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"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?

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3  
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
6  
Are we to assume you have random friends? –  James Jan 31 '12 at 15:45
1  
@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

1 Answer 1

up vote 16 down vote accepted

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
 [1] 0.0000 0.0045 0.0095 0.0220 0.0210 0.0395 0.0570 0.0835 0.0890 0.1165
[11] 0.1480 0.1770 0.1955 0.2265 0.2490 0.2735 0.3105 0.3350 0.3910 0.4165
[21] 0.4690 0.4560 0.5210 0.5310 0.5745 0.5975 0.6240 0.6430 0.6950 0.7015
[31] 0.7285 0.7510 0.7690 0.8025 0.8225 0.8280 0.8525 0.8645 0.8685 0.8830
[41] 0.8965 0.9020 0.9240 0.9435 0.9350 0.9465 0.9545 0.9655 0.9600 0.9665
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7  
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
2  
@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
2  
(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
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@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
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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

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