| birthday {stats} | R Documentation |
Computes approximate answers to a generalised “birthday paradox”
problem. pbirthday computes the probability of a coincidence
and qbirthday computes the number of observations needed to
have 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 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.
This formula is approximate, as the example below shows. For
coincident=2 the exact computation is straightforward and may be
preferable.
qbirthday |
Number of people needed for a probability prob that k of
them have the same one out of classes equiprobable labels.
|
pbirthday |
Probability of the specified coincidence |
Diaconis P, Mosteller F., “Methods for studying coincidences”. JASA 84:853-861
## the standard version qbirthday() ## same 4-digit PIN number qbirthday(classes=10^4) ## 0.9 probability of three coincident birthdays qbirthday(coincident=3,prob=0.9) ## Chance of 4 coincident birthdays in 150 people pbirthday(150,coincident=4) ## Accuracy compared to exact calculation x1<- sapply(10:100, pbirthday) x2<-1-sapply(10:100, function(n)prod((365:(365-n+1))/rep(365,n))) par(mfrow=c(2,2)) plot(x1,x2,xlab="approximate",ylab="exact") abline(0,1) plot(x1,x1-x2,xlab="approximate",ylab="error") abline(h=0) plot(x1,x2,log="xy",xlab="approximate",ylab="exact") abline(0,1) plot(1-x1,1-x2,log="xy",xlab="approximate",ylab="exact") abline(0,1)