Roulette I've seen a new absurd claim that the gambler's fallacy isn't.....

after 490 spins played...

 

You see that 34 out of the 37 numbers returned to their 1/37 as they should! only 3 numbers left. But i'm 100% certain that over time given enough spins, these 3 will also reach their normal average point of 1/37 FACT!
And hey, we also made a nice profit of 2622 is units along the way.

have a nice day.

 

All this is about how random works, not how you should play. please keep this in mind before you test things. it's just given as an example.

 

Is anyone interested in solving the problem I posed?
If we bet for at least x wins in y spins, how do we bet so as to keep our bets at a minimum?

I'm even thinking of offering a small prize to who can solve this, what would be appropriate?

 

So it seems some of you are trying to make an argument for the Binomial distribution helping you get an edge. But there's a contradiction in your argument; the BD assumes that outcomes are independent (which you acknowledge). It also doesn't claim to predict the ORDER of outcomes (again, you recognize this). e.g. in 20 spins the probability that there will be at least 4 reds is 99.81%. Good enough? Not really. So let's say you see 12 blacks in a row and figure that there are 8 spins left in which 4 of them must be red. Why is this flawed thinking in the light of what you know about the BD? Is there a paradox?

 

No paradox from what I see, there is a defined probability for the desired number of occurances in N independent trials.. It is blind to order or events. So in a small sample we expect the certainty to be low.

There is nothing mystical here, the underlying probability provides the basis of expectation and nothing else. As you know flip a coin 100 times and it's very unlikely they will be all heads or tails or even "nearly all heads or tails" because that's such a rare event for a fair coin. Yes it can happen and that's EXACTLY what the binomial distribution calculation that tells us. So if you flipped a coin 15 times recorded the results, and then you flip another 15 times when we add both of those up we expect the 30 fair coin flips will only be all heads 1 in 1073741824 times. It cares not past or future.

So if we said, ok I want to target roulette even chance bets. Say we look at probability of at least 40 EC events in 120 spins which will hold true 4056 times for every time it fails. You may notice in that first 60 spins there is a deficit (less than 20) of reds/blacks, or a deficit of say 5 odds/evens, or a deficit of highs/lows. That means the remaining spins will have at least 20+deficit (e.g. 25) events and this will hold true 4056 times for every time it fails. The inverse of this is also true, you shouldnt be expecting more than 80 of any EC to occur in 120 events but you will witness that once every 4056 times you watch 120 spins.

So how does that help? Well given there is 60 decisions and 25 wins are known to exist in the remaining 60 spins with a certainty of 4056 to 1 or 99.975% confidence. Those 25 or so events in 60 can be played with a labby fairly confidently knowing your win rate is higher than 1/3.

 

I'm afraid not. The probability applies to the whole sequence, not the remainder of the sequence when some spins have already occurred. This is the fallacy I see time and time again on forums, and yes, it's classic gambler's fallacy. The 99.975% confidence you have applies to the 120 spins, NOT the 60 spins remaining. This is because the probability of the 60 spins which have already appeared is 1.0. To find the probability of the remaining 60 spins you need to recalculate your chances of getting 25 wins in 60 spins. You're assuming that the initial probability (for 120 spins) carries over to the last 60 spins. It doesn't. By making this assumption you're violating the principle of independence, which the binomial distribution assumes.

 

@ Median Joe, that is so funny. You are trying to tell the king he is naked.

Try this one on for size you fallacy dependent number crunchers: if the outcome on red/black is balanced for the past 120 spins then when will singles on the weak side happen in the next 60 spins?

And my favorite part, knowing it does not matter because you can't predict the future, not even with magical math. So it all boils down to the suspension of disbelief or just another inconvenient truth. Anyway that you add it up you are bound to get the short end of the stick and some sequence of death moment that balances the algorithm.

 

I remember in the old days we would argue this.

So 10 reds in row have a 1 in 1,376 chance of happening.
9 reds in a row just happened..... the next spin's chances of being red are still 50/50
even though it being the 10th red has a 1 in 1,376 chance of happening, it's still 50/50.
I remember those days. @mr j

 

But it doesn't have a 1 in 1,376 chance of happening, it IS still 50:50. The same logic applies whether you're talking about the very next spin or the next 100 spins. The "probability" of all wins which have already happened is 1 (certain), it's only the probability of future events which are uncertain. It seems obvious, so how come so many system players make the mistake? Don't they notice after their calculations that the reality doesn't match up? They never seem to learn.

 

Median Joe is, of course, correct.
That's how I should have put it, the past spins have a probability of 1 (it has already happened) and the rest need to be recalculated.
Now, knowing this, can we work out how to bet for x ECs in y future spins? We can find the probability, but how about the betting method?

