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[deepceleron]

A series of bets with an infinite possible number of bets will always win a single bet unit.

I guess that's the question.  Isn't it possible to lose forever?

If the probability is greater than zero, I can always move forward in the series of bet outcomes until I find a win. In an infinite series of wagers, there are an infinite number of wins yet to be won.

[organofcorti]

I've derived the expectation for any two roll sequence, house edge included. I'm assuming in this two roll sequence that different bet amounts (a1 and a2) and different probability games (p1 and p2) are used for each roll.


Expected profit, 1 roll sequence
Roll 1, win:  0.99*a/p-a   with probability = p
Roll 1, lose:  -a   with probability = (1-p)

Expected profit = (0.99*a/p-a)*p - a*(1 - p)
                = - 0.01*a ..... (1)

Expected profit, 2 roll sequence.

Roll 1, win:  0.99*a1/p1-a1   with probability = p1
Roll 1, lose:
   roll 2, win: 0.99*a2/p2-a2 - a1  prob = p2*(1 - p1)
   roll 2, lose: -(a1 + a2)   prob = (1 - p2)*(1 - p1)

Expected profit = (0.99*a1/p1-a1 )*p1+ (0.99*a2/p2-a2-a1)*p2*(1 - p1) -(a1 + a2)*(1 - p2)*(1 - p1)
                = - 0.01*a1 - 0.01*a2*(1-p1) ........ (2)


Since a1, a2 are always positive, and p1 is less than one and positive, the expected profit for a two roll sequence in the presence of a house edge will always be negative, and does not depend on the probability of the second roll.

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!

[narayan]

dooglus, please change the way you calculate the house edge to 2%% of the WINNINGS instead of 1% of the whole bet.

[vlees]

2%% of the WINNINGS

So it's 0.00002 * winnings?

(2%%)

[DomenicoRomano]

...
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.

[organofcorti]

...
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?

[DomenicoRomano]

...
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?

[dooglus]

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?

[DopeFish]

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.

[dooglus]

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.

[organofcorti]

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

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.

[Dabs]

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.

[organofcorti]

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.

[🏰 TradeFortress 🏰]

dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them

[mechs]

dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them

What exactly is the advanyage of inputs over any other web wallet?

[🏰 TradeFortress 🏰]

dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them

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.

[mechs]

dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them

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? 

[VTC]

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)

[PrintMule]

dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them

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.

[HollowIP]

dooglus: implement Inputs withdraws, pretty please?

People keep asking me about them

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.