How 999dice.com is stealing your coins, and exactly why you won't believe me

[dooglus]

So, on any given roll, you are risking about 11.15 BTC to win 1 satoshi. Of course your chances of losing are 1 in 1.28 BILLION, so it's "impossible".

Don't be so sure.

On Just-Dice, at 49.5% chance of winning, we have seen a winning streak of length 30, and a losing streak of length 32.

Those are 1-in-3-billion and 1-in-1.5-billion chances respectively.

Billion-to-one shots do happen. (Especially when you have over a billion bets on the site)...

>>> 1 / 0.505**32
3,123,747,346

>>> 1 / 0.495**30
1,451,590,214

Also, are you sure about your math there? If you have only a 1-in-1.28 billion chance of losing, then you can expect to win 1 satoshi around 1.28 billion times before you lose your 11.15 BTC.

There are 100 million satoshis in 1 BTC, so 1.28 billion satoshis is 12.8 BTC, and so your expectation is positive (win 12.8 BTC for every 11.15 BTC you lose).

[dooglus]

because they want to win every time and they forget that the casinos were not invented for that. you cannot win against them.

that's the point..we can't. win aginst them but.they are still addicted to some.techniques who.they belive will work aginst them

People win
against
casinos
every
day.

The trick is
to quit while
you're ahead.

If you play
long enough
you will lose
in the end.

[trixter]

because they want to win every time and they forget that the casinos were not invented for that. you cannot win against them.

that's the point..we can't. win aginst them but.they are still addicted to some.techniques who.they belive will work aginst them

People win
against
casinos
every
day.

The trick is
to quit while
you're ahead.

If you play
long enough
you will lose
in the end.

haiku or hai about no?

[dooglus]

that's the point..we can't. win aginst them but.they are still addicted to some.techniques who.they belive will work aginst them

People win
against
casinos
every
day.

The trick is
to quit while
you're ahead.

If you play
long enough
you will lose
in the end.

haiku or hai about no?

Arnab went as far as to use a table to lay out his weird post. I've no idea why anyone would do that.

[trixter]

Arnab went as far as to use a table to lay out his weird post. I've no idea why anyone would do that.

meh its ok.  I just couldnt think of anything funny to say so I thought a pun would have to do.

[dooglus]

Arnab went as far as to use a table to lay out his weird post. I've no idea why anyone would do that.

meh its ok.  I just couldnt think of anything funny to say so I thought a pun would have to do.

What is a haiku?
I'm just splitting my text up
like Arnab above.

[Jarx]

What is a haiku?
I'm just splitting my text up
like Arnab above.

haiku is traditional
and short form of
Japanese poetry.

[dooglus]

What is a haiku?
I'm just splitting my text up
like Arnab above.

haiku is traditional
and short form of
Japanese poetry.

Thanks for the info.
Haiku seems pretty easy.
I was worried there may have been some kind of silly rule about syllable counts.

[trixter]

Thanks for the info.
Haiku seems pretty easy.
I was worried there may have been some kind of silly rule about syllable counts.

That depends if you are into traditional art or modern art which throws out all the rules to wake up the sheeple.  

[keepinquiet]

So, on any given roll, you are risking about 11.15 BTC to win 1 satoshi. Of course your chances of losing are 1 in 1.28 BILLION, so it's "impossible".

Don't be so sure.

On Just-Dice, at 49.5% chance of winning, we have seen a winning streak of length 30, and a losing streak of length 32.

Those are 1-in-3-billion and 1-in-1.5-billion chances respectively.

Billion-to-one shots do happen. (Especially when you have over a billion bets on the site)...

>>> 1 / 0.505**32
3,123,747,346

>>> 1 / 0.495**30
1,451,590,214

Also, are you sure about your math there? If you have only a 1-in-1.28 billion chance of losing, then you can expect to win 1 satoshi around 1.28 billion times before you lose your 11.15 BTC.

There are 100 million satoshis in 1 BTC, so 1.28 billion satoshis is 12.8 BTC, and so your expectation is positive (win 12.8 BTC for every 11.15 BTC you lose).

Well don't misquote me then say I'm wrong I go on in that post to show that if you roll the dice often enough, you WILL hit that impossible 1-in-1.28-billion failure. During my tests (and betting) I hit a 37-in-a-row loss on a 49.95% bet... twice I think.

As for my math on that, the 1 satoshi profit and the 11.15 BTC cost were estimates. No way to know the 11.15 BTC for sure, because in X number of rolls, the profit won't always be the same, as it doesnt account for losses "eating" some of those rolls.

