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1  Economy / Gambling discussion / Re: The reason why Crash Games usually crash at lower values on: November 04, 2021, 02:18:52 PM
If I take multiplier=1.05, should I have P(X<=1.05) = 1/33 + 32/33*(0.01 + 0.99*(1-1/1.05)) = 8.57%, which is pretty low, it doesn't mean if I bet on 1.05x each time, only 8.57% of chance it crashes with multiplier less than 1.05? I am quite confused. If I win 1.05x, deduct the bet=1 I made, I still have 0.05 win, isn't it?

So 857 times out of 10 000 times it will crash below 1.05 (on average). So lets say that you are going to play 10 000 games putting 1$ in each game cashing out at 1.05. In this situation you will win 9143 times winning 9143*0.05 = 457.15$ and lose 857 times losing 857 * 1$ = 857$ ending up with 399.85$ net loss.

I'll check rest of your calculation in free time
Yes, I think it sounds right to me. What I am going to say is the math may guarantee for gambler will lose for the long term (the return is just 96% for all possible multipliers taken into account); however, if you bet the small multiplier always, since the probability of getting the crash multiplier less than your betted multiplier is smaller than 10% (if my math is correct), it will not increase the return above 100% but it will gain a bit advantage to players?
2  Economy / Gambling discussion / Re: The reason why Crash Games usually crash at lower values on: November 03, 2021, 10:52:15 PM
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!).

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.

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

"theoretically" Huh?


I read the reply a couple of times and recall what I learnt from the college stat., I think it really gives me good information on how everything is being controlled. I am planning to write an app to reproduce this game for fun. I did some research and I see the following formula to approach the cumulative probability function for the multiplier given as  crash_multiplier = (100 * e - h) / (e - h) / 100

P(X<=x) = 1 - 1/x

where X is a random multiplier and x is the certain cap of the multiplier, it is saying that the chance to see any multiplier less than a given value of x is the complement of 1/x. So when x=2, it is 1-1/2 = 0.5 or 50% of chance to have any multipliers less than 2x. Here I am confused about the gameplay. What should I do to pick the crash_multiplier for each game? I saw that P(X<=x)=1-1/x is derived in the assumption that h is a uniform random number so I guess   I should just pick a uniformly random number (h) between for [0, e-1] as the crash_multiplier for each game and let the player bet on it and wait for the player's cashing out? And that is all?

What about the incremental of the multiplier in time, could the multiplier increase at any interval?




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!).

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.

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

"theoretically" Huh?


After reading this, I read more online and I found this one https://github.com/MindingTheData/Crash-Analysis/blob/master/Crash.ipynb
 it uses a similar formula for multiplier but better approximation for the cumulative prob function, I run the code and add all the expectation values for multiplier from 1 to 1000

Code:
e = 0
N = 0
for multiplier in np.arange(1, 1000, 0.01):
  N = N+1
  e = e + 1 + ((1/33) + (32/33)*(.01 + .99*(1 - 1/(multiplier-.01))))*-1 + (multiplier-1)*(1 - ((1/33) + (32/33)*(.01 + .99*(1 - 1/(multiplier-.01)))))

e/(N/1.0)

it ends up with a return of about 96%. If I take multiplier=1.05, should I have P(X<=1.05) = 1/33 + 32/33*(0.01 + 0.99*(1-1/1.05)) = 8.57%, which is pretty low, it doesn't mean if I bet on 1.05x each time, only 8.57% of chance it crashes with multiplier less than 1.05? I am quite confused. I don't understand why all the discussions in this thread are about winning the double of the bet (2x). If I win 1.05x, deduct the bet=1 I made, I still have 0.05 win, isn't it?

I try the following simulation

Code:
e = 2**52
N = 20000
M = 100000
win = 0
for n in range(M):
    h = np.random.randint(0, e-1, size=N)
    crash_mul = (((100 * e - h) / (e-h)) // 1) / 100.0
    crash_mul = (np.random.uniform(size=N)>=(1/33.0))*crash_mul
    bet_mul = np.random.uniform(1.05, 100.06, size=N)
    win = win + bet_mul[bet_mul<crash_mul].sum()

win/(N*M)
which gives something close to 96% but if I fixed the bet_mul to 1.05, I get about 96.9%. Why is it?

If I understand it correctly, the advantage the casino took is on the 1/33 chance to crash instantly, but the bet it still has 32/33 of chance that fixing the 1.05 may end up with a higher return in the simulation, I don't know what is wrong in my code.


[moderator's note: consecutive posts merged]
3  Economy / Gambling discussion / Re: The reason why Crash Games usually crash at lower values on: November 03, 2021, 04:31:50 PM
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?
4  Economy / Gambling discussion / Re: The reason why Crash Games usually crash at lower values on: November 03, 2021, 03:34:44 PM
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.


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
PossibilitiesMultiplier
H=01
H=11.11
H=21.24
H=31.42
H=41.66
H=51.99
H=62.48
H=73.31
H=84.96
H=99.91


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.
5  Economy / Gambling discussion / Re: The reason why Crash Games usually crash at lower values on: November 03, 2021, 01:40:31 PM
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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