Does clipping off the outliers...

...make any real difference?

Let's say you have a bell curve representing the distribution of results over time, for a negative EV game. The highest point of the curve would represent expectation, or the mean of all results (and I'm not certain that the preceding statement is strictly correct). Now let's say that you artificially limit the occurence of outliers, such as by setting loss limits and/or win/quit points.

Now, by definition, in a -EV game, that mean result is to the left of 0 on the X-axis, and therefore more of the area under the curve is in negative than positive territory. In other words, the total magnitude of losses is greater than that of wins.

(I then got to wondering: a game can be -EV even if you win more often than you lose; what matters are the aggregate results. I then wondered what the bell curve would look like for a high-variance game where you won considerably more when you won than you lost when you lost--say, betting on eleven at the crap table.)

It would seem that loss limits chop off a greater portion of the bell curve (on the left) than win limits chop off from the right side of the curve. This is because the curve is taller on the left than on the right in a negative EV game, for any value of X, where X is the absolute difference between 0 and the cutoff point. This makes sense, because a loss limit is most likely to be beneficial in a negative EV environment. But what does that do to the FREQUENCY of losses? Are they reduced as well? That would seem not to be the case, as quitting at a stop-loss point ENSURES a loss, where playing on would have swung the game back into the positive, at least SOME of the time.

I've been thinking about this kind of analysis in environments where the play session is artificially limited, as in loss-rebate promotions, tournaments, and short-lived promotions.

I'm not sure.
I think all sessions are limited as opposed to the theoretical results obtainable over a great time period.
I think all curves have the tails lopped off by practicalities of time, sobriety, etc.

Curves are mentally shifted by near-misses or other people's recent wins but in reality everyone really faces the central data rather than the outliers. Point-Seven-Out is common but its not the most likely. Those roulette numbers may not trend in the way assumed from the annunciator but its not going to be some 17 reds in a row for most people. Its going to be that central area under the curve with the extreme tails lopped off. Most of the time!

Interesting question. The easy answer is that it doesn't matter; all expectation is just edge*action, but that wouldn't do justice to your problem. A win limit will cut off the right side "tail," but will also make the peak of your curve farther to the left of zero. That is, you will hit your win goal and leave a lot of the time, but those times when you don't leave you will lose more because you won't have those tail-end winning sessions to balance them out. A loss limit will cut off the left side of curve but will shift the top of the bell to the left, again (I think). It doesn't really matter because in the long run it will all blend together.
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You came to the board late but there was a really good discussion of this type of thing in the arena of craps. Search goatcabin's posts and blog.

Quote: mkl654321

(I then got to wondering: a game can be -EV even if you win more often than you lose; what matters are the aggregate results. I then wondered what the bell curve would look like for a high-variance game where you won considerably more when you won than you lost when you lost--say, betting on eleven at the crap table.)

It would seem that loss limits chop off a greater portion of the bell curve (on the left) than win limits chop off from the right side of the curve.

Not if the right side is measured from the mean, rather than 0. The thing to remember is that Bernoulli processes always converge on the normal curve over time (assuming 0<p<1). The question is just how long that takes. The higher the variance, the longer it takes. Yo bets take longer to converge on the mean than pass bets. Basically every slot game is also like this - win frequencies are nowhere near 50% but conditional payoff amounts average many times the wager.

Clipping or limiting one side of a curve does indeed have the effect you suggest - it guarantees a result at the limit, eliminates results beyond that limit -- but also reduces the sequences which would have ended up on the other side of the mean. So a loss limit eliminates all losses beyond the limit, but also increases overall loss frequency and reduces win frequency. The magnitude of these changes depends on how close to the mean you set the limit(s).

No win limit, loss limit, or any betting system can increase nor decrease the house edge. I recommend studying the Central Limit Theorem if you're not convinced.

Quote: Wizard

No win limit, loss limit, or any betting system can increase nor decrease the house edge. I recommend studying the Central Limit Theorem if you're not convinced.

Well, of course not. The question I asked was whether win/loss limits increased or decreased the frequency and magnitude of winning and losing sessions, respectively.

Quote: mkl654321

Well, of course not. The question I asked was whether win/loss limits increased or decreased the frequency and magnitude of winning and losing sessions, respectively.