And as it turns out, my math in the above examples was wrong anyway. I'll explain more on that later - I was forgetting to use binomial distribution on the first part too.

[keepinquiet]

Side note, a Haiku isn't just 3 lines, it should be 5/7/5 syllables. I also believe a "traditional" haiku is about nature or something similar. Nowadays it's just a 5/7/5 poem.

 Don't Lose Your Bitcoin
In No Way Provably Fair
     999dice Scam

See? Haiku

[dooglus]

And as it turns out, my math in the above examples was wrong anyway. I'll explain more on that later - I was forgetting to use binomial distribution on the first part too.

I don't think you need the binomial distribution.

Lets use 999dice's numbers: Betting 95% [...]

To recoup your losses on 999dice, you need to bet 19.38x your last bet.

Your error is that you need to multiply your bet by 20.3877551, not 19.38x. If you multiply by less than 20 your net profit will be smaller the longer your streak, and it will be negative if the streak is long enough.

To calculate how much you need to increase your bet to keep the same net profit:

>>> edge = 0.001
>>> chance = 0.95
>>> (1 - edge) / (1 - edge - chance)
20.3877551020408

[trixter]

To calculate how much you need to increase your bet to keep the same net profit:

>>> edge = 0.001
>>> chance = 0.95
>>> (1 - edge) / (1 - edge - chance)
20.3877551020408

I think the problem here is that 999dice asks for a percentage increase which is always 100% lower than actual because your last bet is added.  I did not verify the math though to see if 19 or 20 was used I just know that you have to subtract 1 from what is entered on the webpage.

In short it was not a devious clever calculated number designed to cause people to lose money over time but a more simple mistake which I think is far more common

[keepinquiet]

Oops, you're right. I had 19.38 in my head because thats the base # and 999dice used funky multipliers. On loss increase bet by 100% actually DOUBLES your bet, which when you think about it, makes sense  when using the term "increase your bet by" where I am more inclined to think of 100 as 100% of your bet - 50 would be half, 200 would be double.

So for 999dice, if you enter 1938 in the box, it IS increasing it x20.38.

As for the binomial, yes, you do. I wasn't thinking clearly. Here's why:

Say you are making 20 "runs". Not rolls, runs. A run is where you bet until you max out and quit, or win. Much easier to do that math, as if you are doing rolls, you could have 1 chance at maxxing out, or 15-20 depending on how many rolls until you do. And you wouldn't want to quit in the middle of a run, you're losing by doing that.

So, say your chance of maxxing out and losing big is only .28% per run - your multiplers and bankroll is big enough that you can fail 21 times before you max out. .28% or 1 in 357 odds.

When you are doing something like flipping a coin, rolling a die, etc, that has a set chance for success or failure, the math to calculate your odds over time is not simple division. While if I do that once, my chances are 1 in 357 (.28%), if I do it 20x, my chances of it happening at least once can be much greater, or much less, than it appears.

Look at it this way... you have a 100 sided die. The chance of it landing on 1 is 1% - 1 in 100 odds.

What if you roll it 100 times? Your odds of getting a 1 are NOT 1 in 1. 1 in 1 odds means you WILL get a 1. You could roll it 500 times and not get a single 1. You could roll it 14 times and get 3.

Seeing something with a 1% chance of failure, and assuming you have a 50/50 shot at it if you do it 50 times is wrong. It doesnt work like that.

You need binomial distrbution - which is a statistical method to calculate probability.

In truth, if you can fail 21 times before losing big, and the chances of that happening the first time you drop the dice or click the button, and commit to the run are .28%, and you do it 20 times, for the total of 20 runs, you chances of failing 1 or more times is actually about 5.6%. Double it to 40 runs and the chance increases to 10.88% - not 11.2%. It's not straight math. It's a stupid complicated forumula that needs to be calculated over and over and totalled up to get an accurate number.

Its why a lot of gamblers who calculate odds fail, because they don't look at it right. And I completely goofed when posting my examples in this thread, forgetting that important part.

[dooglus]

You need binomial distrbution - which is a statistical method to calculate probability.

To simplify things, suppose we're doing a 50/50 bet, and we bust if we lose 3 times in a row. Let's say it pays out 2x, so there's a 0% house edge too.

Like you say, we might play 1000 times without hitting 3 losses in a row, or we might lose the first 3 plays and bust straight away.

What we're interested in here is what is the probability of any particular martingale sequence busting.

Well, it busts if we get 3 losses in a row, and each play is independent and loses with probability 1/2, so the probability of busting is 1/8. Every time we play, we have a 1/8 chance of busting, and a 7/8 chance of winning. So 7 times out of 8 we win 1 unit, and 1 time out of 8 we lose 1+2+4 = 7 units. That all sounds right, since total losses = total wins.