 

I hear you. So that means you'll be accepting the following bet: Take any random source (random.org, Excel, RS, RX, collected spins from any BM casino etc.). Choose one side of 120 consecutive events (R/B, E/O, H/L, Head/Tail - your choice). For each time your choice shows less than 40 times I pay you 1000 EUR. If the opposite happens you pay me 10 EUR. Fair, isn't it?

 

I learned it. I use math to rule out the possibility that math can predict the future.

If the odds on an EC bet are near 50/50 then the only useful tool is coincidence. I even coined a phrase for it, "coincidental change." It's based on the "variable change" used to quantify counting cards. Only with coincidental change you jump in blind to any advantage using the same exact strategy nevertheless. It works like this. You speculate on any advantage that is in a current state of executing 100% perfect. Once it breaks from being 100% perfect you stop using it. That takes situational awareness and depends on a 100% confirmation of an existing condition. MathZombies, that tend to be xenophobic regarding conditional awareness, have a fixation on projecting prediction as a kind of super powerful suit of armor. So they are pleased with themselves at the image in the mirror, Spandex Boy, the new savior of gamblers.

 

If you read carefully I am referring to the entire set of decisions (120 spins) once X decisions are already known the remaining spins plus what we have already seen will contain what the distribution calculation says (minimum of 40 total) this will hold true 4056 times for every time it fails due to the underlying event probability, even if you take every second spin result or even if you decide to include or exclude a future spin totally randomly.

The binomial distribution function is not some dubious add on to probability theory, it IS the calculation that tells us how confident we can be to expect the event probability to hold true. It tells us that we actually need a lot of events to have certainly that the distribution counts approach the theoretical probability of the event we are interested in. This is why there is so much variance in a small set of trials (exactly what the binomial distribution function says) and why with millions of trials it becomes almost certain to the point the probability of failure is too small to compute with tools like excel.

The order of the decisions is also irrelevant as they are all independent events and this is a distribution calculation, NOT the probability calculation of a Markov chain. It is NOT the probability of the next set of events. It is the probability of at least 40 EC occurances in a set of 120 spins collectively.

You cannot say the known events invalidate the future event given the probability itself provides the confidence for ANY set of independent events, past present or future.

If you say it's not true then you are saying probability has no value and that probability distributions are invalid. It would make a mockery of quality assurance processes, reliability of aircraft for travel could not be relied upon based on defect and failure rate probabilities (which are engineered to provide nine 9's reliability). They all use the same equations to predict future failure events within acceptable tolerances, yes FUTURE events and that is why flight hours are strictly monitored as the successful flight hours budget eats into the reliability the more hours an airplane flies, even when the probability of a failure is a billion to 1. Fly a 1000 hours and it will make a difference.

So if you think you can flip a coin 1000 times and they all come up heads with the same probability as getting a rough 50/50 distribution of heads vs tails then you don't understand what probability is.

The probability distribution calculation doesn't even care if you have a large set of historical independent events from a fair coin or wheel and you randomly select a subset of those past results (as one might do using a montecarlo simulation) or even if you randomly select future spins. They will still obey the distribution of the underlying event probability with the degree of certainty provided by the binomial distribution function.

So why don't you get a large set of historical data or a reliable random source and walk forward through it and count all the times where there are more than 40 reds in 120 spins vs less than 40. I guarantee the ratio will be very close to 4056:1 on a European wheel.

Then repeat the above plucking numbers randomly from your historical data. Then do it again taking every 2nd number, then again every 3rd number, every 4th number. Or randomly decide to either pick a past number from a large set or the result of the next spin or randomly decide to include or exclude it before the event. It will make no difference as all events are independent and the probability of the underlying event is the only thing that matters and it alone through the binomial distribution calculation tells us how confident we can be on the number of occurrences we can expect to see.

 

He is of course wrong.

You are both confusing a Markov process (a chain of dependent events) with a probability distribution calculation. They are not even close to the same thing, but I can compute the probability of Markov chain reaching the desired state and then compute the probability distribution on that probability to understand how many occurances I can expect to land in that state (or not) within N independent trials.

I'm sorry if you feel I am repeating myself but there is a level of mathematical illiteracy going on and I'm having to explain the difference between a verb and a noun.

The question I have is if the binomial distribution function DOES NOT tell me the probability of witnessing X occurrences within N trials then please explain how you compute that. I guarantee if you do it the long way working out all the possibilities and add them up you will have re-invented the binomial distribution calculation.

Example: how confident can I be to see 1 head in 3 flips with a fair coin ? (Probability 0.5)

Ok so we want to find the probability of not seeing at least one head. I have the following 8 sequences ALL equally likely (0.5 x 0.5 x 0.5 = 0.125):
TTT
TTH
THT
THH
HTT
HTH
HHT
HHH

So to find the probability of seeing any 3 flips containing at least one head we can see that 7 of the sequences contain a head and the top one is all tails.

