Hi,
Thanks for taking the time to explain a little more your example. I
believe now I understand your concerns better.
Most people, of course, want to make money gambling and, if they
gamble a lot, they want to make sure they profit. We all know only a
few games can satisfy that goal, games such as Sports betting if you
have certain information, blackjack under certain conditions, video
poker under certain conditions. I believe most video poker sites are
geared towards people with the play for profit goal.
I agree also that many people play negative games; I'm sure the great
majority of people do. I certainly do, and my playing strategy, if I
had to characterize it according to what is on the vpFREE Dogma,
would be of the type "anecdotic evidence" strategy.
You have chosen to be more careful and to devise a strategy such as
to maximize your chances of achieving a goal when risking a certain
amount. In the example occupying your last posts, you want to risk a
dollar, make $36, and you have to decide between two games with
different rules and paybacks. You choose the game that gives you the
highest probability of achieving your goal.
Your example, of course, has practical interest for a guy who goes to
a casino every Sunday and wants to gamble $1, and quit when he either
wins $36, or loses the $1. The joint he goes to only has craps and
roulette. If he were to choose the game with highest ER (I'm using
Dogma terminology), he would choose craps, but you have determined
that would be wrong; by choosing roulette, at the end of the year he
would be better off. This is completely true.
How can that happen? You can think of your two games as two machines.
Roulette is a machine that takes a dollar and either gobbles it or
gives you thirty-six dollars. Craps takes the dollar and also either
gobbles it or gives back thirty-six dollars. Your risk and goal
conditions have effectively changed the ER of the craps machine. For
your risk and goal conditions, the ER of roulette is the same,
94.74%, and that of craps is now 92.6%. After you determined what the
modified ER is, your strategy then consists of choosing the game with
the highest ER.
I do admit I have a great deal of confusion with the terminology used
in these posts. I believe now I know what people mean by ER and EV in
the posts I have read, but in a future post, here or elsewhere, when
those terms are used, if the author does not make it clear from the
content what he is talking about, I might misunderstand him since I
don't have telepathy and my guess might be wrong. The terminology
used for strategies is less clear to me. Max-EV strategy seems to
also mean sometimes to play as many hands as possible, other times
just choosing the best machines as far as ER, or the best holds of
cards. I guess now optimal is to play the best effective ER to lose
less money, and favorable to play the best ER to win more money. They
both coincide on positive machines. I'm groping in the dark on the
last few sentences. Anyway, in your example, since you choose the
game with the best effective ER, you are using max-EV strategy in
your unfavorable, optimal strategy, my guess.
I agree the type of problem exemplified by the roulette and craps
problem could be useful if worked out for video poker situations. As
you said, a lot of people play negative machines. Most visitors on
the Strip play negative machines. An analogous problem in video poker
to the craps one would be stated this way, for example. A gambler has
a choice among a few negative video poker machines of different
paybacks and volatility. He will only play a hundred dollars on an
evening. He will stop playing if he either loses the hundred dollars,
or wins a thousand dollars. What are his best choices? The working
out of the optimal solution to this problem is more complicated here
than in your example, not only because there are not two different
machines but many, but because the outcome of a video poker bet is
not simple as in the case of craps and roulette, but multiple, you
have many outcomes with different probabilities to consider. As your
example illustrates, the gambler would have to work out
the "effective ER" of the many choices. I believe players do this in
an intuitive way, not systematically, and thus in an actual casino
situation they probably often make the wrong choices. The problem of
course is relevant to a visitor, who would be a likely person to have
the stated risk limit and goal.
Thanks again for having taken the time to explain; it was, for me,
educational.
E
"Generally speaking, the most important questions in life are only
problems of probability." Laplace.
> Steve Jacobs,
>
> Pardon my stupidity, but I still fail to see the importance of
your
> roulette-craps example. Here is my main observation.
>
> A max-EV strategy should not be used on the pass bet in craps. The
> pass bet in craps has a negative return. As you said, the pass bet
> has a house edge of 1.4%. If you use a max-EV strategy on the pass
> bet, you will lose your money completely in the "long term".
