The reason why Crash Games usually crash at lower values - page 5.

Tytanowy Janusz

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Quote from: gin.zhe on November 03, 2021, 11:31:50 AM

ah, now I understand. Many thanks.

If a player decides to bet $1 at a time but chooses to cash out when it is 1.05x reaches and repeat this strategy thousands of times, will it benefit the player to have positive gain? Why not and how does the math forbid that to happen?

with 1.05x you need to bet and win 20 times to double your 1$. When you lose you lose your 1$. The problem with this strategy is that crash (and every casino game) is calculated in a way that casino has statistic advantage over you. So in this case you will lose once each 19 times (ON AVERAGE! You can have 100 wins strike but sooner or later your luck will end). So each time you will win 95 cents with this strategy, you will lose 1$ and be short 5 cents (ON AVERAGE!).

Quote from: gin.zhe on November 03, 2021, 08:40:31 AM

Usually you would have to avoid betting continuously since it would make your chances of winning go down and I heard a lot of players say how fixing the seeds again works for them, so therefore I usually use crash after resetting my seeds and not continually at the same time.

Each bet on provably fair casino is completly random. So no ... resetting seed, "avoid betting continuously", wearing talismans does not work.

Quote from: gin.zhe on November 03, 2021, 08:40:31 AM

Mathematics is always theoretically correct but an interesting point to analyze here.

"theoretically" Huh

?

fiulpro

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Most of the time I don't really play crash really realizing all the things and this is honestly the first formula that I saw in my life regarding these games.
When we talk about this formula how would all the casinos follow the same one ?
Crash games for me are really one based on luck but seeing your formula I feel like I should get more into see how it works and how it would be beneficial for the player to encash every once in a while when they start.

Quote from: Tytanowy Janusz on November 03, 2021, 10:31:44 AM

Quote from: gin.zhe on November 03, 2021, 08:40:31 AM

Besides, I saw someone said there is at least 50% of chance the multiplier is 2 or less. If I bet 1 dollar at a time, each time I cash out if saw the multiplier raised to 1.1x, will it give me more chance to win something small each time? If no, why not?

No. There is always the same risk/reward ratio. There is no difference in strategy you pick. There is no diifference in multiplier you aim to hit. You always have ~90-98% from your bet back to your account from each bet, on average. This 2-10% is casino house edge. There is no way to outperform other, not even saying about outperforming casino. The only way to win is by having luck.

Quote from: gin.zhe on November 03, 2021, 08:40:31 AM

crash multiplier = (100 * e - h) / (e - h) / 100

no matter what h is, the crash multiplier is always ONE?

Its not 1. OP gave few examples. You had to make a mistake in the order of performing mathematical operations

Quote

Possibilities	Multiplier
H=0	1
H=1	1.11
H=2	1.24
H=3	1.42
H=4	1.66
H=5	1.99
H=6	2.48
H=7	3.31
H=8	4.96
H=9	9.91

Usually you would have to avoid betting continuously since it would make your chances of winning go down and I heard a lot of players say how fixing the seeds again works for them, so therefore I usually use crash after resetting my seeds and not continually at the same time.

Mathematics is always theoretically correct but an interesting point to analyze here.

gin.zhe

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Quote from: Tytanowy Janusz on November 03, 2021, 10:43:12 AM

Quote from: gin.zhe on November 03, 2021, 10:34:44 AM

Thanks. But I think that example is valid when E=10, what I am confused about is the casino chose E=2^52, even I choose H=1000 or 1E8, the formula still gives me 1 instead.

Maybe because E=2^52 = 4 503 599 627 370 496 so H=1000 is close to 0, even 1E8 is close to 0 in this scale, because comparing to E=2^52 your pick - H=1E8 is 50 million times smaller. Use H=2^50 or something like this (or 2^52-1) to check if you calculate everything correct.

ah, now I understand. Many thanks.

If a player decides to bet $1 at a time but chooses to cash out when it is 1.05x reaches and repeat this strategy thousands of times, will it benefit the player to have positive gain? Why not and how does the math forbid that to happen?

