A standard, thoroughly shuffled 52-card deck is dealt one at a time to 5 players (players 1 - 5) in standard fashion, until the deck is exhausted.

Using non-brute force methods, show which player is most likely to be dealt the last Diamond in the deck.

The probability is not affected by whether the remaining cards after the last diamond are dealt out, so let's assume all 52 cards are dealt out.  Player 2 will get the last card; player 1 the next to last card; player 5 the one before that, etc.

The probability is not changed by dealing the last card (bottom of the deck) first to player #2, the next bottom to player 1 etc. and then see who gets what in this direction is the first diamond.

Player 2 then has a 1/4 chance of getting a diamond, which then is in fact the diamond in question. That's just on this first go round (equivalent to last in the original methodology).  Player #1 then gets a chance: having 3/4 chance player #2 didn't get it, and then a 13/51 chance of getting that first (last) diamond. Overall that gives player #1 (the "second" to get a card) a 13/68 probability of getting the diamond in question, a lower probability than player #2.  Similarly players 5, 4 and 3 have increasingly lower probabilities of getting the last diamond on that "first" round.

If none of the first five (players 2,1,5,4,3 in that order) get that diamond, another round is gone through, in which again, player 2 has the best chance.  So player 2 has the best chance overall.

Posted by Charlie on 2004-05-25 08:39:17