help? continous automatic shufflers

Ive used this strategy 4 times and have won every time but im new to the game and if someone could check this theory out for me i would greatly appreciate it im about to head up to the riverboat in a day.

If using the Basic strategy against a continous automatic shuffler your odds of winning should be 50% so your chances of losing for your
1st hand 50% 1 unit
2nd hand 25% 2 units
3rd hand 12.5% 4 units
4th hand 7% 4 units

11 units lost if you lose 4 hands in a row

5th hand start over at 1 unit ect...

statistically shouldnt you have won much more units than you lose over a period of time. Plus you will win extra money off of blackjacks, doubling, and splitting. Also doesnt the fact that you play more hands faster make your time more worth while because you are winning. This system is based off of the martingale system and it seems to work but if people could please tell me any errors they see i would appreciate it.

in blackjack you win 43% of your hands and lose 47% of your hands and push the rest. No system of the sort you describe will defeat the house edge built into the game.

As bigplayer pointed out, the odds of winning a hand of blackjack aren't 50%. They are 43% with ties factored in. With ties factored out, the odds of winning a hand are about 48%. But there is more.

You postulated that if the odds of losing one hand were 50%, then the odds of losing the second hand are 25%, the third 12.5% and the fourth 7% (don't know how you came up with the 7% figure). That is wrong. The odds of losing the second, third, or fourth hands are the same as the odds of losing the first hand. Although the cards have memory, it is not so strong that it significantly alters the odds of winning and losing, except at very high or very low counts. For the purposes of a martingale system, blackjack can be seen as a series of independent trials, much like roulette or craps.

If you started out with the first hand and were looking ahead, you could say the odds of losing four hands in a row are .52x.52x.52x.52 or 7.31%. But that is looking ahead. Once you lose the first hand, your odds of losing the second hand are still the same 52% and so on. That is how one comes up with the 7.31% figure in the end.

Short answer: any system that bases its bets on previous hands won or lost is doomed to failure. One may win short-run trials more often with such systems, but in the end, the losses will outweigh the wins by the percentage of the house advantage times the total amount bet.

> 1st hand 50% 1 unit
> 2nd hand 25% 2 units
> 3rd hand 12.5% 4 units
> 4th hand 7% 4 units

The prob of winning 1 *OR MORE* in a row is 0.5, but remember the average # of consecutive wins is ~0.90476.

The prob of losing *exactly* 1 in a row is 0.2375.
The prob of losing *exactly* 2 in a row is 0.1247.
The prob of losing *exactly* 3 in a row is 0.06546.
The prob of losing 4 *OR MORE* in a row is 0.07235.

In the cases of losing *exactly* 1, 2 & 3 in a row, this means that the following hand, by definition, must be a win.

In the case of losing 4 or more in a row, the next hand could be a win (~47.5% of the time) or a loss (~52.5% of the time).

Where the "house" (and the odds) are going to catch up with you is in the "losing 4 or more in a row" category. You lose your 11, then lose another 1, another 2, another 4, another 4, another 1, etc. If you think it's not possible to lose 8, 10, 12 or more hands in a row, you haven't been playing very long.

Remember this, the average # of consecutive losses is ~1.10526. This "extra" 0.10526 of a loss is the equivalent of a "tax" put on every sequence of wins. To illustrate, say you're flat betting $10. If you win exactly 1 in a row, it will be followed by an average of 1.10526 losses. The average win (playing b.s.) on a $10 bet is $12.046, and the average loss on a $10 bet is $10.99. So winning exactly one in a row means a win of $12.046 followed by 1.10526 losses @ $10.99, or -$12.1468. This is a net loss of 0.1009, or about 1%. Without the "tax" of 0.10526 * $10.99 or $1.1568, the win-loss sequence would have netted you $12.046 - $10.99 or $1.056. This "tax," unfortunately, is unavoidable.

The numbers will change for different numbers of consecutive wins, but the best you can ever do is get an overall sum for all sequences of about -0.9%. For example, the best progression within the parameters that the last bet cannot be higher than 10 times the first bet and that no bet after the 2nd can exceed the sum of all previous bets) is 10-30-40-45-75-100, which yields an avg-net/avg-bet of -0.898092%

The probs and the "tax" guarantee that NO method of varying bets (whether it's up-on-lose or up-on-win or some combination of the two) can ever create a positive result.

