Ask The Wizard #105

Thank you for the compliment. Assuming the dealer hits a soft 17 you are adding 0.012% to the house edge by playing 4-8 deck strategy in a two deck game. Playing double deck strategy in a 6 deck game costs 0.008%. To take this question further I wondered about a more extreme case of playing 4-8 deck strategy for the dealer standing on soft 17 in a single deck game where the dealer hits a soft 17. In this situation the incorrect basic strategy adds 0.038% to the house edge.

According to Cecil Adams the term originates from restaurants and soda fountains of the 1920s. He says it started out meaning to be out of something and then became an expression to drive off a customer.

p.s. In December 2004 another reader wrote with another explanation. According to inspirationline.com the term originates from a restaurant named Chumley’s at 86 Bedford Street in Greenwich Village, New York City. It started as a warning to leave the building because the police were coming and evolved to mean to get rid of something.

Yes, the probabilities are the same. Seven random cards out of 52 have the same odds regardless of how they are taken out of the deck or whom you share them with.

(12/52)*(11/51)*(10/50)*(9/49)*(8/48) = 0.00030474, or about 1 in 3282.

The reason is if the dealer gets a 22 and you have 21 or less then the hand pushes. This works strongly to the dealers favor and should be a disincentive to put more money on the table by doubling or splitting.

It is my policy not to count in Las Vegas. Since I live here I don’t want to make any enemies out of prospective clients. So I am allowed to play blackjack at all but two casinos locally. However last January I went to Reno and Lake Tahoe for a few days and was told not to play blackjack at four different casinos.

The probability of a four of a kind in any given hand is 13*48/combin(52,5) = 0.0002401. Let’s assume in two hours you can play 120 hands. The probability of exactly two four of a kinds would be combin(120,2) × 0.0002401² × (1-0.0002401)¹¹⁸ = 0.000400095 = 1 in 2499.41.

Generally speaking 50 and 100 play machines have lousy pay tables and thus should be avoided. However assuming you did find a decent pay table ask yourself what you would play on single play and then divide that by 50 or 100. For example if you play the $1 single line machines then you should play 2 cent 50 line or 1 cent 100 line games.

No. Some might argue that it would take an infinite number of losses to lose in this situation, which would be impossible. The truth is that 0.5^infinity approaches 0 but does not equal zero. If this did happen you would lose $2^infinity. The expected return of this strategy is thus 1- $2^infinity * 0.5^infinity = $1 - 1 = 0. Another more graceful way to look at is that as your bankroll increases the expected value still remains unchanged at zero. So the limit of the expected value as the bankroll approaches infinity is zero. In other words an increasing bankroll doesn’t help your odds, even if it goes to infinity.

Griffin is of course correct. The expected value of hitting is -0.552613 and standing is -0.535787. Some plays I don’t list because they are either so obscure I didn’t find them or so unlikely I didn’t bother to list them.

Yes! If the player had the advantage a betting system could not help but work in the long run. The reason is the house/player advantage is immutable. Betting systems can not change it.