So 7 x 0.125 = 0.875 or 87.5% probability of seeing at least 1 head in three flips.

Are you going to argue that is not the case?

You just did the equivalent of a Binomial distribution calculation for a simple case and funny enough the binomial distribution function provides the identical answer we got above.

So we have an 87.5% chance that on ANY three coin flips past present or future that I will see AT LEAST 1 head. You can take any random set or mix of past present or FUTURE events and that holds true 87.5% of the time and it is guaranteed to fail 12.5% of the time because it is based on the inherent probability of the coin being a 50/50 outcome.

How about at least 2 heads? Well that's much less probable as there are only 4 sequences with 2 or more heads so we get 4*0.125 = 0.5 or 50%

Say we wanted to be super confident? Increase the trials or decrease the minimum expected heads. In 20 flips I can be 99.9979972839355% confident of seeing at least 2 heads.

No mysteries or fallacy. If you disagree then the onus is on YOU to show how you compute the probability of witnessing X events in N independent trials.

And if you dare say the future events are not somehow independent trials and can't be included in the distribution calculation then you are now saying they the future events care about the past and the fallacy is on you.

 

Sure you can. Where are the throngs of mathBoyz from the past that missed that one?

 

No. That's gambler's fallacy as I've already explained. I'm not sure you understand the meaning of independence. Let me remind you of what you said in your earlier post :

Let's start from the beginning. You calculate the probability of at least 40 reds (assume you're betting on red) in 120 spins using the BD. This is 99.975% as you correctly state. Now, suppose you see only 15 reds in the first 60 spins. You then conclude that there must be 25 reds (at least) in the remaining 60 spins with probability 99.975%. In order to conclude this you must have subtracted the number of reds in the previous 60 spins (ie, 15) from 40, in order to get 25 which is the expected number of reds (with probability 99.975%) over 120 spins. Do you see how this violates the assumption of independence?

Definition :

Two events are independent, statistically independent, or stochastically independent if the occurrence of one does not affect the probability of occurrence of the other (equivalently, does not affect the odds).

Notice that there are two events in the definition. You are referring to one event (that of getting 15 reds in 60 spins) in order to calculate the probability of occurrence of the other event (getting at least 25 reds in 60 spins), which is a big no no. This is exactly what independence says makes no difference. To get the true probability of the event that you will get 25 reds in the next 60 spins you must recalculate. Using the BD again gives a probability of 88.74%. If you're not convinced, think about other numbers, but use the exact same logic.

To use Turbo's example, suppose there are 10 spins. The probability of at least one red in the sequence is 99.873%. Now suppose that the first 9 spins are black, does this allow you to conclude that the probability of the last spin being red is 99.873%? Of course not, that would be absurd. And yet the logic is exactly the same as your example, which only seems more plausible because the numbers are more reasonable.

Correct, but completely irrelevant to the point you're trying to make, which is that you can use past results to predict future results using the BD. What you need to do is find sets of 60 spins where there are no more than 15 reds, and then see how many times there are at least 25 reds in the next 60 spins. Then compare your results with a set of 60 spins completely ignoring past results. You'll find that there is no difference in the probability whatsoever.

 

Median Joe,

You're talking way above them. They can't comprehend what you're writing. The big problem that people like Twoup, Luckyfella, and Turbo have is they keep looking back at the probability of the whole sequence, even though the spins have already happened, rather than the probability of hitting on the next spin or series of spins. The random walk is something that they will never grasp. In their gambler's fallacy world, they believe that numbers are due and they think that they can use math to beat the random game.

 

Doc,
You're probably right, but it's strange how TwoUp obviously understands the binomial distribution and can compute probabilities, yet seems unable to grasp the much more basic notion of independence. He uses the word quite often, but isn't applying the concept. Maybe it's because these guys desperately WANT there to be winning systems and it makes them blind to certain concepts. After all, if outcomes were dependent there would be endless system fun, and systems could actually work, but without it, none of them do. Bet selection for games of independent outcomes is an oxymoron.

A lot of people have trouble with the idea of independence, and I can kind of understand it if you're a newbie, but some of these guys have been using systems for YEARS. They must just make certain incorrect assumptions and then never check whether the results actually confirm or deny them. It's weird.

 

Take Reading Randomness for an example. It has nothing to do with prediction fallacies. It's checking each spin to see if you are still winning. Your bet selection is nothing more than a guess. It has no magical powers. When you are winning, by coincidence alone, your trend appears to be working. As soon as the trend or pattern loses then the appearance of the trend or pattern no longer works. The secret is getting good at finding coincidences that last longer than others that don't. People give up on trends and patterns because they want them to work all the time. When they discover that these things don't work all the time then they call foul on it all. It's like their brains shut down. If they can't have it working like an ATM machine then they don't want it. So they go down the road in search of systems like math can predict the future.