I want to generate 30,000 random number decisions (spins) but only the numbers 1, 2, 3, and 4. then I want the computer to find the numbers 1, 2, and 3 anytime they follow each other in any order, in 3 decisions (spins). then I want to know how many times the number 4 follows anytime 1 2 and 3 hit in 3 spins in any order (anotherwords, there is no repeat of 1, 2, or 3 on the 4th decision.) I want to know what probability the computer comes up with for event 4 to hit on the 4th spin, after events 1, 2, and 3 follow each other successively in any order, in 3 spins.
This is the basic question I am trying to get an answer to:
We have 4 possible outcomes, which are randomly generated and each outcome always has a 1 in 4 chance to appear. If in 4 trials, on the first trial outcome A appears, and on the second trial outcome B appears, and on the third trial outcome C appears, or outcomes A, B and C appear in any order in the first three trials, what is the probability that outcome D will appear in the fourth trial?
The simple answer would seem to be that outcome D has the same 1 in 4 chance to appear on the fourth trial that it did at the first trial. However, I have been looking at something that indicates that in fact outcome D now has a less than 25% chance to appear on the 4th trial, due to the fact that the 3 previous outcomes A, B, and C also each have a 1 in 4 chance to re-appear, which ends up reducing outcome D's 25% chance to occur that it had at the first trial.
I would like to know what computer results show, and if it indicates that the fourth outcome showed less than 25% immediately following the other 3 outcomes, then that is very good news.
Let's try it without a program, just an algorithm.
On bet number 1 you have a 75% chance of hitting step number one. D is the only loss.
On bet two you have a 50% chance because whatever hit in step one (plus D) is opposite half the choices remaining.
On bet three, step three, you have a 25% of hitting the remaining condition while avoiding the other two (plus D)
I think this last step is right. You have a 25% chance of hitting D on the forth step.
So you do this for four in a row:
75% X 50% X 25% X 25% = 2.34%
If I wrote a simulation that did millions of spins then it would be this outcome, if I did the algorithm right.
hmmm, I did not really follow that but are you saying that the probability for D hitting on 4th spin following A-B-C in any order in 3 spins is only 2.34%? No, you know that's not right, it is somewhere between 17% and 20 or 25%
what I am asking is, if we divide the wheel into four 9 number sections, and 3 sections hit without repeat in 3 spins, what is the probability that the 4th section will hit on the 4th spin?
I'll try it this way:
Step one is 3/4
Step two is 2/4
Step three is 1/4
Step four is 1/4
The formula for the odds for four heads or tails in a row is:
(.5) times (.5) times (.5) times (.5)
Half of .5 = .25, (25%)
Half of 25% is 12.5%
Half of 12.5% = 6.25%
So there is a 6.25% chance for 4 heads in a row.
If there were a four headed coin and on each step one more was off limits until the fourth flip then that alone would even be more rare than a two sided coin. Yes, my algorithm is correct. Were you going to be rich?
what is your final answer for the probabilty for the event I am asking about?
the payouts make it always a wash, with 4 possible outcomes, we can bet on the one vs the other three, or bet on all 3 vs the one (similar to betting 2 dozens against one dozen, or one dozen against two), it is always a wash and the house edge prevails. BUT, if the events I am asking about are not a wash because the expected probability of the 3 events become greater and the expected probability of the one event becomes smaller, than yes you have a winning bet.
Quote from: simon on October 31, 2009, 01:52:42 PM
what is your final answer for the probabilty for the event I am asking about?
Do I get a "lifeline?"
This isn't Slum Dog Millionare --but, my final answer is 2.34%
that was a great movie. so you're saying that with 4 possible events A-B-C-D, the probability for event D to hit on the 4th spin following events A-B-C hitting in any order in the 3 previous spins has been reduced from the 25% probability it had at spin 1, to only 2.34% probability on the 4th spin after A, B, and C hit on the 3 previous spins? Don't you think then, that if you divide the wheel up into 4 nine number sections, all you would have to do is wait for 3 sections to hit without repeat in 3 spins, and then bet on all 3 sections to hit on the following spin, since your probability to hit one of them would be so great, vs the very small probability of the final event to hit on the 4th spin?
Hi Simon,
Gizmo's analysis is correct regarding that particular sequence (that of one of each of 1,2,3, in any order, followed by by a 4) but to answer your question - the probability is 25%, no more or less. It has to be, because you are talking about independent trials, so whatever sequence has come before the arrival of the 4 will make no difference.
If you were to try the bet thousands and thousands of times then you would win 97%+ times with a nine unit payoff for each. In 100 bet groups you would win 97 X 9 = 873. You would lose three times 27 for 81 lost. Yes, it looks very good. I'm going to have to write a new sim. It's an interesting problem looking for the rarity of an exact situation over four steps. The problem is that you will only see that precondition a little less than 10% of the time. I'm going to find a rarity that happens more commonly, and that has similar long shot odds. 27 numbers out of 37/38 is a good bet. It's funny that the single bet odds are 27% to lose that bet, but the conditional bet is 3% to lose, if you round the numbers a little.
