I promised that I do the counting for him. ;-) BTW second raffle is complete moneywise.
Leelou... that looks like a doable solution for the next (third) raffle ... but somehow I still have the impression that it is not mathematically perfectly balanced out, because one still would have to set a +1 +5 +10 +15 +25 offset (in this example for 5 winners) after the modulo operation by hand. So it would rule out certain winner combinations and thus not being truly random. In my example two consecutive participants could never win simultaneously (for example ticket number 5+6).
I created the 2nd Participation thread here.