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Topic: Just-Dice.com : now with added CLAMs : Play or Invest - page 171. (Read 454769 times)

full member
Activity: 210
Merit: 100
blockchain.info keeps your private key on the browser, they never know it.
inputs.io does not.
You actually don't need to trust blockchain.info as long as you have their extension that does not allow changing the JavaScript code.
That interesting, make it more hacker-proof than even coinbase then.  But that means if your computer crashes you lose the private keys?
full member
Activity: 210
Merit: 100
dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them Tongue
What exactly is the advanyage of inputs over any other web wallet?
You don't need to wait for confirmations (and no TX fees) if you're paying offchain. It's also the most secure web wallet, with 2FA, GPG auth, location based authorization, session listing & management, and automatic tor/proxy block if you have enabled that.
you really think it is more secure than blockchain.info? 

It is when you cast aside any possible trust issues, like a thought of a situation,where site owners run away with your btc.
I don't know who the owners of blockchain.info are anymore than I do inputs.io.  I personally use coinbase for my web wallet.  Incorporated in US, $5M in venture funding and owners are well known. Only problem is lately they seem to have problems pushing btc transfers onto the blockchain without hour long delays.
sr. member
Activity: 518
Merit: 250
dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them Tongue
What exactly is the advanyage of inputs over any other web wallet?
You don't need to wait for confirmations (and no TX fees) if you're paying offchain. It's also the most secure web wallet, with 2FA, GPG auth, location based authorization, session listing & management, and automatic tor/proxy block if you have enabled that.
you really think it is more secure than blockchain.info? 

It is when you cast aside any possible trust issues, like a thought of a situation,where site owners run away with your btc.

I don't really see the need for inputs withdraws. It shows up in your account in minutes and confirms shortly after. It's probably healthy to have to wait 10 minutes for your funds anyways.
hero member
Activity: 980
Merit: 500
FREE $50 BONUS - STAKE - [click signature]
dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them Tongue
What exactly is the advanyage of inputs over any other web wallet?
You don't need to wait for confirmations (and no TX fees) if you're paying offchain. It's also the most secure web wallet, with 2FA, GPG auth, location based authorization, session listing & management, and automatic tor/proxy block if you have enabled that.
you really think it is more secure than blockchain.info? 

It is when you cast aside any possible trust issues, like a thought of a situation,where site owners run away with your btc.
VTC
member
Activity: 84
Merit: 14
Doog: I not that fond of the new reconnect feature you've implemented. I like to have the site open so I can check up on the bet history and chat. Would it be possible to let the filtered bets and chat stream (don’t have to be live like usual)? Or let us  download it on reconnect.

+1 for looking at the chat stream

Or add a "don't disconnect on idle" as an option to the account settings.  Most users won't enable it probably so not much of an increase in server load (I'm guessing that's why you started to disconnect idling clients)
full member
Activity: 210
Merit: 100
dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them Tongue
What exactly is the advanyage of inputs over any other web wallet?
You don't need to wait for confirmations (and no TX fees) if you're paying offchain. It's also the most secure web wallet, with 2FA, GPG auth, location based authorization, session listing & management, and automatic tor/proxy block if you have enabled that.
you really think it is more secure than blockchain.info? 
vip
Activity: 1316
Merit: 1043
👻
dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them Tongue
What exactly is the advanyage of inputs over any other web wallet?
You don't need to wait for confirmations (and no TX fees) if you're paying offchain. It's also the most secure web wallet, with 2FA, GPG auth, location based authorization, session listing & management, and automatic tor/proxy block if you have enabled that.

Sending coins also requires entering a PIN that you have to click on the pinpad. Standard keyloggers won't catch that - only a screen capturing malware can. They'd also have to do the request through your IP address otherwise you'd get an email alert.
full member
Activity: 210
Merit: 100
dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them Tongue
What exactly is the advanyage of inputs over any other web wallet?
vip
Activity: 1316
Merit: 1043
👻
dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them Tongue
donator
Activity: 2058
Merit: 1007
Poor impulse control.
Instead of talking about infinite bankrolls or infinity, let's talk about the max bet instead on the site. Currently 1% of the invested is 460, so that is the max profit. The max bet depends on the chance to win.

