Ask the Wizard! (No. 5)
March 4, 2000 column
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?
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.
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.
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?
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?
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?
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.
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