Topic: Dooglus "AMA" - page 2. (Read 2895 times)

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 ?

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?

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% ?

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.

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?

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://bitcointalksearch.org/topic/m.7141055

cool! i didn't know there was a 4 dec limit on chance.  good to know.

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

Only EV plus games are PVP games where players can win and not the house. 

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.  

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)?

I hesitate to mention this, but if you carefully pick your martingale strategy it is more effective than a single large bet at achieving your goal (whatever that goal is).

Yes and yes.
With "1% house edge", statistically speaking, you should expect to lose 1% of your total wagered amount.

So , if you are betting in an attempt to win something , it is best to bet all you can afford to bet at once ?

Does this imply that the martingale strategy is worse than a normal strategy ?

It's because when you accept those odds and gamble you effectively reduce your bet by 1% (this is true over an n case scenario where n approaches infinity). Hence, mathematically it is better to bet in smaller amounts as you 'lose' less from the edge. You don't notice this as in comparison to the number of bets you make the losses you get from the 1% are tiny. Maybe after something like 50 million bets that 1% would be clearer but over the short term variance wins out.

Well , I first expected it to still work , because you could just bet at 90% with 0.1 BTC over and over.

0.1 * (1.1)^25 = 1.083 BTC

So , taking the same percentage of 90% chance of winning,

(90/100)^ 25 = 7.17% of winning

What you said makes sense now that I have calculated it , but why does this happen ?

You have a lesser chance of winning and also a lesser amount of winnings , does this mean a bet with lesser chances of winning is always better ?

Game rules:
1000 players bet 1000 times in a game where they have a 50% chance of winning each bet (actually I think they have a 49.999999999% chance or whatever - Idunno exactly how random works, tbh). They each have a starting balance of 1 unit and must bet 1 unit. (changed from original so there's less 0s to write)

Set rules:
One game must be played 50 times.

Results of 10 sets: https://docs.google.com/spreadsheets/d/1KR_Wd9Z7K1YfWCsOgiq0KyRF7lD4dYaBnbW99BkSbSY/pubhtml

Conclusion:
House ended up very slightly (each game and set, the house seemed likely to lose, but rarely won huge) - nothing abnormal. Theory's definitely bunk, and I've now paid my debt to rationality by manually typing all those numbers in.

Script run (only "real" change is to print "casino profit" after each game so I don't have to think):

Yay! Thank you, dooglus. I'll run it 50 times, average what the house earns, then do that set of fifty 9 more times. Tonight's my last night of free time for a while. Time well spent, IMO. ETA: Actually, I'll do EV neutral, since there's still a fair chance I'll get some doubt benefit if variance is with me.

OK, still too easy for the player.  What if we don't let them cash out until they've made a million bets, and still only let them start with enough to make a single bet?

Still no good.  It takes a few more players before one goes to the moon, but it still happens, and when it does he breaks the bank: