bustabit.com -- The Social Gambling Game

[ranlo]

"
!prob >= 2.06
Shiba: Probability of ≥2.06: 47.814538%

"

So this means in 1000 games 478 should be wins and 522 looses. This is a 44 difference (522-478=44)
In 1686 games this means that 806 (805.9 to be exact) should be wins and 880 should be looses. This is a 74 difference.

The reality:

I played 1686 consecutive games with 2.06 cashout point.
The results: 777 games won and 909 games lost. This is a 132 difference.

The difference of 132 is nearly double as it should been, 74.

This is pure matematics and probabilities. The deviation is way to big. A 10% - maxim 20% could call normal because of the variance, but nearly double?? I have made many tests with randon numbers and this is not normal.

The conclusion: I think the game it's NOT FAIR.

Edited: By the way, the last game ID thai I played is 1933142, and the first I guess is 1933142-1686.




-edited-
I had an error when recording data from game.

It takes 100 million+ rolls to get near the estimated ratio, which is all just part of statistics. Some won't accept them if they're under 1b. This is like rolling 15x, all being fails at 50%, and saying that it's clearly broken, while ignoring the people winning 50+ times in a row.

[dooglus]

The difference of 132 is nearly double as it should been, 74.

This is pure matematics and probabilities. The deviation is way to big. A 10% - maxim 20% could call normal because of the variance, but nearly double?? I have made many tests with randon numbers and this is not normal.

Could you share your tests with us? I understand that you later realised that you had somehow failed to "count" correctly, but even if you had learned to count correctly what makes you feel that you're justified in saying that a difference of 132 is "not normal"?

According to my tests, when you play 1686 times with a chance of winning of 47.814538%, you will see a difference of (losses-wins) equal to 132 or more around 8% of the time, and so for it to happen once out of one trial is entirely normal.

Here's the code for my test. I'd like to compare it with yours:

Code:

#!/usr/bin/env python

import random

win_chance = 47.814538 / 100
games = 1686
trials = 100000
results = {}

trial = 0
while trial < trials:
    c = 0
    wins = 0
    while c < games:
        if random.random() < win_chance:
            wins += 1
        c += 1

    losses = games - wins
    diff = losses - wins
    if not results.has_key(diff):
        results[diff] = 0
    results[diff] += 1
    trial += 1

keys = results.keys()
keys.sort()
sum = 0
for k in keys:
    sum += results[k]
    print k, sum * 100.0 / trials

[Coef]

The number of games with crashpoint >= 2.06 in a total of 1686 games should follow a binomial distribution of  N = 1686 and p = 47.814538%. So the chance to have less than or equal to 777 wins is 0.081149924, which is very close to the number dooglus obtained by simulation.
An easy way to calculate it is to use "=BINOMDIST(777, 1686, 47.814538%, 1)" in Excel.

[ranlo]

The number of games with crashpoint >= 2.06 in a total of 1686 games should follow a binomial distribution of  N = 1686 and p = 47.814538%. So the chance to have less than or equal to 777 wins is 0.081149924, which is very close to the number dooglus obtained by simulation.
An easy way to calculate it is to use "=BINOMDIST(777, 1686, 47.814538%, 1)" in Excel.

This is a pretty cool function I wasn't aware of. So if we wanted to see the chance of 35 losses in a row at 49.5%, would that be:

"=BINOMDIST(35, 35, 49.5%, 1)"

?

[Coef]

The number of games with crashpoint >= 2.06 in a total of 1686 games should follow a binomial distribution of  N = 1686 and p = 47.814538%. So the chance to have less than or equal to 777 wins is 0.081149924, which is very close to the number dooglus obtained by simulation.
An easy way to calculate it is to use "=BINOMDIST(777, 1686, 47.814538%, 1)" in Excel.

This is a pretty cool function I wasn't aware of. So if we wanted to see the chance of 35 losses in a row at 49.5%, would that be:

"=BINOMDIST(35, 35, 49.5%, 1)"

?

When the last field is "1", it is to calculate the CDF. And when the last field is "0", it is to calculate the PDF.

If you want to see the chance of having exactly 30 wins in 35 bets at 49.5%, you can use "=BINOMDIST(30, 35, 49.5%, 0)" and will get 0.000735%.
If you want to see the chance of having exactly 35 wins in 35 bets at 49.5%, you can use "=BINOMDIST(35, 35, 49.5%, 0)" and will get 0.0000000020473% which is exactly 0.495^35.

