Re: Problem with playing the wrong way

On 26 Oct 2005 00:21:08 -0700, "Eddie Inside Numbers"
<eddie_inside_numbers@xxxxxxxxx> wrote:

>All I know is that when someone begins a sentence with "Huh?" it says
>to me that he can't believe what he just read, no matter what his
>reasons. So, I kind of figured you and I were on the same page about
>this scheme, which is why I couldn't understand why you were giving me
>a ration of grief.
>
>I'm sorry if you feel you were misquoted or taken out of context.
>
>I still say you and I agree that this scheme is not such a hot idea.
>Maybe we're not on the exact same page, but I think we are reading from
>the same hymnal.

See my other post, after I simulated it.
>

>> >> *You* are
>> >> concocting a scenario to suit your dislike of this scheme, aren't you?
>
>(Not to be picky, but that last sentence is one that you wrote to me,
>not the other way around.)

Right.

>> If I have time, I am going to put this into WinCraps and compare it
>> with just a straight DP in an amount that results in approximately
>> equal bet handles. You will see that this reduces variance.
>
>While I'm right here, is that good or bad? Seriously. I don't speak
>statistics as well as other languages. You always mention variance, but
>I've never understood whether it's something I want or not.

An excellent question. A few months ago, there was a show on the
History Channel about different ways to "beat the house." It had
segments on blackjack counting, "dice influencing," poker, etc.

Mostly, it was crap, but they had a guy on who teaches casino
management at UN Reno. He made one of the most intelligent statements
I've ever heard about gambling. He said that, when people gamble, they
are "buying variance." What this means is that, although over millions
and millions of bets the casinos are sure to win and the players,
taken as an aggregate, are sure to lose, variance, a characteristic of
independent, random events, gives any player a chance to win, even
long term. The probability of any player being ahead decreases with
the number of bets, all other things being equal. If there were no
variance, then no one would win. Not only that, it would be pretty
damn boring if you lost 1004 of every 1980 passline bets, never more,
never fewer. The variance of a bet is determined by its probability of
winning and the payoff.

A passline bet has almost a 50% chance to win, and the payoff is even
money, so the variance is low. Since the payoff is even, the only way
you can get way ahead or behind is to win or lose way more than 50% of
your bets.

A bet on the 12 has a low probability of winning, but pays 30 to 1 (or
29 to 1. Because of high payoff, it doesn't take much deviation from
the expected W-L percentage to make a large difference.

The wonderful program WinCraps, on the "Advantage" tab of the
"Configure" screen, shows the expected value and standard deviation
for every bet, on a per-roll and per-decision basis. Technically, the
standard deviation is the square root of the variance. I use the term
"variance" in its more general meaning, i.e. "the degree by which the
distribution of outcomes is expected to spread out from the mean." Or
s.t. like that.

So, it's "good" that there is variance, so people can have a chance to
win. The question is, how much is good for you? That's a question each
player has to answer for him/herself. If you want to play for 3 hours
on $100, you don't want very much variance. High variance means you
have a chance to win more and lose more, that your results may go far
afield from the mean expectation.

The probability of breaking even is a function of the expected loss
divided by the standard deviation. Think of the expected loss as a
handicap. To overcome this handicap, you need some good luck. How
much? The ev/SD tells you what fraction of a standard deviation you
need to come out BETTER than the mean expectation in order to overcome
the handicap. For example, I posted the results of a simulation of a
session of 60 passline bets.

mean net: -$4.61
SD: $114

So, in this case, in order to overcome the expected loss of $4.61, the
dice have to select an outcome for you that is 4.61/114 = .04 of a
standard deviation better than expectation. How likely is that? Every
statistics book has a "table of Z values" or "Areas under the standard
normal curve." The part of a standard deviation (or multiple) is the
"Z" value, which is usually in the left-hand column, and you look up
the probability along the row to the right. For .04, the probability
listed is .484.

Because of this principle, betting pass or DP and taking or laying
odds in high multiples gives you the best chance of breaking even or
better, because the expected loss is minimized and the odds bets give
you pretty high variance. This minimizes the value ev/SD, so you don't
have to be as lucky to win. The down side of this, of course, is that
it also increases the probability of a large loss, since you are just
as likely to be unlucky as lucky.

So, people who want to win a lot of money, and are willing to pay a
lot for this chance, choose high-variance types of play.

How much variance do you want to buy, and how much are you willing to
play?
Cheers,
Alan Shank
.

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