o_e_l_e_o
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February 06, 2020, 10:58:21 AM |
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-snip- 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.
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deisik
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February 06, 2020, 11:47:28 AM |
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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 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
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o_e_l_e_o
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February 06, 2020, 02:17:52 PM |
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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 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.
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shield132
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February 06, 2020, 03:41:00 PM |
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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.
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sana54210
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February 06, 2020, 05:04:39 PM |
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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.
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". luck is luck. You can't get luck by strategy but you can get it by accident.
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.
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deisik
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February 08, 2020, 08:30:39 AM |
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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 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?
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o_e_l_e_o
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February 08, 2020, 09:09:56 AM Last edit: February 08, 2020, 11:07:33 AM by o_e_l_e_o |
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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. 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. 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. 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 0.5 10, 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 0.5 5 which is 1 in 32. By losing one toss, the player's probability of winning drops by two percentage points. With 5 losses and 11 rolls remaining, the probability of winning drops to around 0.5 (50%). The probability of at least one win does not increase after a series of losses; indeed, the probability of success actually decreases, because there are fewer trials left in which to win. The probability of winning will eventually equal the probability of winning a single toss, which is 1/16 (6.25%) and occurs when only one toss is left. 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.
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Sadlife
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February 08, 2020, 09:28:39 AM |
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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.
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deisik
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February 08, 2020, 10:42:00 AM |
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...
Okay, let's do some quoting here 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 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
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hosseinimr93
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February 08, 2020, 02:07:09 PM Last edit: February 08, 2020, 02:23:14 PM by hosseinimr93 |
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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 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 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 0.5 10. If you lose the first roll, chance of hitting 10 reds in a row will increase to 0.5 9. If you lose the second roll, chance of hitting 10 reds in a row will increase to 0.5 8. 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.
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o_e_l_e_o
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February 08, 2020, 02:29:07 PM |
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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 This is not what you say now This is exactly what I am saying now: 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. 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. 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.
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deisik
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February 08, 2020, 02:51:40 PM |
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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 This is not what you say now This is exactly what I am saying now: 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 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")
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o_e_l_e_o
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February 08, 2020, 03:58:50 PM |
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Simply put, you can't reevaluate the probability of the whole series of flips after you have made some flips Yes, you absolutely can. Please read about conditional probability as I suggested above. 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") 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.
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Zicadis
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February 08, 2020, 05:49:07 PM |
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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.
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Kyraishi
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February 08, 2020, 05:57:56 PM |
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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 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.
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Yabes
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Gula membunuhmu.
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February 08, 2020, 05:59:30 PM |
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The round is randomness, why you still thinkin about after red must be green?
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deisik
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February 08, 2020, 09:00:15 PM |
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Simply put, you can't reevaluate the probability of the whole series of flips after you have made some flips Yes, you absolutely can. Please read about conditional probability as I suggested above Could you explain it in a couple of sentences? 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") 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 Yes, the odds of getting 10 heads in 10 flips yet to be made are 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 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
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o_e_l_e_o
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February 08, 2020, 09:11:52 PM Last edit: February 08, 2020, 09:30:27 PM by o_e_l_e_o |
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The odds of getting 10 heads in 10 flips yet to be made are 1/1024 regardless of what has already happened 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. but it doesn't mean that the odds of hitting 10 heads have suddenly became 1/2 after you flipped heads 9 times before. That's exactly what it means, for that given series of 10 flips. To repeat, the odds are applicable only to future events, not the ones that already happened 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. 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. 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!"
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fullhdpixel
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February 09, 2020, 03:23:40 AM |
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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. 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.
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deisik
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February 09, 2020, 07:22:02 AM |
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The odds of getting 10 heads in 10 flips yet to be made are 1/1024 regardless of what has already happened 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. but it doesn't mean that the odds of hitting 10 heads have suddenly became 1/2 after you flipped heads 9 times before. That's exactly what it means, for that given series of 10 flips 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) 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!" 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
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