Topic: (Bustabit) How often do long streaks of red come? (Read 624 times)

Well, I don't really think that is your intention

To make it abundantly clear, I don't say you are deliberately implying (read, insinuating) that the past rolls or flips somehow determine the outcomes of future bets since we both seem to agree with the assumption that the bets are independent of each other. With that said, however, if you don't imply something, it may still follow from what you say. As you can see, it doesn't necessarily have anything to do with my English skills


Generally speaking, you are trying to apply a very abstract (thus basic) concept to a real-life situation but without adding back the factors which have been removed when making this concept abstract. The example with teams A and B clearly reveals the numerous pitfalls of this approach, its deficiency and inadequacy

If you are going to keep on arguing against a point that I am not making and have never made, there is no point trying to help you understand. My statement does not in any way imply that flipping 5 heads somehow facilitates flipping 5 more heads, and if you still think it does despite me explaining this several times then that's an issue with your English skills, not my logic skills.

I am going to make 10 flips.
Each flip has a probability of 0.5 to be heads.
Before I start, I have a 0.5¹⁰ (1 in 1024) probability of my next 10 flips being heads.
After I've made 1 flip, I have a 0.5⁹ (1 in 512) probability of my next 9 flips being heads.
After I've made 2 flips, I have a 0.5⁸ (1 in 256) probability of my next 8 flips being heads.
After I've made 3 flips, I have a 0.5⁷ (1 in 128) probability of my next 7 flips being heads.
After I've made 4 flips, I have a 0.5⁶ (1 in 64) probability of my next 6 flips being heads.
After I've made 5 flips, I have a 0.5⁵ (1 in 32) probability of my next 5 flips being heads.
After I've made 6 flips, I have a 0.5⁴ (1 in 16) probability of my next 4 flips being heads.
After I've made 7 flips, I have a 0.5³ (1 in 8) probability of my next 3 flips being heads.
After I've made 8 flips, I have a 0.5² (1 in 4) probability of my next 2 flips being heads.
After I've made 9 flips, I have a 0.5¹ (1 in 2) probability of my last flip being heads.

Now, if all my flips do end up as heads, then as I progress along my run of heads from having made 0 flips to having made 10 flips, the probability of my reaching 10 heads in a row becomes more likely.

This is going to be my last post on the subject, since I am running out of ways and examples to explain this very basic concept.

Yes, team B is more likely to win the game

But this is not the same as "as you progress along a run of losses, reaching 10 losses in a row becomes more likely, not less". Since we assume that the teams are of equal strength (read, it was a lucky kick for team B), losing the first half doesn't in the least mean that the team A has become closer to losing the second half, but this is what follows from the original example, i.e. losing the first half somehow facilitates losing the second half, i.e. losing all the halves, no matter how many there could be. Actually, your example is sooner proving my point rather than refuting it, as well as making it more clear overall. And this is definitely not semantics

I wonder why some people argue and grasp the concepts differently!

For me, my English is poor but I can say this with true certainty. The more you play, the more likely you are to experience streaks. Whether it is red or green, just by playing much much more you can increase your likelihood to see this. It does not mean you WILL!

Every person who diced millions of hits can understand this;)

@deisik

Assume that there's a football match between team A and Team B. Both teams are exactly equal in strength and the chance of winning is exactly 50% for each of them
After first half ends, all players forget what happened in the first half. So what happened in the first half has no effect on the second half.

Team A lose the first half.
Now the second half is going to be started.
All players have forgotten what happened in the first half.

Do you agree that Team A has a smaller chance to win the game but both teams have equal chance for winning the second half?

Let's come back to the flip game.

First half = 5 first flips
Second half = 5 second flips
Game = 10 flips

You are going to flip a coin 10 times and you will win the game if you win at least one of 10 flips.
You lose the first 5 flips. Do you agree that your chance to win the game has decreased comparing to the time you hadn't flipped any coin?
I am not talking about the next flip. I am talking about the game which is winning one flip out of 10 and you have already tried your luck 5 times.

Which is a statement of fact.

You seem unable to separate the probability of individual flips from the probability of a series of flips as a whole. The probability of any one flip doesn't change. The probability of the series changes are you progress along that series. Those are both simply facts.

You can argue semantics about the way I have phrased the question or heads, tails, flips, rolls, wins, losses, etc., as much as you want, but it is entirely irrelevant. Since you have refused to answer my question, let me answer it for you:

If you have flipped 8 heads in a row, then you have 2 flips left in your series of 10. You have a 0.5² (1 in 4) chance of reaching 10 heads in a row.
If I have flipped 3 heads in a row, then I have 7 flips left in my series of 10. I have a 0.5⁷ (1 in 128) chance of reaching 10 heads in a row.
As you have progressed further along a run of heads, you are more likely than me to reach 10 heads in a row.

None of this depends on a "factor of time" or pays any attention whatsoever to "past events in terms of probabilities". This answer looks exclusively at flips which are yet to happen - 2 in your case and 7 in mine.

If you want to argue against my math then feel free. If you want to argue semantics for no reason then I'm out.

But this is what directly follows from your assertion that "as you progress along a run of losses, reaching 10 losses in a row becomes more likely, not less"


Let me rephrase your question

So, did you stop beating your wife every morning? Yes/No. This question is meaningless if you don't have a wife, obviously, but it is as meaningless if you had a wife and never beat her (barring the situation when you actually have a spouse and beat her regularly). But even the latter condition is itself somewhat suspicious and dubious as it assumes the possibility that you were physically able to beat your wife as it might in fact be the other way around, i.e. your wife beating you, regularly or otherwise

Your question (a set of questions) is of the same type, so forcing me to reply to it (pick up one of the replies), you are as well forcing me to involuntarily admit the fact that "as you progress along a run of losses, reaching 10 losses in a row becomes more likely, not less" in more specific terms and changed ones (heads instead of losses, flips instead of rolls). But it doesn't in the least change or challenge the underlying idea, that you can't speak of past events in terms of probabilities. All in all, you discard a factor of time which is quintessential to the idea of probabilities

I have never claimed otherwise.

As above.

I'm not asking about individual flips. I'm asking about completing the series of 10 flips. There are three possible answers to my question:

You, having already flipped 8 heads in a row, are more likely to reach 10 heads in a row.
Me, having already flipped 3 heads in a row, are more likely to reach 10 heads in a row.
We are both just as likely to reach 10 heads in a row.

Please pick one.

I never meant it that way


I intentionally decided not to answer this question

Because you are complicating matters beyond what is necessary here (30 seconds, two players instead of one, etc). Regardless, let's just say you are only adding another layer of complexity which doesn't add anything in terms of clarity and understanding. Everything is the same as before as far as the original example is concerned, i.e. the fact that I flipped a coin 8 heads in a row doesn't make it more likely that I'm going to continue flipping heads (provided the coin is fair, of course, and my flips are entirely random, i.e. no skill is involved)

Put differently, my chances of flipping heads in the remaining 2 flips haven't and couldn't have changed since both you and I proceed from the assumption that flips are independent of each other. So these chances remain the same as they were before you or I started the series (whatever they might be), apart from being a purely abstract construct (if we talk about a series of events). It is this seeming connection which leaves you susceptible to the Gambler's Fallacy reversed. In other words, your new example is as meaningless as the first one since it can be reduced to the latter

Just lol. Lets just insult me rather than address your misunderstanding?

Please answer this question:

And so what?

If there are fewer flips remaining, how does it make these fewer flips more dependent on your previous flips if there is any dependence at all (because there is none, which you seem to agree with)? As I can see, well, as you can see, your frustration is in fact directed not so much at me but rather at yourself

Basically, you are trying to apply an abstract idea, that of odds in a series of events like coin flipping to a real life situation (that is, coin flipping, huh), and arrive at impossible inferences (which you dare not speak out loud) like previous rolls or flips affecting the remaining ones. But I understand your cognitive collapse

We can't precisely predict when does 15-20 red streak come and go, because that is gambling we can not predict the accurate result or outcome of every game, we will only hope for our luck on that day, but there are some ways to know what would be the output of the game, based on what I had read before a man won in the lotto because he observed and analyze the process of the draw lots, he is a statistician, so by doing some statics of the lotto for four months, he came up with the correct output. He is now a millionaire because of that. You can make statistics In every gambling that uses a machine to operate the game because the same process will be applied to start the game. The machines are already program to start the game with the same process, so if you will make ways and if you are going to do some research, then probably you will win the game. Gambling is better if you win the game, so for you to win the game, you should do some research and analyze the process of the game. It is better if you are going to make a way to win than to wait for you to win.

i've ever got 4 streak losses at 1.1x multipler. And more than 10 at 2x multipler in long run. Get 5 streak losses is very normal happened on bustabit.

Just how martingale only works if you have unlimited funds, j think that for the sake of clarity we should never involve the law of averages with gambling. Exlecially for the newbies, it must be made clear that their possibilities of their wagers are not connected. Everything else is just fluff.

Can you please be more specific, 5 loosing streak at what odds? Because, bust-a-bit is not a dice game so can't assume you mean at 49.5%. If you mean immediate crash that is at 1.0 then I have to admit you have been really unlucky. For me I never got more than 2 at max.

Indeed and talk of any strategy they cannot turn the odds in your favor, some strategies tend to be slow starters and explode like martingale does while some explode initially and gets smaller as it goes on - an example would be something like betting on 10% win chance.

I was reading through and was wondering the same and you probably said almost what I would. I would just like to add the fact that if you are talking about loosing streak on a expected cashout at 2.00 odds then I can tell you that you may even have bigger streaks than 15 because it's similar to dice and I have seen streaks of 13-14 often at primedice and had a streak of 9 myself not so long ago.

For years, people are researching about how to overcome the continuous losing streaks and unfortunately no one is coming up with any possible solution. It shows, it cannot have any solution but it is the solution for houses to beat us when try to beat them through martingale.

That is a common mistake for new gamblers ( that bolded one). That they actually believe that gambling could be some of sort mathematics calculations, making winning patterns and use in gambling is not actually it works but some gamblers did it.

Definitely, here in gambling there is no 10/10 winning streak and it might happen that 0/10( that is near to possibility). This is means that luck is much needed in gambling to win not that mathematical calculation.

This is not a simple question

What bets are you talking about, flipping a coin, rolling dice or spinning a roulette? What do you mean exactly by winning or losing here? Without defining all these specific conditions, your question is meaningless on its own as there can be every possible answer like losing 10 bets in a row less likely than losing 5 bets in a row

Because you have fewer flips remaining! This is a very basic concept that you are struggling to grasp.

You and I both have a fair coin, and we are both going to flip it 10 times in a row. Every flip has a 1 in 2 chance of being heads. After 30 seconds, you have flip 8 heads in a row. I'm slower at flipping than you, so I've only flipped 3 heads in a row. Who is more likely to flip 10 heads in a row? You or me?

---

Thank you! I feel like I'm banging my head on a wall here.

As I said before, any single flip has the same chance of being heads as any other single flip, but a series of 10 flips all being heads becomes more likely as you flip 1, 2, 3, 4, etc. heads because there are fewer flips remaining. This is simply a fact. I don't understand how we can still be arguing it.

I ask you a simple question.
Which one is more likely? Losing 10 bets in a row or losing 5 bets in a row?

Assume that you are going to make 10 bets and your purpose is to win at least one of them.
Before the game starts, you are allowed to try your luck 10 times.
You lose 5 first bets. Now you have lost 5 of your bets and you can try your luck only 5 times. It's clear that your chance has been decreased. Because you are no longer allowed to make 10 bets. You are allowed to make only 5 bets.

