Dooglus "AMA"

[Kluge]

Hold on, before you jump on me for asking something ridiculously idiotic...

I was specifically thinking of the tendency for people to continue gamble until they're no longer able, especially on sites where you have to pre-fund accounts (you need to deal with the bother and, with BTC, confirmation wait times of withdrawal and possibly a withdrawal fee). I don't know of any figures which'd be relevant, but I'd be surprised if a very simple coinflip site, where there's always a 49% chance of the player winning, did not actually haul a 5% averaged profit or more due to gambling fallacies (letting it ride, then believing you can recover a loss by betting more, eventually losing their entire roll with a bad series of gambles).

-So, then, why not offer a casino where the player has an edge per play (maybe even significant), but still has the cards stacked against them simply by gambling psychology? The only concern, then, is with bots, but perhaps you could just raise the stakes high enough where bot-users would be hesitant or maybe even demand some kind of flat fee to play the game. Alternately, perhaps you could have it where players need to "unlock" that "EV+" game, perhaps through a large number of losses/bets with EV- games in the casino.

[1715669570]

1715669570

[a1choi]

Hold on, before you jump on me for asking something ridiculously idiotic...

I was specifically thinking of the tendency for people to continue gamble until they're no longer able, especially on sites where you have to pre-fund accounts (you need to deal with the bother and, with BTC, confirmation wait times of withdrawal and possibly a withdrawal fee). I don't know of any figures which'd be relevant, but I'd be surprised if a very simple coinflip site, where there's always a 49% chance of the player winning, did not actually haul a 5% averaged profit or more due to gambling fallacies (letting it ride, then believing you can recover a loss by betting more, eventually losing their entire roll with a bad series of gambles).

-So, then, why not offer a casino where the player has an edge per play (maybe even significant), but still has the cards stacked against them simply by gambling psychology? The only concern, then, is with bots, but perhaps you could just raise the stakes high enough where bot-users would be hesitant or maybe even demand some kind of flat fee to play the game. Alternately, perhaps you could have it where players need to "unlock" that "EV+" game, perhaps through a large number of losses/bets with EV- games in the casino.

Kluge, you seem to be a senior member here, but you really do have to brush up on your math and fallacies yourself.  I'll respond to this assuming you're not trolling all of us here.  

The math dictates that with large enough sample size, results will tend to the odds of the game.  Thus if the game has a 1% house edge, the house's profits will tend to 1% of the total wagered volume.  This holds true as players bet more.

If players bet more, they will tend to lose more relative to their overall wagered volume.  This is why a bankroll of a certain value will certainly go to zero as you play more, and that most wise people advise you to leave the tables and never play again when you are up.  As you play more, your total wagered will increase, and your expected loss (1% of that growing number) will definitely outgrow your total balance.

The profit of the house (house edge) can NOT increase to 5% as people play more.  It can fluctuate to 5% because of variance, but it will always tend to 1% as more bets are wagered/hands played.

A casino will definitely go out of business offering a +EV game because as more play is generated, their profits will fall in line with the math.  Their bankroll will dwindle, on average, with every bet.

What you might be thinking about is their actual profit.  Their total winnings minus operational costs.  That profit margin is definitely going to be very high, and not related to the house odds (1%)

Edit:  to address your point about players that lose their entire bankroll, roll to ruin so to speak.  If you have a lot of players, some will win and some will lose.  there will be enough winners that cancel out the losers, and still average back to the 1% in the long run.

[dooglus]

I'd be surprised if a very simple coinflip site, where there's always a 49% chance of the player winning, did not actually haul a 5% averaged profit or more due to gambling fallacies

This itself is a fallacy.  It doesn't matter why people play, or what they're thinking.  In a game with a 2% edge the house will average 2% profit on turnover in the long run(*).  Even if every player played until they bust, that doesn't change anything.  Some would get to play though their deposit more than 50 times over before busting, and some less.  On average they'd play through 50 times and so lose 2% of their total wagered amount.

-So, then, why not offer a casino where the player has an edge per play (maybe even significant), but still has the cards stacked against them simply by gambling psychology?

Because that makes no sense at all.  In a game of no skill, psychology plays no part.  Actual returns approach expected returns over the long run.  Offer a +EV no-skill game and quickly go bust.

(*) I know, Just-Dice has a 1% edge but only 0.33% profit.  That's because "the long run" is longer than you think.

[a1choi]

Because that makes no sense at all.  In a game of no skill, psychology plays no part.  Actual returns approach expected returns over the long run.  Offer a +EV no-skill game and quickly go bust.

There must be some site in the existence of the internet that has tried to fight the math and done a +EV game.  haha.

actually, +EV games happen in land casinos.  Not intended of course, but usually due to some math error combined with game rules/dealer error (and perhaps some unscrupulousness on the players part).  There are a lot of APs (advantage players) who try to find these games and take advantage of them before the casinos find out.  This was how Phil Ivey won the amount away from the casinos playing baccarat.  Edge sorting.