Quote from: Tangram on October 31, 2009, 02:24:28 PM
Hi Simon,
Gizmo's analysis is correct regarding that particular sequence (that of one of each of 1,2,3, in any order, followed by by a 4) but to answer your question - the probability is 25%, no more or less. It has to be, because you are talking about independent trials, so whatever sequence has come before the arrival of the 4 will make no difference.
I think it is not quite that simple, because you have 3 events each of which can repeat on the fourth decision, vs one event that has not shown yet. In any case I would love to see what the computer would show.
Quote from: Gizmotron on October 31, 2009, 02:34:39 PM
The problem is that you will only see that precondition a little less than 10% of the time. I'm going to find a rarity that happens more commonly, and that has similar long shot odds. 27 numbers out of 37/38 is a good bet. It's funny that the single bet odds are 27% to lose that bet, but the conditional bet is 3% to lose, if you round the numbers a little.
......... no it happens over and over, I have several ways I divide the wheel up into nine number sections for each of the even chance factors, and I can keep track of all of them as they hit and the sequence I am looking for comes in frequently (especially when playing a rapid roulette virtual machine), and often simultaneously under the 3 factors the way I code them.
Your exact condition of hitting three of four in any order for three spins without the forth position hitting and any of the other three not repeating is a little less than 10% You can assume that any breach of the condition must allow four steps to occur before the next sequence of four starts. I agree that on each spin a new four step sequence can be tested for. I wonder what that does to the odds. If you have a new sequence starting on each spin then I would think that it increases the odds of losing by 25%. You might lose 27% of the time. That is in line with the true single event odds. I think you must let all four spins happen before checking for a new precondition to occur.
I check on a continuous basis not divided up, anotherwords if any 3 previous spins anywhere in the cycle of spins I am tracking shows 3 of the 4 events without repeat, that is the trigger to bet on the next spin for a repeat of any of the 3 events which just hit. I guess you are saying that is a problem for what I am trying to do? this is why I just want a computer to do what I originally asked, so I can see what it comes up with and if a greater or lesser than 25% comes out for the event I want to track.
The condition is a four spin sequence. The math is for exactly four completed spins. If you do it a thousand times then it reaches the statistical normality of 2.34% for it to occur if a thousand four spin sequences happen back to back. If you go by each spin then the odds are 25%.
So let's look at it from another point of view. Every four spins you will have a 2.3% of the D slot will hit on the fourth spin after each one of the other three have hit. After the first three have hit there is a 25% chance that the D section will hit. I think the sim will produce that 25%. I'm not sure what is missing here. I think it's an individual spin and so it must have an individual spin's odds. Tangram is correct. There is no point in producing this sim.
hmmm, that's not exactly what I want to hear but I suppose it's true. the thing is I was looking at a very unique formula that shows that something a little different happens then the 25% expectation for the 4th event after the first 3 hit, although maybe you are right and you can't just look for any 3 spins that show 3 of the 4 events, the 4 spins have to be divided up back to back. which is not a problem, I just want to know if the formula I was looking at has any validity. if possible maybe I can scan the page of the book to show it to you. basically the author says that if we watch 4 spins, and A hits then B hits then C hits, it is a better bet to bet on A or B or C, then D, and he has an interesting explanantion and formula to show why. I just want to know what the computer shows.
Before each four spin segment the odds are 2.3%. After the first spin the odds are a little better. Each spin works step by step toward the 25%,
later on I will see if I can scan and upload a couple pages from the book I am referring to. it is definitely something you have not seen before. It should be read carefully and with an open mind, before concluding if the formula is valid or not.
I doubt that anything in a book will change my mind. On step one, there is a 2.34% chance that all the conditions will come true in all four steps. If step one comes true then there is a 3.12% chance that the next three things will take place. After that it will take 6.25% chance that the last two things will happen. Finally it will take 25% for the last step to occur. I think I did that right.
here are five pages from an interesting out of print book. I have been developing a system based somewhat on the author's remarks. I would like an opinion if his formulas on pages 5 and 7 regarding 3 events and 4 events have any validity or not. thanks.
p.3
p.4
p.5
p.6
p.7
Quote from: Tangram on October 31, 2009, 02:24:28 PM
the probability is 25%, no more or less. It has to be, because you are talking about independent trials, so whatever sequence has come before the arrival of the 4 will make no difference.
I think Tangram is correct here simon, sorry.
That said, I bet four groups of nine all the time, and one of my favoured scenarios is betting the fourth group after three spins with no group repeats, but I bet that group at least twice, and up to four times. Stopping after any win of course.
