SQL Binomial Distribution/Arithmetical Overflow - sql

Question

I created a function for a cumulative binomial distribution. It works well for extremely modest sample sizes, but I get a arithmetical overflow on larger samples.
The largest culprit is the n!. In excel 170! = 7.3E+306. 171! = #NUM!
Excel has an internal function that calculates binomial distribution, and it works with ns much, much larger than 170.
Is there something I can do to limit the magnitude of the #s generated?
EDIT: I played with this
SET #probout = 2*3*4*5*6*7*8*9*10*11*12
Worked fine
SET #probout = 2*3*4*5*6*7*8*9*10*11*12*13/10000
Resulted in overflow
Function below.
ALTER FUNCTION [dbo].[binomdist_cumulative]
(
#n int
,#k int
,#p float
)
RETURNS float
AS
BEGIN
-- Local Variable Declarations
-- ---------------------------
DECLARE #kfac float
,#nfac float
,#nkfac float
,#i float
,#f int
,#probout float
SET #i = 0
SET #f = 0
SET #nfac = 0
SET #kfac = 0
SET #nkfac = 0
SET #probout = 0
WHILE #i <= #k
BEGIN
--k!
SET #f = #i-1
SET #kfac = #i
IF #kfac > 0
BEGIN
WHILE #f > 0
BEGIN
SET #kfac = #kfac*#f
SET #f = #f -1
END
END
ELSE
BEGIN
SET #kfac = 1
END
--n!
SET #f = #n-1
SET #nfac = #n
IF #nfac > 0
BEGIN
WHILE #f > 0
BEGIN
SET #nfac = #nfac * #f
SET #f = #f -1
END
END
ELSE
BEGIN
SET #nfac = 1
END
--(n-k)!
SET #f = #n-#i-1
SET #nkfac = #n-#i
IF #nkfac > 0
BEGIN
WHILE #f > 0
BEGIN
SET #nkfac = #nkfac * #f
SET #f = #f -1
END
END
ELSE
BEGIN
SET #nkfac = 1
END
--Accumulate distribution
SET #probout = #probout + #nfac/(#kfac*#nkfac)*POWER(#p,#i)*POWER(1-#p,#n-#i)
SET #i = #i+1
END
RETURN #probout
END

Answer 1

Let me give you a hint.
If you calculate the full factorials, you are quickly going to get overflows. If you do an incremental calculation, the you won't.
For instance, instead of calculating (5 // 3) as (5*4*3*2*1) / ((3*2) * (3*2*1)), calculate it as: (5 / 3) * (4 / 2) * (3 / 3) * (2 / 2) * (1 / 1). . . oh, wait, you can see that the last three terms are all "1".
To be clear, you want to calculate the product of:
((n - i) / (n - k - i)
For i between 0 and k - 1. That is, you are dividing the product of k consecutive numbers ending in n with k consecutive numbers starting with 1.
You'll see that this incremental approach will forestall the issues with overflow.