[Kluge]

Thanks. I have to reconsider this. I was assuming people are more likely to let things ride until their bankroll's out, at which point they're unable to recover, even if it's EV+. The premise is that gamblers have already written off what they gamble, and should they be up, they'll continue gambling, but if they're down, they'll also continue gambling. Compared to the average user, the casino has an unlimited bank roll. They can wait out the user until luck turns against them for a while and wipes their relatively tiny bankroll out (assuming it eventually does, which is almost guaranteed, especially with high stakes where it takes far fewer bets to wipe the player out). JD doesn't do pre-funding (AFAIK), but that figure's enough to make me assume I'm wrong.

Someone have a simulator to help me fully realize I'm wrong? I dunno how to create the formula for doing this quickly. 1,000 users have a $1,000 bankroll and must make 1,000 $25 bets with a 51% chance of 2x return, 49% chance of 0x return. The casino has an unlimited bankroll. The casino only loses if someone cashes out their winnings, which in this simulator, requires someone to make 1,000 $25 bets without their bankroll ever hitting $0. They get to keep whatever's left at the end of that.

[dooglus]

Thanks. I have to reconsider this. I was assuming people are more likely to let things ride until their bankroll's out, at which point they're unable to recover, even if it's EV+.

The problem with this is that while most of your players will play until they lose, one of them will never go below zero.  He will win enough that he'll never run out of funds, and the positive edge you've given him will allow him to use basic bankroll management techniques to safely keep playing for bigger and bigger stakes until he breaks you.

This guy was playing on Just-Dice today, betting at 90% chance to win (generally a pretty bad idea) and winning.  Notice his profit at no point went negative.



Now imagine he was playing with a significant positive expectation...

JD doesn't do pre-funding (AFAIK), but that figure's enough to make me assume I'm wrong.

I don't know the term.  What's pre-funding?  And what figure?

[Kluge]

I don't know the term.  What's pre-funding?  And what figure?

Where you have to deposit something on the site to bet with. US land casino, you take $1,000 in and get chips. You can't use dollars, must use chips.

I updated my second post here. My stupid's getting in the way, here. I'll write up a script to help me out if there isn't one pre-made. I was thinking about what you ended up posting (thanks)... I'm just unsure if there's really no advantage to having an "unlimited" bankroll. The casino would definitely be screwed out of a lot if someone ended up themselves getting an "unlimited" bankroll from winnings, where they're relatively safe from going bust.

(eta: "figure" was just a reference to the % you posted on what JD "should" earn vs what it does)

[dooglus]

There must be some site in the existence of the internet that has tried to fight the math and done a +EV game.  haha.

actually, +EV games happen in land casinos.  Not intended of course, but usually due to some math error combined with game rules/dealer error (and perhaps some unscrupulousness on the players part).  There are a lot of APs (advantage players) who try to find these games and take advantage of them before the casinos find out.  This was how Phil Ivey won the amount away from the casinos playing baccarat.  Edge sorting.

I'm not sure Ivey admits to using edge sorting.  The story I heard suggests that the dealer was colluding with him, placing the high cards upside down in the deck for him.  That would probably be considered closer to "cheating" than "advantage play" wouldn't it?

Poker is the most obvious +EV game in land casinos.  You're playing against other customers with the casino taking a rake, so as long as you're a good enough players that you can beat your peers and beat the rake, you make a profit.

Blackjack is another obvious example if you can find a game where you can count cards.  I used to count cards at Bitcoin casino strikesapphire.com until they changed the penetration and banned me.  Made a few hundred dollars, withdrew into five dollar Bitcoins and hedl.

And a third example is sign-up bonuses.  If a casino offers you 2x your deposit as a sign-up bonus, it's often +EV if you play the right games.  They often make you play through the bonus 20 times or something, in the hope that that's enough to make you lose it, but unless the edge is 5%, your expectation is positive even after playing through it 20 times.  You need to do the math carefully, but it's possible to find +EV promotions.

Some places offer "no zero" roulette, where all bets pay out their actual fair odds.  You get 36x for a single number for example.  That's a zero house edge.  I've seen it suggested that it's +EV for the house because people play until they bust, but I don't believe it.  I think it's offered as a loss-leader to get new customers, and then withdrawn in the hope that they'll play "proper" 2.7% edge roulette.

[dooglus]

Where you have to deposit something on the site to bet with. US land casino, you take $1,000 in and get chips. You can't use dollars, must use chips.

You have to deposit to JD before you can play.  All bets are off-blockchain.  You play with your balance, and withdraw when you're done.  In a sense you don't play with Bitcoins, you play with JD-IOUs.  When you're done, you click 'withdraw' and hope that we send you real Bitcoins in exchange for your IOUs, in much the same way you hope the casino is going to give you dollars for the clay chips you won.  They might not, but they always have so far.

I'm just unsure if there's really no advantage to having an "unlimited" bankroll. The casino would definitely be screwed out of a lot if someone ended up themselves getting an "unlimited" bankroll from winnings, where they're relatively safe from going bust.

As far as the house is concerned, isn't everyone the same player?  All the site sees is an endless stream of bets which either win or lose.  In a sense that's indistinguishable from a single player with an unlimited bankroll.  If we can handle an endless stream of finite bankrolls, can't we also handle a single player with infinite funds?

