Problem suggested by Alan Murray

[George Berzsenyi]

Two people are playing basic 21 with a known sorted deck, i.e. the cards always follow the sequence ...,A,2,3,4,5,6,7,8,9,10,J,Q,K,A,2,3,4,... indefinitely. (By basic, I mean no naturals or splitting or insurance or doubling down, just highest score <=21 wins.) If the second player (the Dealer) must hit on 16 or lower and stand on 17 or higher, show that the first player has the advantage and determine what fraction of hands that player can be guaranteed to win. If the Dealer is free to play any strategy, show that the Dealer has the advantage and determine what fraction of hands the Dealer can be guaranteed to win. 

[David Ash]

If the dealer is forced to draw on 16 and stand on 17, and I'm assuming there are no exceptions for soft 17's, then I believe the best strategy is for the player to play A-2-3-4-5-6 for a score of 21. This forces the dealer to bust w/7-8-9 leading to a push with 10-J for the player and Q-K for the dealer. The cycle repeats so the overall fraction of victory, with one win and one push each time through, is 3/4.

If the dealer is free to play any strategy, I'm going to assume that truly does mean free to play any strategy, including not to draw any cards at all if he knows he will lose anyways and any draw would be disadvantageous.

Best play is for player to start with 7. Dealer wins at least the next two hands e.g. 9-10 beats 8-7 and then K-A beats Q-J. On the next hand the dealer draws up to the '6' card, not drawing at all if necessary--losing that hand and then setting the player back to starting with 7 and repeating the cycle. Overall dealer wins 2/3 of the time.