In a card game for four players, a pack of fifty-two cards is dealt round so
that each player receives thirteen cards. A hand that contains no card greater than
nine is called a yarborough. How many deals are necessary for the probability
of at least one hand being a yarborough to be greater than (1/2) ? (Ace ranks high)
1+1+1+1+1+1+1+....=S_n
S_n =1+1+1+1+1+1+1+....
2S_n = 2 + 2 + 2+.......
.......... subtracting
−S_n =1−1+1−1+1−1+1−1+1−1+....
−S_n =(1/2)
S_n =−(1/2)
I have found this while experiment . I know the sum diverges
but is it pretty cool?
Kindly rectify me if there is any fault on this non rigorous
process
I have found some Ramanujan proof
S_n =1+2+3+4+5+6+7+...
4S_n = 4+ 8 + 12+...
−3S_n =1−2+3−4+5−6+7−8+......
−3S_n =(1/4)
S_n =−(1/(12))
Ramanujan had done this on his notebook