Just-Dice.com : now with added CLAMs : Play or Invest

[infested999]

Here goes my design

Just some notes.

- The "grey" bar behind all the tabs is unnecessary and doesn't fit in, better just delete it.
- The "Chat" tab has a different background color than the outline of the chat box does.
- The max profit does not fit next to the wins/losses at all.

For reference, here is what I got on Firefox:


You have a separator between the "chat" page and the chat tab itself.

[dooglus]

Here goes my design, still under development. I am also making a theme switcher.

I tried it, and it looks quite good to me.

I have a problem with the roll buttons being on a separate line from the input fields at the zoom level I use:

[templerunAI]

• a theme that changes color based on how well you are doing?
• a theme without betting buttons (for the investors)

To do.

Here goes my design

Just some notes.

- The "grey" bar behind all the tabs is unnecessary and doesn't fit in, better just delete it.
- The "Chat" tab has a different background color than the outline of the chat box does.
- The max profit does not fit next to the wins/losses at all.

For reference, here is what I got on Firefox:
<img snip>
You have a separator between the "chat" page and the chat tab itself.

Check out, v0.1.1

The tabs have a little lighter shade of background, feels good.

Here goes my design, still under development. I am also making a theme switcher.

I tried it, and it looks quite good to me.

I have a problem with the roll buttons being on a separate line from the input fields at the zoom level I use:
<img snip>

Fixed!


For everyone, I will update my primary post with everything!

https://bitcointalk.org/index.php?topic=238613.msg2840579#msg2840579

[organofcorti]

........  I am interested in both the probabilities and the loss runs, now that you have defined it. So you mean to say, that at 87.7779% chance to win, I need to get (on average) 300,000 loss runs? Meaning, 300,000 martingale sequences before I hit a 6 loss streak?

Not quite. The expected number of sequences until you exactly 5 losses in a row is 41772.04. The expected number of loss runs until you hit more than 5 losses in a row 3e05.

Keep in mind that the variance is large. If the expected number of loss runs is 1/3e05, then 95 times out of a hundred that you experiment, you'll hit a hit loss streak of 6 or more at between 7,595 and 1,106,661 sequences.

So, a loss run is any martingale sequence? Not including just 1 loss. That means at least 2 losses in a row? Or does loss run mean any sequence of losses, including a single loss?

"Losses in a row" can be any positive integer from zero up: 0, 1, 2, 3 .... For example loss, loss, win, loss, win, win is 2 losses in a row, followed by a one loss sequences, followed by a zero loss sequence.

According to the same formulas (and even other formulas from other websites) the probabilities of a 7 loss streak or run at this chance to win is extremely low, which I took and interpreted as, the 6th roll in these two sequences are most probably the last to lose, the next one (the 7th) will most likely win.

That was true with the first time. On this second time, I still don't know yet if it will win since I have not yet made the 7th roll. However, since the average number of loss runs until the next greater than 6 run of losses in a row is about 2.4 million, and I've only rolled 1.3 million times, I think I'm not going to lose the next roll.

I also understand that, they keep saying all rolls are independent and dice have no memory, and assuming that, I still have an 87.7779% chance the next roll will win.

They do keep saying that and it is true. I absolutely guarantee you that as long as the site is fair you will have an 87.7779% chance to win every time you play the 87.7779% game. I also think you wont lose the next roll, in fact I'm 87.7779% sure you won't lose. More info on memoryless probability distributions: http://en.wikipedia.org/wiki/Memorylessness




@ infested999

If I have bankroll of 0.512 btc and I martingale with min bet 0.001 with the only purpose of doubling that 0.512 ... <snip>

I realised that the probability of having exactly doubled your money in exactly 512 rolls is extremely low - less than 0.25%. This is because as well as calculating the probability of 9 losses in a row, you need to also include the possibility of the last roll being a loss and only having 0.511 btc. Also, you need to include the probability of the last two rolls being a loss, the last three rolls and so on to the last 8 rolls being a loss.

Most people would then keep playing past 512 rolls until a win, but then the original probabilities calculated are incorrect. However if you make your calculations based on loss runs this is no longer a problem.

Would "how much bank do I need to have a 90% chance of a positive balance in 512 rolls" or something similar be a better question? Or maybe "I want to win x btc on probability p and before y loss runs, how much bank do I need?".