If I play 87.7779% the max bet would be 3,598 BTC.

If I start a martingale sequence to survive 7 consecutive losses and win the 8th, it would look like this.

0.00086500
0.00763092
0.06731904
0.59388037
5.23914032
46.21905808
407.73890358
3,597.02296846 = just a little bit below the max bet.

On average, I should hit an 8 loss streak in about 20 million rolls.

If I bet 0.00086500 I will win 0.00011058 per sequence. After 100k rolls, I'd have profited 11.06, after 1 million rolls 110.58, and after 10 million rolls 1,105.80. I can stop at any time before then so I never hit the 8 loss streak. Or I can hit it once while betting low amounts so I will probably not hit it again for another 20 million rolls, on average.

I'd need a bankroll of only about 4057. It will probably take me at least 2 to 3 months to do 10 million rolls, if I can manage to do 100k rolls per day.

Does that make sense? Are my odds or probabilities wrong? I just used the formulas provided by dooglus and organofcorti.

Okay, maybe this doesn't make sense, so you guys go back to talking about infinite amounts of money and time.


I just had a quick look - it seems to make sense. You're forgetting about variance though - 99% of the time you'll hit an 8 loss streak sometime after the first ~ 200000 wins (~ 230000 rolls).

The expected maximum number of losses in a row is:

Code:
Expected maximum number of losses in a row ~ Hn/(− log(1−p)) - 1/2

where n = number of wins and
         Hn is the nth Harmonic number, Hn = sum( 1/1 + 1/2 + 1/3 + ... 1/(n-1) + 1/n )

The the 10millionth roll will come after about 8.77 million wins, so Hn = sum( 1/1 + 1/2 + 1/3 + ... 1/8.77e06)


  Hn = sum( 1/1 + 1/2 + 1/3 + ... 1e-07) = 16.56406

So if you make 10 million rolls, the expected maximum losses in a row will be ~ 16.56406/(− log(1−0.877)) - 0.5 =  7.404318

I think, given these points, there's a good chance you'll hit 8 losses in a row before 10 million rolls.


Edit:

Alternatively, you can estimate the max losses in a row as:

Code:
log(n/p)/log(1/(1-p)) - 0.5772/log(1-p) - 0.5

n = rolls, p = probability to win. In your case:

Code:
log(1e07/0.877)/log(1/(1-0.877)) - 0.5772/log(1-0.877) - 0.5 = 7.529575


Much the same as the above result, and probably more accurate. However I prefer using 'wins' rather than 'rolls' since it's possible to derive the distribution of the maximum which allows easy calculation of confidence intervals. There's no easy way to do the same with maximum wins per roll.
legendary
Activity: 3416
Merit: 1912
The Concierge of Crypto
Instead of talking about infinite bankrolls or infinity, let's talk about the max bet instead on the site. Currently 1% of the invested is 460, so that is the max profit. The max bet depends on the chance to win.

If I play 87.7779% the max bet would be 3,598 BTC.

If I start a martingale sequence to survive 7 consecutive losses and win the 8th, it would look like this.

0.00086500
0.00763092
0.06731904
0.59388037
5.23914032
46.21905808
407.73890358
3,597.02296846 = just a little bit below the max bet.

On average, I should hit an 8 loss streak in about 20 million rolls.

If I bet 0.00086500 I will win 0.00011058 per sequence. After 100k rolls, I'd have profited 11.06, after 1 million rolls 110.58, and after 10 million rolls 1,105.80. I can stop at any time before then so I never hit the 8 loss streak. Or I can hit it once while betting low amounts so I will probably not hit it again for another 20 million rolls, on average.

I'd need a bankroll of only about 4057. It will probably take me at least 2 to 3 months to do 10 million rolls, if I can manage to do 100k rolls per day.

Does that make sense? Are my odds or probabilities wrong? I just used the formulas provided by dooglus and organofcorti.

