Dooglus "AMA"

[a1choi]

You'll be limited by real-life barriers, like the invisibility 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.  

Also, can you explain more about the invisibility of the satoshi?  never heard that term until now.

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.

man i was stoked to hear about some new harry potter shit with math and stuff..  but alas, you had to turn all stodgy and math textbooky on me.  =p

yeah earlier on, i was trying to see if i could take advantage of rounding errors by betting small at other payouts..  how do you make sure those don't exist?  Always round down or only allow payouts/chance which provide "round" win values based on initial bet?

[Salmon1989]

Say, my target is to double my initial deposit.
Is it really possible to use martingale (or whatever strategy) to achieve the goal with higher than 49.5% success rate (one single bet)?

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 have done some calculation after reading your example.
If I make a series of bets in the following way:
Bet 1: Amount: 0.1 btc; Multiplier: 11x; Stop and get 2 btc if I win, proceed with bet 2 if I lose.
Bet 2: Amount: 0.1 btc; Multiplier: 12x; Stop and get 2 btc if I win, proceed with bet 3 if I lose.
Bet 3: Amount: 0.1 btc; Multiplier: 13x; Stop and get 2 btc if I win, proceed with bet 4 if I lose.
...
Bet 10: Amount: 0.1 btc; Multiplier: 20x; Stop and get 2 btc if I win.

I can get 2 btc with a chance of 0.496394553

If I split the 1 btc into 100 bets in similar way, the chance to get 2 btc is 0.496509733
If I split the 1 btc into 100000 bets in similar way, the chance to get 2 btc is 0.496522213

The win chance does get closer to 0.5 with more steps, but it seems it cannot be made arbitrary close to 0.5.

I am so glad that I asked you the question. Thanks a lot.

[jambola2]

Say, my target is to double my initial deposit.
Is it really possible to use martingale (or whatever strategy) to achieve the goal with higher than 49.5% success rate (one single bet)?

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 have done some calculation after reading your example.
If I make a series of bets in the following way:
Bet 1: Amount: 0.1 btc; Multiplier: 11x; Stop and get 2 btc if I win, proceed with bet 2 if I lose.
Bet 2: Amount: 0.1 btc; Multiplier: 12x; Stop and get 2 btc if I win, proceed with bet 3 if I lose.
Bet 3: Amount: 0.1 btc; Multiplier: 13x; Stop and get 2 btc if I win, proceed with bet 4 if I lose.
...
Bet 10: Amount: 0.1 btc; Multiplier: 20x; Stop and get 2 btc if I win.

I can get 2 btc with a chance of 0.496394553

If I split the 1 btc into 100 bets in similar way, the chance to get 2 btc is 0.496509733
If I split the 1 btc into 100000 bets in similar way, the chance to get 2 btc is 0.496522213

The win chance does get closer to 0.5 with more steps, but it seems it cannot be made arbitrary close to 0.5.

I am so glad that I asked you the question. Thanks a lot.

I have more questions on this , but before that can you clarify what "Luck %" of overall users on JD is ?
If everyone followed this strategy , wouldn't it technically break the house edge of 1% , making it more like 0.7% ?

[dooglus]

But in theory you can get arbitrarily close to 0.5.  I think.  

The win chance does get closer to 0.5 with more steps, but it seems it cannot be made arbitrary close to 0.5.

I am so glad that I asked you the question. Thanks a lot.

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?

[dooglus]

I have more questions on this , but before that can you clarify what "Luck %" of overall users on JD is ?
If everyone followed this strategy , wouldn't it technically break the house edge of 1% , making it more like 0.7% ?

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).

[jambola2]

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).

This is even more confusing.

Tell me if I've got it wrong , but this is what I think you are saying :-

House edge is the percentage of the amount you risk that you would expect to lose to the house on average.

That means that the strategy you previously outlined made you risk less but still expect to win the same amount.


The method you stated gets you closer and closer to 0.5 probability , but what would happen if you started with a hypothetical EV neutral game ? What would your EV tend to ?

[dooglus]

That means that the strategy you previously outlined made you risk less but still expect to win the same amount.

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.

The method you stated gets you closer and closer to 0.5 probability , but what would happen if you started with a hypothetical EV neutral game ? What would your EV tend to ?

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

[jambola2]

That means that the strategy you previously outlined made you risk less but still expect to win the same amount.

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.

Okay , what about the EV of the martingale strategy ?
If I want to get from 1 BTC to 2 BTC , would martingale betting or betting one single time be better ?

I wonder if it is just a myth propagated by casino owners

[Light]

Yes.

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

Everyone else just tells me I'm wrong.  

I cannot believe I'm saying this but I think you might be right.

I went ahead and played around with a case where the player starts at 1 wants to move to 2 by making two bets (of any size between 0-1 inclusive). Hence I went to go graph the function to see if it was true and I got this:

https://www.desmos.com/calculator/obkfifgbnl

Where y = probability of succeeding
and d = the value of the first initial bet

Notice how at both d = 0 and d= 1 the probability is 0.495 as expected (as you are either betting nothing then 1 or 1 then nothing and both are equivalent cases). And in between you get a probability higher than the 0.495 offered for the single bet.

I've tried a few set values for cases where you split your value up to more than two and you do get a better result. I can only theorise that this is because as your bet size approaches 0 with the number of bets approaching infinity your expectation approaches 1.

However, what I do not understand at the present is why this is so. I almost fell out of my chair when the numbers came out (I checked like 6 times), as it's inferring that you can get better than what the house 'technically' offers. The problem with this is that your expectation is better than just flat betting and logically that doesn't make sense. Both should have the same expectation.

