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

[omahapoker]

very impressed with $23 million invested at just-dice

keep up the good work

[coincolonel]

very impressed with $23 million invested at just-dice

keep up the good work

And its value is increasing along with recent bullish of bitcoin price.. Let's all pray, to the moooon... 

[Gatekeeper]

very impressed with $23 million invested at just-dice

keep up the good work

Amazing isn't it?  #1 dice site by a long way and the only site i would ever trust and use.

dooglus does an amazing job 

[freedomno1]

very impressed with $23 million invested at just-dice

keep up the good work

Amazing isn't it?  #1 dice site by a long way and the only site i would ever trust and use.

dooglus does an amazing job  

I agree it's one of the safest places to store Bitcoins (in relation to trust) and from someone who I trust personally
Good old Satoshidice charts and a new way of building a gambling site
Go Dooglus and his sock puppet army

[Light]

very impressed with $23 million invested at just-dice

keep up the good work

Only sad thing is that there are no whales (that I can see) who play with anywhere close to the max limit now offered. Also overall bet volume (in terms of BTC) has seemed to be following a downward trend in the last couple of months. Wish there were more players.

[Acidyo]

I've been trading doge for btc recently. I have maybe 30 Mdoge left.

Make it sound like thats not a lot...thats at least 10k USD I believe. Feel free to donate it around! Or if I had a suggestion, donate it to Sean's Outpost or that BTC charity foundation that donates to worthy causes.

Do a do-good thing, dooglus thing haha

Why do people like to tell others what to do with their money?

[dooglus]

Do a do-good thing, dooglus thing haha

Why do people like to tell others what to do with their money?

https://bitcointalk.org/index.php?topic=631601 

[theskillzdatklls]

Do a do-good thing, dooglus thing haha

Why do people like to tell others what to do with their money?

https://bitcointalk.org/index.php?topic=631601 

such a face palm.

[dooglus]

such a face palm.

Is it just me or is your signature barely legible?

[organofcorti]

such a face palm.

Is it just me or is your signature barely legible?

It's just you. The signature is actually completely illegible.

[dooglus]

It's just you. The signature is actually completely illegible.

Hey it's you!

You might be able to help me with this...

If someone was to say "ha there's no way there's ever been a losing streak of 28 or more bets on any dice site ever at 49.5%" and I say "oh yeah?  I bet it has happened on Just-Dice alone!" and then someone else says "I bet you 49 BTC that it didn't happen" and I say "OK then!"...

What are my odds of winning?

Suppose that there are 357,937,388 bets at 49.5% and that they were all made by the same person.

What's the odds that it contains a losing streak (p(loss) = 0.505) of length 28 or more?

We know that the odds of a losing streak of length 28 happening in your first 28 rolls are 1 in 1/0.505**28 = 203,161,501.  That makes me think I have a greater than 50% chance of winning.

I'm told that the expected number of rolls before you see a 28 loss streak is around 514 million, but I don't know why.  That makes me think I have a lesser than 50% chance of winning.

So what are the odds of winning?

"Some of this actually happened."

[BayAreaCoins]

such a face palm.

Is it just me or is your signature barely legible?

I can't read it either. They have been like this for a while from the copy n paste from JD/DD. I said fuck it and changed the colors on mine to get it visible.

[organofcorti]

It's just you. The signature is actually completely illegible.

Hey it's you!

You might be able to help me with this...

If someone was to say "ha there's no way there's ever been a losing streak of 28 or more bets on any dice site ever at 49.5%" and I say "oh yeah?  I bet it has happened on Just-Dice alone!" and then someone else says "I bet you 49 BTC that it didn't happen" and I say "OK then!"...

What are my odds of winning?

Suppose that there are 357,937,388 bets at 49.5% and that they were all made by the same person.

What's the odds that it contains a losing streak (p(loss) = 0.505) of length 28 or more?

We know that the odds of a losing streak of length 28 happening in your first 28 rolls are 1 in 1/0.505**28 = 203,161,501.  That makes me think I have a greater than 50% chance of winning.

I'm told that the expected number of rolls before you see a 28 loss streak is around 514 million, but I don't know why.  That makes me think I have a lesser than 50% chance of winning.

So what are the odds of winning?

"Some of this actually happened."

I haven't given any though to the distribution of longest runs since this: https://bitcointalk.org/index.php?topic=80312.msg2338366;topicseen#msg2338366

I'll have to try and spin some of the old neurons up and see what happens.


EDIT: Can you explain this in more detail?

I'm told that the expected number of rolls before you see a 28 loss streak is around 514 million, but I don't know why.  That makes me think I have a lesser than 50% chance of winning.

[dooglus]

EDIT: Can you explain this in more detail?

I'm told that the expected number of rolls before you see a 28 loss streak is around 514 million, but I don't know why.  That makes me think I have a lesser than 50% chance of winning.

I get confused between the expected number of rolls for something to happen and having a 50% chance of that thing happening.  It feels as if the (alleged) fact that you expect to have to roll 514 million times on average to get a 28 losing streak means that since we have less than that number of bets then we have a less than 50% chance of having seen a 28 losing streak.

