Something i've noticed lately, mostly from one system-seller's site to another, is the supposed effects of "stop-win/losses" (, or win/loss limits per shoe).

"Our system beats 88% of the shoes...", turns out to mean that each shoe is played only until some relatively-small number of wins is churned out... or that not nearly all of the shoe is played. But is an 88% game-edge over the house required to achieve such results shoe to shoe (, even neglecting the commission and/or in-built banker-bet's edge).

The best way to answer this is to work from the random-walk theory already laid out at https://acrobat.com/#d=dUaZa*i6TYme*urC8pGwNw . That work culminates with the function g(n) for the probability of no "catch-up" after 2n games... with the first outcome, whichever, setting the side to be caught-up to.

Continuing on from that work, let G be the chance of a catch-up after 2n games. Then G = 1 - g(n). Also, let G1L be the chance of a catch-up after a loss on the first game; and G1W be that for a fall-back after a win. (In general, the integer m in the GmL or GmW shall stand for the number of outcomes to be [caught-up/fallen back] to, after either side has followed/led, respectively, by a current total of m. Either side may and shall follow/lead by 1 win/loss, respectively, after one game. Furthermore, G = G1L = G1W because, eg, the ratio of catch-ups (beginning with an L) to the non-catch-ups beginning with an L (over the 2n games) remains the same as for G1W, and the latters combine to make up the former.)

Now suppose we want to know the chance of reaching 2 W's from the outset (, staying ahead the whole time), to quit the shoe before or at 72 P-B decisions (or n=36), on average. To do this, pretend that we had an L-outcome before the single (inherent) L to be caught-up with by G1L. This chance may now be expressed as G2L.

If we lose on the next outcome, roughly 1/2 of the time, we'd be left with a chance of G3L to succeed; but if it comes a W for us, we'd be left with a chance of G1L to succed. Which means that G2L(n) = 1/2 X [G1L(n-1/2) + G3L(n-1/2)]. Because this then holds for any integer m, G1L(n) = 1/2 [G0L(n-1/2) + G2L(n-1/2)]. Therefore, given that G0L(of any n) = 1 because no losses (existed/)remained to be caught-up with (, as we won the second game where G0L first occurred above), we may write GL(n) = 1/2 [1 + G2L(n-1/2)] ---> G2L(n-1/2) = 2G1L(n) - 1 = 2G(n) - 1 = 2(1 - g(n)) - 1 = 1 - 2g(n). Plugging the numbers into the approximated function for g(n), at the link provided, we find that the chance of leaving a shoe as soon as ahead by two wins (after up to 80 games) is 2(1 - g(36)) - 1 = about 0.8121 . (Note that the chance of leaving right away with only one win ahead is G1L(n) = G(n) = 1 - g(n), hence about 0.9094 for n=36.)

One may easily continue all this, putting 1/2 X [G1L + G3L] equal to 2G1L - 1, because both are G2L, to arrive at G3L = 1 - 3g(n); then G4L = 1 - 4g(n); etc... until realizing that likely, GmL = 1 - mg(n). (Anyone want to write out the general case?)

Anyway, we now see that on average, anyone can win 91% of the shoes if using a win-limit/shoe of 1 , and no loss-limit/shoe. And long as the loss-limit remains more than the win-limit, and the goals aren't too-lofty... anyone can "win more shoes than lose" by the averages alone.

Beyond that though, the above is a powerful tool by which to gauge many of the otherwise-tedious qualitative aspects of gambling, egs, when one is beginning to fall behind too far by "flat betting" (to reasonably expect to catch-up over a given period of play), and graylove's idea of how far ahead either side should be to remain in the lead throughout a shoe. And more importantly, a means to an even more-indepth understanding of gambling, in general.
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I'll add it to the thread, "Gambling" theory... , if it passes intact here to some further results.

For they who have plodded through some of the numbers above, here are a couple of things i omitted in my haste to write something about this subject. (Not too surprising that Ellis couldn't fill in some my oversights for me, lol.)

Sometimes when "pulling seemingly-unrelated bits of information out of thin air", you miss one. I prefer that sort of approximation-approach to most sorts of problems, as opposed to grinding things out the non-analytical way... the latter which often leads to so much more of the tedious, non-elegant "show your work" stuff.

1. First, though it's true that, "If we lose on the next outcome, roughly 1/2 of the time, we'd be left with a chance of G3L to succeed; but if it comes a W for us, we'd be left with a chance of G1L to succed. Which means that G2L(n) = 1/2 X [G1L(n-1/2) + G3L(n-1/2)]..." ... to avoid the linearity of that which won't play out at the higher levels of m, note that the distances between the Gm-pairs on the righthand side of that equation become closer by the ratios 0/2, 1/3, 2/4, 3/5, 4/6, and so on to 1.

2. Second, G1L(n) = [1 - g(n + 1/2)] (, not [1 - g(n)],) because if 2 X (n - 1/2) games could result in ending up ahead by one win (, for the "equivalent" catch-up over 2n games), then the full n for G1L(n) shall be better reflected by the full g(n + 1/2) -value.

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Hopefully, this will allow for a fuller conclusion of the actual numbers, and some other applications... as i go.

Garnabby. If I'm reading this right, basically you are saying that given a stop-win of 1 and no stop-loss you will succeed around 9/10 times.

Say I stop at 1 for 9 shoes in a row.

I am now +9 all up (not including commission)

On the 10th shoe I lose a huge amount (for example: more than 9) and thus I lose overall. Is that what you are saying?

Plasia,

What Garn is attempting to explain, in his own simple words (and a tad of mathematical obsfucation), is why some people from BTC, who "claim" to have won 15 or 19 or 20 shoes in a row, most probably did it by winning 1 unit and calling it a winning shoe.

Of course his premis is all wrong, as is his motive, but what he appears to be trying is to tell everyone that only by making 1 unit could you possibly run 20 shoes in a row as winners.

I'd would beg to differ, but then I'd be on the losing side of any argument with "The Appointed One" as anything I would say on the matter would be total lies.

The best I could come up with was 19 out of 20. I can assure the G man that it was more than 1 unit per shoe.

As I now tend to just ignore G's message posts, he'll have to talk to you or himself about it.

AD

Yes, playing by a system w/o any statistical basis, on average you will lose the amount bet X the house-edge on each game.

And incidentally, over many games and hence plays of the same bet-money, the house's edge will be compounded... until it only seems as though that edge must have been higher, to have reduced one's session-BR by so much more, perhaps by cheating, etc. (Another concept which Ellis "doesn't get".)

Beyond that, these sorts of calculations are/will be useful in maximizing the benefits of any logically-progressing and grounded-to-begin-with systems.

The operative words here are "statistical basis." Of course G is absolutely correct. Particularly he is correct about the compounded house edge making the edge "seem" higher. In terms of a single finite bankroll one might see that the house edge is 100%!

The above is not to say that a player can go an awful long time picking spots and winning. An awful long time.

It also assumes random trials but that is another thread.

A