I'm going to mull it over.

[Simon8x]

That means that the strategy you previously outlined made you risk less but still expect to win the same amount.

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.

Okay , what about the EV of the martingale strategy ?
If I want to get from 1 BTC to 2 BTC , would martingale betting or betting one single time be better ?

I wonder if it is just a myth propagated by casino owners

If you are talking about the strategy of making 2x bets and doubling the wager upon loss, it is worse as the total wagered amount is HUGE if you want to get from 1 BTC to 2 BTC.

[Light]

Okay , what about the EV of the martingale strategy ?
If I want to get from 1 BTC to 2 BTC , would martingale betting or betting one single time be better ?

I wonder if it is just a myth propagated by casino owners

If your solely looking for move from 1 to 2 or n to k then split bets should give you a better expectation than a single massive bet (if the math in my previous post is right). I'm not sure how this maths works out with the idea that martingale is bad, I'm not sure whether it's bad because of expectation or because your more probable to go bust.

[Mitchell]

Well, what you guys are talking about isn't really marginale, since that is about doubling your bet after each losing bet so that your next bet will 'make up' for all of your losses to that point. I think that "Risk Spreading" is a better name for the method described by dooglus.

Correct me if I am wrong

[DomenicoRomano]

Here is an example of a multi-bet strategy that improves your odds over a single bet strategy for a fixed bankroll.

Strategy 1: Bet 3 BTC, p = 0.7425         - get 4 BTC with p = 0.7425
Strategy 2: Bet 1 BTC, p = 0.495,
                if lose bet 2 BTC, p = 0.495. - get 4 BTC with p = 0.744975

Strategy 2 has better odds, by p = 0.002475 or 0.3%.  Strategy 2 is martingale starting bet of 1, exit of 4 or bust.

This is explained by the fact that Strategy 2 doesn't bet 3 BTC unless the first bet loses whereas Strategy 1 bets 3 BTC.  Your expected losses are proportional to the total amount bet.  The average amount bet with Strategy 2 is 2 BTC.

This is an example of the principal that: The less you bet the less you lose.

It is essential for each strategy that you stop betting.  The amount you lose grows with each bet.

[Light]

Well, what you guys are talking about isn't really marginale, since that is about doubling your bet after each losing bet so that your next bet will 'make up' for all of your losses to that point. I think that "Risk Spreading" is a better name for the method described by dooglus.

Correct me if I am wrong

No, I'm pretty sure your right. See the thing is that your idealised aims are different - in martingale you are looking to make a profit and so you continue to play until you either make that profit or go bust. In comparison in this case you play to reach a certain specified goal. The problem I have in my head, is that probability doesn't give a crap about your aims it's irrespective of it so your expectation shouldn't change with differing betting patterns. But the maths clearly contradicts this. So right now I'm not really too sure what to tihnk.

[001sonkit]

hey guys, back from being hermit for a very very long time (meh schoolwork)
anyway

Just being late on some topic, I'd like to brought back up the topic of "+EV" game from Kluge
I think it is possible to have some +ev game, in my opinion, it should act as a loyalty program for the people who bets. After all, the aim of casino is "sustainably draws money out of the player's pocket". And as you guys have previously mentioned, sometimes it is just about how the player controls their logical mind and stop betting anymore (like me, up and away).

If having a +EV game or must win game, unlocked by reaching a certain threshold. This thing could just stimulate their irrational brain to continue to bet, until they reach the threshold, since they just won't like to lose the golden opportunity even if they are currently in the green. They will eventually touch the red when they bet more and it would be a feedback loop to have them incentivised to continue to bet until they are finally up (and continue to chase for the +EV game... )

The money used for holding the +EV game comes from the bettors themselves at the end of the day. (In which all of the above are taught by Dragons.tl, that 3D gaming thing for Bitcoin)

But anyway, I know Doog will never change stuff, coz it is Just Dice.

Time to go voyage for the next whale.

[KingOfSports]

Kluge, look at bitcoinvideopoker, they have growing jackpots last I checked and once a game become +EV due to the progressive jackpot being so high, a ton of bots play until it is hit. It is pretty interesting to watch.

[Light]

They will eventually touch the red when they bet more and it would be a feedback loop to have them incentivised to continue to bet until they are finally up (and continue to chase for the +EV game... )

That is a fallacy unfortunately. There is a reasonable probability that they never in fact do get to losing margins and hence all you do is end up behind (as the house). If they do get to a certain margin and then give them +EV, they will eventually gain (over the long term) back their losses unless you remove the +EV. In effect all you're doing is cutting your profit margin and the only advantage to that would be if you gained enough users to overcome those losses.

@doog: Do you have an explanation for why bet splitting works? I can't think of the logic behind it.

[dooglus]

@doog: Do you have an explanation for why bet splitting works? I can't think of the logic behind it.

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.

[dooglus]

Well, what you guys are talking about isn't really marginale, since that is about doubling your bet after each losing bet so that your next bet will 'make up' for all of your losses to that point. I think that "Risk Spreading" is a better name for the method described by dooglus.

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.

[dooglus]

If you are talking about the strategy of making 2x bets and doubling the wager upon loss, it is worse as the total wagered amount is HUGE if you want to get from 1 BTC to 2 BTC.

Right, but what if you make an r-1 BTC bet with chance 99/(r*(r+1)), and if you lose, bet the rest at the same chance, where r is the square root of 2?

That has the effect of getting from 1 BTC to 2 BTC if you win either the first or second bet, and does it more reliably than putting the whole 1 BTC on at 49.5%.