For a week or so before JD launched, we ran the site in "testnet" mode.  People could deposit testnet coins and play with them.  I also made up "fractional reserve" testnet coins out of thin air and gave them away generously.  The site was full of whales with huge bankrolls.  They didn't manage to get anywhere close to bankrupting the site.  Not that that proves anything.  Nakowa came closer with a few real coins than the testers did with their millions of fake testnet coins.

(eta: "figure" was just a reference to the % you posted on what JD "should" earn vs what it does)

This is a chart showing the actual and (offset) expected profit since nakowa finished his insane run.  Notice how close the two lines run to each other:



There's an offset of some 36k BTC which nakowa "should" have lost but didn't.  We don't expect to ever make that back again.  The expectation is that we stay 36k below 1%, but that it becomes less and less significant over time.

By my estimation in about 100 years if we keep the same volume as this year and have average luck, the site profit will be around 0.99%.  

Edit: and what you should reply is: look at how the black line goes up twice as steeply as the green line since the start of March.  that shows the site has been keeping 2% of turnover when it should only keep 1%.  That's because players play until they bust.  I don't have an answer for that, except that the site has been consistently almost twice as lucky as expected since March.  Weird.

[littlebox]

yes, by being a game creator at CrowdDice.com you get a slight edge (EV+) over the game players.

[Light]

I don't have an answer for that, except that the site has been consistently almost twice as lucky as expected since March.  Weird.

Well the answer is the same answer as pretty much every single question about why people lost. It's just variance. People are foolish to think that variance occurs over a set period of days, it can occur for any period in the 'short term' which is any point in time that is non-infinite.

So far the responses in this thread have been pretty much spot on - no matter what psychology you're using to explain it, it should be irrelevant as your profit will trend towards your house edge eventually. Not to mention the idea of having 'pre-funded' accounts might not take into account people like myself who if actually given +EV would be more than happy to pound the casino in the long term for profits. If I've actually got an edge no reason you couldn't just abuse it via botting all day long and eventually approaching your 1%. Unlocking +EV after losing a certain amount is interesting, but once again unless you cap the +EV such that it ends after a certain period it's completely open to abuse.

[Kluge]

Edit: and what you should reply is: look at how the black line goes up twice as steeply as the green line since the start of March.  that shows the site has been keeping 2% of turnover when it should only keep 1%.  That's because players play until they bust.  I don't have an answer for that, except that the site has been consistently almost twice as lucky as expected since March.  Weird.

Dammit. I'll write a script next week, because now I have absolutely no idea if I'm being stupid or not. I've gotta write it out. What you say makes complete sense, though. Fucking statistics, HOW DO THEY WORK?! Thanks. I'll go lose some money, now. ETA: Yay, winrar.

[a1choi]

I'm not sure Ivey admits to using edge sorting.  The story I heard suggests that the dealer was colluding with him, placing the high cards upside down in the deck for him.  That would probably be considered closer to "cheating" than "advantage play" wouldn't it?

http://apheat.net/2013/05/11/phil-ivey-won-7-8-million-pounds-by-edge-sorting-baccarat/ From what i heard, the rules of the game were agreed upon before hand which included that his requests to the dealer to flip certain cards would be upheld.  The casino must have thought it was a superstitious ritual and not know that it could allow him to perceive small defects in the cards.  

Poker is the most obvious +EV game in land casinos.  You're playing against other customers with the casino taking a rake, so as long as you're a good enough players that you can beat your peers and beat the rake, you make a profit.

yes agree!  but this game has an element of skill which affects EV.

Blackjack is another obvious example if you can find a game where you can count cards.  I used to count cards at Bitcoin casino strikesapphire.com until they changed the penetration and banned me.  Made a few hundred dollars, withdrew into five dollar Bitcoins and hedl.

also agree!  counting on online gambling sites is just too easy as well.  You could set up a bot to give you the exact count and EV and perfect bet amount based on kelly.  i'm not surprised they quickly changed their game after that and banned your IP.  you're too smart for them!

And a third example is sign-up bonuses.  If a casino offers you 2x your deposit as a sign-up bonus, it's often +EV if you play the right games.  They often make you play through the bonus 20 times or something, in the hope that that's enough to make you lose it, but unless the edge is 5%, your expectation is positive even after playing through it 20 times.  You need to do the math carefully, but it's possible to find +EV promotions.

Some places offer "no zero" roulette, where all bets pay out their actual fair odds.  You get 36x for a single number for example.  That's a zero house edge.  I've seen it suggested that it's +EV for the house because people play until they bust, but I don't believe it.  I think it's offered as a loss-leader to get new customers, and then withdrawn in the hope that they'll play "proper" 2.7% edge roulette.

definitely have seen +EV signup bonuses, sometimes used as loss leaders or just management ignorant of the math behind the scenes.  I always feel great when i can find a good deal.  even if its just for a hundred or so dollars.  haha!

and I don't believe 0% roulette could ever be +EV for the house, at least not from the winnings directly from that game.  I'd hesitate to offer that the variance of the game would wreak more havoc than the winnings would provide comfort.  But overall, as long as its doing its job as loss leader, and bringing players to play other games, it might be worth it.  Just keep the limits lower on that 0% table.  I remember reading some scientific paper that if the bankroll of the player is bigger than the house, the player will bankrupt the house playing 0% odds.  i'll try to search for that paper.