As the game goes and you lose the bets, you get closer to completing 10 losing streak.

Too many people seem to believe that past results are an indicator of future occurrences.

This is exactly why anybody that doesn't have at least a basic understanding of probability or mathematics should absolutely stay away from any type of gambling.

Otherwise, they're just going to get rekt eventually. Know your game BEFORE you play, not AFTER.

I kinda agree but if we actually see every bet independent, shouldn't we say it's actually 1/2 since for every bet the odds are half only. It could have been 9 heads over 9 tails and then the chances of hitting 9 heads is 1/2 as well.

Indeed it has nothing to do with what has come before

However, that is in direct contradiction to what you have been saying earlier, i.e. "as you progress along a run of losses, reaching 10 losses in a row becomes more likely, not less". Really, how can it become more likely if previous outcomes are in no way determining future ones? I'm just rephrasing what follows from your own words

Anyway, I think you should stop oscillating between these two mutually exclusive points of view, don't you think? And I already explained where your reasoning fails and why you are, or should be, confused. In simple terms, because there is neither more nor less likely in respect to events that have already occurred

No, it doesn't. The 1 in 2 chance of the final 10th flip being heads says absolutely nothing about what has come before. The chance for the 10th flip would be 1 in 2 regardless of if we had just flipped 9 heads, or 9 tails, or literally any combination of heads or tails. The odds that those 9 flips in a row would all be heads would have been 1 in 512 (0.5⁹) before we started flipping, but now they have happened so the odds are irrelevant.

You keep making this statement that I am applying probabilities to events in the past when that is the exact opposite of what I am doing. I am specifically excluding past events from the calculation of probabilities for future events.

We do seem to be going round in circles here.

It seems that we have entered a circular argument here. But you can still see how fundamentally incorrect your assumption is if you take a peek at the other part of the equation. Okay, you flipped 9 heads in a 10 flip series, and now you say that the odds of hitting 10 heads have become 1/2 for that series. So far so good, and everything seems fine and quite in line with elementary logic, right?

But that necessarily means the probability of hitting 9 heads in that series of 10 flips has always been 1/2 all the time. Don't you feel something is inherently wrong here? If you don't, I'll explain (once more). What is wrong here is you judging past events from a probability point of view which you can't do. There is no such probability as 1/2 with respect to 9 flips in this series as it is and has always been whatever it is for any other first 9 flips in a series of 10

Note, that it doesn't in the least mean that the odds of flipping tails after 9 heads in a row have suddenly become 1023/1024 (or whatever)


I don't challenge this. I challenge the opposite view, which is in fact not very far from the Gambler's Fallacy in its roots

Agreed although there is law of averages that tends to event out things but each roll and in case of Bustabit each round is independent of the previous round and there is an absolute certainty that you can have a large number of crashes at 1.00 although I don't recall having a crash at 1.00 more than 2 times at max. It's almost like loosing a bet on 99% in dice.

From my experience in bustabit most of times you will see the rise from 1.00 to 1.5 mostly and then if it passes 3.00 then it sometimes goes really big, but again that's my experience and someone else would have their own numbers.

That's obviously not what I've been talking about, though. I'm talking about the odds of a series of 10 flips all being heads as your progress through that series.

That's exactly what it means, for that given series of 10 flips.

I'm not applying odds to past events. That's the exact opposite of what I'm doing. I'm intentionally and specifically not including past events when calculating the odds of future events.

If I have already flipped 5 heads, then that is a fact. The odds are irrelevant. The probability is irrelevant. Any calculations are irrelevant. It has happened.
The odds of my next 5 flips being heads are 1 in 32.
Therefore, the odds of me completing my run of 10 flips and all of them being heads is currently, at this present moment in time, 1 in 32.

Not recalculating odds as you progress down a run is the very essence of the Gambler's Fallacy, which is the point I originally made. A gambler at the start of the run of 10 thinks "I only have a 1 in 1024 chance of all 10 flips being heads." After flipping 5/6/7/whatever heads, because he thinks at that moment in time he still has odds of 1 in 1024, he erroneously thinks "A tails will definitely come up soon!"

Could you explain it in a couple of sentences?


Yes, the odds of getting 10 heads in 10 flips yet to be made are 1/1024 regardless of what has already happened


Since you are posting this, it means that you are still confused about what I'm talking about

Your next flip has 1/2 chance of being heads (or tails, for that matter) as any other single flip you make, but it doesn't mean that the odds of hitting 10 heads in a given series have suddenly became 1/2 after you flipped heads 9 times before. To repeat, the odds are applicable only to future events, not the ones that already happened, so your inference would be fundamentally incorrect as far as probabilities as such are concerned

It may seem like nitpicking or some useless mental gymnastics, or whatever, but in reality it is fundamental to our understanding how odds actually play out in real life. Being able to calculate them purely arithmetically (as you do) without being aware of the fact that you are dealing with what can be loosely called "undefined behavior" exposes you to a bunch of pretty ugly mistakes and misjudgments

The round is randomness, why you still thinkin about after red must be green?

There is no definitive number. If there was, then everyone would be drowning in money.

I don't get people who are trying to find out how long red streaks are in this regard. There is no limit to them. There is always the tiniest odds that you'll be completely overwhelmed by some black swan event. And since expected value is negative in the long run, I don't see the point of preparing for these streaks.

Have some fun and relax. Don't take gambling so seriously as if it's a profession.

If you're going to get a huge streak of reds, then it's important to cut your losses sometimes.

One of the most common issues with the mentality of most players is that because they've already seen a huge streak of red, that the next bet is almost certainly going to be green.

These people then up their wager and end up getting ruined when the red streak continues. As such, sometimes it's better to exit on a small loss, rather than risk a huge one.