Okay, maybe this doesn't make sense, so you guys go back to talking about infinite amounts of money and time.
donator
Activity: 2058
Merit: 1007
Poor impulse control.
Anyway, he's right. the Martingale sequence must end in a win or it's not a Martingale sequence.

I found a few posts on Wizards of Odds:

If I had an infinite amount of money and time, and the casino would take any bet, then could I ensure a profit by playing the Martingale (doubling after every loss until I win) on a fair bet on the toss of a coin?
— Anonymous

No. Some might argue that it would take an infinite number of losses to lose in this situation, which would be impossible. The truth is that 0.5^infinity approaches 0 but does not equal zero. If this did happen you would lose $2^infinity. The expected return of this strategy is thus 1 - $2^infinity * 0.5^infinity = $1 - 1 = 0. Another more graceful way to look at is that as your bankroll increases the expected value still remains unchanged at zero. So the limit of the expected value as the bankroll approaches infinity is zero. In other words an increasing bankroll doesn’t help your odds, even if it goes to infinity.

I think he's wrong with his "0.5^infinity approaches 0 but does not equal zero" - he's treating infinity like it's a number.  In fact 0.5^x tends to 0 as x tends to infinity, but the amount risked, 2^x tends to infinity as x tends to infinity - but I think he's right that the expectation remains at zero.

The posts immediately above and below that one on that page are similar too.

I see your point and I raise you one Smiley

If you just consider the amount won per sequence and then investigate the limit as number of rolls approaches infinity:

limit n->infinity 2^n - (2^n - 1) = 1

which is a positive expectation for the sequence as it approaches infinity. AFAICT the sequence can never actually be infinite since then it would not end, by definition not be a completed sequence and could not have an expectation. Ignore that - not useful to consider when the expectation actually sums to infinity.

I can't access that website from work - are the other posts more rigourously presented?


Edit: I should have just let Wolfram Alpha do all the heavy lifting.

The expectation of the total return per sequence n , for n -1 losses in a row = 0, 1, 2, ...., n-1, with a win on the nth roll:

E(return) =  sum(n=0 to n = Inf) (1-0.5)^(n-1) (2^n-(2^n-1))  = 2

Since the initial cost of the sequence is 1, E(profit) = E(return) - 1 = 1

If Wolfram alpha is to be believed and I have the expectation definition correct, the expected profit is still 1.

Seem logical? Personally, I don't believe in infinity. I've never seen one.

legendary
Activity: 2940
Merit: 1333
Anyway, he's right. the Martingale sequence must end in a win or it's not a Martingale sequence.

I found a few posts on Wizards of Odds:

If I had an infinite amount of money and time, and the casino would take any bet, then could I ensure a profit by playing the Martingale (doubling after every loss until I win) on a fair bet on the toss of a coin?
— Anonymous

No. Some might argue that it would take an infinite number of losses to lose in this situation, which would be impossible. The truth is that 0.5^infinity approaches 0 but does not equal zero. If this did happen you would lose $2^infinity. The expected return of this strategy is thus 1 - $2^infinity * 0.5^infinity = $1 - 1 = 0. Another more graceful way to look at is that as your bankroll increases the expected value still remains unchanged at zero. So the limit of the expected value as the bankroll approaches infinity is zero. In other words an increasing bankroll doesn’t help your odds, even if it goes to infinity.

I think he's wrong with his "0.5^infinity approaches 0 but does not equal zero" - he's treating infinity like it's a number.  In fact 0.5^x tends to 0 as x tends to infinity, but the amount risked, 2^x tends to infinity as x tends to infinity - but I think he's right that the expectation remains at zero.

The posts immediately above and below that one on that page are similar too.
newbie
Activity: 23
Merit: 0
Doog: I not that fond of the new reconnect feature you've implemented. I like to have the site open so I can check up on the bet history and chat. Would it be possible to let the filtered bets and chat stream (don’t have to be live like usual)? Or let us  download it on reconnect.
legendary
Activity: 2940
Merit: 1333
dooglus, please change the way you calculate the house edge to 2%% of the WINNINGS instead of 1% of the whole bet.

