We were given an example like this one in class on 9/22/08.
First bar purchased = 1 coupon
$ X_2 $ = # of extra bars for a different coupon
$ X_2 $ is geom$ (\frac{n-1}{n}) $
$ X_3 $ = # of extra bars after 2nd coupon to get 3rd coupon
$ X_3 $ is geom$ (\frac{n-2}{n}) $
$ X_4 $ is geom$ (\frac{n-3}{n}) $
$ X_n $ is geom$ (\frac{1}{n}) $
Avg. # of coupons
E[# needed]=$ \sum_{i=1}^n E[Xi] $