Yes, you absolutely can. Please read about conditional probability as I suggested above.

If you don't recalculate the odds like you claim you aren't allowed to, then what you are stating is that the odds of 10 flips all being heads is 1/1024 regardless of what has already happened. So if I flip 9 heads, by your logic, my next flip only has a 1/1024 chance of being heads and a 1023/1024 chance of being tails, which is obviously nonsense.

You obviously don't take into account the factor of time (which is crucial to the whole idea of a random event)

As you are evaluating the probability a series of flips at the moment before you started flipping (i.e. with 10 flips still ahead of you) after you have already made some flips. In other words, your inferences are correct but only when you remove the factor of time from the equation (which you can't). If you don't, the probabilities of past events no longer exist. Note, it is not like they are 100% as the idea of probabilities itself makes no sense in respect to events that already happened with their outcomes known to us

When talking about probabilities we are talking about events that will either happen in the future or their outcomes are still unknown to us. Simply put, you can't reevaluate the probability of the whole series of flips after you have made some flips and already know their outcomes as this violates the definition of probability as a description of how likely an event is to occur or how likely its certain outcome is. In this case, the event in question will be the outcome of 10 flips like all being heads, so you can't reevaluate it after you already flipped a few coins ("becomes greater")

This is exactly what I am saying now:


No, it isn't. The chance of any single flip being heads is exactly the same as any other single flip being heads, but the chance of all 10 flips eing heads becomes more likely as you progress through the series with all previous flips being head.

I am not wrong here, so I would suggest instead of arguing against me you try to find out where your misunderstanding is. As I stated above, it's a common misconception. This is called Kolmogorov-type conditional probability, if you want to look it up. If there are two possible events, called A and B, then the conditional probability of A given B, (i.e. the probability of A occurring if B also occurs), is given by the following equation:

P(A|B) = P(A∩B)/P(B)

P(A|B) is the conditional probability of A occurring given B occurring
P(A∩B) is the probability both A and B occur
P(B) is the non-zero probability of B occurring

Lets take A to mean "flipping 10 heads in a row".
Lets take B to mean "flipping 5 heads in a row".
P(A) is therefore 1/1024.
P(B) is therefore 1/32.

As P(A∩B) is the probability of both happening, this therefore becomes the same as the P(A) (the probability of A alone), because to flip 10 heads in a row, then you must already have flipped 5 heads in row.

P(A|B) therefore becomes P(A)/P(B). P(A) is 1/1024, and P(B) is 1/32. P(A|B) therefore becomes (1/1024)/(1/32), which is 1/32.

So, although the chance of flipping 10 heads in a row is 1/1024, if you have already flipped 5 heads, then your chance of reaching a run of 10 heads in a row is now only 1/32.

This is not what o_e_l_e_o meant.
Reading previous posts, it's completely clear that o_e_l_e_o knows that different rolls are independent from each other.

When you start a game, the chance of hitting 10 reds in a row is 0.5¹⁰.
If you lose the first roll, chance of hitting 10 reds in a row will increase to 0.5⁹. If you lose the second roll, chance of hitting 10 reds in a row will increase to 0.5⁸.

What makes you get o_e_l_e_o wrong is that, you are talking about next 10 rolls following the final red you have already hit, but o_e_l_e_o is talking about 10 rolls including previous losses.

If you lose 5 rolls in a row, the chance of hitting 10 reds in next 10 rolls doesn't change. That's still 1 in 1024. But the chance of losing 10 rolls in a row (5 previous rolls + 5 next rolls) increases.

Okay, let's do some quoting here


This is not what you say now

Indeed, people get it wrong thinking that when they progress along a streak of losses (reds), their chances of hitting a green become more likely as they remain the same as ever. But it is as wrong to assume that these chances become less likely (the opposite of more likely) for the whole streak of 10 losses in a row. Why you can't do that I explained in my previous post. You implicitly evaluate and take into account chances of past events where there are no more chances, to begin or end with

I guess i'd consider forfeiting in gambling and accept the losses rather than continue playing than chasing losses we all know that when you are emotional and desperate, you commit series of bad decisions.

But if you still want to make bets, maybe bet on a higher percentage then roll and slowly generate revenue to also create a new strategy.

That's not what I'm saying. The entire point I'm making is that his chances of hitting a red on any individual roll are exactly the same, regardless of what happened before.

Emphasis mine. This is the point I am making. Take this example. You are going to flip a fair coin 10 times. You correctly calculate the odds of flipping 10 heads to be 0.5¹⁰, which is 1 in 1024. You begin flipping. After 5 flips, you have flipped 5 heads. Your odds of reaching 10 heads is not still 1 in 1024 when you have already flipped 5 heads and only have 5 flips remaining, because the first 5 flips now have odds of 1 in 1 because they have already happened. These past events have no bearing whatsoever on your current odds. To still say you have odds of 1 in 1024 to flip 10 heads, when you've already flipped 5 heads, is completely wrong. Because you now only need to flip 5 heads, and precisely because there are no odds in respect to past events, your odds are now only 0.5⁵ which is 1 in 32.


The probability of any one flip is identical, regardless of what came before.
The probability of a series of flips all being heads becomes greater as you work along the series, because there are fewer flips left to make.

You are still confused

And let me explain where and why exactly. But before we proceed, let me also say that you are right about the Gambler's fallacy. But you are in fact using the same logic and reasoning as that insane gambler, though reversed and inverted. The fallacious Gambler thinks (as you correctly note) that the farther he goes down the losing streak, the higher are his chances to hit a green, which is rightfully wrong (pardon this choice of words but it seems to be confusingly appropriate)

However, you essentially make the same mistake by claiming that the farther the poor mate goes down the losing streak, the higher are his chances to hit a red. But the truth is his chances are exactly the same. And here's the root and source of your confusion. You recalculate the odds of hitting 10 reds in a row after you already hit 5 reds, i.e. in  hindsight, which you must not do as there are no odds in respect to past events. Isn't it essentially the same mistake as the fallen Gambler's opposite assumption?

