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.

Let L(n) be the probability of the nth card being the last diamond.

We have to consider sums of L(n) to anwer the question,

i.e., which is the largest

S(2) = L(52) + L(47) + L(42) + ... L(17) or,  //player 2's chance of last diamond
S(1) = L(51) + L(46) + L(41) + ... L(16) or,  //player 1's chance of last diamond
S(5) = L(50) + L(45) + L(40) + ... L(15) or,  //player 5's chance of last diamond
S(4) = L(49) + L(44) + L(39) + ... L(14) or,  //player 4's chance of last diamond
S(3) = L(48) + L(43) + L(38) + ... L(13).     //player 3's chance of last diamond

Fortunately, there are the same number of L(n) terms for each player. And, the heirarchy is:

1/4 = L(52) > L(51) > L(50) > L(49) ... L(13) > L(12) = L(11) = ... = L(1) = 0.

If we accept the heirarchy, then

S(2) > S(1) > S(5) > S(4) > S(3).

All that remains is to prove the heirarchy (which has always been the essence of this problem).

L(m) = 1/4 * (1 - (L(52) + L(51) + ... + L(m+1))).

Since all the terms L(13) .. L(52) are nonzero, as m gets smaller then the sum has more terms and L(52) + L(51) ... + L(m+1) gets bigger. So then as m gets smaller, L(m) gets smaller.

There for the heirarchy is true and player 2 has the best chance
of getting the last diamond.

 

Posted by vectorboy on 2004-05-27 11:54:33