If you want to see the chance of having less than or equal to 10 wins in 35 bets, you can use "=BINOMDIST(10, 35, 49.5%, 1)" and will get 0.976982%.

Edit: Fixed typos.

[dooglus]

The number of games with crashpoint >= 2.06 in a total of 1686 games should follow a binomial distribution of  N = 1686 and p = 47.814538%. So the chance to have less than or equal to 777 wins is 0.081149924, which is very close to the number dooglus obtained by simulation.

I didn't post my results, but they were like this:

Code:

-122 0.0001
-110 0.0002
-108 0.0005
-106 0.0007
-104 0.0008
-102 0.0011
-100 0.0015
[...]
-6 2.7609
-4 3.0792
-2 3.4406
0 3.8267
2 4.2453
4 4.7016
6 5.207
[...]
68 45.466
70 47.3885
72 49.3342
74 51.2791
76 53.2283
78 55.1682
[...]
130 91.8775
[...]
268 100.0

So 91.8775% chance of a difference of 130 or less, and so a 100 - 91.8775 = 8.1225% chance of a difference of 132 or more.

That's really pretty close the 8.1149924% you calculated.

[RHavar]

Thanks guys! Nice seeing some solid statistic analysis =)

[ranlo]

The number of games with crashpoint >= 2.06 in a total of 1686 games should follow a binomial distribution of  N = 1686 and p = 47.814538%. So the chance to have less than or equal to 777 wins is 0.081149924, which is very close to the number dooglus obtained by simulation.
An easy way to calculate it is to use "=BINOMDIST(777, 1686, 47.814538%, 1)" in Excel.

This is a pretty cool function I wasn't aware of. So if we wanted to see the chance of 35 losses in a row at 49.5%, would that be:

"=BINOMDIST(35, 35, 49.5%, 1)"

?

When the last field is "1", it is to calculate the CDF. And when the last field is "0", it is to calculate the PDF.

If you want to see the chance of having exactly 30 wins in 35 bets at 49.5%, you can use "=BINOMDIST(30, 35, 49.5%, 0)" and will get 0.000735%.
If you want to see the chance of having exactly 35 wins in 35 bets at 49.5%, you can use "=BINOMDIST(35, 35, 49.5%, 0)" and will get 0.0000000020473% which is exactly 0.495^35.

If you want to see the chance of having less than or equal to 10 wins in 35 bets, you can use "=BINOMDIST(10, 35, 49.5%, 1)" and will get 0.976982%.

Edit: Fixed typos.

Thanks for that, . I've noted these down so I can use them in the future. Is there a similar way to calculate odds of hitting certain cards in poker, as well? Like let's say you have two hearts and two came out on the flop. There's two cards left to draw, and there are 9 hearts left that you can't see. Is there a quick/easy way to calculate odds using the # of outs per draw, rather than doing two separate calculations?

[WEBcreator]

Thanks guys! Nice seeing some solid statistic analysis =)

But this is just some analysis? There is no proof that this will take effect on game right? 

[RHavar]

But this is just some analysis? There is no proof that this will take effect on game right? 

I'm not really sure what you mean. What they're doing is figuring out the answer to questions like: "I made 1686 bets @ 2.06x, but only won 777 of them. How unlucky was I?"

If you want to know if you were cheated or not, really what you need to do is look at the provably fair system (like the analysis I posted a page earlier). In the end, you can hit a 1-in-a-million bad luck run, but that doesn't mean you were cheated at all.

Or if you want to know the probability of a single outcome, you can use shiba's command

!prob <condition>


like:

!prob >= 2.23x
!prob < 1.3x
!probt = 3.4x
etc.

[dooglus]

Is there a similar way to calculate odds of hitting certain cards in poker, as well? Like let's say you have two hearts and two came out on the flop. There's two cards left to draw, and there are 9 hearts left that you can't see. Is there a quick/easy way to calculate odds using the # of outs per draw, rather than doing two separate calculations?

I don't think the binomial distribution helps you with that specific case, because the odds of drawing another heart change on the turn and river - it's 9/47 on the turn, and either 8/46 or 9/47 on the river (depending on whether you drew a heart on the turn).

The binomial distribution is useful when the individual probabilities don't change - ie. when the events are independent of each other - like dice rolls, or bustabit games.

Edit: having said that, you can get pretty close if you ignore the fact that the probabilities change slightly as you draw.

Assuming the probability of drawing a heart is 9 in (52-5), the chances of drawing 0, 1, or 2 more hearts in 2 cards are:

0   2   19.15%   0   65.37%
1   2   19.15%   0   30.96%
2   2   19.15%   0   3.67%

A rule of thumb is that you double the number of outs to get the percentage chance of drawing on the turn: 9 outs, 18% on turn, 18% on river, 36% total.

