Topic: How 999dice.com is stealing your coins, and exactly why you won't believe me - page 5. (Read 41936 times)

2.5 million doge is a really big sum of money already so if in fact they let new players win means that you can create new accounts and win in all of them

Winning 2.5 mil DOGEs doesn't mean the site is not rigged. It would make sense for such site to let new players win to earn their trust.


This. Accusing site of being a scam and promoting them in your sig at the same time looks pretty weird.

But you won 2.5 million doge so the casino is not riggef after all if it lets you win that amount, the fact that you lost afterwards doesnt mean shit

Maybe take their URL out of your signature then?

I'm fucking pissed off I won 2.5 mil doge over 3 days there and now I just lose lose lose lose in the most retarded ways.  This site is definately a scam, I can't even break even and I've been trying for months.  I've lost over a thousand dollars here easily just trying to break even.  Doesn't seem to matter if I play safe or bet big, either way I lose.

no, but now you know how to make sure not to get scammed from them and you know that they are lying when they say its provably fair -> its not (this lie is a scam in itself - though it COULD be an accident)

So after all of this there is still not 100% proof that they scam

Right. I don't think anyone is counting rolls. We're counting "martingale sequences" - start at your base, play until you win or bust, that's 1 sequence.


I don't think that's true.

If I make 6 attempts and each one has a 1/8 probability of failing, the probability that at least 1 of them fails is 1 minus the probability that none of them fail, ie.:

1 - (1 - 1/8)^6
= 1 - 0.875^6
= 1 - 0.4488
= 0.5512

No BD in sight. No factorial, etc.


First off, you made a couple of mistakes:

1) I think you meant ".05% chance" (since that is 1 in 2000 : 100 / 0.02 = 5000 and 100 / 0.05 = 2000)

2) Also, did you mean "1 in 20" not "1 in 200" (since 2000/100 = 20)?

Anyway, the truth is that when the number of runs you're making (100) is significantly lower than the odds against failure happening (2000) then the naive guesstimate ("2000/100 = 20, so it's 1 in 20") is pretty accurate, and a little conservative. It tells you your chance of failure is HIGHER than it really is.

In fact when you make 100 runs and each has a 1 in 2000 chance of failure, your overall chance of failure is lower than 1 in 20. It's more like 1 in 20.5 - to be precise, it's 1 in:

1 / (1 - (1 - 1/2000)^100)
= 20.49916711818098

Like you said, if you make 2000 runs with a 1 in 2000 chance of failure per run, the overall chance of failure isn't 1 in 1. The actual chance of failure is lower than the naive runs/odds calculation would lead you to believe. In this case it's 1 in 1.58:

1 / (1 - (1 - 1/2000)^2000)
1.58174652400438

very good find.
i wont say they are cheating but they are not a provably fair system then

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.

Except there is absolutely zero proof of that. None.

Keepingitquiet,

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

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

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.


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.

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.

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

I don't think you need the binomial distribution.


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

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

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.