Quote:DocIsn't variance by definition a positive number? I now try to refer to positive deviation -- is there an issue of correctness here? I'm not really sure.

Deviation is a positive number too :)

I didn't think that was necessarily so. If I am considering numbers within a distribution of mean µ, then an individual observation x would have a deviation from the mean equal to x-µ, which could be either positive or negative. In computing variance and standard deviation, these individual deviations are squared before they are added, so that both variance and standard deviation are positive numbers, but not individual deviations necessarily. Positive deviation implies an observation higher than the mean, which in the context of a gambling session suggests winning money or losing less money than expected on average.Quote:weaselmanDeviation is a positive number too :)

Quote:goatcabinI am objecting to mkl's assertion that "you will lose". Here is his statement: "The more you bet, the more you will lose (odds bets excluded)."

That is clearly incorrect.

Semantics again, Alan. As you approach certainty, there becomes no meaningful distinction between absolute (mathematical) certainty and practical certainty. A 99.99999% chance of winning (or losing) is good enough for me.

Quote:goatcabinWhy not? How would that hurt you?

It would hurt me to stop and assess my results every 400 hands because I would be wasting my time gathering useless information. It's the same reason why the casino doesn't halt all play and count all the money every thirty minutes.

Quote:DocI didn't think that was necessarily so. If I am considering numbers within a distribution of mean µ, then an individual observation x would have a deviation from the mean equal to x-µ, which could be either positive or negative. In computing variance and standard deviation, these individual deviations are squared before they are added, so that both variance and standard deviation are positive numbers, but not individual deviations necessarily. Positive deviation implies an observation higher than the mean, which in the context of a gambling session suggests winning money or losing less money than expected on average.

Right - *standard deviation* is a positive number since it's defined as the square root of variance (and not vice-versa), but any specific deviation of an outcome from the mean can be negative. Since we're talking about specific outcomes, that all makes sense. However, there's still a difference between what people think of as "good luck" and what is technically a positive deviation. For example, after playing $10,000 in handle on the passline, you're down $40. That's a positive deviation (the mean is -$141) but most people wouldn't think losing $40 in craps is "good luck".

Quote:weaselmanso, if he added "likely" after "will", you would agree then?

No, the word "will" is inappropriate, as is "the more you bet" without distinguishing between betting more, once, and betting many times. If I bet $1000 on the pass line, my probability of winning is .4929. OTOH, if I bet $10 100 times, my probability of coming out ahead is more like .44. etc. etc. This is because the standard deviation goes up more slowly than the expected loss, as I'm sure you, and mkl, are aware. "You will lose" means zero probability of not losing.

Quote:weaselmanThat's simple misunderstanding then. I used "get lucky" as a synonym to "win". If I lose, I don't consider myself lucky even if I lost less than EV.

With this correction, do you agree that we are saying the same thing now?

Yes, I believe so.

Quote:weaselmanNot really. If the distribution is symmetric, increasing SD raises your chances of winning as much (actually, slightly less) as your chances of loosing. Yes with a huge SD you can win a lot, but you can lose a lot as well, so on average, it's the same.

Yes, really. Compare betting the pass line for $5 with single odds against 3, 4, 5X odds. With the higher SD, same ev, the probability of a winning or breakeven session is increased. For 100 bets, the ev's are -$7.07, but the SD's are $98 and $246, so a smaller degree of "positive deviation", to use Doc's term, is required to overcome the expected loss. That is all I am saying. We all know that variance works both ways, and I am constantly pointing that out -- high variance spreads the graph out, but it doesn't change the mean. That doesn't make it "the same". High variance means that more players using that strategy will be ahead at any point in time. It's almost like insurance, in a weird way. The players at -1 SD are losing more than low-variance players at -1 SD, balancing the fact that the +1 SD high-variance players are ahead, while the + 1 SD low-variance players are not.

Quote:weaselmanIt's misleading though for the reason I mentioned above. If SD is large, you may not need as much "luck" to win, but at the same time it doesn't take much "lack of luck" to go bankrupt either. If you consider both possibilities, you'll have to conclude that you need about the same total amount of luck.

The same total amount of luck for what? All I am saying is that a larger SD requires less luck to come out ahead. Of course it increases "risk of ruin"; of course the same degree of bad luck results in larger losses than with low variance. The fact that it works both ways does not change the fact that high variance increases one's chance of winning.