[ranlo]

Is there a similar way to calculate odds of hitting certain cards in poker, as well? Like let's say you have two hearts and two came out on the flop. There's two cards left to draw, and there are 9 hearts left that you can't see. Is there a quick/easy way to calculate odds using the # of outs per draw, rather than doing two separate calculations?

I don't think the binomial distribution helps you with that specific case, because the odds of drawing another heart change on the turn and river - it's 9/47 on the turn, and either 8/46 or 9/47 on the river (depending on whether you drew a heart on the turn).

The binomial distribution is useful when the individual probabilities don't change - ie. when the events are independent of each other - like dice rolls, or bustabit games.

Edit: having said that, you can get pretty close if you ignore the fact that the probabilities change slightly as you draw.

Assuming the probability of drawing a heart is 9 in (52-5), the chances of drawing 0, 1, or 2 more hearts in 2 cards are:

0   2   19.15%   0   65.37%
1   2   19.15%   0   30.96%
2   2   19.15%   0   3.67%

A rule of thumb is that you double the number of outs to get the percentage chance of drawing on the turn: 9 outs, 18% on turn, 18% on river, 36% total.

The chance would be less than 36%, though, so that's an over-exaggeration (important detail when playing something that's odds-based). But thanks! I think I need to take a statistics class on Coursera or something to better understand all these things, .

[RHavar]

This is how crazy things have been:

[ranlo]

This is how crazy things have been:

Any chance you can add a line for expected profit, just to see a comparison between where it is and what it would average?

[RHavar]

Any chance you can add a line for expected profit, just to see a comparison between where it is and what it would average?

Yeah, that would be quite interesting. One thing about BaB is that it uses a dynamic house edge, but not all the information is known (when someone would have cashed out). But it could be conservatively estimated by assuming they would've cashed out at +0.01 of what they busted.  (And an over-estimate could be calculated by assuming they were exclusively relying on their autocashout)

I'll try get around to that tomorrow, as I'm actually interested in this very much =)

[ranlo]

Any chance you can add a line for expected profit, just to see a comparison between where it is and what it would average?

Yeah, that would be quite interesting. One thing about BaB is that it uses a dynamic house edge, but not all the information is known (when someone would have cashed out). But it could be conservatively estimated by assuming they would've cashed out at +0.01 of what they busted.  (And an over-estimate could be calculated by assuming they were exclusively relying on their autocashout)

I'll try get around to that tomorrow, as I'm actually interested in this very much =)

Maybe a trick would be to mathematically determine the "average" somehow? Like if you can easily ping the database to pull the information and craft the house edges / total bets, you can get an "average" house edge to use that would encapsulate everything.

[RHavar]

Maybe a trick would be to mathematically determine the "average" somehow? Like if you can easily ping the database to pull the information and craft the house edges / total bets, you can get an "average" house edge to use that would encapsulate everything.

You could estimate, but it's impossible to do it perfectly. If you imagine the game busts at 0x, you really have no idea what people were going to cashout. You can make a guess, but it'll still be a guess. Especially the whales tend to just play on gut feelings, so that's not something easy to predict and it makes a big difference if you assume they were going to cashout at 1x (0% house edge)  or a super high autocashout (normally ~0.9%).

Anyway, a few people have been asking for it, so here's our updated chart:

[ranlo]

Maybe a trick would be to mathematically determine the "average" somehow? Like if you can easily ping the database to pull the information and craft the house edges / total bets, you can get an "average" house edge to use that would encapsulate everything.

You could estimate, but it's impossible to do it perfectly. If you imagine the game busts at 0x, you really have no idea what people were going to cashout. You can make a guess, but it'll still be a guess. Especially the whales tend to just play on gut feelings, so that's not something easy to predict and it makes a big difference if you assume they were going to cashout at 1x (0% house edge)  or a super high autocashout (normally ~0.9%).

Anyway, a few people have been asking for it, so here's our updated chart:

Nice rebound there! Looks like one of the whales had a turn-around on luck.

[RHavar]

Our biggest win yet:

https://www.bustabit.com/game/1954550


A 15 BTC win, on a game that went over 5000x!

[ranlo]

Our biggest win yet:

https://www.bustabit.com/game/1954550


A 15 BTC win, on a game that went over 5000x!

Talk about risky though. A 1 BTC bet and holding out until 16x...