[a1choi]

I remember reading some scientific paper that if the bankroll of the player is bigger than the house, the player will bankrupt the house playing 0% odds.  i'll try to search for that paper.

I'm thinking about this paper which doesnt corroborate with what i said earlier..  http://www.library.unlv.edu/center-gaming-research/2008/09/article-impact-finite-bankroll-even-money-game.html.

but still i think the math stands up that if the player has a larger bankroll than the house, player has a larger chance to bankrupt the house than the house has to bankrupt the player on a 0% game

and this is where i calculated the ruin %..  if you input the odds of the house win as 0.505 and loss at 0.495, at 100x bankroll compared to max bet, the chance of ruin is 14%...  which is really high.  i wonder how they came up with that formula.
http://www.urbino.net/digital-books.cfm?dID=13&p=3&sID=87  bottom of the page.

[dooglus]

but still i think the math stands up that if the player has a larger bankroll than the house, player has a larger chance to bankrupt the house than the house has to bankrupt the player on a 0% game

It seems kind of obvious that in a 0% edge game whoever has the biggest bankroll has the smaller chance of going bust...  Not to say that everything is intuitive with probability stuff, but in this case I think it makes sense.

and this is where i calculated the ruin %..  if you input the odds of the house win as 0.505 and loss at 0.495, at 100x bankroll compared to max bet, the chance of ruin is 14%...  which is really high.  i wonder how they came up with that formula.

Pretty sure they're assuming the max bet is a constant.  Casinos generally don't reduce the max bet as players win...

When the bankroll is crowd-sourced and changes dramatically over time like it does at JD it makes sense to set the maximums dynamically so they change with the bankroll.  That has a side-effect of vastly reducing the risk of ruin.

[a1choi]

Pretty sure they're assuming the max bet is a constant.  Casinos generally don't reduce the max bet as players win...

When the bankroll is crowd-sourced and changes dramatically over time like it does at JD it makes sense to set the maximums dynamically so they change with the bankroll.  That has a side-effect of vastly reducing the risk of ruin.

makes lots of sense.  i guess the best way to calculate risk of ruin is to go through a bernoulli process then.  Assuming a reasonal bankruptcy point > 0.  maybe 30-40% of original bankroll.  since going to 0 would be infinite time.

[dooglus]

makes lots of sense.  i guess the best way to calculate risk of ruin is to go through a bernoulli process then.  Assuming a reasonal bankruptcy point > 0.  maybe 30-40% of original bankroll.  since going to 0 would be infinite time.

Maybe consider the risk of losing half the bankroll - a kind of a 'half life' thing.

Also, what if someone came with enough coins to bet the maximum at 98% over and over.  Each time they have a very good chance of winning 0.5% of the house bankroll.

If they win 138 times in a row, they've won half the bankroll, and they can do so with 6% chance:

>>> 0.995 ** 138 = 0.5007
>>> 0.98 ** 138 = 0.06154

If the max profit per roll was a full 1% of the bankroll, they only need to win 69 times in a row, and they can do so with 25% chance:

>>> 0.99 ** 69 = 0.4998
>>> 0.98 ** 69 = 0.24808

So by cutting the max profit from 1% to 0.5% we've made it ~4 times harder to win half the bankroll by max-betting at 98%.

[a1choi]

So by cutting the max profit from 1% to 0.5% we've made it ~4 times harder to win half the bankroll by max-betting at 98%.

i remember halving the kelly lead to quartering the risk.  something to do with a squared of the denominator thing...

[jambola2]

This guy was playing on Just-Dice today, betting at 90% chance to win (generally a pretty bad idea) and winning.  Notice his profit at no point went negative.

What does generally a pretty bad idea mean ? As far as I can understand , the house edge should be equal at all levels and that would just be a lot less fun (as you have no chance of winning something substantial) , with the same amount of risk.

Is my understanding wrong ?

[a1choi]

This guy was playing on Just-Dice today, betting at 90% chance to win (generally a pretty bad idea) and winning.  Notice his profit at no point went negative.

What does generally a pretty bad idea mean ? As far as I can understand , the house edge should be equal at all levels and that would just be a lot less fun (as you have no chance of winning something substantial) , with the same amount of risk.

Is my understanding wrong ?

its probably just the risk of gambling so much to win so little.  I dont think Doog means that only 90% is a bad idea.  maybe he has a range of 85-95% thats bad in his mind..  or perhaps 90% up to 98.99% is a bad range..  dunno.  i think its subjective.

[dooglus]

What does generally a pretty bad idea mean ? As far as I can understand , the house edge should be equal at all levels and that would just be a lot less fun (as you have no chance of winning something substantial) , with the same amount of risk.

Is my understanding wrong ?

If you want to win 1 BTC, you could:

a) risk 0.1 BTC at 9% for an 11x payout, or you could
b) risk 10 BTC at 90% for a 1.1x payout.

In each case you expect to lose 1% of the amount you risk, so b) costs about 100 times more than a) in terms of the amount you expect to lose, while only having a 10 times better chance of success.

That makes 90% seem like a bad deal compared to 9%, even though they both have a 1% edge.

Does that make sense?

