disprove this probability of a net gain

[simon]

If we have a mathematical expectation of winning and losing a certain amount of bets within a given series of outcomes, and we can start the series with what would be some of the "losing" bets, that is, bets that we know would have resulted in a loss, why then can we not assume we can achieve a net gain within the given series?  That is, if probability theory states that within a certain amount of spins you can expect that for example 50% will win and 50% will lose, and we have a way to eliminate some of the losing bets within the series, why can we not expect to win more bets than we lose within the entire series?

I have a system which seems to me cannot be disproved as a solution for overcoming the house edge, if one is to use probability theory.  So in fact it could be proven that at least a one unit gain should be achieved within approximately 80 continuous spins, if the outcomes conform to expected
mathematical probabilities.  the one unit gain may be achieved much faster, and may occur within just a few spins, at which point the series stops and you would start a new session.  But according to probability theory, this one unit gain should be achieved within 80 spins.  unfortunately, the improbable always seems to be happening at the wheel which never seems to deliver up what is supposed to occur  (I have witnessed many triple repeats of a single number,
and once witnessed a column hit 15 times in a row), which is why I cannot have faith that the system will win at the casino, though it should win according to probability theory.

So I'm wondering why my theory is incorrect (it must be, because no one can beat the wheel mathematically, right?) so something must be wrong, and I want a "math guy" or somebody to tell me what that is.  The system I have devised is more complicated then can be desribed here, but it is perhaps based on a flawed assumption, and I could present a much simpler example here, and if this simple example is flawed, then so is the whole thing, my more complex system, that is.

So, for example, suppose we will play a pre-determined amount of spins, let's say 38 to simplify things, and we will back one of the even chances and stick to it throughout.  Now according to the laws of probability, we should expect to win half our bets, and lose half our bets, and lose twice to the zeroes.  So, if that is the case, then if we have a way to eliminate some of our losing bets,
then we should come out with a net gain (perhaps just one unit) by the end of the series, according to probability theory.  So, how we do eliminate some of the losing bets?  By waiting for the losing bets to occur at the beginning of the series. 

So, suppose we wait for any combination of successive losing bets, anotherwords a zeroe hits plus any streak of any even chance, and now we will back the other even chance for the rest of our series.  I know this immediately sounds like a basic and naieve way of betting, but I want to know how it can be disproved mathematically that it would not work.  Because this is how I can prove mathematically that it does work:  We begin our count of 38 spins when let's say the zeroe and/or five of any even chance (lets say red) hits successively.  Our bet now for the rest of the spin cycle would be the other even chance (black.)  Six of the decisions that would defeat our bet on black have been eliminated.  we will bet now for 32 more spins.  If the mathematical expectation is that we will lose twice to the zeroes and 18 times to red within a 38 spin cycle (and I am only talking probability theory, not what tends to actually happen at the wheel), then red has already hit 5 times and the zero has already hit once and we did not lose these bets; now we should expect another 13 reds and another zeroe (at a double zeroe wheel) and 18 black within the 32 spins left to bet (which would put us ahead +5.)  This is ofcourse over simplified but in fact does follow what probability theory tells us will happen.  DOESN'T IT??

So to summarize, my question is, if the expected probability is that we will win a certain amount of bets and lose a certain amount of bets within a certain amount of spins, and we did not incur a loss from some of the "losing" bets because they were at the beginning of the spin cycle and we did not play them, why then can we not expect a net gain by the end of the cycle (if not before), according to the laws of probability??

Just thinking out loud... but if anyone can give me the answer to this question, that would be great...

[Lohnro]

So to summarize, my question is, if the expected probability is that we will win a certain amount of bets and lose a certain amount of bets within a certain amount of spins, and we did not incur a loss from some of the "losing" bets because they were at the beginning of the spin cycle and we did not play them, why then can we not expect a net gain by the end of the cycle (if not before), according to the laws of probability??

Just thinking out loud... but if anyone can give me the answer to this question, that would be great...

I cannot give you mathematical reasons, but common sense tells me this:

1. You may miss the losing bets at the "begining" of the series, but you also miss the winning ones.
2. The table does not know or care when "your" series started, each spin is the start of a "new" series.

[Lohnro]

Also check out alarian's post:

nolinks://vlsroulette.com/alarian's-school/no-more-progressions-please!!/

[ikarianman]

i believe probability will not work for roullete,betting spin after spin or betting once a month is the same thing,except once a month you will lose or win slower.where probability comes is that you witnessed a column appear 15 times.then,though its not certain but almost certain that you will win in the next 3 spins,i dont know if there is a record of 18 repeats of a column.in such extreme events i think it worths betting heavily because it hasnt happend yet for the time roulette exists so why happen to you?,otherwise i dont bother playing at all.

also,results have no a starting point and an ending point,so how can you limit them in some thousands spins? for 1000 spins the 900 could be  losing spins or wining spins,you never know.

[TwoCatSam]

simon

Your idea has legs.  However, the term "probability" is what will trip you up.  You could even use the word "should".  This is what "should" happen.  And many times it will;  many times it won't.

There is an erroneous school of thought which says reds and blacks should "even out" over a period of time.  They will get within a certain percentage, but the period of time could be huge.  While the percentage of the difference decreases, the actual number of reds vs blacks may stay exactly the same. 

Also, William Feller in 1958 discovered that in any series, be it flips of a coin or whatever--the side that gets ahead usually stays ahead for the full trot.  He talks of 20,001 flips of a fair coin.  If heads leads at the beginning, there is a huge chance the lead will never change.  This does not mean you will not win unit along the way as they can get very close.  And, yes, the lead does change so you could start out red and switch to black.

I have used Random.org and flipped a few coins myself.  I was amazed at how often the "leading" side stayed ahead.

Talk to /Compa about the standard deviation.  He's the guy who can help you.

Sam

[MATTJONO]

hi simon,

I have no answer for you rearly mate. all I can say is in the past I always used to look at the previous results spun as many as possible, to make decisions on bets to make.

one time I deposited. £5 on betfred casino.1p bets. I started playing a system and after 45mins or so I was on £6.50p. which I felt happy with. however I had around sixty odd spins history that betfread allowed you to see. (the time loggen in)

anyway as I was playing my system slow/boring/however still making profit.

I seen this.while playing.
B,B,R,R,B,B,B,B,R,B,R,B,R,B,R,B,R,B,B,B,B,B,B,R,B,B,R,B,R,B,R,B,B,R,B,R,B,R,B,R,B,B,B,B,B,B,B,B,B,B,R,B,R,B,B,B,R  AS YOU CAN SEE I SEEN 16 TIMES IN A ROW WE SEEN RED NEVER REPEAT AND ALWAYS landed on black after 16 reds.

this is where I placed £6 on red and 50p on the 0

0 span in and was now on £18.
I started to bet randomly what I always end up doing, trying to win big. however this time I was always betting alot of chips on red when red appeared, as I thought the probibility would change and reds would begin to repeat.

anyway I found myself on £97 from just £6.50p  (I was chuffed at this point) I was chuffed reds was going mad runs of 5 in a row, runs of 8 in a row, blacks where nowhere to be seen  .

HOWEVER I DID NOT LET MYSELF STOP. I COULD SAY THAT PROBIBILTY HELPED MY OUT IN GETTING £6 TO £97 +£91 in 30mins or so.

BUT,BUT,BUT THE probiblity dd not help me out in going mad I looked again at the history spins, (I) didnt see alot of repeating colums and did not see alot of this and hardly any of that, none of that, wow no 1,2,3s. wow no repeating numbers.wow,wow,wow. THE demon (GAMBERLER) was well and truly let lose I was betting on everything everywhere on the board. stupid bets amount even missing out numbers with fifty odd quid on the board layout.

anyway in 15-20mins I was a gonner and was on 0.73p and it was a £1 minimum bet so I was feeling stupid.

some of you will know exactly what im on about here.

I go on sometimes forgive me everyone.



cheers,

mattjono

[TwoCatSam]

MATTJONO

"I was chuffed"

How does know he is "chuffed".  I may have been and not known it!   

Sam

[simon]

Thank you for your responses.... the system I am working is based on what is supposed to happen when there are 3 possible outcomes (the dozens or columns) and all possible outcomes are easily illustrated and identifiable.  If we break down the outcomes into what I will call a "set", and the set involves 3 spins/3 decisions, without going into every possible event and example, there are 27 possible outcomes which break down as follows:  9 sets will start same (A-A-?), and 18 sets will start different (A-B-?).  Of the 9 sets that start same, 3 will start same and end same (A-A-A) and 6 will start same and end different (A-A-B).  Of the 18 sets that start different, 6 sets will start different and end with first decision (A-B-A), 6 sets will start different and end with second decsion (A-B-B), and 6 sets will start different and end with 3rd decision (A-B-C).   I suppose this is hard to follow if you are not involved with it, but my system is based on betting on how these sets will end, that is, I wait for the first two decisions and then bet on the third decision, and start again, and I have a way of effectively eliminating the bets that I would normally lose based on the betting strategy and based on these probability statistics.  I guess the problem is that within 83 spins you cannot expect these 27 sets to occur plus two zeroes to hit, which is what I am basing my strategy on.  So the question is, how CLOSE to the mathematical expectation would these 27 possible events come to happening within 83 spins?  Not EVEN close??  And how many spins then would it require for all these possible events to occur in a percentage equal to what the expectation is?  Or do these 27 possible events not even relate to what the probability of them happening is?  Thanks for any help with these questions!