If we look into the statistics provided by the houses, it will be always tempting anyone to try like the top one, that is the reason always all that gambling houses are providing easy access to leaderboards. I believe gambling industries are surviving by marketing like "see how other people are winning and why not you try to win like that".

That is true. Strategies cannot bring luck but it may help to maximize the chances of being lucky. Mathematically strategies could win but our luck may play "destroying role" against those calculations too.

I can't understand what do you mean exactly, streaks of red when? Anything above 1.00 is green (1.01 is already profit) and I have never seen 1.00 (immediate crush) to be shown in a row for 15 times and higher, haven't even seen it 2-3 times in a row personally. Will be glad if you make your question more detailed. If you are going to build some strategy, look at bustabit's Leaderboard and check profiles of users you would love to see and there you'll see statistics of them, how much they bet, when, what was the result and etc.
Good luck but in overall strategy isn't very beneficial, luck is luck. You can't get luck by strategy but you can get it by accident.

That's exactly the point I am making.

The chance of rolling 10 reds in a row is 1 in 1024. If you have already rolled 5 reds in a row, those odds for those 5 rolls no longer count because they have already happened. They have a probability of 100%. You only have 5 rolls left, and so your chance of rolling 5 more reds is no longer 1 in 1024, but now only 1 in 32 because the previous rolls are irrelevant. Past rolls have no bearing on future rolls.

People make the mistake of not realizing that as you progress through a run of reds, your chances of reaching 10 (or whatever) in a row becomes more likely, not because previous rolls have any bearing on future events, but precisely because they don't. They have a probability of 100%, and so can be excluded from any probability calculation.

Well, I tend to disagree with you here, at an entirely fundamental level

It could be said that you are basically making the same mistake as the proverbial fallacious Gambler, even though in reverse, obviously. Look, chances make sense only with respect to future events, right? So before you started to roll the chances of hitting 5 reds in a row in a series of no matter how many rolls longer than that are 1 in 32 (or whatever), but after you rolled 5 times and actually hit 5 reds, the odds of that event no longer make sense as it already happened

And it is not like they are 100% for the simple reason there are no more odds as the event has already transpired. What I mean to say is that the probability of a series of independent events is an entirely abstract idea, with the implication being that you should consider it as such (i.e. as only a theoretical concept), and only when dealing with future outcomes (read, not in the middle of a series of rolls). But I understand your confusion

I wouldn't say it make "no sense", but I can see where you are coming from. Perhaps if people thought "With each of my 10 rolls, I have a 50% chance of losing" as opposed to "I only have a 0.1% chance of losing 10 times in a row", they would be less likely to follow a silly strategy like Martingale.

What people also seem to get wrong is not understanding that as you progress along a run of losses, reaching 10 losses in a row becomes more likely, not less. Lots of people think, "Well, I've just rolled 5 reds, I'm bound to get a green soon", when actually after already rolling 5 reds the odds of rolling 5 more has changed from 1 in 1024 to only 1 in 32.

To make things clear, I do not deny that rolls are independent of each other

Buy if we assume that (a correct assumption anyway), then, as I see it, we shouldn't calculate the odds of hitting 20 reds (or blacks, or whatever) as such calculation (and probability thus obtained) makes no sense in the real world. And that gives us a clue why so many people fall for this infamous Gambler's fallacy

People understand (purely mathematically) how to calculate the probability of hitting 20 consecutive losses, and that makes them feel that the rolls and their outcomes are somehow linked to each other. In other words, if it were explained that this probability is only an abstraction, people would be more cautious with martingale and similar strategies 

Ahh, I see where you are coming from now. My apologies. If you reset to the beginning of a new set of 10 every time you win and essentially ignore the remaining rolls from the original set then yes, your equation is correct. Each set begins with the first roll after a win, and can have number of rolls from 1 to 10 in it. As you say, y then becomes the number of wins, rather than the number of sets of 10.

Let's say I lose the first roll and the second roll and win the third roll. Now I can say that I have won one of 10 first rolls. As all rolls are independent from each other I can say that I skip 4th-10th rolls and the next roll is 11th one. Then I lose 11th-20 rolls and my balance is emptied.

(1-x)^y doesn't give the chance of not having 10 losses in a row in y consecutive rolls. But it gives the chance of winning when you use martingale strategy and your purpose is to win y rolls.

I think there were some problems with my wording.

This statement is correct, that the run of 10 obviously resets when you win, but that's not what your (1-x)^y equation above calculates. I'll try to give examples below.

Your equation of (1-x)^y assumes each set of 10 is entirely independent of each other set of 10. In 1000 rolls, there will be exactly 100 sets of 10 consecutive rolls.

___________	__________________________
First set	01 02 03 04 05 06 07 08 09 10
Second set	11 12 13 14 15 16 17 18 19 20
Third set	21 22 23 24 25 26 27 28 29 30
___________	__________________________

And so on. This is obviously not how Martingale works, because if you win on roll 3 and 17, and lose on all the others, then you will have a >10 loss streak, but your calculation doesn't take that in to account. Your calculation looks at the chance of losing 10 rolls out of 10, over 100 completely separate sets of 10 rolls which are independent of each other.

What actually happens is that any consecutive 10 numbers can make up their own set, as such:

___________	__________________________
First set	01 02 03 04 05 06 07 08 09 10
Second set	02 03 04 05 06 07 08 09 10 11
Third set	03 04 05 06 07 08 09 10 11 12
___________	__________________________

And so on.

