bustabit.com -- The Social Gambling Game - page 54.

Erza

legendary

Activity: 1078

Merit: 1000

Quote from: ranlo on October 16, 2015, 04:41:56 PM

Quote from: ?? on ??

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...

He is a whale holding 1 btc that long will never scare him Grin

. Btw last time I saw him winning around 20btc after loss-60 btc that means profit 80btc and last night saw him already -60btc again. That is crazy for him

joksim299

legendary

Activity: 2198

Merit: 1014

Bitdice is scam scam scammmmmmmmmmmmmmmmmmmmmmmmmm

Quote from: ?? on ??

Our biggest win yet:

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

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

Unbelievable win, sadly he is still 125BTC in the negative.

ranlo

legendary

Activity: 1988

Merit: 1007

Quote from: ?? on ??

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...

ranlo

legendary

Activity: 1988

Merit: 1007

Quote from: ?? on ??

Quote from: ranlo on October 13, 2015, 12:49:56 AM

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.

ranlo

legendary

Activity: 1988

Merit: 1007

Quote from: ?? on ??

Quote from: ranlo on October 12, 2015, 11:46:17 PM

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.

ranlo

legendary

Activity: 1988

Merit: 1007

Quote from: ?? on ??

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?

ranlo

legendary

Activity: 1988

Merit: 1007

Quote from: dooglus on October 11, 2015, 05:20:50 AM

Quote from: ranlo on October 10, 2015, 10:17:28 PM

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,

.

dooglus

legendary

Activity: 2940

Merit: 1333

Quote from: ranlo on October 10, 2015, 10:17:28 PM

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.

WEBcreator

hero member

Activity: 840

Merit: 1000

Quote from: ?? on ??

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? Grin

ranlo

legendary

Activity: 1988

Merit: 1007

Quote from: Coef on October 10, 2015, 09:23:45 PM

Quote from: ranlo on October 10, 2015, 09:16:54 PM

Quote from: Coef on October 10, 2015, 09:10:50 PM

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?

dooglus

legendary

Activity: 2940

Merit: 1333

Quote from: Coef on October 10, 2015, 09:10:50 PM

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.

Coef

hero member

Activity: 882

Merit: 1000

Exhausted

Quote from: ranlo on October 10, 2015, 09:16:54 PM

Quote from: Coef on October 10, 2015, 09:10:50 PM

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.

ranlo

legendary

Activity: 1988

Merit: 1007

Quote from: Coef on October 10, 2015, 09:10:50 PM

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

hero member

Activity: 882

Merit: 1000

Exhausted

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.

dooglus

legendary

Activity: 2940

Merit: 1333

Quote from: jonsi on October 10, 2015, 02:45:00 PM

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

ranlo

legendary

Activity: 1988

Merit: 1007

Quote from: jonsi on October 10, 2015, 02:45:00 PM

"
!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.

jonsi

legendary

Activity: 1397

Merit: 1019

Thank you.

I'll make sure to double and triple check before posting anything so I don't end up embarrassing myself again.

jonsi

legendary

Activity: 1397

Merit: 1019

Quote from: ?? on ??

Quote from: jonsi on October 10, 2015, 03:43:15 PM

Well, I just won 777 games, not 790.

So now, not only that it doesn't makes sense, but I think you should check your calculation and compare it with my bet history of my last 1686 games, bacause I had 777 wining games and 909 loosing games. I don't know zhere you come up with the 790 wining games bacause the reality is different.

That's not true. The 1686 games you're talking about is between 1931456 and 1933142 (inclusive), which means you should have won:
https://gist.github.com/RHavar/f5d1cdc76d916e15688a

Please check them all, because I guarantee you won them all, and I can guarantee there's 790 of them. I checked both the provably fair, and the database. Every single game you should have won, you won. All 790 of them.

I checked my script again and I found an error on the counting part, I have missed something. Also from the total number of games some wining games were not recorded.

I apologize that I rushed on some conclusion based on wrong data.

jonsi

legendary

Activity: 1397

Merit: 1019

~~Well, I just won 777 games, not 790.~~

So now, not only that it doesn't makes sense, but I think you should check your calculation and compare it with my bet history of my last 1686 games, bacause I had 777 wining games and 909 loosing games. I don't know zhere you come up with the 790 wining games bacause the reality is different.

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

jonsi

legendary

Activity: 1397

Merit: 1019

Quote from: ?? on ??

Quote from: jonsi on October 10, 2015, 02:45:00 PM

"
!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.

I don't quite understand, you've made a net profit of 4.9 bitcoins, and never had any problems with withdrawing them, and still think it's unfair? I must be a really, really shitty scammer.

Anyway, the hash of game 1933142 was efe51340f3f8744d90d3c9ac349745a4aeddee2852eddbddcc4fee47c7d1a79c, as you probably saw on your screen. So why don't you try plug that into this 3rd party provably fair verifier:

https://jsfiddle.net/1L1uqcgv/6/embedded/result/

And you'll notice, the outcomes will have matched what you saw.

This is because I don't gamble. I just use mathematics. The conclusion is based on probabilities. If it was a deviation by let's say 20%, I could call it normal, but a deviation of 89,18%!! It's way way way to big to call it normal.

By the way, the games were played to test a script I wrote. And I'm talking about all the games, not just one game.

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

Topic: bustabit.com -- The Social Gambling Game - page 54. (Read 293938 times)