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Ask the Wizard #107

I heard the other day that if you’re playing a 6:5 game (or the even money game at Rio), you should double down when dealt a natural to help offset the lower payout. Is this correct? What would the expected loss be for that play?

"Anonymous" .

This would be a terrible play. For example if you doubled on a blackjack against a 5 (six decks dealer stands on soft 17) your expected gain would be 0.622362, according to my blackjack appendix 9I. So even in an even money game this would still be an error costing about 38% of the bet.

Wong states in Professional Blackjack on page 23 the following, "If you get to twelve by 10-2 or 2-10 (where 10 means any 10-count card), and two or fewer decks are being used (or seven or fewer if the dealer stands on soft seventeen), you should hit." Is that correct? I can see it for a one or two deck game where composition-dependent strategy has a certain amount of value to it, but he’s saying that you should hit a 10-2/2-10 when SEVEN decks are used (S17)! That doesn’t sound right to me.

"Anonymous" .

Wong is referring to a player 12 against a dealer 4 and is quoting The Theory of Blackjack, page 176, by Peter Griffin. Yes, he is right. In a seven deck game the expected value by hitting is -0.210820 and standing is -0.211106, so hitting is higher. However with eight decks hitting is -0.2111161 and standing is -0.211100, so standing is higher. This is such a borderline play that the number of decks does make a difference between seven and eight. Here is an even better example. With A-4 against a 4 you should double all the way through 26 decks but hit with 27 or more.

I go to Vegas once a year and enjoy playing Pai Gow poker there because all casinos near me do not allow banking. My question is: How big of an etiquette breach is it for players to pull away bets and not play a hand when someone decides to bank. This has happened to me often (usually at smaller casinos, Sahara, etc...) and usually players say "If I wanted to give another player my money I’d play in the poker room" This really bothers me and I just wanted your thoughts.

"Anonymous" .

This would make me furious too. While not banking it should not make any difference who is banking. I have never heard of an etiquette rule written about this situation but it falls under a breach of common courtesy in my opinion.

Hello, I just wanted to say that I love your site, and I have a question. I just got back from a trip to Vegas and noticed that some people bet way too high for their available bankroll, for example they will join a $5 minimum roulette table with $20. This greatly increases the house edge, because if that person loses the first 4 games, they have lost the entire bankroll and can no longer play. Using the standard deviation of a $5 game, can you calculate the minimum bankroll needed, so that say 95% of the time you can cover natural losing streaks? You did something similar on your betting systems page when you put a cap on the max bet for a Martingale better. How does the house edge change when a regular bettor only starts with $20, or $40, or whatever?

"Anonymous" .

Thanks for kind words. The house edge is always the same in any game given the same rules and skill level of the player. The bankroll and betting strategy do not matter. Even if I sat down at a $5 game with $5 with the goal of winning $1,000,000 the house edge would still be the same. Although my probability of succeeding is low my worst casino scenario is nothing compared to the best case scenario.

How many "successful" card counters, i.e. the ones who do it right, do you think are in a casino on any given night?

"Anonymous" .

In would say in a large Strip type casino the number of counters who know what they are doing on a given night is in my best guess one half of a single person (or two casinos would have one person). The reason I think it is this low is in my many hundreds of hours at the blackjack tables I only spotted other counters twice.

In the initial two cards can you tell me what the odds are of receiving 7 hands of Ace King or better at hold’em in 35 hands?

"Anonymous" .

The probability of receiving ace/king is (8/52)*(4/51) = 0.012066. The probability of receiving any pair is (3/51) = 0.058824. So the probability of a pair or better is 0.07089. The probability of receiving exactly seven hands of ace/king or better is combin(35,7)*(.07089)^7*(1-.07089)^28 = 0.00772. To work out the probability of 7 or more we would have to go through a total of 7 to 35 one at a time. This adds up to 0.010366551.