Part a) Has a recursive structure. Let An be the average after the nth card has been pulled. A5=3; An=(4An-1+n)/5.
This can be written explicitly as An = n - 4 + 22n-9/5n-5
A(10)=6+211/55 = 6.65536
A(100)=96+2191/595 = 96.000000001243308091
Part b) Call L(n) the expected lowest card after pulling card n.
L(n)=6n-5/5n-5 holds only up to n=10.
L(10)=65/55 = 7776/3125 = 2.48832
But L(11) is only 46536/15625, slightly less than 46656/15625.
My best attempt at L(100) involved Stirling Numbers of the Second Kind and using excel I know it is very close to L(100)=89.58333334
Not exactly 89 7/12 because the chances of the smallest card being 4,3,2, or 1 are slightly different than what I actually did. Charlie gave the exact rational result. |