[jjc326]

I do understand OP and it's worth a shot.  You're thinking that even though people might be "winning" with an edge of like 1% or something, they'll keep playing until they lose.  It is risly but if you could make it owrk that'd be great.  You would have to stop bots because they wouldn't have "emotion" so that was a great point too.

[dooglus]

Someone have a simulator to help me fully realize I'm wrong? I dunno how to create the formula for doing this quickly. 1,000 users have a $1,000 bankroll and must make 1,000 $25 bets with a 51% chance of 2x return, 49% chance of 0x return. The casino has an unlimited bankroll. The casino only loses if someone cashes out their winnings, which in this simulator, requires someone to make 1,000 $25 bets without their bankroll ever hitting $0. They get to keep whatever's left at the end of that.

I'll run that simulation and let you know how it goes.

[dooglus]

Someone have a simulator to help me fully realize I'm wrong? I dunno how to create the formula for doing this quickly. 1,000 users have a $1,000 bankroll and must make 1,000 $25 bets with a 51% chance of 2x return, 49% chance of 0x return. The casino has an unlimited bankroll. The casino only loses if someone cashes out their winnings, which in this simulator, requires someone to make 1,000 $25 bets without their bankroll ever hitting $0. They get to keep whatever's left at the end of that.

I'll run that simulation and let you know how it goes.

OK, maybe I misunderstood your specification, but here are the results:

I coded a simulation, ran it 6 times.  The first time I ran it the first player busted and the 2nd player made 1000 bets without busting.  The other 5 times the first player made 1000 bets without busting:

player    1  won bet    1.  balance = 1025
player    1  won bet    2.  balance = 1050
player    1 lost bet    3.  balance = 1025
player    1 lost bet    4.  balance = 1000
player    1  won bet    5.  balance = 1025
player    1 lost bet    6.  balance = 1000
player    1 lost bet    7.  balance =  975
player    1  won bet    8.  balance = 1000
player    1  won bet    9.  balance = 1025
player    1 lost bet   10.  balance = 1000
[...]
player    1  won bet  991.  balance = 2475
player    1 lost bet  992.  balance = 2450
player    1 lost bet  993.  balance = 2425
player    1 lost bet  994.  balance = 2400
player    1 lost bet  995.  balance = 2375
player    1  won bet  996.  balance = 2400
player    1 lost bet  997.  balance = 2375
player    1  won bet  998.  balance = 2400
player    1  won bet  999.  balance = 2425
player    1  won bet 1000.  balance = 2450
player 1 made 1000 bets without busting

Here's the simulation code:

#!/usr/bin/env python

import random, sys

players = 1000
bets_per_player = 1000
starting_balance = 1000
stake = 25

def roll():
    return random.random() < 0.51

player = 1
while player <= players:
    balance = starting_balance
    bets = 0
    while True:
        bets += 1
        if roll():
            balance += stake
            print "player %4d  won bet %4d.  balance = %4d" % (player, bets, balance)
        else:
            balance -= stake
            print "player %4d lost bet %4d.  balance = %4d" % (player, bets, balance)
            if balance < stake:
                print "player %d bust after %d bets" % (player, bets)
                break
        if bets == bets_per_player:
            print "player %d made %d bets without busting" % (player, bets)
            sys.exit(0)

    player += 1

print "all the players busted"

[a1choi]

What does generally a pretty bad idea mean ? As far as I can understand , the house edge should be equal at all levels and that would just be a lot less fun (as you have no chance of winning something substantial) , with the same amount of risk.

Is my understanding wrong ?

If you want to win 1 BTC, you could:

a) risk 0.1 BTC at 9% for an 11x payout, or you could
b) risk 10 BTC at 90% for a 1.1x payout.

In each case you expect to lose 1% of the amount you risk, so b) costs about 100 times more than a) in terms of the amount you expect to lose, while only having a 10 times better chance of success.

That makes 90% seem like a bad deal compared to 9%, even though they both have a 1% edge.

Does that make sense?

i guess my risk preferences are more risk averse than yours.  i'd gladly risk the 100x for the 10x more chance of winning.  90% just seems a much better shot than 9%.

but personally i'd be limited to only acting on the 0.1BTC bet because i don't have the bankroll for 10BTC bets.

[dooglus]

I coded a simulation, ran it 6 times.  The first time I ran it the first player busted and the 2nd player made 1000 bets without busting.  The other 5 times the first player made 1000 bets without busting:

OK, so let's give the casino a better chance.  Each player starts with just $25, and has to flat-bet $25 over and over until they either bust or make 1000 bets.

Now the casino is bound to win, right?