Quote:weaselmanPut it another way, consider a single bet with extremely large SD. You can win a lot with relatively higher probability, that's correct. But at the same time, you can lose even more with even higher probability. And, if that happens, you are going to need to make many more subsequent bets to dig you out of the hole (again, risking to lose even more on each of them). You may need only a little bit of luck to win one bet, but overall you need a lot of it to come out ahead.

What do you mean by "overall"? What is "a lot" of luck? The amount of luck you need to come out ahead, as I've stated over and over, is a function of ev/SD; the ev increases directly with the number of bets, the SD with the square root thereof, which is exactly why it becomes less likely to win the more you play a negative ev game. A gambler needs to understand what he/she is gambling for: minimizing loss rate to stay at the table? - low variance; willing to risk losing the bankroll for the chance for a big win - high variance -- both may have the same ev. 98 Steps' method introduces a high measure of skew - lots of winning sessions, risk of a big loss, but nobody except Krebs himself believes that this produces a positive expectation.

Cheers,

Alan Shank

Woodland, CA

Quote:mkl654321Semantics again, Alan. As you approach certainty, there becomes no meaningful distinction between absolute (mathematical) certainty and practical certainty. A 99.99999% chance of winning (or losing) is good enough for me.

But I don't think any individual player ever plays enough to approach certainty. It's not just semantics; I think there is an important, material difference between saying:

You can't beat craps.

AND

You can't win at craps.

You can't beat craps, in the sense of achieving a positive expectation by making different bets in different amounts at different times.

You can win at craps, even over a lifetime's play, but the more you play the less likely that is.

Cheers,

Alan Shank

Woodland, CA

Quote:goatcabin

The amount of luck you need to come out ahead, as I've stated over and over, is a function of ev/SD

Well, I just have to say it again, that you seem to have a very unconventional definition of luck.

Suppose you bet $100 on a number in roulette. You expectation is about -$5.26, and SD is ~$576.

By your definition, the "amount of luck" you need to win is 0.00913.

Now consider betting $100 on black. The expectation is still -$5.26, but the SD is only $100. By your definition, this bet requires 0.0526 of luck to get ahead - almost 6 times more than the other one!

Yet, as we all know, you will (yes, will :) ) win a lot more often betting on red, then on a number (about 18 times more often to be precise - 47% of time vs. 2.6%).

Quote:weaselmanWell, I just have to say it again, that you seem to have a very unconventional definition of luck.

Suppose you bet $100 on a number in roulette. You expectation is about -$5.26, and SD is ~$576.

By your definition, the "amount of luck" you need to win is 0.00913.

Now consider betting $100 on black. The expectation is still -$5.26, but the SD is only $100. By your definition, this bet requires 0.0526 of luck to get ahead - almost 6 times more than the other one!

Yet, as we all know, you will (yes, will :) ) win a lot more often betting on red, then on a number (about 18 times more often to be precise - 47% of time vs. 2.6%).

Oh, please, weaselman, when you are talking about one bet, neither the ev nor the standard deviation has any meaning. You cannot lose $5.26, and your results cannot vary by $576, or $100. On the one hand, you win $3500 or you lose $100, a difference of $3600; on the other, you win $100 or lose $100, a difference of $200. Also, you know full well that I am not talking about the probability of winning a single bet.

Let's compare 100 $10 bets for each. They each have an ev of -$52.63, while the standard deviations are about $100 for the black vs. about $576 for the single number. So, the black has an ev/SD of about .53, while the single number's is about .09. I ran a quick simulation of 10,000 sessions of each. For the black, there were 2784 winning sessions, 448 broke even and 6768 lost. For the single number, 4968 sessions won, none broke even and 5032 lost. Those figures are quite close to what the ev/SD ratios indicate -- .53 SD associates with a probability of .298, while the single-number set came out considerably better than the .464 for .09 SD. Let's say that those 10,000 sessions were by 10,000 different players - about 28% of the "black" players came out ahead, while almost half of the single-number players did. OTOH, if you look at it from the point of view of one player of each type playing 10,000 sessions, at the end of the 10,000 sessions they are going to be about equal.

What I'm getting at here is that individual players are not "doomed" to lose at craps, although the players, taken as an aggregate, are.

Cheers,

Alan Shank

Woodland, CA