Can you give a simple example of what you mean, what it would change, and why that would be good?

It seems to me that it either means the investors win less (and I promised them the house edge would be 1%) or the players lose more (in which case wouldn't they play at the competition?).

But I'm willing to at least try to understand your suggestion.  I guess my main stumbling block at the moment is "WINNINGS".  If I bet 1 BTC and receive 2 BTC back (2x payout) did I win 1 BTC (the profit) or 2 BTC (the payout)?  How are you defining WINNINGS?
newbie
Activity: 33
Merit: 0
...
When will the expectation for the multi-roll strategy be more than for the single roll?



 - 0.01*a  <  - 0.01*a1 - 0.01*a2*(1-p1)
           a  >  a1 + a2*(1-p1)


So, compared to a single roll, if the amount bet on the first roll plus the amount bet on the second roll multiplied by the probability of a loss on the first roll is less than the amount bet on any game, then you can expect to lose less.

For a non-martingale two roll sequence I guess you're right, dooglus!

...


This is a simple consequence of the house edge being proportional to the bet size and the same for each individual bet.  It is true for all bets not just the subset of bets where the 1 roll has the same chance of winning as the 2 rolls.

Here is the same information:

Code:
a1 = bet 1
a2 = bet 2
Which has a higher average amount lost, a1 or a2?
It can be shown: When

      a1 > a2

For a bet sequences, s1 and s2, where each sequence contains any bets:

   If sum(s1 sequence) > sum(s2 sequence), then s1 has a higher average loss.



Can you show the

Code:
It can be shown:

part?


I just did.  Do you not see it?
donator
Activity: 2058
Merit: 1007
Poor impulse control.
...
When will the expectation for the multi-roll strategy be more than for the single roll?



 - 0.01*a  <  - 0.01*a1 - 0.01*a2*(1-p1)
           a  >  a1 + a2*(1-p1)


So, compared to a single roll, if the amount bet on the first roll plus the amount bet on the second roll multiplied by the probability of a loss on the first roll is less than the amount bet on any game, then you can expect to lose less.

For a non-martingale two roll sequence I guess you're right, dooglus!

...


This is a simple consequence of the house edge being proportional to the bet size and the same for each individual bet.  It is true for all bets not just the subset of bets where the 1 roll has the same chance of winning as the 2 rolls.

Here is the same information:

Code:
a1 = bet 1
a2 = bet 2
Which has a higher average amount lost, a1 or a2?
It can be shown: When

      a1 > a2

For a bet sequences, s1 and s2, where each sequence contains any bets:

   If sum(s1 sequence) > sum(s2 sequence), then s1 has a higher average loss.



Can you show the

Code:
It can be shown:

part?
newbie
Activity: 33
Merit: 0
...
When will the expectation for the multi-roll strategy be more than for the single roll?



 - 0.01*a  <  - 0.01*a1 - 0.01*a2*(1-p1)
           a  >  a1 + a2*(1-p1)


So, compared to a single roll, if the amount bet on the first roll plus the amount bet on the second roll multiplied by the probability of a loss on the first roll is less than the amount bet on any game, then you can expect to lose less.

For a non-martingale two roll sequence I guess you're right, dooglus!

...


This is a simple consequence of the house edge being proportional to the bet size and the same for each individual bet.  It is true for all bets not just the subset of bets where the 1 roll has the same chance of winning as the 2 rolls.

Here is the same information:

Code:
a1 = bet 1
a2 = bet 2
Which has a higher average amount lost, a1 or a2?
It can be shown: When

      a1 > a2

For a bet sequences, s1 and s2, where each sequence contains any bets:

   If sum(s1 sequence) > sum(s2 sequence), then s1 has a higher average loss.

full member
Activity: 196
Merit: 100
2%% of the WINNINGS

So it's 0.00002 * winnings?

(2%%)
member
Activity: 98
Merit: 10
I do not sell Bitcoins. I sell SHA256(SHA256()).
dooglus, please change the way you calculate the house edge to 2%% of the WINNINGS instead of 1% of the whole bet.
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