We were given an example like this one in class on 9/22/08.
First bar purchased = 1 coupon
X2 = # of extra bars for a different coupon
X2 is geom[(n-1)/n]
X3 = # of extra bars after 2nd coupon to get 3rd coupon
X3 is geom[(n-2)/n]
X4 is geom[(n-3)/n]
Xn is geom(1/n)
Avg. # of coupons
E[# needed]=