Tytanowy Janusz

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Quote from: gin.zhe on November 03, 2021, 10:34:44 AM

Thanks. But I think that example is valid when E=10, what I am confused about is the casino chose E=2^52, even I choose H=1000 or 1E8, the formula still gives me 1 instead.

Maybe because E=2^52 = 4 503 599 627 370 496 so H=1000 is close to 0, even 1E8 is close to 0 in this scale, because comparing to E=2^52 your pick - H=1E8 is 50 million times smaller. Use H=2^50 or something like this (or 2^52-1) to check if you calculate everything correct.

gin.zhe

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Quote from: Tytanowy Janusz on November 03, 2021, 10:31:44 AM

Quote from: gin.zhe on November 03, 2021, 08:40:31 AM

Besides, I saw someone said there is at least 50% of chance the multiplier is 2 or less. If I bet 1 dollar at a time, each time I cash out if saw the multiplier raised to 1.1x, will it give me more chance to win something small each time? If no, why not?

No. There is always the same risk/reward ratio. There is no difference in strategy you pick. There is no diifference in multiplier you aim to hit. You always have ~90-98% from your bet back to your account from each bet, on average. This 2-10% is casino house edge. There is no way to outperform other, not even saying about outperforming casino. The only way to win is by having luck.

Quote from: gin.zhe on November 03, 2021, 08:40:31 AM

crash multiplier = (100 * e - h) / (e - h) / 100

no matter what h is, the crash multiplier is always ONE?

Its no 1. OP gave few examples:

Quote

Possibilities	Multiplier
H=0	1
H=1	1.11
H=2	1.24
H=3	1.42
H=4	1.66
H=5	1.99
H=6	2.48
H=7	3.31
H=8	4.96
H=9	9.91

Thanks. But I think that example is valid when E=10, what I am confused about is the casino chose E=2^52, even I choose H=1000 or 1E8, the formula still gives me 1 instead.

Tytanowy Janusz

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Quote from: gin.zhe on November 03, 2021, 08:40:31 AM

Besides, I saw someone said there is at least 50% of chance the multiplier is 2 or less. If I bet 1 dollar at a time, each time I cash out if saw the multiplier raised to 1.1x, will it give me more chance to win something small each time? If no, why not?

No. There is always the same risk/reward ratio. There is no difference in strategy you pick. There is no diifference in multiplier you aim to hit. You always have ~90-98% from your bet back to your account from each bet, on average. This 2-10% is casino house edge. There is no way to outperform other, not even saying about outperforming casino. The only way to win is by having luck.

Quote from: gin.zhe on November 03, 2021, 08:40:31 AM

crash multiplier = (100 * e - h) / (e - h) / 100

no matter what h is, the crash multiplier is always ONE?

Its not 1. OP gave few examples. You had to make a mistake in the order of performing mathematical operations

Quote

Possibilities	Multiplier
H=0	1
H=1	1.11
H=2	1.24
H=3	1.42
H=4	1.66
H=5	1.99
H=6	2.48
H=7	3.31
H=8	4.96
H=9	9.91

gin.zhe

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Thanks for the great post. I am studying the game for fun and happen to see the math in the forum. It is the first time I understand how it works with my limited college knowledge in statistics. I do understand the formula to produce the multiplier as follows

crash multiplier = (100 * e - h) / (e - h) / 100

e is the maximum multiplier one could have, but from other replies, it mentioned e will be 2^52 in javascript, does it mean the multiplier could go up that that high? I don't know javascript but I try to implement the formula (with e=2^52) in c++, no matter what h is, the crash multiplier is always ONE?

Besides, I saw someone said there is at least 50% of chance the multiplier is 2 or less. If I bet 1 dollar at a time, each time I cash out if saw the multiplier raised to 1.1x, will it give me more chance to win something small each time? If no, why not?

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Quote from: webtricks on August 01, 2020, 02:18:25 AM

Quote from: Saint-loup on July 31, 2020, 04:03:31 PM

I don't know where you've found those formulas but they seem more complicated than RHavar's code.

