birthday {stats}  R Documentation 
Computes answers to a generalised birthday paradox problem.
pbirthday
computes the probability of a coincidence and
qbirthday
computes the smallest number of observations needed
to have at least a specified probability of coincidence.
qbirthday(prob = 0.5, classes = 365, coincident = 2) pbirthday(n, classes = 365, coincident = 2)
classes 
How many distinct categories the people could fall into 
prob 
The desired probability of coincidence 
n 
The number of people 
coincident 
The number of people to fall in the same category 
The birthday paradox is that a very small number of people, 23, suffices to have a 50–50 chance that two or more of them have the same birthday. This function generalises the calculation to probabilities other than 0.5, numbers of coincident events other than 2, and numbers of classes other than 365.
The formula used is approximate for coincident > 2
. The
approximation is very good for moderate values of prob
but less
good for very small probabilities.
qbirthday 
Minimum number of people needed for a probability of at least

pbirthday 
Probability of the specified coincidence. 
Diaconis, P. and Mosteller F. (1989). Methods for studying coincidences. Journal of the American Statistical Association, 84, 853–861. doi: 10.1080/01621459.1989.10478847.
require(graphics) ## the standard version qbirthday() # 23 ## probability of > 2 people with the same birthday pbirthday(23, coincident = 3) ## examples from Diaconis & Mosteller p. 858. ## 'coincidence' is that husband, wife, daughter all born on the 16th qbirthday(classes = 30, coincident = 3) # approximately 18 qbirthday(coincident = 4) # exact value 187 qbirthday(coincident = 10) # exact value 1181 ## same 4digit PIN number qbirthday(classes = 10^4) ## 0.9 probability of three or more coincident birthdays qbirthday(coincident = 3, prob = 0.9) ## Chance of 4 or more coincident birthdays in 150 people pbirthday(150, coincident = 4) ## 100 or more coincident birthdays in 1000 people: very rare pbirthday(1000, coincident = 100)