[simon]

Anotherwords, are the 27 POSSIBLE outcomes of 3 possible decisions (or what I am referring to as "sets"), be PREDICTIVE in any way and if so, how so?   Specifically, can these 27 possible outcomes be predictive of what will occur if we track a certain amount of sets as they occur successively within x amount of spins...

[Tangram]

Hi Simon,

I think it may be easier to bet on trends rather than the expectation that the patterns will all materialize within a given number of spins. Imbalance is much more common than balance. There's an interesting thread by Arteinvivo on this approach which you might be interested to look at. It's based on the even chances but you could adapt it easily to dozens or columns.
nolinks://vlsroulette.com/arte's-studio/variations-on-patterns-distribution-and-combination/

How did you come up with a figure of 83? I made it 105 spins before you get all the patterns (on average).

[simon]

hmm, that's quite the thread... well...

with 3 possible outcomes, there are 27 possible sets.  set= 3 decisions (A-B-C, or A-B-A, or C-C-C, etc...)  If every set that is possible came in one after the other, this would involve 3 x 27 = 81 spins (right?) (plus am adding 2 spins for the zeroe and double zeroe-- technically they should show twice in 38 spins, but in this strategy they also would have to hit on the third decision of the set to cause a loss, and not every set is even bet on, so I am assuming/guestimating will need to overcome 2 losses to the zeroes in 83 spins.)

as mentioned previously there are 9 possible sets that start the same (A-A...) and I do not bet on these, I am only betting the 18 possible sets that start different (A-B...)  According to the statistics, if I bet the first or second decision of these sets to be the outcome of the third decision, I will be right half the time and wrong half the time (technically, I would lose 6 bets and win 6 bets) and I will lose to the 6 possible sets that start different and end different (A-B-C, or C-B-A, etc.)  So if I am betting one dozen or column, it's a wash (win 6 x 2 = +12, and lose 6 and 6 = -12.)  But I have a way to eliminate some of the bets that would have caused a loss with the bet selection that will be employed on all the successive spins of the series, thus hopefully giving me an edge within this series of spins, if a continuous series of spins would reflect results that are similar to the possible outcomes of 3 possible decisions.  also if probability doesn't deliver the expected sets, then I am prepared to make a 9 bet progression on a single dozen that won't kill me if it busts out (at least you don't need as much capital to progress on a single dozen as you would to progress on an even or, heaven forbid, a two dozen bet.)

[Tangram]

I just realized I made a mistake, I got the figure of 105 from here - nolinks://vlsroulette.com/reference-area/some-math-and-the-'law-of-the-third'/

There's a formula at the end of the post which you can use to calculate the number of spins before a full set of events come in. I used N = 27 and the formula gave me 105. But then I realized each pattern is 3 spins long, so this number has to be multiplied by 3 which gives 315. 

I think I'll check this by writing a computer simulation, it seems like an awful lot of spins. But when you think about it, the way you calculated it means that you are expecting one pattern to come in after another, and this would be like expecting all 12 streets (or all 37 numbers) to come in in 12 spins (or 37 spins), and the odds against this happening are huge. There are always many repeats, which is why on average it always takes longer than you'd expect to get the whole set of events.

[simon]

yup that's an interesting post, it will take me a while to digest it.  in the mean time I have 75 continuous spins I personally recorded from six different double zeroe wheels and I am just going to go through them all as I have time and see how the percentages hold up as concerns the dozens and columns, as compared to the percentages of all possible set combinations.  these spins can be divided into 25 sets which is pretty close to what I am basically looking at for a playing session.  so far I have run the dozens of the first wheel and I find that the percentages are not far out of whack with the percentages of all possible set combinations, in fact they are pretty close, very close in some cases (I am looking at 7 different percentages as concerns the possible outcomes of these sets.)  also I will be looking at longest losing streak for the bets I would be making.  but now I gotta get back to the real world.  thank you much for the links and your input.

[simon]

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