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
Jer: by hoodat)
The 126 possible hands are not all equally likely.
Each of the first 5 cards has 5 opportunities at which it may be discarded, for a survival rate of (4/5)^5 = .32768. The probabilities of each survival are not independent so, most easy calculations can't itemize the probabilities, but one is easy: If the 1 survives, it will be the lowest. It is .32768 probability of surviving, so out of the 126 possible hands, .32768 * 126 = 41.28768 are only expected to have a 1 (and therefore a 1 low) (rather than 56). The fractional nature of this expected value is due to the fact that order of selection does have an effect on the probability of survival of a given card. The other numbers are also at a disadvantage compared to the 6, which has (4/5)^4 = .4096 probability of survival. The effect is enough to raise the expected value of the lowest card left to 2.48832 for part b and 6.65536 for part a. The latter (part a) is also affected by the greater probability that numbers higher than 6 have an even higher probability of survival than the 6.
Edited on July 24, 2007, 10:51 am

Posted by Charlie
on 20070724 10:44:24 