For example, the best progression within the parameters that the last bet cannot be higher than 10 times the first bet and that no bet after the 2nd can exceed the sum of all previous bets) is 10-30-40-45-75-100, which yields an avg-net/avg-bet of -0.898092%

I don't want to start something here, but by what definition of "best" is that progression better than 10-10-10-10-10-10?

>...avg-net/avg-bet of -0.898092%

>I don't want to start something here, but by what definition >of "best" is that progression better than 10-10-10-10-10-10?

As is indicated in the first line the criterion is: Average Net (i.e., the sum of the extensions of all probs times net results) divided by Average Bet (i.e., the sum of the extensions of all probs times the bet amount). If memory serves, the quotient for flat betting is about -0.924%. There are *many* progressions which outperform flat betting, but none of them by much and none which get you anywhere close to positive territory.

This is not the same as "ROI" (which deals with totals instead of averages) which is never effected by a progression. But different progressions produce different average results and different average bets, and the criterion above is one way to measure those differences.

"This is not the same as "ROI" (which deals with totals instead of averages) which is never effected by a progression. But different progressions produce different average results and different average bets, and the criterion above is one way to measure those differences."

So if you aren't using "ROI", what makes one set of average results or different average bets "better" than another?

What exactly is your criterion measuring, and what makes one value "better" than another? What is your definition of "better" and why?

Sounds like the old lose a little bit on every bet you make, but make up for it on volume.

The prob of winning 1 *OR MORE* in a row is 0.5, but remember the average # of consecutive wins is ~0.90476.

These two statements contradict each other. If p is the probability of winning and q=1-p is the probability of losing then the average number of consecutive wins is (conveniently) p/q. If we're talking about an ubiased coin toss here, then the average number of consecutive wins is exactly 1. If we're talking about blackjack, then the probability of winning 1 or more is more like 0.475.

>The prob of winning 1 *OR MORE* in a row is 0.5, but remember the average # of consecutive wins is ~0.90476.

These two statements contradict each other. If p is the probability of winning and q=1-p is the probability of losing then the average number of consecutive wins is (conveniently) p/q. If we're talking about an ubiased coin toss here, then the average number of consecutive wins is exactly 1. If we're talking about blackjack, then the probability of winning 1 or more is more like 0.475.<

There is no contradiction and the numbers are correct.

Using your p/q formula and plugging in .475 and .525, which are the approximate values, the quotient is 0.90476, just as I stated.

And, yes, we're talking about blackjack, where the probability of winning one or more in a row is EXACTLY 0.5 and the probability of losing one or more in a row is EXACTLY 0.5.

If you don't believe it, write a program to count sequences (ignoring push).

> Using your p/q formula and plugging in .475 and .525, which are the approximate values, the quotient is 0.90476, just as I stated.

Actually I used your 0.90476 to come up with my 0.475.

> And, yes, we're talking about blackjack, where the probability of winning one or more in a row is EXACTLY 0.5 and the probability of losing one or more in a row is EXACTLY 0.5.

> If you don't believe it, write a program to count sequences (ignoring push).

Did we not just agree that it was approximately 0.475/0.525? You don't need to count sequences. The only way to win "one or more in a row" is to win the next hand. Likewise, the only way to lose "one or more in a row" is to lose the next hand.

>Did we not just agree that it was approximately 0.475/0.525? You don't need to count sequences. The only way to win "one or more in a row" is to win the next hand. Likewise, the only way to lose "one or more in a row" is to lose the next hand.<

I labored under that same incorrect assumption for a number of years until I wrote a program to actually count the sequences.

If you want to learn the true math behind the occurance of sequences of wins and losses, I suggest you do the same.

In the meantime, consider this awkward fact.

Under your theory, the prob of winning *EXACTLY* N in a row is PW ^ N (where PW is the prob of winning a single hand), and the prob of losing *EXACTLY* N in a row is PL ^ N (where PL is the prob of losing a single hand).