Hi Bombus, I just posted above five pages from a book, did you read them? (carefully?)
Simon,
There's nothing wrong with the maths, but the mistake comes from confusing a sequence of outcomes with a single outcome. For example, on page 5, the author correctly gives the probability of not getting a repeat in 3 spins, but then goes on to say:
It would certainly be logical to place bets that have already been selected because the odds are in favor of repeats
Implying that you can somehow get an advantage by waiting for no repeats in the first 2 spins and then betting for a repeat on the 3rd, which is a fallacy because the odds of a repeat haven't changed on any particular spin, it only seems as though they have when you look at the sequence as a whole. It's exactly this kind of thinking which leads people to believe that a black is more likely on the next spin because you have just seen 10 reds in a row.
Quote from: Tangram on November 01, 2009, 06:29:04 AM
It's exactly this kind of thinking which leads people to believe that a black is more likely on the next spin because you have just seen 10 reds in a row.
well believe it or not a math major at a prestigious university personally told me that if red hit 10 times in a row he would definitely bet black and I said "but isn't that the well known gambler's fallacy?" and he basically told me that it's not a fallacy and there is still some debate about this because everything has to even out in the end, which was really surprising and confusing for me to hear, and yes I know about the law of large numbers, etc.
Be that as it may, it is the author's specific statements and formula on p.7 under the heading THE FORMAL STATEMENT OF THE IDEA that I want to focus on and find out if the statements make sense and if the formula is valid.
Asking the question - "are spins independent?" goes to the heart of the matter because if they truly are under all circumstances, then no system or method based on past results can possibly work. There does seem to be some evidence to the contrary but it's all anecdotal - no objective test has ever been made that shows that spins are
not independent. However it's not always easy to dismiss the arguments of those who say there is a degree of non-independence, despite the common-sense view that there can't be because every spin is a new event.
Quoteif every table game result is an independent event, how can we ever expect any particular number to come up at all? We can't, because there would be nothing to stop the wheel from selecting a different number, every time. And yet, the same people who say that these numerical events are immaculately independent, expect the numbers to conform with the probabilities. But if such events were truly independent, there would never be a moment, or even a sustained period, when any number could be expected to show up.
There is a causative force that compels numerical events to seek their legitimate place within their assigned probabilities. Whether the dice or wheel have a memory is irrelevant. The influence originates from the effects of statistical propensity, the authority that governs the probabilities of random numerical events.
The key to getting a clear handle on this lies in seeing the difference between viewing table decisions one at a time, or in groups. On a one-by-one basis, it is true that there will never be a time when any number is mandated to appear or not appear. But even in a sampling as small as 3000 spins, you will never see what might be regarded as a catastrophic deviation from the statistical expectation. There's not an unbiased roulette table on earth that can make it through that many spins without our number 8 coming up at least two or three dozen times times.
nolinks://nolinks.thegamblersedge.com/propensity.htm (nolinks://nolinks.thegamblersedge.com/propensity.htm)
There
does seem to be a contradiction here, and I've never heard a satisfactory explanation for it.
Anyway, going back to the "formal statement":
QuoteNow, if we consider the random process from the standpoint of not repeating a selection we see that the situation is the same as that involved in a series of trials without replacement.
I disagree with this. It isn't the same thing at all. Just because the calculations happen to be the same it doesn't mean that the physical situation is too. The only way you could make roulette a game of "no replacement" is by covering each pocket on the wheel every time the ball lands in the corresponding number, analogous to removing cards from the deck in Blackjack.
p.1
I am posting pages 1 and 2 so the other pages are not taken out of context and you can read the author's comments on randomness, etc. and see that he seems to have his head on straight.
(I trust that these pictures of the document are readable to the viewer and that after clicking on them once to enlarge, they can be clicked again to enlarge to the correct size.)
p.2
As I understand it what the author is saying is that if you isolate a 4 spin sequence, and 3 of 4 possible outcomes hit in the first 3 spins, then because these 3 outcomes are eligible to repeat again, the combined possibility of each of these 3 different outcomes hitting again on the 4th spin reduces the probability of outcome 4 to hit on the 4th spin, in the same way that you will never see 36 numbers hit in 36 spins, because once a number hits it is eligible to hit again and the repeats are inevitable.
I am confused about the formula on p.7 which seems to indicate that there is only about a 10% chance for event 4 to hit on spin 4 after the first 3 events hit in the first 3 spins (9.37% chance-- but *anything* less than 25% would be great news.)
Simon, that better chance you are hoping to confirm only happens before all four events take place. You are hoping to have those better odds on the fourth spin. They only exist on spin 1 of the four. I read that paper. You can't combine replacement odds with non-replacement odds to formulate a true equation. Roulette never has a condition of non-replacement. It is ridiculous that the author attempted to pass this off as a valid argument.
