 0.3910 0.4165
 0.4690 0.4560 0.5210 0.5310 0.5745 0.5975 0.6240 0.6430 0.6950 0.7015
 0.7285 0.7510 0.7690 0.8025 0.8225 0.8280 0.8525 0.8645 0.8685 0.8830
 0.8965 0.9020 0.9240 0.9435 0.9350 0.9465 0.9545 0.9655 0.9600 0.9665

• Indeed, one can show via Schur convexity, that for any nonuniform distribution of birthdays, the probability of a match is at least as great as in the uniform case. This is Exercise 13.7 of J. Michael Steele, The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities, Cambridge University Press, 2004, pg. 206. – cardinal Jan 31 '12 at 13:16
• @Xi'an: Indeed. Now, if only I knew someone who did book reviews for a high-quality, high-readership stats magazine, I'd suggest they review it to give it higher visibility to statisticians...but where to find such a person... – cardinal Jan 31 '12 at 13:51
• (For those who may be wondering about my immediately preceding comment, it references the fact that @Xi'an is the newly appointed book reviewer for Chance.) – cardinal Jan 31 '12 at 19:36
• @Xi'an, check this out and see what you think: table(replicate(10^5, max(tabulate(sample(1:365,360,rep=TRUE))))). – whuber Jan 31 '12 at 20:12
• It's probably not clear, except to R cognoscenti, that the code in previous comments by @Xi'an and myself simulates the OP's situation. Running it establishes that the chance of 9 or more people sharing a birthday, out of 360 randomly chosen from a uniformly distributed population, is only around 40 out of 100,000. The most likely value for the maximum number of shared birthdays is 5. – whuber Jan 31 '12 at 20:41