Since were talking about winning *EXACTLY* N in a row in each case, there is no overlap; so the sum of all probs should equal 1.0. Five minutes on a spreadsheet will demonstrate to you that the sum of those probs do NOT total 1.0 (or anything close to it)! Ergo, my previous and your current assumption is WRONG!

Once you understand the correct math governing the sequences of wins and losses, you'll see that the sum of all probs DOES exactly equal 1.0, and you'll kick yourself for not "seeing" the true relationship years ago.

If you want more detail or some C code to help you figure this out, feel free to Email me.

> Under your theory, the prob of winning *EXACTLY* N in a row is PW ^ N (where PW is the prob of winning a single hand), and the prob of losing *EXACTLY* N in a row is PL ^ N (where PL is the prob of losing a single hand).

These are the probabilities of winning or losing N hands in N trials. You may or may not win or lose the (N+1)th trial so they are the probabilities of winning/losing N *OR MORE* in a row. The probability of winning *EXACTLY* N in a row would be PW^N*PL; similarly for losing *EXACTLY* N in a row: PL^N*PW.

> Since were talking about winning *EXACTLY* N in a row in each case, there is no overlap; so the sum of all probs should equal 1.0.

No, of course not. For N > 1 there is the possibility of winning some of the trials and losing some.

Getting back to your original post, there are two things that bother me. First, if you calculate your expectation for a series of independent trials over all possible outcomes, your result should always equal your average bet times your advantage - regardless of your betting scheme. Second, in blackjack, you lose more hands than you win and (consequently) you have longer losing streaks than winning streaks on average. I believe you'll have a hard time trying to dispute these facts.

> If you want more detail or some C code to help you figure this out, feel free to Email me.

If you want some help debugging it feel free to send it. :-) I'm sorry, but as far as I can tell your results contradict the mathematics. And my money is riding on the mathematics.

> The probability of winning *EXACTLY* N in a row would be PW^N*PL; similarly for losing *EXACTLY* N in a row: PL^N*PW. <

You are correct. I misspoke (mistyped?), writing it on the fly. Those are the formulas I formerly used when I thought as you think. It�s been a long while since I used those old, incorrect, formulas.

> No, of course not. For N > 1 there is the possibility of winning some of the trials and losing some. <

What?!!! When looking at runs and losses of EXACTLY N in a row, these are mutually exclusive events each with its own unique probability. There is NO possibility of overlap.

I�ll say this again: go to your spreadsheet program, use your incorrect formulas, total up the probs for all events (within reason � say 100 of each) and observe the inescapable fact that the sum is nowhere near 1.0.

But, when you use the CORRECT math -- math that is confirmed by numerous trials -- the sum of all probs always equals exactly 1.0, as it should and as it MUST.

> Getting back to your original post, there are two things that bother me. First, if you calculate your expectation for a series of independent trials over all possible outcomes, your result should always equal your average bet times your advantage - regardless of your betting scheme. <

That is correct. The fact that exactly half or your runs will be winners and exactly half of your runs will be losers has NO effect on ROI. Your error is assuming that it must.

BUT, you can only understand how one progression could outperform another when you know the CORRECT math governing the occurrence of runs of various lengths. This is not to say any progression can ever achieve a positive outcome. It�s just that some progressions are slightly less negative than others.

> Second, in blackjack, you lose more hands than you win and (consequently) you have longer losing streaks than winning streaks on average. I believe you'll have a hard time trying to dispute these facts. <

That�s funny, I don�t recall disputing those facts. In fact, I actually cited them (remember the 0.90476 discussion?). Your error is in assuming that there is an inconsistency here.

Think about it:
(1) Equal number of winning and losing runs;
(2) Average length of winning runs shorter than that of losing runs.

Ergo, more hands lost than won and NO conflict and NO inconsistency.

> If you want some help debugging it feel free to send it. :-) I'm sorry, but as far as I can tell your results contradict the mathematics. And my money is riding on the mathematics.<

Nobody likes a smartass :-) But I�m afraid both your ignorance and your arrogance are showing. Sad that an otherwise intelligent human being would be so close minded. I will send code on request only.

