Consider a deck of 10 cards numbered in order from 1 to 10. Pick up the first five cards (1 to 5). Randomly discard one and take the 6. Randomly discard one again and take the 7. Continue until the 10 has just been taken.

a) What is the expected average of the five cards in the final hand?
b) What is the expected value of the smallest card in the final hand?

Recompute parts a) and b) where you still hold 5 cards, but go all the way through a 100 card deck.

(In reply to re: Some starting thoughts by Jer)

Your methodology may be in error.  There are 126 different combinations of hands that can be made [9!/(5!4!)].  Each of these hand have equal probability.

The sum of these 126 combinations comes to 3780.  Thus, the average value of these 126 hands will equal 30.

For the lowest digit, the number of occurrences for these values are as follows:

1 -- 56

2 -- 35

3 -- 20

4 -- 10

5 --  4

6 --  1

The average expected value for this lowest card is exactly 2.

Posted by hoodat on 2007-07-24 09:37:06