The system I described doesn't double the bet, because it doesn't need to; it's not betting at 49.5%.  But it does increase the bet such that winning the 2nd bet will make up for the loss of the 1st bet and cause you to end up with exactly the same profit as if you had won the first bet.

Maybe it's only strictly martingale if you're playing for a 2x multiplier, and double your stake on a loss, I don't know the script definition.  But the system I'm proposing is very martingale-ish in that it has the "making up for losses" part.

When you bet your whole bankroll in a single bet, you expect to lose 1% of it.

When you split it up and bet the pieces in order from smallest to biggest, and stop when any bet wins you often don't end up betting the whole bankroll, and so you expect to lose 1% of less than the whole bankroll.

By splitting it up you reduce the amount you expect to bet, and so you reduce the amount you expect to lose.

I deposited 11 BTC (or 7? I forget) into BitZino once and lost it all to a bad streak of losing blackjack hands.

Later, my girlfriend deposited 0.01 BTC (or 0.1?  I forget) and won 32.768 BTC by getting a royal flush in video poker.

I conclude that sometimes people are lucky and sometimes they're unlucky.

If you're concerned about the fairness of the site they offer you the ability to check it by publishing hashes and algorithms and stuff.  Check it out!

No, your expectation increases.  Your potential profit is the same, your potential lose is the same, but chance of winning increases, so your expected profit increases.


In a 0 EV game your EV is and tends to 0.

I'm not sure what to say about that.  Where to even begin?

The JD luck statistic is unrelated to the house edge, or payout multipliers.  It simply indicates whether players have won more (>100%) or less (<100%) bets than expected, based on the chance they played at.

Currently the overall luck stat across all users is 100.06%.  It should be closer to 100%, but early on a couple of players made a point of brute-forcing a 0.0001% win.  One hit it playing 'hi', and one playing 'lo'.  Both hit it in about 500k rolls, twice as quickly as expected, and this skewed the global luck statistic upwards dramatically.  It's effectively a million bets with twice the expected 'luck'.

The other 1231 million bets with mostly average luck dilute the effect so the global luck is now not much over 100%.

Here's a simulation of how the global luck approaches 100% as the number of millions of bets increases:

>>> x = 0; 100.0 * (2.0+x)/(1+x)
200.0

>>> x = 1; 100.0 * (2.0+x)/(1+x)
150.0

>>> x = 2; 100.0 * (2.0+x)/(1+x)
133.33

>>> x = 10; 100.0 * (2.0+x)/(1+x)
109.09

>>> x = 500; 100.0 * (2.0+x)/(1+x)
100.19

>>> x = 1000; 100.0 * (2.0+x)/(1+x)
100.099

>>> x = 1231; 100.0 * (2.0+x)/(1+x)
100.08

This strategy I'm describing doesn't change the house edge.  The house edge is a constant 1%.

What it does do is allows people to expect bet less, and so expect to lose less (they still expect to lose 1% of the amount they risk, but they risk less).

Yeah, hence the "I think. ".  I'm not sure where it converges.

I was thinking of using the following strategy to split any single bet into a pair of more effective bets:

Suppose we have H and want to gain G.

We could make a single bet to attempt to do that
The payout multiplier would need to be (G+H)/H
The probability of success would be 0.99H/(G+H)

So we would bet H with chance 99H/(G+H) and if we win we end up with H*(G+H)/H = G+H and we've gained G.

Alternatively,

Define A = sqrt(G(G+H)) - G
Define B = H-A
Define P = (A+G)/A
Define C = 99/P = 99A/(A+G)

Then we can make an equivalent pair of bets:

bet A at chance C and if it loses bet B at chance C.  The payout multiplier is P for both bets.

If the first bet wins, we have (H-A) left that we didn't risk, and get A*P = A+G back, so we end up with G+H

If the 2nd bet wins, we have lost A in the first bet, and get B*P

BP = (H-A)(A+G)/A
= (HA - AA + HG - AG)/A
= (H.sqrt(G(G+H)) - HG - G(G+H) + 2G.sqrt(G(G+H)) - GG + HG - G.sqrt(G(G+H)) + GG)/A
= (H.sqrt(G(G+H)) - G(G+H) + 2G.sqrt(G(G+H)) - G.sqrt(G(G+H)))/A
= (H.sqrt(G(G+H)) - G(G+H) + G.sqrt(G(G+H)))/A
= ((G+H)sqrt(G(G+H)) - G(G+H))/A
= ((G+H)(sqrt(G(G+H)) - G))/A
= (G+H)A/A
= G+H

So whether we win the first or 2nd bet, we end up with G+H, as required.

And we have a (1-C)(1-C)/1e4 probability of success.


Can we use that recursively, to split 1 bet into 2, then 2 into 4 into 8, etc. indefinitely?  And if so, what does that do to the overall success rate?

Oh, I meant indivisibility.  As is there's nothing between 0 BTC and 1 satoshi.  Sorry - it's been a long day...

I expect at some point you're going to be wanting bet 0.01 satoshi at 0.00001% if you keep breaking down the martingale into ever more steps.  Neither of which (tiny stake nor tiny chance) is possible.

Except https://bitcointalk.org/index.php?topic=639787.msg7141055#msg7141055

It doesn't really matter.  Whether intentionally or not he's helping the scammer to run his scam.

I'm not sure which is worse - scamming on purpose or scamming but not being able to see what you're doing.

Yes.

And you're the first person who responded in such an open manner.

Everyone else just tells me I'm wrong.  

The trick is to split up your bet (the amount you were going to risk in a single bet) into a series of amounts which sum to a the same, and which form a sequence such that you can bet the smallest amount, and if it wins, you make the same as if you bet the whole amount at 49.5% (so you'll be betting with a smaller chance, and higher payout multiplier).  And if it loses, you want betting the 2nd amount to cover the first loss and make the same net profit.  Etc.

If you can find such a sequence (and you always can, though it can involve some hairy math depending on the length of the sequence you're looking for) then the amount you expect to risk is less than your whole amount (since there's a non-zero chance that you will win before the last bet, and stop at that point), and so the amount you expect to lose, being 1% of the amount you risk, is less than when you make the single bet.

Here's a very simple example:

you have 1 BTC and want to double it.

* you could bet it all at 49.5%, and succeed in doubling up with probability 0.495

* or you could bet 0.41421356 BTC at 28.99642866% with payout multiplier 3.41421356x, and if you lose, bet the rest at the same chance.  If you win either bet, you double up, else you lose.  Your chance of doubling up is 0.4958492857 - a little higher than the 0.495 you have with the single bet.

Cool, huh?

That's breaking the single bet up into a sequence of length 2.

If you break it up into more, smaller bets, then the probability of success increases further.

The more steps, the closer to 0.5 your probability of success gets.

You'll be limited by real-life barriers, like the invisibility indivisibility of the satoshi, and the limit of 4 decimal places on the chance at JD.  But in theory you can get arbitrarily close to 0.5.  I think.  

I don't mind if you PM me, but I often don't notice PMs.

It's better if we talk here, because the watchlist thing works most of the time, so I'll notice your messages quicker here.

I think it's safe to assume that almost everyone reading this thread knows how to convert BTC values to their own local currency.  And not safe to assume they care about US dollars in particular.

This is true, but the greater threat is that once the bankroll has halved a few times nobody trusts that the site is ever going to recover its losses and so the remaining bankroll evaporates as people pull out.

Hi.  Thanks for the PM asking me to check out your game.

I do see a few things worth commenting on:

1. What are the $ signs for?  Is this a Bitcoin or dollar game?  The two change in price relative to each other constantly.

2. The house edge seems pretty high and inconsistent:

Low Donation
0.0004 per player
10 players, 0.004 total
0.00365 paid out   (+ 0.002 0.0008 0.00045 0.0004)
8.75% house edge   (* (- 1 (/ 0.00365 0.004)) 100)

High Donation
0.01 per player
10 players, 0.1 total
0.095 paid out      (+ 0.05 0.02 0.015 0.01)
5% house edge      (* (- 1 (/ 0.095 0.1)) 100)

You're charging 8.75% on one and 5% on the other.  Is that intentional?  Your payout multipliers are the same in both games, so I guess it's an error.

Note that:

0.0004 * 1.12 = 0.000448 (not 0.00045)
0.01 * 1.12 = 0.0112 (not 0.015)

Maybe you meant 1.125x for the 3rd place payouts, not 1.12x.  In which case only the "high" game is wrong.

3. How do we know you will pay anyone at all?

4. How do we know you won't enter and win yourself?

5. What if less than 10 people enter?  Is there an end date after which you refund everyone?

... and that's only if you're betting at 50% chance to win.

People typically play with a slightly lower chance to win, like 49.5%, to cover the house edge and still get paid out 2x on win.

The chance of losing 10 times in a row at 49.5% is 1 in 927

So you're saying the deposit amount was the total of three separate deposits you made?

That makes more sense then.

Congrats.  You're one of the lucky ones.  The people whose coins were used to pay you out might be lucky too.  But at some point it will stop, and there will be large amounts outstanding.

I'm not accusing him of anything.

I don't know what's going on with his account.

Nobody else seems to have any problems though.

I don't think it's a good idea.

Why would you want a program to bet for you on a site where there's no winning strategy?

If you dislike playing enough that you do want to actually do it, perhaps you should stop playing.

You can just send your coins directly to my donation address in the FAQ if you want to get rid of them quickly. 

If you deposited 1.1505 BTC, why is he sending you 0.002 BTC, when he promised 2%?


2% of 1.1505 BTC is 0.02301 BTC.

Does OP suck at math or do you suck at shilling for him?

Only because you deleted the next line that I wrote:


The whole point I am making is that there is a difference between "the expected number of rolls to see a 6" and "the number of rolls you need to have a 50% chance of seeing a 6".


Read about what "expected value" means.

It sounds like you think it means "most likely value", but it doesn't.

Take your 1000 sided die.  Let's play a game.  If you roll a 6, you win a million dollars, but if you roll anything else, you lose a dollar.

What's the expected value of rolling the die?  Almost every time you roll (99.9% of the time), you lose a dollar.  So you 'expect' to lose almost every time.

But that's not what expected value means.  Your expected value for each roll is +$999.001:

(1 * 1e6 - 999 * 1) / 1000 = 999.001