[superresistant]

According to the same formulas (and even other formulas from other websites) the probabilities of a 7 loss streak or run at this chance to win is extremely low, which I took and interpreted as, the 6th roll in these two sequences are most probably the last to lose, the next one (the 7th) will most likely win.
That was true with the first time. On this second time, I still don't know yet if it will win since I have not yet made the 7th roll. However, since the average number of loss runs until the next greater than 6 run of losses in a row is about 2.4 million, and I've only rolled 1.3 million times, I think I'm not going to lose the next roll.
I also understand that, they keep saying all rolls are independent and dice have no memory, and assuming that, I still have an 87.7779% chance the next roll will win.

They do keep saying that and it is true. I absolutely guarantee you that as long as the site is fair you will have an 87.7779% chance to win every time you play the 87.7779% game. I also think you wont lose the next roll, in fact I'm 87.7779% sure you won't lose. More info on memoryless probability distributions: http://en.wikipedia.org/wiki/Memorylessness

Interesting article. Now the question is :
Is just-dice a Markov chain characterized as memoryless or not ? How can you verify it ?

[dooglus]

Interesting article. Now the question is :
Is just-dice a Markov chain characterized as memoryless or not ? How can you verify it ?

That is the question.

I wonder what the answer is!

[dooglus]

TRY IT! CSS:
v0.1.1 : http://pastebin.com/raw.php?i=cZWurEt7

I tried the new one.  It still puts the bet buttons on a new line.  In fact it's a little worse now.  It only used to do that if I zoomed in, but now it does it at 100% zoom too:



(chromium on linux)

Overall I like the new design.  It just has a few wrinkles to iron out.  

[g83]

TRY IT! CSS:
v0.1.1 : http://pastebin.com/raw.php?i=cZWurEt7

I tried the new one.  It still puts the bet buttons on a new line.  In fact it's a little worse now.  It only used to do that if I zoomed in, but now it does it at 100% zoom too:



(chromium on linux)

Overall I like the new design.  It just has a few wrinkles to iron out.  

Looks like the fontsize ist 0.5 too high for roll

[wolverine.ks]

Custom theme ideas mentioned in chat...

• change cursor to harpoon when whale starts betting
• change background to ocean waves when whale starts betting
• flash 'all bets' tab when whale starts betting

[organofcorti]

According to the same formulas (and even other formulas from other websites) the probabilities of a 7 loss streak or run at this chance to win is extremely low, which I took and interpreted as, the 6th roll in these two sequences are most probably the last to lose, the next one (the 7th) will most likely win.
That was true with the first time. On this second time, I still don't know yet if it will win since I have not yet made the 7th roll. However, since the average number of loss runs until the next greater than 6 run of losses in a row is about 2.4 million, and I've only rolled 1.3 million times, I think I'm not going to lose the next roll.
I also understand that, they keep saying all rolls are independent and dice have no memory, and assuming that, I still have an 87.7779% chance the next roll will win.

They do keep saying that and it is true. I absolutely guarantee you that as long as the site is fair you will have an 87.7779% chance to win every time you play the 87.7779% game. I also think you wont lose the next roll, in fact I'm 87.7779% sure you won't lose. More info on memoryless probability distributions: http://en.wikipedia.org/wiki/Memorylessness

Interesting article. Now the question is :
Is just-dice a Markov chain characterized as memoryless or not ? How can you verify it ?

No idea. Why is that important?

[Dabs]

Not quite. The expected number of sequences until you exactly 5 losses in a row is 41772.04. The expected number of loss runs until you hit more than 5 losses in a row 3e05.

Keep in mind that the variance is large. If the expected number of loss runs is 1/3e05, then 95 times out of a hundred that you experiment, you'll hit a hit loss streak of 6 or more at between 7,595 and 1,106,661 sequences.

So, a loss run is any martingale sequence? Not including just 1 loss. That means at least 2 losses in a row? Or does loss run mean any sequence of losses, including a single loss?

"Losses in a row" can be any positive integer from zero up: 0, 1, 2, 3 .... For example loss, loss, win, loss, win, win is 2 losses in a row, followed by a one loss sequences, followed by a zero loss sequence.

So "Losses in a row" is basically how many wins, because that includes zero loss sequences, except if there are "Wins in a row" which are counted as one up to the next loss. Correct? This translates to approximately 87.7779% of the number of rolls? Or the other way around, about 13% of the number of rolls is the losses, therefore, the "Losses in a row" sequences.

Also, 95 times out of a hundred, I will hit a loss streak of 6 or more after 7,595 sequences, but before 1.2 million sequences. Which means I'm almost guaranteed to hit the loss streak after a million rolls.

I think the debate now is whether just-dice is really memoryless or not, but that's another issue. We can all agree that it might indeed be actually memoryless. Or maybe that's just Gambler's Fallacy whispering in my ear again ...