Wrong:

player    1 lost bet    1.  balance =    0
player 1 bust after 1 bets
player    2 lost bet    1.  balance =    0
player 2 bust after 1 bets
player    3 lost bet    1.  balance =    0
player 3 bust after 1 bets
player    4 lost bet    1.  balance =    0
player 4 bust after 1 bets
player    5 lost bet    1.  balance =    0
player 5 bust after 1 bets
player    6  won bet    1.  balance =   50
player    6 lost bet    2.  balance =   25
player    6 lost bet    3.  balance =    0
player 6 bust after 3 bets
player    7  won bet    1.  balance =   50
player    7 lost bet    2.  balance =   25
player    7  won bet    3.  balance =   50
player    7 lost bet    4.  balance =   25
player    7  won bet    5.  balance =   50
player    7  won bet    6.  balance =   75
player    7 lost bet    7.  balance =   50
player    7 lost bet    8.  balance =   25
player    7 lost bet    9.  balance =    0
player 7 bust after 9 bets
player    8 lost bet    1.  balance =    0
player 8 bust after 1 bets
player    9  won bet    1.  balance =   50
player    9  won bet    2.  balance =   75
player    9  won bet    3.  balance =  100
player    9  won bet    4.  balance =  125
player    9 lost bet    5.  balance =  100
player    9  won bet    6.  balance =  125
player    9 lost bet    7.  balance =  100
player    9  won bet    8.  balance =  125
player    9  won bet    9.  balance =  150
player    9  won bet   10.  balance =  175
player    9 lost bet   11.  balance =  150
player    9 lost bet   12.  balance =  125
player    9 lost bet   13.  balance =  100
player    9 lost bet   14.  balance =   75
player    9 lost bet   15.  balance =   50
player    9  won bet   16.  balance =   75
player    9 lost bet   17.  balance =   50
player    9 lost bet   18.  balance =   25
player    9  won bet   19.  balance =   50
player    9 lost bet   20.  balance =   25
player    9 lost bet   21.  balance =    0
player 9 bust after 21 bets
player   10 lost bet    1.  balance =    0
player 10 bust after 1 bets
player   11  won bet    1.  balance =   50
player   11 lost bet    2.  balance =   25
player   11 lost bet    3.  balance =    0
player 11 bust after 3 bets
player   12  won bet    1.  balance =   50
player   12  won bet    2.  balance =   75
player   12 lost bet    3.  balance =   50
player   12 lost bet    4.  balance =   25
player   12  won bet    5.  balance =   50
player   12 lost bet    6.  balance =   25
player   12 lost bet    7.  balance =    0
player 12 bust after 7 bets
player   13 lost bet    1.  balance =    0
player 13 bust after 1 bets
player   14  won bet    1.  balance =   50
player   14 lost bet    2.  balance =   25
player   14  won bet    3.  balance =   50
player   14  won bet    4.  balance =   75
player   14  won bet    5.  balance =  100
player   14  won bet    6.  balance =  125
player   14  won bet    7.  balance =  150
player   14  won bet    8.  balance =  175
player   14 lost bet    9.  balance =  150
player   14  won bet   10.  balance =  175
[...]
player   14  won bet  224.  balance =  125
player   14 lost bet  225.  balance =  100
player   14 lost bet  226.  balance =   75
player   14  won bet  227.  balance =  100
player   14  won bet  228.  balance =  125
player   14 lost bet  229.  balance =  100
player   14 lost bet  230.  balance =   75
player   14 lost bet  231.  balance =   50
player   14 lost bet  232.  balance =   25
player   14 lost bet  233.  balance =    0
player 14 bust after 233 bets
player   15  won bet    1.  balance =   50
player   15  won bet    2.  balance =   75
player   15  won bet    3.  balance =  100
player   15 lost bet    4.  balance =   75
player   15  won bet    5.  balance =  100
player   15  won bet    6.  balance =  125
player   15 lost bet    7.  balance =  100
player   15  won bet    8.  balance =  125
player   15  won bet    9.  balance =  150
player   15 lost bet   10.  balance =  125
[...]
player   15 lost bet  991.  balance =  850
player   15 lost bet  992.  balance =  825
player   15 lost bet  993.  balance =  800
player   15  won bet  994.  balance =  825
player   15  won bet  995.  balance =  850
player   15  won bet  996.  balance =  875
player   15  won bet  997.  balance =  900
player   15 lost bet  998.  balance =  875
player   15 lost bet  999.  balance =  850
player   15 lost bet 1000.  balance =  825
player 15 made 1000 bets without busting

[dooglus]

OK, so let's give the casino a better chance.  Each player starts with just $25, and has to flat-bet $25 over and over until they either bust or make 1000 bets.

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:

player    1 lost bet    1.  balance =    0
player 1 bust after 1 bets
player    2  won bet    1.  balance =   50
player    2  won bet    2.  balance =   75
player    2  won bet    3.  balance =  100
[...]
player    2  won bet   18.  balance =  225
player    2  won bet   19.  balance =  250
player    2  won bet   20.  balance =  275
player    2 lost bet   21.  balance =  250
player    2 lost bet   22.  balance =  225
[...]
player    2 lost bet   51.  balance =   50
player    2 lost bet   52.  balance =   25
player    2 lost bet   53.  balance =    0
player 2 bust after 53 bets
player    3 lost bet    1.  balance =    0
player 3 bust after 1 bets
player    4  won bet    1.  balance =   50
player    4 lost bet    2.  balance =   25
player    4 lost bet    3.  balance =    0
player 4 bust after 3 bets
player    5  won bet    1.  balance =   50
player    5 lost bet    2.  balance =   25
player    5 lost bet    3.  balance =    0
player 5 bust after 3 bets
player    6 lost bet    1.  balance =    0
player 6 bust after 1 bets
player    7 lost bet    1.  balance =    0
player 7 bust after 1 bets
player    8  won bet    1.  balance =   50
player    8 lost bet    2.  balance =   25
player    8  won bet    3.  balance =   50
player    8 lost bet    4.  balance =   25
player    8  won bet    5.  balance =   50
player    8 lost bet    6.  balance =   25
player    8  won bet    7.  balance =   50
player    8 lost bet    8.  balance =   25
player    8 lost bet    9.  balance =    0
player 8 bust after 9 bets
player    9  won bet    1.  balance =   50
player    9 lost bet    2.  balance =   25
player    9  won bet    3.  balance =   50
player    9 lost bet    4.  balance =   25
player    9 lost bet    5.  balance =    0
player 9 bust after 5 bets
player   10 lost bet    1.  balance =    0
player 10 bust after 1 bets
player   11 lost bet    1.  balance =    0
player 11 bust after 1 bets
player   12  won bet    1.  balance =   50
player   12  won bet    2.  balance =   75
player   12  won bet    3.  balance =  100
player   12  won bet    4.  balance =  125
player   12 lost bet    5.  balance =  100
player   12  won bet    6.  balance =  125
player   12 lost bet    7.  balance =  100
player   12  won bet    8.  balance =  125
player   12  won bet    9.  balance =  150
player   12  won bet   10.  balance =  175
player   12  won bet   11.  balance =  200
player   12  won bet   12.  balance =  225
player   12  won bet   13.  balance =  250
player   12 lost bet   14.  balance =  225
player   12 lost bet   15.  balance =  200
player   12 lost bet   16.  balance =  175
player   12 lost bet   17.  balance =  150
player   12 lost bet   18.  balance =  125
player   12  won bet   19.  balance =  150
player   12  won bet   20.  balance =  175
player   12 lost bet   21.  balance =  150
player   12 lost bet   22.  balance =  125
player   12 lost bet   23.  balance =  100
player   12  won bet   24.  balance =  125
player   12  won bet   25.  balance =  150
player   12 lost bet   26.  balance =  125
player   12 lost bet   27.  balance =  100
player   12 lost bet   28.  balance =   75
player   12 lost bet   29.  balance =   50
player   12 lost bet   30.  balance =   25
player   12  won bet   31.  balance =   50
player   12 lost bet   32.  balance =   25
player   12  won bet   33.  balance =   50
player   12 lost bet   34.  balance =   25
player   12 lost bet   35.  balance =    0
player 12 bust after 35 bets
player   13 lost bet    1.  balance =    0
player 13 bust after 1 bets
player   14 lost bet    1.  balance =    0
player 14 bust after 1 bets
player   15 lost bet    1.  balance =    0
player 15 bust after 1 bets
player   16  won bet    1.  balance =   50
player   16  won bet    2.  balance =   75
player   16 lost bet    3.  balance =   50
player   16 lost bet    4.  balance =   25
player   16 lost bet    5.  balance =    0
player 16 bust after 5 bets
player   17 lost bet    1.  balance =    0
player 17 bust after 1 bets
player   18 lost bet    1.  balance =    0
player 18 bust after 1 bets
player   19 lost bet    1.  balance =    0
player 19 bust after 1 bets
player   20  won bet    1.  balance =   50
player   20 lost bet    2.  balance =   25
player   20  won bet    3.  balance =   50
player   20 lost bet    4.  balance =   25
player   20 lost bet    5.  balance =    0
player 20 bust after 5 bets
player   21 lost bet    1.  balance =    0
player 21 bust after 1 bets
player   22  won bet    1.  balance =   50
player   22  won bet    2.  balance =   75
player   22 lost bet    3.  balance =   50
player   22  won bet    4.  balance =   75
player   22 lost bet    5.  balance =   50
player   22 lost bet    6.  balance =   25
player   22  won bet    7.  balance =   50
player   22  won bet    8.  balance =   75
player   22 lost bet    9.  balance =   50
player   22 lost bet   10.  balance =   25
player   22  won bet   11.  balance =   50
player   22 lost bet   12.  balance =   25
player   22  won bet   13.  balance =   50
player   22 lost bet   14.  balance =   25
player   22 lost bet   15.  balance =    0
player 22 bust after 15 bets
player   23 lost bet    1.  balance =    0
player 23 bust after 1 bets
player   24  won bet    1.  balance =   50
player   24 lost bet    2.  balance =   25
player   24  won bet    3.  balance =   50
player   24  won bet    4.  balance =   75
player   24  won bet    5.  balance =  100
player   24  won bet    6.  balance =  125
player   24  won bet    7.  balance =  150
player   24 lost bet    8.  balance =  125
player   24 lost bet    9.  balance =  100
player   24 lost bet   10.  balance =   75
player   24 lost bet   11.  balance =   50
player   24 lost bet   12.  balance =   25
player   24 lost bet   13.  balance =    0
player 24 bust after 13 bets
player   25  won bet    1.  balance =   50
player   25 lost bet    2.  balance =   25
player   25 lost bet    3.  balance =    0
player 25 bust after 3 bets
player   26 lost bet    1.  balance =    0
player 26 bust after 1 bets
player   27 lost bet    1.  balance =    0
player 27 bust after 1 bets
player   28 lost bet    1.  balance =    0
player 28 bust after 1 bets
player   29 lost bet    1.  balance =    0
player 29 bust after 1 bets
player   30 lost bet    1.  balance =    0
player 30 bust after 1 bets
player   31  won bet    1.  balance =   50
player   31  won bet    2.  balance =   75
player   31  won bet    3.  balance =  100
player   31  won bet    4.  balance =  125
player   31 lost bet    5.  balance =  100
player   31  won bet    6.  balance =  125
player   31 lost bet    7.  balance =  100
player   31 lost bet    8.  balance =   75
player   31 lost bet    9.  balance =   50
player   31 lost bet   10.  balance =   25
[...]
player   31 lost bet 999991.  balance = 502750
player   31 lost bet 999992.  balance = 502725
player   31 lost bet 999993.  balance = 502700
player   31 lost bet 999994.  balance = 502675
player   31 lost bet 999995.  balance = 502650
player   31  won bet 999996.  balance = 502675
player   31  won bet 999997.  balance = 502700
player   31 lost bet 999998.  balance = 502675
player   31  won bet 999999.  balance = 502700
player   31 lost bet 1000000.  balance = 502675
player 31 made 1000000 bets without busting