You can: (A) keep your head in the sand and hold to your faulty assumptions; or (B) actually LEARN something. Your choice.

> The fact that exactly half or your runs will be winners and exactly half of your runs will be losers has NO effect on ROI.

It's true (assuming an even number of runs) that exactly half of your runs will be winning streaks and exactly half will be losing streaks. Why? Because each winning streak is terminated by a losing streak and vice versa. Does this mean that "the probability of a winning streak is exactly 0.5"? Yes and no.

If you select a streak at random and ask "what is the probability that this is a winning streak?" the answer is 0.5. If you happen to forget whether you are currently on a winning streak or a losing streak and ask "what is the probability that this is a winning streak?" the answer is more like 0.475. The difference is in considering a streak picked at random from the set of streaks, and considering a streak containing a trial picked at random from the set of trials.

When you go from saying "probability of a winning streak" to "probability of winning 1 or more in a row" you implicitly stop talking about streaks and start talking about trials. You can't interpret that phrase in such a way that the correct answer is 0.5. So is "the probability of winning 1 or more in a row exactly 0.5"? No, but at least now I see what you meant.

At this point you may be screaming "that's what I've been trying to tell you!" Well, I'm sorry, but that's not what you said.

As for your analysis of progressions at the start of this thread, I see that you are analysing the particular case where a winning streak is regularly followed by a slightly longer losing streak. Allow me to suggest a basis for winning progression for this case:

1. Bet the minimum until you win a hand. You are waiting for a winning streak.
2. Bet the maximum until you lose a hand. Your winning streak is now over. Let n be the length of your winning streak.
3. Bet the minimum until you win a hand or for at most n more hands. Your losing streak is now over. Go to step 2.

Ironically, to use this progression strategy you need to keep track of the "count" n by incrementing it during winning streaks and decrementing it during losing streaks.

> It's true (assuming an even number of runs) that exactly half of your runs will be winning streaks and exactly half will be losing streaks. Why? Because each winning streak is terminated by a losing streak and vice versa. <

So far, so good. You�ve stated what is more or less obvious, at least when some thought is devoted to the subject.

> Does this mean that "the probability of a winning streak is exactly 0.5"? Yes and no.

If you select a streak at random and ask "what is the probability that this is a winning streak?"...etc. <

The point you have not addressed is the fact that if you sum the probabilities of all events it MUST be exactly equal to 1.0. If you use your Pw^N*Pl and Pl^N*Pw formulas and add up all those probs the sum does NOT equal 1.0 or even come close to it. What more proof do you need to convince you that those formulas are wrong?

Thus, all of that �if you�re in or if you�re not� stuff is irrelevant. When it comes to analyzing and/or designing a progression, you must know the exact probability of the occurrence of each event. Your formulas, because they are just plain flat wrong, do not and cannot tell you that.

Interestingly, the strategy you have suggested runs exactly contrary to the actual math; so would perform worse than simple flat betting.

I�ve written a very simple, very fast program which will demonstrate the correct math for you. It does not use a blackjack engine due to complexity/run time considerations. Instead, it just counts the number of winning and losing streaks of each length for the pass line bet in craps, and reports the raw counts and prob of occurrence of each. Still, you have to let the program run for 80 to 100 million passes before the numbers �settle down� enough to show the true relationships between them; so it�s really an �overnighter.�

I will be happy to email both the source and executable to you upon request. Or, you can write your own program, if you wish.

But I�ll tell you this much now, just to pique your curiosity. Partial list of probability rankings:

1st: Tie between winning and losing 1 or more in a row.
2nd: Winning exactly 1 in a row (i.e., a win followed by 1 or more losses).
3rd: Losing exactly 1 in a row (i.e., a loss followed by 1 or more wins).
4th: Tie between winning and losing exactly 2 in a row.

Using your formulas, the two sequences WL and LW would have identical probabilities (e.g. 0.475 * 0.525 and 0.525 * 0.475). The fact is the probabilities are significantly different.