Quote from: RHavar on January 22, 2018, 07:05:18 PM

Code:

99 / (1 - X)

X is uniformly distributed on [0,1] because it comes from the seed
The result is then divided by 100 to get the crash multiplier

I have derived formula on my own. It's basically the same. What you are showing are the steps to find the result from coding point-of-view. But in compact form it's exactly same as the second formula I derived in the OP, let me show you how:

Step 1: Starting point -> 99 / (1-X)

Step 2: Inserting the value of X in the formula:
-snip

Yes you're right X=H/E

Quote from: RHavar on January 22, 2018, 07:05:18 PM

Code:

// 2. r = 52 most significant bits
  seed = seed.slice(0, nBits/4)
  const r = parseInt(seed, 16)

  // 3. X = r / 2^52
  let X = r / Math.pow(2, nBits) // uniformly distributed in [0; 1)

  // 4. X = 99 / (1-X)
  X = 99 / (1 - X)

Quote from: webtricks on August 01, 2020, 02:31:01 AM

Quote from: Insanerman on July 31, 2020, 10:59:43 PM

Quote from: webtricks on July 31, 2020, 12:47:57 PM

~

But I don't think that Roobet uses the old algorithm as there are already huge number of players that win a big amount in playing Roobet's crash game. And how are you even sure that they are using RHavar's code and not their own? I think only those famous betting platforms with crash games back then only uses that, and new innovating ones are making their own adjustments with the code.

Just because Roobet is growing successfully, that doesn't mean it is a pioneer of Bitcoin Gambling Industry! Stop believing gambling sites blindly. Here's the proof that Roobet's crash game is based on the same algorithm (taken right from the 'Fairness' page of Roobet):

Do you see a red box? Can you read the formula inside? Isn't it same to the one in OP?

Don't think that I will write random things anywhere on forum without any knowledge. If I was sure enough to include that line in OP, it does mean that I am sure about the fact. Period!

You're right it's the v1 version
https://github.com/Dexon95/Bustabit/blob/master/gameserver/server/lib.js

But in this version the house edge is computed like that as far as I understand :

// In 1 of 101 games the game crashes instantly. if (divisible(hash, 101)) return 0; // Use the most significant 52-bit from the hash to calculate the crash point var h = parseInt(hash.slice(0,52/4),16); var e = Math.pow(2,52); return Math.floor((100 * e - h) / (e - h));

While in the Roobet version, 101 has been replaced by 25.
It means there is an instant crash every 25 game rounds, ie 4 times out of 100.
So we can assume the house edge is at 4% on Roobet... Undecided

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Quote from: iamsheikhadil on August 02, 2020, 06:29:34 AM

While crash shows that it's hitting many times under 2x, the one or two times we also see it hits thousands or even millions of wins. I think it's evenly spread out and even in other games, the chances of hitting 2x is exactly the same of what is in crash. Thanks for the nice explanation of how it originated and how it's calculated!

If it hits many times above 2x or more then what will be the benefit of the house who is hosting the crash game ? Ofcourse as with every other game, you have very few chances to get more out of gambling houses and crash game is not any different. You can only hit big in a crash games if you have money to take risk in many bets before you land on a big milestone.

iamsheikhadil

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While crash shows that it's hitting many times under 2x, the one or two times we also see it hits thousands or even millions of wins. I think it's evenly spread out and even in other games, the chances of hitting 2x is exactly the same of what is in crash. Thanks for the nice explanation of how it originated and how it's calculated!

Tytanowy Janusz

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Quote from: webtricks on July 31, 2020, 12:47:57 PM

So yesterday a forum member asked me why do Crash Games usually crash at lower multiples (mostly lower than 2x)? Does that mean these games are not fair? Does that mean owner of the site manipulated the game so lower multipliers appear more often than the higher multipliers? I thought other forum members may have these doubts too so I decided to create this thread.

