Ask the Wizard #37
I was studying your appendix 1 on the blackjack site, and came across something odd. Basic strategy in a six-deck game (with double after split, dealer stays on soft 17 etc. - all the usual Strip rules), dictates that an A,2 against a dealer 5 is a double-down. Yet, on the appendix, the player's expected return is HIGHER if you hit, rather than double (compare .1334 for hitting, vs. .126 for doubling). The same is true for an A,4 v. a dealer showing a 4 (compare .0593 for hitting with .0584 for doubling down). All of the other splits and doubles work out. What's up with these two examples? Thanks in advance.
Fred from San Diego, USA
Appendix 1 is based on an infinite deck. Both hands you mention are borderline plays and the number of decks affects which play is better. For example, A-4 against a 4 favors doubling with 26 decks and hitting with 27 decks. A-2 against a 5 also crosses over somewhere between 8 and an infinite number of decks.
Where is the best casino in Las Vegas to play Spanish 21?
Michael from Philadelphia, USA
The Venetian. To the best of my knowledge they are the only casino in Las Vegas which stands on a soft 17 in Spanish 21, lowering the house edge from 0.76% to 0.40%.
Update: The Venetian later switched to hitting a soft 17. As of this update (May 14, 2013) the best Spanish 21 game is at the D, which allows re-doubling.
I was wondering if you thought continuous shufflers have an effect on basic strategy? I know they speed up the number of hands per hour which is usually bad for the player, but is basic strategy still effective in this instance? Doesn't basic strategy slightly change depending on the number of decks?
Danny from Mission Viejo, California
I first addressed this topic in my December 1, 2000, newsletter. For those who missed it I just added blackjack appendix 10 to my site, which explains the effect on the house edge under both a cut card and continuous shuffler game. To answer your question, no, the basic strategy does not change. Basic strategy is always developed based on a freshly shuffled shoe, which is always the case when playing against a continuous shuffler.
My question has to do with the House edge and element of risk calculations for Casino War for the Casino Niagara Rules (i.e., 3-1 pay out on raise and lose the original wager). How did you come up with these numbers I am currently trying to calculate them? I am having trouble. Thanks for you help.
Mark from Vancouver, Canada
Let's let d be the number of decks. The probability of a tie on the first round is (4*d-1)/(52*d-1)= 0.073955. The probability of a tie in the second round is 12*4*d/(52*d-2)*(4*d-1)/(52*d-3)+(4*d-2)/(52*d-2)*(4*d-3)/(52*d-3) = 0.073974. Lets call p1 the probability of a tie in the first round and p2 the probability of a tie in the second round. Then the player return is p1*(2*p2 +(1-p2)/2*(1-2))= -0.023301. Multiply by -1 and you have the house edge of 2.33%. I hope I didn't go over this too quickly.