Displaying 1 result from an estimated 1 matches for "pbirthday2".
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pbirthday
2020 May 22
1
pbirthday() for larger number of classes
...se 0.
For large c, sum(log1p(-(1:(n-1))/c)) is close to 0, so a more robust
version would be
to return if(n >= 2) -expm1(sum(log1p(-(1:(n-1))/c))) else 0 in the
default case 'coincident = 2'
(internally: if (k == 2) ...).
## Auxiliary function *just* considering 'coincident = 2'
pbirthday2 <- function(n, classes = 365) {
c <- classes # as pbirthday()
if(n >= 2) -expm1(sum(log1p(-(1:(n-1))/c))) else 0 # return value
suggested for the case 'if (k == 2) ...'
}
pbirthday (3, classes = 2) == pbirthday2(3, classes = 2) # identical
pbirthday (3, classes = 2^53) ==...