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"
?
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"
?
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
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
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
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]
