(Bustabit) How often do long streaks of red come?

[o_e_l_e_o]

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 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.

[Darker45]

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 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.

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.

[ralle14]

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.

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!

Having 15-20 red in a row is not something extraordinary but first, be specific if this is all about dice or roulette.

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

[libert19]

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.

[o_e_l_e_o]

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.

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.

[alani123]

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!

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.

[hosseinimr93]

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.

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%

[Ailmand]

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.

[TheUltraElite]

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.

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 (<took me some time to type that username without copying ) 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://bitcointalk.org/index.php?topic=5223146.msg53771714#msg53771714

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

[imstillthebest]

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.

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  .

[janggernaut]

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.

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  .

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

[XenoFever]

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!

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.

[o_e_l_e_o]

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%

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 4x and you got over 10 streak losses is very normal to happen.

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.

[abel1337]

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!

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.

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.

[RealMalatesta]

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!

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.

[hosseinimr93]

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%

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.

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.

[o_e_l_e_o]

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.

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%

[hosseinimr93]

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.

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.

[o_e_l_e_o]

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.

[deisik]

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

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