Not all crash game shows actual odds of winnings. It is not roulette or dice where you can calculate your odds and house edge easily. Formula with E and H is shown only in fairness section and only for few gambling sites (big part of "provably fair" casinos are "provably fair" only for 1-2 games, rest games "provably fair" section is under construction forever). House edge on these games can be set to 20% and you don't know it. Does it mean that output is "manipulated" and "not fair"? To me, As long as casino share formula how crash is calculated it is fair. It's gambler fault that did not check how high casino set house edge.

But back to OP. Great explanation how crash is calculated. Good work.

Formula with H and E is very clever because it provides consistent commission for the casino no matter where players positions their bets.

You want to double? You have 4x% to do so

Quote

Hence, in 5 out of 9 case, value is lower than 2x.

You want to tripple? you have <33% to do so - 3/10 above 3

You want to do 10x? you have <10% to do so

and so on and on. No matter where you put your bet. Math take care that you have same risk/reward ratio and casino have it's share.

webtricks

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Quote from: Insanerman on July 31, 2020, 10:59:43 PM

Quote from: webtricks on July 31, 2020, 12:47:57 PM

~

But I don't think that Roobet uses the old algorithm as there are already huge number of players that win a big amount in playing Roobet's crash game. And how are you even sure that they are using RHavar's code and not their own? I think only those famous betting platforms with crash games back then only uses that, and new innovating ones are making their own adjustments with the code.

Just because Roobet is growing successfully, that doesn't mean it is a pioneer of Bitcoin Gambling Industry! Stop believing gambling sites blindly. Here's the proof that Roobet's crash game is based on the same algorithm (taken right from the 'Fairness' page of Roobet):

Do you see a red box? Can you read the formula inside? Isn't it same to the one in OP?

Don't think that I will write random things anywhere on forum without any knowledge. If I was sure enough to include that line in OP, it does mean that I am sure about the fact. Period!

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Quote from: Insanerman on July 31, 2020, 10:59:43 PM

Quote from: webtricks on July 31, 2020, 12:47:57 PM

~

Does algorithm really matter? It's not just the crash game that usually has low values and less possibility of winning, it is also in every game in gambling, and in any platform. But I don't think that Roobet uses the old algorithm as there are already huge number of players that win a big amount in playing Roobet's crash game. And how are you even sure that they are using RHavar's code and not their own? I think only those famous betting platforms with crash games back then only uses that, and new innovating ones are making their own adjustments with the code.

I think that the algorithm is what matter the most in software controlled games.In slot machines it is the algorithm that controls everything and although most of the providers say each spin is independent of each other we all know that getting on the reels the top combination happens rarely because the algorithm has decided so bu being programmed by the developers of the game to behave that way.Same with crash games and other software related games.

webtricks

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Quote from: Saint-loup on July 31, 2020, 04:03:31 PM

I don't know where you've found those formulas but they seem more complicated than RHavar's code.

Quote from: RHavar on January 22, 2018, 07:05:18 PM

Code:

99 / (1 - X)

X is uniformly distributed on [0,1] because it comes from the seed
The result is then divided by 100 to get the crash multiplier

I have derived formula on my own. It's basically the same. What you are showing are the steps to find the result from coding point-of-view. But in compact form it's exactly same as the second formula I derived in the OP, let me show you how:

Step 1: Starting point -> 99 / (1-X)

Step 2: Inserting the value of X in the formula:

where R is a random number taken from first 52-bits of the hash and E is 2⁵²

Step 3: Taking LCM in denominator:

R = H

Step 4: Taking 'E' to the numerator:

Step 5: Finally dividing the result by 100 to get multiplier value:

I just missed one step, if the resulting value of multiplier is below 1 then return 1. Rest of the formula is completely correct.

Insanerman

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Quote from: Wexnident on July 31, 2020, 11:59:59 PM

~
I guess Op just wanted to prove that Crash games are provably fair, and is not manipulated by the gambling sites. Well, the core idea of the algorithms of crash should closely follow what Op said and shouldn't deviate that much. They may change it up to increase their house edge or vice versa, but the idea remains the same. Besides, with the amount of crash matches that happen every day, someone winning big every now and then isn't even that weird or odd.