That's why it is actually far more likely that Martingale will result in your going bust than your calculation would suggest.
(1-0.0009766)¹⁰⁰ (100 separate sets of 10 rolls) gives a chance of one of those sets of 10 being all losses of 9.31%
The real chance of rolling 10 losses in a row in 1000 consecutive rolls is 38.55%

I completely understand you. I thought about this when I was calculating the probability.

Assume that I am using martingale strategy. Whenever I lose a bet and, I double the bet amount and whenever I win a bet, I return to the base a bet.
Given my balance, after I return to the base bet, I can lose 10 times in a row.

After I win a bet and return to the base bet, I must win at lease one of 10 following bets.
If I have a win on the third bet, the first set of 10 is finished and now I am on the second set. Because in the first set, I need at least one win and I have already made that. Now I need to win one of the 10 bets of the second set.
Once I win a bet, the set is finished. I can skip the remaining bets of the set and go to the next set.
As all bets are independents from each other, I think my assumption is true. Maybe I am wrong.

Any idea this community will be giving you out, definitely will not save you. If someone says once in 10 attempts and when you will be gambling it may occur for every 5 attempts. Because, your bad luck may work quicker than you could imagine. Even you will be setting are low base bet amount be preparing for withstanding against 50 continuous red streak, you may face 55 continuous red streak. This is how, our fate will work. You cannot escape.

I am actually sorry for my discouraging words but what I am saying is 100% from my own experience. One day or other, you will be beaten in gambling. Yes, we cannot escape with profits or we will not find an ending to our gambling until facing running short of bankroll.

I think It's not a new game. I played it before around 2016 and It is one of my favorite gambling game on my whole record on playing different gambling games.

I've experienced a red streak on playing on Bustabit before, I think it is depending on the odds or multiplier you are playing with, In my case, as I remember I am targetting 1.5x on every play I make and I win 8/10, I am skip playing, It means I play for a while then take a rest on playing then come after. I don't play straight on Bustabit.

The chance of getting 10 losses in a row in such a large run is actually far higher than that.

Your calculation gives the odds of not getting a run of 10 losses when considering separate sets of 10 rolls each. Provided there is a win at some point in those 10 rolls, then that set doesn't count as an overall loss and you look at the next set of 10 rolls. That's not how Martingale works though. Martingale is a continuous set of rolls, and isn't subdivided in to sets of 10. In your calculation, if the first set of 10 had a win on the third roll, and the second set of 10 had a win on the seventh roll, that isn't accounted for despite there being 13 losses in between the two wins.

For example, the chance of flipping heads 10 times in a row is 0.5¹⁰ = 0.0009766
Using your equation (1-0.0009766)¹⁰⁰ looks at 100 different sets of 10 flips, and probability of getting 10 heads in any one set is 0.0930861.
If instead of looking at 100 sets of 10 flips we look at 1000 continuous flips, then the chance of getting 10 heads in a row is now 0.38545, which is obviously far higher.

---

Payout of 4x is a 24.8% win chance, and so a 75.2% loss chance. So 0.752¹⁰ = 5.78% chance of happening. So yes, pretty common.

There is no way for you to predict when the 15-20 red streak will come and go, Bustabit is a new gambling website with a revolutionary protocol to take your online gaming experience to the next level. Upon creation of a Bustabit account, the user will have a Bitcoin address with which they can deposit bitcoins to begin playing. Gambling is very unpredictable. If you won the game, it was a matter of luck. There is no winner in every gambling website because most of the time, you already spend so much money because of continuous losing then, if you win, it won't reach how many bets you previously spent. The casino or the website is always the winner in every gambling site.

Payout 4x and you got over 10 streak losses is very normal to happen. It's same when you played dice game with25% win chance, and how long streak losses you get?

Huge balance isn't enough to be able survived in long run with martingale strategy, you need unlimited balance to make it happens

when i was new on crypto i never tried bustabit but i have played a games simillar to bustabit . getting 5 streak loss is verry normal  , i rememer i got over 10 red streaks when i set the payout to times 4  . the results are random but some says its on the seed that affects the game  but im not convinced yet though i observed that red streaks come often after you hit alot of green streaks for a long period of time   . try to keep an eye out of it next time you play and see if works on you too  . lastly the martingale  . well you need to have huge balance to make this effective in the long run  and you better dont afk it because i think the system knows if your afk or not , and will give you tons of reds  .

I tried to explain this to the OP in my previous post, but seems like they dont wish to acknowledge the posts nor do they want to discuss the same.

o_e_l_e_o () has also shown some important pointers about this topic and I wished the OP would actually come back to reply to some of these posts.

He has opened another thread here - https://bitcointalksearch.org/topic/m.53771714

I guess the OP is desperate to make some profit from these games.

I have played Bust-a-bit when I was new in cryptocurrency and the worst that I had come across so far was 5 streak in a row. I usually stop playing for a few rounds when this occurs. I tried figuring out when it usually occurs but I guess this is random since each run is generated by a random algorithm.

Also, martingale doesn't work long-run which everyone is already aware of. Have tried this several times before and if you will keep repeating it for the long run you will end up bankrupt.

To add what o_e_l_e_o said, 0.11% is probability of losing 10 times in a row when you are going to play only 10 times.

As the chance of losing 10 times in a row is that low, some people think that they can use martingale strategy and get rich.
Here is how martingale strategy decreases the chance of winning.
Assume that you have 1 BTC. The chance of doubling your balance with one wager is 49.5%
Some people think that they can wager 1 mBTC instead of 1 BTC, use martingale strategy and double their balance easily.
In this case for doubling the balance you must win 1000 times and you can lose up to 10 times in a row. As o_e_l_e_o said, the chance of losing 10 times in a row is only 0.0010787. But the chance of not happening this 1000 times is (1-0.0010787)¹⁰⁰⁰ =  0.34 = 34%

So, the chance of losing 10 times in a row depends on the total games you are going to play too.