BTW, your author has no consideration for the short term effect of the characteristics of randomness in the forms of trends, dominances, and patterns. I believe that most people are stuck on attempting to see a mathematical weakness in multiple spins where one will never be found. If each spin is independent then there is no force or math that prevents patterns and trends from forming from observance of past spin results. They are never predictive but they are always either a continuance or they aren't a continuance.
hmmm, bummer. well I will mull over what you said. thank you for your input.
hmmm, I'm sorry, don't hit me for being so thick but I don't get it. you say "You are hoping to have those better odds on the fourth spin. They only exist on spin 1 of the four." so what has changed just because I am tracking the sequences and wait till spin 4? I guess this is similar to predicting that the next 4 spins won't be all black, which is not the same as predicting that the 4th spin won't be black after the first 3 were? are you saying that the only way to set up the bet correctly is to bet that spin 1 will be A, spin 2 will be B, spin 3 will be C, and spin 4 won't be D and there is absolutely no other way to bet (with advantage) that spin 4 won't be D, and that skipping the first 3 bets instead of playing them achieves nothing? yes that's probably what you're saying, but if for example we track 18 nos without repeat, and if it were feasible to bet on all these 18 numbers till one repeated within 36 spins, wouldn't that be a good bet? It's not feasible to bet on 18 numbers to repeat within the next 18 spins, but the bet I am trying to work is feasible (if the logic was, anyway.)
Regarding the prestigious mathematician, things do tend to conform to a normal distribution, but it doesn't happen every ten or twenty spins....that is a classic misconception. It tends to happen over many more spins, which is why there can never be an advantage in waiting for a specific sequence and then betting with or against it. Even if you bet black after 10 reds, your chances of winning do not change, and the probability of getting another red only becomes smaller if you were betting for black from the first spin. Therein is the fallacy, there's nothing debatable about it.
I'll round the numbers so you can see it better:
spin 1 of 4 = 2% odds
spin 2 of 4 = 4% odds
spin 3 of 4 = 6% odds
spin 4 of 4 = 25% odds
On spin 1 all four spins must come true for the condition to be true all the way. For you to get those odds you must bet all starting points and hope that every one will last to all 4 spins being true. Only that does not happen. In fact, only 25% of the time does the last condition come true after the other three have occurred.
So look at this for the 18. If you bet the other 18 for the next 18 spins you expect to win half of them in the long term averages. If you bet one same number for 18 spins then you expect to win half of those in the long term averages. You are stuck on multiple spins = creative combinations. If you walk up to a table that has had thirty reds in a row you should not expect the number three hitting on the next spin because of the number of times the first street has not hit in the past tens spins. There is no logic in combining spins into multiple combinations of separate events. That is the primary flaw. You can't play multiple bets on more than one spin at a time and get paid on one spin for another.
ok I will think about what you said, I guess you're right (unfortunately.) Six I'm glad you saw this thread, as you can read the document I mentioned to you before when you were experimenting with 2 dozen betting and I recommended backing the 2 previous dozens that hit, but I guess that idea is getting shot down here fast.
Yes, I'll read it. I personally am not saying there isn't merit in what you're talking about, but it sounds a bit fallacious. I actually scrapped the two dozen betting for reasons that I think one of us mentioned backed in the divisor thread. It was based on a similar principle and by default you would bet the last dozen out along with another selected ad hoc. It didn't really work. I have concluded that it's unwise to bet more then 18 numbers (not because the 2 dozen bet selection failed, but because of the wager compared to the return) therefore I concentrate on the ECs now.
Quote from: Number Six on November 01, 2009, 12:05:40 PM
I have concluded that it's unwise to bet more then 18 numbers (not because the 2 dozen bet selection failed, but because of the wager compared to the return) therefore I concentrate on the ECs now.
I have found that I'm better at 2 dozen / 1 dozen betting as a balancing act versus the EC's. If I see a lot of singles I bet the other 2 dozens and if I see a lot of repeats (doubles in series) I bet the same dozen to repeat. It's a slow moving transition when things tend to go one type or the other with consistency. The only thing that kills that is when it's very choppy. At times doubles dominate and at times singles dominate very strongly. If you are not married to two dozens at a time, 24 numbers, then you can adjust to following the last 12 numbered size group, whatever it may be. I track 4 different sets of 36/38 numbers for each group. Two are based on the table layout and two are based on the inside numbers. I just do way better with this than I do with the EC's. During rapid chop I search for something else in other groups. Sometimes I go to the EC's because they are in powerful domination conditions. I just keep looking for styles that I like. So in the end what really counts is session management. I use start BR limits in order to limit downturns. I use win limits to get out of a session while in an up condition. It all runs very closely to balanced. I let all win streaks to go to the first loss.