But IMO, there are a lot of gambling sites out there that uses their own algorithm and had their limits or was being manipulated. I've encountered a lot of issues in many gambling platforms I've tried back then and read many accusations of them being either too much fair that they made others win drastically big, or they made many loss too much. And that leads me to another idea that algorithms doesn't matter nowadays, as many enters gambling mostly not as their profit making habit, but a time to enjoy playing gambling games.

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Quote from: Wexnident on July 31, 2020, 11:59:59 PM

Can't really expect the casino to create an algorithm that would prove to be an advantage to players now, can we. I've never really seen crash games being unfair tbh, I mean, even my common sense would indicate that if they didn't crash at lower values, they'd probably be losing out quite a lot.

But then if they would ever made an unfair one, it would simply make a huge impact in their reputation as the possibility of loses are bigger than the wins. I think they are making their own algos that would fit their categories such as how many wins can a user have in their crash games together with the chances they could have from time to time.

But I do disagree with Insanerman, that algorithms do have impact in gambling and knowing it would make a gambler be attentive when to bet and what strategy to use. If we didn't knew algorithm, we'll soon be losing our money as we only know how to make fun and be satisfied.

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Can't really expect the casino to create an algorithm that would prove to be an advantage to players now, can we. I've never really seen crash games being unfair tbh, I mean, even my common sense would indicate that if they didn't crash at lower values, they'd probably be losing out quite a lot.

Quote from: Insanerman on July 31, 2020, 10:59:43 PM

Does algorithm really matter? It's not just the crash game that usually has low values and less possibility of winning, it is also in every game in gambling, and in any platform. But I don't think that Roobet uses the old algorithm as there are already huge number of players that win a big amount in playing Roobet's crash game. And how are you even sure that they are using RHavar's code and not their own? I think only those famous betting platforms with crash games back then only uses that, and new innovating ones are making their own adjustments with the code.

I guess Op just wanted to prove that Crash games are provably fair, and is not manipulated by the gambling sites. Well, the core idea of the algorithms of crash should closely follow what Op said and shouldn't deviate that much. They may change it up to increase their house edge or vice versa, but the idea remains the same. Besides, with the amount of crash matches that happen every day, someone winning big every now and then isn't even that weird or odd.

Insanerman

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Quote from: webtricks on July 31, 2020, 12:47:57 PM

~

Does algorithm really matter? It's not just the crash game that usually has low values and less possibility of winning, it is also in every game in gambling, and in any platform. But I don't think that Roobet uses the old algorithm as there are already huge number of players that win a big amount in playing Roobet's crash game. And how are you even sure that they are using RHavar's code and not their own? I think only those famous betting platforms with crash games back then only uses that, and new innovating ones are making their own adjustments with the code.

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Thanks for sharing the valuable information, math has been always the language of universe. The probabilities are not in favor of player but using different money management technics can reduce the chances of getting busted sooner than hitting profit target. The Limbo game in Stake has the similar odds but this line alone explain everything why gamblers tend to lose in the long term:

Quote

Hence, in 5 out of 9 case, value is lower than 2x

+ House Edge

Interesting stats by bustabit:

Saint-loup

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I don't know where you've found those formulas but they seem more complicated than RHavar's code.

Quote from: RHavar on January 22, 2018, 07:05:18 PM

Code:

99 / (1 - X)

X is uniformly distributed on [0,1] because it comes from the seed
The result is then divided by 100 to get the crash multiplier

So we can understand that we need to have (1-X) below 0.5 to get a crash multiplier above 2
99/0.5=99x2=198
=>CM=198/100=1.98

So we can conclude that less than 50% of (1-X) values give crash multipliers above 2 and thus less than 50% of X(ie seed) values.
Or more than half of game rounds give crash multipliers below x2

Topic: The reason why Crash Games usually crash at lower values - page 5. (Read 1259 times)