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.

Part a) For 10:    6.65536

Part a) For 100:  96  [more accurately 96.000000001]

Odds of having the 10:  1.00

Odds of 9:  0.8       = 0.8

Odds of 8:  0.8 ^ 2 = 0.64

Odds of 7:  0.8 ^ 3 = 0.512

Odds of 6:  0.8 ^ 4 = 0.4096

Remaining odds of 1 - 5: 5 - (1.00+0.8+0.64+0.512+0.4096) = 1.6384 chances in 5

Individual odds of 1,2,3,4,5 = 1.6384 / 5 = 0.32768

Part a) Weighted average values:

[10 x 1.00 + 9 x 0.8 + 8 x 0.64 + 7 x 0.512 + 6 x 0.4096 + (1+2+3+4+5) x 0.32768] / 5 = 6.65536

Part a) for 100 is similar and was done on a spreadsheet.

100: 1.00    99: 0.8   98: 0.8^2  n: 0.8^(100-n)

(1,2,3,4,5): 0.8^95 (same odds)

Edited on July 23, 2007, 3:14 pm

Posted by Leming on 2007-07-23 14:39:02