I didn't have to use the binomial distribution anywhere - I just raised the probability of losing a single play by the number of plays it takes to bust me. 1/2 ^ 3 = 1^8.

If you were to ask "how many times do we expect to win 1 unit before we bust?" or "how many times do we expect to bust if we make 1000 plays?" then the math gets difficult. Note, however, that neither of these questions are what we're looking at. We're just looking at our expected profits from playing the game.

I got mixed up in a bet about this kind of stuff a while ago, betting on the existence or otherwise of a streak of a certain length the JD bet history. I'm still not sure how lucky I was to win that bet given the probabilities involved.

And I completely goofed when posting my examples in this thread, forgetting that important part.

I think the biggest mistake was that you used 19.x instead of 20.x as the stake multiplier, making it look like your bust cost you less than it actually would have. If you use the 20.x figure instead you'll find that your expected losses when you bust are greater than the sum of your expected wins.

[camelson]

Hi this is owner of 999dice https://i.imgur.com/PHC9ex7.jpg Please send email to his job. Thank you. http://back9ins.com/about-us/contact-us/

[mtwelve]

Keepingitquiet,

I've PM'ed you. did you receive it?

[dooglus]

Hi this is owner of 999dice https://i.imgur.com/PHC9ex7.jpg Please send email to his job. Thank you. http://back9ins.com/about-us/contact-us/

To whom it may concern,

One of your employees (I'm not sure which, but he has a smug grin) is running a Bitcoin dice site where you have to click a button to get the server seed hash.

I know, right?!

I trust you can take it from here and deal with the matter appropriately.

Love, etc.

[keepinquiet]

Hi this is owner of 999dice https://i.imgur.com/PHC9ex7.jpg Please send email to his job. Thank you. http://back9ins.com/about-us/contact-us/

Except there is absolutely zero proof of that. None.

[keepinquiet]

You need binomial distrbution - which is a statistical method to calculate probability.

To simplify things, suppose we're doing a 50/50 bet, and we bust if we lose 3 times in a row. Let's say it pays out 2x, so there's a 0% house edge too.

Like you say, we might play 1000 times without hitting 3 losses in a row, or we might lose the first 3 plays and bust straight away.

What we're interested in here is what is the probability of any particular martingale sequence busting.

Well, it busts if we get 3 losses in a row, and each play is independent and loses with probability 1/2, so the probability of busting is 1/8. Every time we play, we have a 1/8 chance of busting, and a 7/8 chance of winning. So 7 times out of 8 we win 1 unit, and 1 time out of 8 we lose 1+2+4 = 7 units. That all sounds right, since total losses = total wins.

I didn't have to use the binomial distribution anywhere - I just raised the probability of losing a single play by the number of plays it takes to bust me. 1/2 ^ 3 = 1^8.

If you were to ask "how many times do we expect to win 1 unit before we bust?" or "how many times do we expect to bust if we make 1000 plays?" then the math gets difficult. Note, however, that neither of these questions are what we're looking at. We're just looking at our expected profits from playing the game.

I got mixed up in a bet about this kind of stuff a while ago, betting on the existence or otherwise of a streak of a certain length the JD bet history. I'm still not sure how lucky I was to win that bet given the probabilities involved.

And I completely goofed when posting my examples in this thread, forgetting that important part.

I think the biggest mistake was that you used 19.x instead of 20.x as the stake multiplier, making it look like your bust cost you less than it actually would have. If you use the 20.x figure instead you'll find that your expected losses when you bust are greater than the sum of your expected wins.

Well two things:
First, you cannot use binomial distribution for events which rely on each other (lose 3 in a row) unless you calculate the odds of losing 3 in a row and use that as a single failure event. Problem is a single failure event "uses up" rolls. Ie: if you are rolling 20 times, but have 3 failures, 6 of the 20 are "used" up as part of the failure run. Hence my use of "runs" instead of rolls.

Second: if you have a 12.5% chance of failure (your 1/8th example above) and you make 6 attempts, what is your chance of failing? It's not 1in8 / 6. The ONLY way to accurately gauge it is the BD(tired of typing that).

One run is 1/8th. Two runs is not 1 in 4. 8 runs is absolutely not 1 in 1 - it's totally possible to do 8 and not bust at all. When I started dice betting I miscalculated the odds and didn't know why until I studied statistics.

Thinking a .02% chance (1 in 2000 chance) of failure, and you make 100 runs, and you'll only have a 1 in 200 chances of losing will make you go broke when you lose a hell of a lot more often.