Using your formulas, the two sequences WW and LL would have significantly different probabilities (e.g., 0.475 ^ 2 * 0.525 = 0.1185 and 0.525 ^ 2 * 0.475 = 0.1309). The fact is that the probabilities for each are identical.

Given the above, perhaps you can see why some progressions outperform others when evaluated on the basis of AVERAGE (not total) NET / AVERAGE (not total) BET.

I've gotten about all I can out of this thread, but since you've been patient with me I'll try to answer your remaining questions.

> The point you have not addressed is the fact that if you sum the probabilities of all events it MUST be exactly equal to 1.0. If you use your Pw^N*Pl and Pl^N*Pw formulas and add up all those probs the sum does NOT equal 1.0 or even come close to it. What more proof do you need to convince you that those formulas are wrong?

At the risk of stating the obvious again... the event space for these formulas is all possible outcomes for N trials. Take N = 2 for example. The possible outcomes are WW, WL, LW and LL. If you sum the probabilities for each of these possible outcomes you will get 1.0. If you ignore all oucomes except WW and LL you will not get 1.0.

> Interestingly, the strategy you have suggested runs exactly contrary to the actual math; so would perform worse than simple flat betting.

I didn't exactly spend a lot of time on it. If that's so, simply reverse the minimum and maximum bets.

> I will be happy to email both the source and executable to you upon request. Or, you can write your own program, if you wish.

I have access to streak data from CVData, thankyou. If I ever decide to delve further into this area I'll be in touch.

> Using your formulas, the two sequences WL and LW would have identical probabilities (e.g. 0.475 * 0.525 and 0.525 * 0.475). The fact is the probabilities are significantly different.

No, these are the the correct probabilities (not withstanding the fact that the expected number of winning streaks of length one is greater than the expected number of losing streaks of length one). If you believe the probabilities should be different please explain why the ordering of two independent trials affects the probability.

> Using your formulas, the two sequences WW and LL would have significantly different probabilities (e.g., 0.475 ^ 2 * 0.525 = 0.1185 and 0.525 ^ 2 * 0.475 = 0.1309). The fact is that the probabilities for each are identical.

Once again, these are correct despite the fact that the expected number of winning streaks of length two is close to the expected number of losing streaks of length two.

I did my best to explain the difference in my last post. Don't you see that even though exactly 0.5 of your streaks are winning streaks, together they constitute only 0.475 of your hands?

> Given the above, perhaps you can see why some progressions outperform others when evaluated on the basis of AVERAGE (not total) NET / AVERAGE (not total) BET.

Yes, I understand now. I thought that by "average net" you meant the average net for all possible outcomes, not the net for the "average" outcome. There's no reason why a progression couldn't give a positive expectation in these circumstances (just as a Martingale gives infinite expectation if you place an upper bound on the number of consecutive losses).

> Don't you see that even though exactly 0.5 of your streaks are winning streaks, together they constitute only 0.475 of your hands? <

That was never in dispute, and was fully discussed.

As you pointed out, when you consider, for example, only a sequence of (for example) two consecutive outcomes, your formulas do, of course, give the correct probabilities for each of the 4 possible sequences.

But that is not the context here. The context is the original post, in which the poster used some incorrect math to arrive at the probabilities of a 1-2-4-4-startover progression. What he needed to know was the correct probabilities of getting:

(a) 1 or more losses in a row;
(b) exactly one loss followed by 1 or more wins;
(c) exactly two losses followed by 1 or more wins;
(d) exactly three losses followed by 1 or more wins;
(e) four or more losses in a row.

I stated the correct probabilities at the time.

> If you believe the probabilities should be different please explain why the ordering of two independent trials affects the probability. <

When I used �WL� and �LW� I was, as should have been clear from the context, referring to the events of (1) a single win followed by one or more losses and (2) and single loss followed by one or more wins.

Similarly, given the context, the �WW� and �LL� referred to events of two wins followed by one or more losses and two losses followed by one or more wins, respectively.

In any case, it�s the �or more� in the above which changes the math, and when dealing with progressions (which is the context) the �or more� has to be taken into consideration.

I apologize for restating the obvious, but some of it had to be said.