[Kluge]

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.

[Kluge]

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):

Code:

#!/usr/bin/env python

import random, sys

players = 1000
bets_per_player = 1000
starting_balance = 1
stake = 1

def roll():
    return random.random() < 0.50

player = 1
while player <= players:
    balance = starting_balance
    bets = 0
    while True:
        bets += 1
        if roll():
            balance += stake
            print "player %4d  won bet %4d.  balance = %4d" % (player, bets, balance)
        else:
            balance -= stake
            print "player %4d lost bet %4d.  balance = %4d" % (player, bets, balance)
            if balance < stake:
                print "player %d bust after %d bets" % (player, bets)
                break
        if bets == bets_per_player:
            print "player %d made %d bets without busting" % (player, bets)
            print "casino profit is %d" % (player-balance)
            sys.exit(0)

    player += 1

print "all the players busted"

[jambola2]

What does generally a pretty bad idea mean ? As far as I can understand , the house edge should be equal at all levels and that would just be a lot less fun (as you have no chance of winning something substantial) , with the same amount of risk.

Is my understanding wrong ?

If you want to win 1 BTC, you could:

a) risk 0.1 BTC at 9% for an 11x payout, or you could
b) risk 10 BTC at 90% for a 1.1x payout.

In each case you expect to lose 1% of the amount you risk, so b) costs about 100 times more than a) in terms of the amount you expect to lose, while only having a 10 times better chance of success.

That makes 90% seem like a bad deal compared to 9%, even though they both have a 1% edge.

Does that make sense?

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 ?

[Light]

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 ?

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.

[jambola2]

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 ?

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.

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 ?

[Salmon1989]

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 ?

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.

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 ?

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

[dooglus]

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 ?

Yes and yes.

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).

[Salmon1989]

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 ?

Yes and yes.

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).

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)?

[dooglus]

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.  

[Rigon]

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

[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. 

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.

[dooglus]

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

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

[dooglus]

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.

[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%.

[Phildo]

Just some thought without any actual math. You are making a series of bets with a hosue edge of 1% so the result can be summarized as a bet with a house edge of 1%.

Let's compare it to roulette where we can actually make the bets at the same time. If you put 10 on red you have a 47.73% chance of winning 10 bucks. That's a hosue edge of 5.26%. If you put 10 on 1-12 and 10 and 13-24 you now have a 63.16% chance of winning 10 bucks but still a house edge of 5.26%.

This isn't a perfect analogue because I increased the wager, but just by guessing (and not doing stats/probability for a while) I would guess the theory might be the same.

[Light]

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.

Ah ok, that makes a lot more sense now. Didn't think of it like that - keep thinking of them as analogous cases which were giving me different results (ie mathematically impossible).

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.

Can you explain why the probability of success increases as you make more smaller bets? Is it for the same reason as well, as you have a better chance of losing less (as your take a risk with only parts of your bankroll and for you to lose it all, all of the bets must fail) it results in an overall better chance of succeeding? 

[Mitchell]

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.

Hmmm, well you are right, it is very martingale-ish, but since it doesn't match the description of a martingale betting system (can be found here) I wouldn't call it that way, even though it's quite similar. But oh well, it doesn't matter. The technique described by you is amazing and I really enjoy reading about it in this thread, even though I fail to see how this is possible.

[zimmah]

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.  

Interesting, are you sure that's correct? As you're basically saying there are ways to nullify the house edge. I doubt this could possibly be true.

[zimmah]

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.

Yes, I'm intrigued as well, I feel this should not happen and it should be wrong.

Beating the house edge should not be possible, that's the whole point...

[Minnlo]

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.

Yes, I'm intrigued as well, I feel this should not happen and it should be wrong.

Beating the house edge should not be possible, that's the whole point...

It is still not going to beat the house.
What doog's strategy does is to lower your expected wagered amount and so lower your expected loss (house edge % * wagered amount) and equivalently increase the successful rate of hitting the target.