For example, if you want to continue to gamble until you win 10,000 times, chance of losing 10 times in a row at least one time is 1- (1-0.0010787)¹⁰⁰⁰⁰ = 99.99%

You mean for the game to continuously crash at zero? That's not very common, but don't be caught up in gambler's fallacy also. Each run has it's own chances of going red, it's not affected by previous runs in a magical cosmos of chance. Not everything is related.

Assuming we are talking about something which has a 50/50 chance of happening (such as flipping a fair coin) here:

You can work out exactly how "close to impossible" it is. The probability of flipping at least 400 heads out of 500 flips is 8.29815×10^-44, so yes, very close to impossible. You can use standard deviations to see the probability of the results not being far from each other, as you put it. At 2 standard deviations from the mean, 95% of the time you would end up in the range of 228 - 272. At 3 standard deviations, 99.7% of the time you would end up in the range of 217 - 283.

The mistake I think you are making here is conflating "all rolls" with "individual rolls". Let's make the numbers smaller to make it easier to follow. Let's say I flip a coin three times. There are exactly 8 possible outcomes:
HHH
HHT
HTH
HTT
TTT
TTH
THT
THH

If you look at all 8 results, you will see there is only 1 which has no heads, so flipping three tails has a probability of 1 in 8, and flipping at least 1 head has a probability of 7 in 8. Now, lets say I have just flipped the coin twice, and flipped two tails. Look at the two I have made bold. I am going to flip a third time. The other 6 out of 8 options are irrelevant at this point, because the probability of any of them happening is now 0, because I have just flipped two tails in a row. That means my next flip only has two possible outcomes - TTT or TTH, therefore a chance of 1 in 2. What happened before is irrelevant.

Now lets go go back to the start of this example and look at my two possible final outcomes - TTT or TTH. From the very start, before I flipped anything, they are both as likely as each other. Both have a 1 in 8 chance of happening. This is the same when we scale things up. Flipping 19 heads in a row followed by another head is exactly as probable as flipping 19 heads in a row followed by a tails.

Not on bustabit but have played on some other crypto casinos, thing is if you play (read: gamble) for hours they are far more frequent then one can imagine.

With not much details mentioned by OP the payout to be assumed here is 2.00 since that's the default settings on most or nearly all bitcoin casinos when you start betting.
Like the others have said predicting long streaks wouldn't help and I agree with them because even if you do know the percentages it could work against you because there can be certain days where these streaks happen more frequent than usual.

It's most likely to be crash since the title mentioned bustabit.

Thanks for the link. And for the lesson.

I cannot argue against what is proven in arithmetic as it is an absolute logic. However, is it not also backed by arithmetic that if you make 500 flips or rolls, for example, at 50% chance of winning, even if the results would not exactly balance out against each other, it will definitely be not far from each other, or the discrepancy will be at the minimal level, right? It is close to impossible that out of 500 rolls, 400 turns out red and only 100 turns out green.

"As long as I keep going, I'll definitely get a green eventually." I guess this is correct, although your balance will have to be able to support every next roll in order for you to eventually arrive at that point. Although it is possible for a fair flip or a roll to give you 100 consecutive red, it is not probable. The main problem with Martingale is that it requires an amount of resources that is almost unlimited.

You are wrong. This is the classic Gambler's fallacy.

You are correct in saying you have to take in to consideration the number of times that you will roll, but only before you have started rolling. Once you have started rolling, previous rolls make no difference whatsoever to future rolls. If you've rolled 19 reds in a row, you might have only had a chance of 0.001% to do that before you started, but there is a 100% chance that just happened, because you just did it. The chance of the 20th roll being red or green is therefore exactly the same as the first roll being red or green.

This line of thinking is common, but completely false. It leads to the creation of betting systems such as Martingale, where people think "As long as I keep going, I'll definitely get a green eventually". Each roll is completely independent of other rolls, and the Martingale system bankrupts people daily.

Every outcome on bustabit is a series of 10 million pre-generated numbers whose median is 1.98.
So, there could be any number of reds or greens as it's totally random. AFAIK, hypothetically, there could be thousands of reds without a green.
But the largest numbers of red we've seen is around 20 or something till date.

Please correct me if I'm wrong. In terms of computing probability you will have to take into consideration the number of times that you will roll, which means that individual rolls, although seemingly independent from each other, are actually somehow connected. Although pre-rolling is a myth to me, the rule of probability will tell you that after rolling 19 times in a row and you got red in all of them, that 20th roll has a very high probability of giving you a green. Even if that next roll has 50% chance of winning, since we take into consideration the 19 consecutive reds in the previous rolls, the probability for a green is high.

You can calculate the odds quite easily. Simply raise the odds of each red happening to the power of the amount of times that you want to account for.

However, one of the more common fallacies in this area is the fact that rolls are not dependent on each other. A lot of people do prerolls and all that in the hopes that they can somehow increase their odds of winning, but what they don't realize is that prerolls don't matter at all because each roll is independent of the other.

That's why I think it's a bit redundant to ask this question to somehow come up with a strategy. Prerolling 19 times doesn't increase the chances of your 20th roll hitting.

It is chance, so you can work it out.

The formula for doing so is simply x^y, where x is your probability of losing an individual roll, and y is the number of rolls you makes.

So for example, lets say you go for go for a payout of "2x". Bustabit has a 1% house edge, meaning you have a 49.5% chance to win, and therefore, a 50.5% chance to lose. The probability of losing is therefore 0.505.

0.505¹⁰ is 0.0010787..., or 0.11%. So at a payout of 2x, you have a 0.11% chance of losing 10 times in a row. That becomes 0.004% for 15 losses in a row, and 0.0001% for 20 losses in a row.

At a payout of "3x", your chance of losing is 0.67, so you have a 0.25% chance of losing 15 times in a row. At a payout of "5x", you have a 3.65% chance of losing 15 times in a row.