Definitely. Set your win limit at $1, and loss limit at $100000, and you will have way more winning sessions than losing sessions.
The question you asked though was if it would make any real difference, and Wiz's answer to that is that it won't, because in the end of the day, if you sum up all your wins and losses, the result will be around the house edge.
If the gap between the loss limit and win limit is huge though (like in my example), it might very well be the case that no losing session shows up in your lifetime, but then again, you can have a 10 in a row strike, starting tomorrow.

The right (non-trivial) question that can be asked here is what is the chance of losing X dollars before making Y. An answer to it may be found on the Wiz's std. deviation page.

While writing this, I ran a quick blackjack simulation, out of curiosity with a win limit of 1 unit, and a loss limit of 100000. After about 5,000,000 hands it has completed 300 sessions (the 5 losing sessions were understandably very long), and was down 29,954 units. Average session length was 16,758 hands (I had to clip the max session lengths at a million hands because I got tired of waiting for results, so some of those 5 sessions lost less than 100000). The loose-to-bet ratio is (who would have thought? :)) 0.6% - just about the house edge.

Quote: mkl654321

Well, of course not. The question I asked was whether win/loss limits increased or decreased the frequency and magnitude of winning and losing sessions, respectively.

Well, of course they do. In a zero house edge game, if the player plays until he wins w or loses l, whichever happens first, then the probability of attaining the winning goal is l/(w+l), assuming the player never overshoots either goal. The house edge increases the probability of the losing goal, according to the house edge and time played.

I may be stepping way out of my math league here, but ...

... when it comes to dollar-amount winnings/losses, is a normal distribution applicable? My thinking is this:

1. No matter how badly you lose a hand of, say, 3CP, you only lose $X. A Q-9-4 to Q-9-3 loss loses you just as much as a A-K-Q mini-royal to Q-6-4.

2. Conversely, it's not by how much you win a hand by, at least as far as the main bet goes. The bonus bet (for lack of a better term) is where the big winnings come (say, Fortune PGP).

Wouldn't a better disstribution be t or Poisson or something other than normal?

Quote: ItsCalledSoccer

I may be stepping way out of my math league here, but ...

... when it comes to dollar-amount winnings/losses, is a normal distribution applicable? My thinking is this:

1. No matter how badly you lose a hand of, say, 3CP, you only lose $X. A Q-9-4 to Q-9-3 loss loses you just as much as a A-K-Q mini-royal to Q-6-4.

2. Conversely, it's not by how much you win a hand by, at least as far as the main bet goes. The bonus bet (for lack of a better term) is where the big winnings come (say, Fortune PGP).

Wouldn't a better disstribution be t or Poisson or something other than normal?

In the long run, all casino gambling propositions converge on the normal curve. Slot games are way, way more volatile than any table game, and the same is true there. Remember, each wagering proposition is just a discrete random variable with a payback range between 0 (total loss) and MAX_AWARD, whatever the jackpot is. In most games, there are only two outcomes, so the passline has a distribution of 0 and 2, with slightly more chances of 0. (You can subtract 1 from everything and put it into "table-games" notation.) Any sequence of variables like this eventually starts looking like the normal curve. The variance of the distribution determines how long that takes.

Quote: MathExtremist

In the long run, all casino gambling propositions converge on the normal curve. Slot games are way, way more volatile than any table game, and the same is true there. Remember, each wagering proposition is just a discrete random variable with a payback range between 0 (total loss) and MAX_AWARD, whatever the jackpot is. In most games, there are only two outcomes, so the passline has a distribution of 0 and 2, with slightly more chances of 0. (You can subtract 1 from everything and put it into "table-games" notation.) Any sequence of variables like this eventually starts looking like the normal curve. The variance of the distribution determines how long that takes.

Thanks. Begging your patience, you obviously know the math better than me ...

This makes sense to me based on the distribution of the result, but I'm missing the part about the distribution of the dollar-amount.

This may be a dumb question, but if the payback range cannot be less than zero, how is a normal distribution of the dollar-amount possible? Wouldn't a distribution where the "answer" can't be <0 be applicable?

PS: Please keep it somewhat simple. I have two engineering degrees and am familiar with mathematics, but not so much with the branches of probability and statistical methods. In other words, I know just enough to sound particularly stupid.