bustabit.com -- The Social Gambling Game

[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.

[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? 

[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, .

[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?

[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.

[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.

[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...

[joksim299]

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.

[Erza]

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 . 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

[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...

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

I guess some people just have enough not to worry, . I was at a local casino and sat at a blackjack table just to watch, and there was a guy that would tip the dealer $20 if he won, $10 if he lost. He ended up tipping well over $1k, and they said it's normal for him to do that (he goes a few times a week and always does $20 on wins, 10 on losses).

[Pony789]

With realhavok going from huge profit zone to huge loss zone in the past days, the site profit has now gone beyond the previous high very fast. Once again, Ryan has the last laugh. 

[ranlo]

BTW, let's see if your theory in https://bitcointalk.org/index.php?topic=1020726.msg12613128#msg12613128 still holds this time hehe.

It took a lot longer than I thought, and was a lot more stressful than I would have imagined it would be, but it seems like my theory did indeed hold =)

Watching chat shows this as well. When there's a lot of busts below 2.0x, it's pretty audible in chat and the number of bettors drops significantly for a while.

[hopenotlate]

Wow, I was away for maybe a month from Bustabit and I just noticed maxprofit is higher than 19BTC 

Someone lost big or Ryan put some liquidity in the pot?

[ranlo]

Wow, I was away for maybe a month from Bustabit and I just noticed maxprofit is higher than 19BTC 

Someone lost big or Ryan put some liquidity in the pot?

I put a lot of extra money (~200 btc) in the bankroll to support higher wins (something the whale wanted) but now that battle has been fought, i'll be shortly pulling money out of the bankroll and reducing my exposure =)

So if you want your chance of a big hit, better be quick (Yesterday someone set a new record, with a >15 BTC profit hit) before i lower it down to about ~10 BTC again

Did they end up losing it again, or running off with it?

[ranlo]

https://www.bustabit.com/user/moeseph?p=5

It's been a while since someone's cashed out >100x with a decent bet size. A smooth 3.13 btc from a 0.03 btc bet

Doesn't look like they were camping for a 100x, either. Just luck, :p.