As far i know long streaks of red situations belongs to dice game and bustabit was crash game which is every rolls will always be crashed but maybe the difference is from the payout itself whether it low or high however talking about losing streaks i think it will depend the percentages of payout itself that if you use low chance of win then you wil met more losing streaks and if you set your payout to high percentages then the losing streaks itself will be so rare

Having 15-20 red in a row is not something extraordinary but first, be specific if this is all about dice or roulette.
If you believed that martingale will work, well it doesn't work. Logically speaking, it will increase your input that continues to reduce your chance of losing everything. The fact, it doesn't make you win and the truth is always the house will win. Don't think too much of profit when you are in gambling, just enjoy the game and if you will win it is better. To be honest, I'm not that kind of gambler who doubling the bets just to cover losses.

If I am not mistaken there are also a code that you could use to see how many percent each of the multiplier would be hit,
For example you want to check the x2 they would show you the percent of provability that it could get that high in every game.
And I also agree that OP should be more specific on which multiplier is he/she is aiming for.

It will depend on what payout you've set on a particular round.Knowing that each bettor would have different selection on when they would cash out.
If you do talk on busting on 1.00 or 1.01 or 1.1 payout then no one knows because everything is random.Theres no such thing about fix or patterns
when it comes to losing streaks but to answer you out it will really just vary on multipliers youve chosen but expect long losing streaks would be
probable on high multipliers ofcourse.

People do not understand mathematics in their head;)

I like martingale but I know I can bust any time and I have in the past. But I do like having a bankroll that can last 20 streaks (not necessarily x2). And then I bet normally and wait for a bad streak, then I start martingale. Such as on roulette. If I see 5 reds in a row, then I start martingale on black.

First thing first, bustabit is a legit site and it is not specific to them. If you create a similar algorithm yourself and try that then also the probability to having a red streak on a certain multiplier is mathematically found but that is again a probable outcome  or the most commonly seen outcomes. It may not be the same when you play and in rare cases you might find that the outcome is not close the calculated result.

I have tried the Seuntjie's dicebot !odds command on Primedice and Stake chatrooms. It gives you an idea of when the red streak will come but again it is not absolute. But I am yet to see any commands from Seuntjie for how long the streak may be.

This according to me dependent on your luck. You may get a nice green streak on a high multiplier (be sure to stop when this happens and cash the profit) OR, you may get a huge red streak on a very small multiplier.

I have been playing enough to witness a streak of win as well as a streak of loss to know now. Well, from my experience and play, I would say, losing streak come more often than winning streak, out of 100, 10 would be winning streak and 20 would be losing streak. That's how it is, generally, because my luck is bad :/

That is a very generic question.The strike of reds can continue to a much higher degree than 20 lost rolls.The reason is because these rolls are controlled by computer software which can be random and can even be 100 consecutive reds although this rarely is the case.Don’t rely on statistics in a game of luck.

There's no way you can measure when a red streak is going to come. If there was a way to cheat the system, casinos would go bankrupt.

The more you want to earn, the smaller your chances are. It goes vice versa. The tricky part is that you can earn 1% of your balance by risking 100% of it. Say you bet 1BTC. You risk 9500$ for a $95 win.

If you lose once $9500, it's bad. Casinos are made for THEM to make profit, not you. So if you go for higher profits, you have to bet dozens, hundreds or thousands of times the same amount in order to win.

If greed takes you over or if you are very unlucky, it could be a huge loss for you in any scenario.

Its been discussed hundreds or even thousands time already. Your bankroll wont help to make martingale works, even if you have unlimited bankroll, it wont make you rich by using martingale. Why? All casinos has their own limit of profit per bet, if you have unlimited bankroll then you hit the max profit per bet while doing martingale, what you can do??
Would you say that it is better to start betting with small amount while you have unlimited bankroll to avoid reaching the max profir per bet? If so, do you think it is worth to play with small amount for martingale while at the same time you have unlimited bankroll?

For sure it will not work since it's been studied strategy and this will not work today but what I mean here is if you got a big bank roll still you can regain back what you lose if you put your bet on good place and you cannot do this again if you balance is so low and drained already that's why we should not take gambling seriously since we might end up broke and just enjoy the game and bet what we can afford to lose since this is much advisable to anyone.

Nahh, martingale strategy never works, and even with a big bankroll, you can get broke LOL. I experience and see in some gambling forum that they are experiencing 20-34 losing streak in 2x odds, that is a lot of money even your base bet will be one satoshi. No strategy works, it's better if you will gamble blindly and trust the luck instead of gambling using any strategy, and in the long term, it will just wipe your bankroll. Just enjoy the game and bet base on your guts.

If there's a people can guess that for sure they became rich for betting on bustabit or in any platform but seriously those red streak is unpredictable that's why it's called gambling and just bet what you can afford to lose if you cannot take to lose in that games but maybe a simple analysis on the game result could help, remember big bank roll can prolong your games and give a high chances of winnings.

 depends on what payout you play  .  lower payout less likely you can exprience those reds but higher payout , higher than 20 is possible for a 20 or more  reds   .  

in a rare occasion  , you can also get longer red strakes on lower payout even 2x .

crazy right  ? but that is possible   . that is called a bad luck   .  

my trick was i only do pre rolls with 0 or a small bet  , then i put my original bet and start hunting the high payout that i want   .

You need to be more specific, bustabit is a crash game where you can choose which payout you are going to play.
15-20 red streak on which payout you are referring to? 2x, 3x, or higher or even lower than 2x?
I'm not a bustabit player as I do not really like crash game, but if you have stake account, then you can get the exact calculation about the odds of what you are looking for by using the chatbot.

Does anyone have any idea how often let's say 15-20 red streak come and go? Would like people who have somewhat of an idea, thanks!