Dung beetles navigate using the milky way galaxy ... (okay, just having some fun here.)

[organofcorti]

Not quite. The expected number of sequences until you exactly 5 losses in a row is 41772.04. The expected number of loss runs until you hit more than 5 losses in a row 3e05.

Keep in mind that the variance is large. If the expected number of loss runs is 1/3e05, then 95 times out of a hundred that you experiment, you'll hit a hit loss streak of 6 or more at between 7,595 and 1,106,661 sequences.

So, a loss run is any martingale sequence? Not including just 1 loss. That means at least 2 losses in a row? Or does loss run mean any sequence of losses, including a single loss?

"Losses in a row" can be any positive integer from zero up: 0, 1, 2, 3 .... For example loss, loss, win, loss, win, win is 2 losses in a row, followed by a one loss sequences, followed by a zero loss sequence.

So "Losses in a row" is basically how many wins, because that includes zero loss sequences, except if there are "Wins in a row" which are counted as one up to the next loss. Correct? This translates to approximately 87.7779% of the number of rolls? Or the other way around, about 13% of the number of rolls is the losses, therefore, the "Losses in a row" sequences.

No. You could think of the number of "loss runs" as the number of wins, though.

Try this. Let "loss" = 0 and "win" = 1. Consider the following 24 rolls:

Code:

0 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0 1 1 0 1 0 1 1 

Now split into the "roll number", "run (or sequence) number" and "losses per run":

Code:

Rolls: 		0  0  0  1	0  1     0  0  1  	1 	1 	0  0  0  0  0  0  1	1 	0  1 	0  1  	1 
Roll number:	1  2  3  4	5  6 	 7  8  9	10	11	12 13 14 15 16 17 18	19	20 21	22 23 	24
Run number:	1		2	 3		4	5	6			7	8	9	10
Losses per run:	3       	1	 2		0	0	6			0	1	1	0

I hope that's clear?

Also, 95 times out of a hundred, I will hit a loss streak of 6 or more after 7,595 sequences, but before 1.2 million sequences. Which means I'm almost guaranteed to hit the loss streak after a million rolls.

I think the debate now is whether just-dice is really memoryless or not, but that's another issue. We can all agree that it might indeed be actually memoryless. Or maybe that's just Gambler's Fallacy whispering in my ear again ...

Dung beetles navigate using the milky way galaxy ... (okay, just having some fun here.)

If the dice rolls at Just-Dice are not memoryless, then the pseudo-RNG it uses would have to be broken. AFAICT, it's not. It's also quite possible to prove (within certain confidence levels) that the RNG is or is not as random as it should be, but I don't really have time for those sorts of shenanigans.

[petrescuerz]

Down for server stuff. Back in 5-10

Deb

[dooglus]

Down for server stuff. Back in 5-10

Deb

And again.  Should be back in 15 minutes max.

[dooglus]

And again.  Should be back in 15 minutes max.

We're back.

There was an issue where 90 investors couldn't view their investment log on the history tab.  It's fixed now.

Sorry for any inconvenience.

[organofcorti]

Something to help you avoid losing everything:

http://organofcorti.blogspot.com/2013/08/145-some-notes-about-just-dicecom.html

3. Summary
So you're probably wondering "How can I use this to help me bet?". Well, it can help you bet safely. If you can budget to withstand a certain number of losses in a row, then you can work out how likely they are to occur before an arbitrary number of wins. The higher the confidence probability you set, the safer you will be.

In the formula below, choose a confidence probability pC (I recommend 0.8 to 0.99, the closer to 1 the safer you'll be) for a game with a probability to win of pX to calculate the number of wins before n losses in a row with confidence pC.


For those wanting to copypaste into a spreadsheet of some sort, try the formula below, although I don't know what the Excel version of the "ceiling" function is:

ceiling(ln(pC)/ln(1 - (1 - pX)^(n + 1))

If you don't want to do that, you can also use the tables from the last section.

Good luck!

[Insanity]

The second just-dice implements video recording to provide provably fairness is the day that I will bet.

[vlees]

The second just-dice implements video recording to provide provably fairness is the day that I will bet.

I like your signature.

[Insanity]

The second just-dice implements video recording to provide provably fairness is the day that I will bet.

I like your signature.

Thanks for your supoprt, feel free to e-mail me at blobsrs@gmail.com if you have any further suggestions or concerns.

[organofcorti]

The second just-dice implements video recording to provide provably fairness is the day that I will bet.

How is video recording provably fair?