It is absolutely true that playing an endless string of unfavorable
wagers
will lead to losing your bankroll. I don't dispute that in the
least, nor
do I dispute the fact that no strategy will turn an unfavorable
game into
a favorable game.
> A max-EV strategy is always wrong on negative games.
> The only time it is right
> is when cashback and comps make the situation positive.
You're confusing the concept of "optimal" with the concept
of "favorable."
For a negative game, optimal means "making the best of a bad
situation."
I don't recommend playing negative games when positive games are
availabe. I also don't recommend playing negative games if you have
the option to simply not play. However, there is value in studying
negative expectation situations and finding optimal strategies for
those situations. If conditions force you to place negative wagers,
then playing optimally allows you to reduce your expected losses
as much as mathematically possible.
> Considering the above, that you will certainly lose totally your
> money in the long term, any money, on the pass bet in craps, you
> could compare a number of bets there and eventually the game will
be
> worse than a single bet on a game with a lower payback. Even Keno
> will be better than the pass bet at craps if you make N pass bets
at
> craps to win what you would at Keno in one bet. N is to be
> determined, but it is a finite number.
>
> Is this right?
Yes, but it isn't what I'm doing in my example. I'm playing to cut
losses to an absolute minimum, and stop play just as soon as the
goal is reached. This is analogous to driving a car in such a way
that you get the best gas mileage. This minimizes the expense of
driving from point A to point B, so that you'll have more money
for playing video poker.
> Or do I still not understand your example?
I think you're not understanding my example, but that is probably my
fault for not explaining it in greater detail (the post was pretty
long as
it was). The craps play wasn't simply a matter of playing a long
string of bets until the craps player inevitably came out worse than
the roulette player. The craps player acts to maximize the overall
probability of reaching the goal to end play as quickly as possible.
This involves two kinds of wagers. WIth a goal of $36, the strategy
is this: if current bankroll is less than or equal to $18, bet the
entire
bankroll. If the bankroll is greater than $18, bet exactly enough
so
that winning this bet will reach the goal of $36. This is
a "maximum
boldness" strategy and gives the player the highest possible
probability of reaching the goal. For this example, the probability
of walking away with $36 is 1/38.872, so the effective return for
the overall parlay is $36/38.872 = 0.926, so the effective house
edge is 8.4% compared to 5.26% for roulette.
This is the best that a craps player can do under the stated
conditions. If the craps player bet $1 each time, hoping to
eventually build up to $36, the probability of making it would
be 1/61.35, much worse than the maximum boldness strategy.
I haven't checked a lot of cases, but I believe that using craps to
parlay would always be better than Keno. The house edge in Keno
is generally about 25% to 30%. If you play "double or nothing"
craps for 20 bets in a row, you will win 2^20=$1,048,576 if you
win all 20 bets. The probability of doing this with the "don't
pass"
···
--- In vpFREE@yahoogroups.com, Steve Jacobs <jacobs@x> wrote:
On Monday 13 October 2003 12:20 am, you wrote:
bet is (949/1925)^20 = 1/1,390,882. This gives an effective return
of 1,048,576/1,390,882 = 0.7539 for a house edge of 24.61%.
By comparison, reducing the effective house edge to 25% by
betting on a single number on the roulette wheel requires a
string of 5 wins and returns $60,466,176 with a probability of
1/79,235,168.
> In any case, the dogma should say that max-EV strategy is wrong in
> negative games, unless comps and cashback make the situation
> positive. I believe vpFREE dogma believes in this (my dogma on the
> vpFREE dogma).
Negative games can be played optimally to minimize losses. I'd
be willing to wager there are vpFREE readers who play negative
games, perhaps on a regular basis. I've played unfavorable VP
games myself, so I guess I'm proof of my own assertion. I'm not
even ashamed to admit it.
In my original example I posed the problem in such a way that the
player was involved in a promotion that made the game positive
overall. I got objections that this wasn't "real" so I pointed out
that it wasn't necessary to make the game positive in order for
the math to work out the way that it does.
Does that help?