But I don't think that feeling is true.

Consider rolling a 6 sided die until you roll a six.  The expected number of rolls to get a six is 6.

But if you only roll 4 times, you have about a 51% of seeing a six, even though you've rolled less than the expected number.

Does that answer your question?  Basically my intuition around this is off (and yet I bet 50 BTC on it)...

[organofcorti]

EDIT: Can you explain this in more detail?

I'm told that the expected number of rolls before you see a 28 loss streak is around 514 million, but I don't know why.  That makes me think I have a lesser than 50% chance of winning.

I get confused between the expected number of rolls for something to happen and having a 50% chance of that thing happening.  It feels as if the (alleged) fact that you expect to have to roll 514 million times on average to get a 28 losing streak means that since we have less than that number of bets then we have a less than 50% chance of having seen a 28 losing streak.

But I don't think that feeling is true.

Consider rolling a 6 sided die until you roll a six.  The expected number of rolls to get a six is 6.

But if you only roll 4 times, you have about a 51% of seeing a six, even though you've rolled less than the expected number.

Does that answer your question?  Basically my intuition around this is off (and yet I bet 50 BTC on it)...

Sorry, I meant how is the 514million figure derived?

[dooglus]

EDIT: Can you explain this in more detail?

I'm told that the expected number of rolls before you see a 28 loss streak is around 514 million, but I don't know why.

No, not really, and apparently I'm misremembering anyway.

I was told "The expected value for 28 or more losses is 410,427,273" without explanation.  I've asked for the reasoning.

[Light]

No, not really, and apparently I'm misremembering anyway.

I was told "The expected value for 28 or more losses is 410,427,273" without explanation.  I've asked for the reasoning.

I believe what you're looking for is described in these two pages.

http://math.stackexchange.com/questions/383704/probability-of-streaks
http://www.askamathematician.com/2010/07/q-whats-the-chance-of-getting-a-run-of-k-successes-in-n-bernoulli-trials-why-use-approximations-when-the-exact-answer-is-known/

Not to be rude, but I'd say the maths is beyond the capabilities of most and as it is a recursive function it would need a computer for large values. Hope this was what you're after. Either way it's interesting reading.

[organofcorti]

No, not really, and apparently I'm misremembering anyway.

I was told "The expected value for 28 or more losses is 410,427,273" without explanation.  I've asked for the reasoning.

I believe what you're looking for is described in these two pages.

http://math.stackexchange.com/questions/383704/probability-of-streaks
http://www.askamathematician.com/2010/07/q-whats-the-chance-of-getting-a-run-of-k-successes-in-n-bernoulli-trials-why-use-approximations-when-the-exact-answer-is-known/

Not to be rude, but I'd say the maths is beyond the capabilities of most and as it is a recursive function it would need a computer for large values. Hope this was what you're after. Either way it's interesting reading.

Yes, it's very complicated and somewhat annoying. I have found that the nicest explanation of the distribution of run lengths is here: http://gato-docs.its.txstate.edu/mathworks/DistributionOfLongestRun.pdf

I'm ok with "expected run lengths for a number of rolls", but I haven't come across "expected number of rolls for a given run length", so I'm interested in the derivation. Can you point it out in the pages you linked?

[theskillzdatklls]

such a face palm.

Is it just me or is your signature barely legible?

Totally possible. I have signatures blocked so I don't even see my own. Might fix this later.

[dooglus]

EDIT: Can you explain this in more detail?

I'm told that the expected number of rolls before you see a 28 loss streak is around 514 million, but I don't know why.

No, not really, and apparently I'm misremembering anyway.

I was told "The expected value for 28 or more losses is 410,427,273" without explanation.  I've asked for the reasoning.

... and was given it:

The expected number of rolls R(N) to get a streak of length N where each roll has probability p is:

R(0) = 0
R(N) = (R(N-1)+1)/p

I tested this via simulation and it does appear to be true.

That gives these expected numbers of rolls for these streak lengths of losing 49.5% rolls (p = 0.505):

 1          1.98
 2          5.90
 3         13.67
 4         29.04
 5         59.49
 6        119.78
 7        239.17
 8        475.58
 9        943.72
10       1870.74
11       3706.41
12       7341.40
13      14539.40
14      28792.88
15      57017.57
16     112908.07
17     223582.32
18     442739.24
19     876713.34
20    1736068.01
21    3437760.41
22    6807448.33
23   13480097.68
24   26693264.72
25   52857951.91
26  104669213.69
27  207265771.66
28  410427272.59

Also, using simulation, I tried to figure out the number of rolls you need to make to have a 50% chance of seeing a streak of length N.  It looks to me like the number is about sqrt(2) times smaller than the above numbers, and so we need to have seen 410M/sqrt(2) = 290M rolls to have a 50% chance of seeing a streak of length 28.

Since we've seen 357M rolls, we presumably have a greater than 50% chance of seeing such a streak.

This is pretty low quality research I did using simulation.  The sqrt(2) thing in particular is very dodgy.