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techsavant
techsavant
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September 23rd, 2015 at 6:37:54 AM permalink
Hi, I was wondering if I could get some insight into this system I was using. It's a variation on Martingale I guess.

I used it on an online bitcoin dice site. Where you choose high or low and what odds you want.

I started with a 6.5 million chip bankroll. (satoshis)

I set the odds to 2.5x (meaning if I bet 10 and win , I get my 10 back plus 15) With 2.5x odds, when you hit a losing streak, the longer it is, more profit you will have when you finally break that streak with a win because the .5x will be your profit on the win.

I choose the low roll every time, but that doesn't really matter. For a low roll to win, you need it to be a 39.6 or lower.

I start out betting 1 chip. I double when I lose. And reset to 1 chip when I win.

I just tested this system a couple hours ago. I stopped when I doubled my 6.5 million chips. I never dropped below my initial 6.5m bankroll amount. It was a constant process of increasing and honestly it seemed like it could go on forever. There was no stressful moments or biting my nails hoping this one hits or whatever, it was actually really boring (and time consuming as the bet amounts are so low, it took thousands of bets sometimes to make just 100 chips).

The difference in this system is you don't just keep plowing ahead when you hit a losing streak. You set a cap on how many losses you can sustain until you call it quits.

So for example. I start betting 1 chip, and say I lose 8 times in a row. (very common), then I will have wagered 128 on my last bet, and also lost 127 in previous bets, so a total of 255. You basically lose double of whatever your last bet was if you don't continue.

I found that if I set my streak cap at 10, so after the 8th loss and betting 128, I would continue with 256 and 512 to try to get the win. If I lose the 10th bet, then I am out 1024 chips. With a 6.5m bankroll, that's really not even a drop in the bucket.

I also found that between these "busts", I would generally win between 4000 and 6000 chips before hitting a losing streak greater than 10. But all I was really looking for was to make more than 1024 chips between 10 loss streaks, which seemed REALLY easy and not hard at all. Like I said earlier, I never dropped below my initial buy-in throughout.

All the while I kept wracking my brain and telling myself, there has to be some catch, there has to be something that can happen to ruin me in this system, it seems to easy, too flawless.

After doubling my 6.5m bankroll, I was just bored with it and didn't even want to continue because there are no exciting moments. You just sit there watching yourself wreck the house.

My question is, was I just running really good ? I use a similar system on diggit.io ( a mine sweeper grid type game where you try to avoid clicking on mines) and it seems to have the same results. As long as I have the discipline to stop doubling and start over when I hit the pre-defined streak cap, I never lose more than I've already won.

What are some pitfalls of this. Is it really possible that I could hit so many 10+ loss streaks in a row that I could bust out a bankroll that's 6,500x larger than my maximum loss (aka 1k after 10 missed doublings) ?

Any input, advice, criticism of this would be appreciated because I'm pretty stoked on this and writing a program to do the wagering for me so I don't have to tend to the computer and just let it run 24/7 and be rich hahahah

I know there has to be some kind of catch. And I know everyone says "betting systems don't work" but I am really curious as to how this one could lead to ruin.

Thanks for Reading :)
Romes
Romes
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September 23rd, 2015 at 7:12:57 AM permalink
Hi techsavant, and welcome to the forums.

I don't think we have enough information... You say you set the odds to 2.5x, well can you set the odds to 10x? What does this do to your odds of winning? What kind of dice are you using where the average roll is 39.6? What's the min and max bets? Basically, you haven't explained the actual game at all, nor how the odds effects the rolls/wins/etc.

Hopefully this doesn't come off as short, but to be honest we don't 'really' need to know the game... unless you want specific EV breakdowns to show how and when you should be expecting to be down. Your system will inevitably fail and bust your bankroll. The fact that you cap your net string losses at 10 might slow this down "the tiniest bit" but it won't change the end all to the game. There's got to be a built in house edge and no amount of betting "differently" or "doubling" will change the house edge and thus overall your Expected Value (EV). EV = NumBets*AvgBet*HouseEdge. An even more simplified version is EV = TotalWagered*HouseEdge. Notice that no where in these mathematical equations does it take in to account "what if I double my bet on the next hand?" If you double your bet on the next hand then you're simply upping your AvgBet/TotalWagered and therefor your eventual, inevitable losses.

Martingale systems, especially when you have such a large bankroll in comparison to your bets, work "most" of the time. It's the times when they don't work that they will punish your bankroll. Say you get 200 rolls low in a row! That took you what, an hour or two? Well in the next 10 min you could have a losing streak and lose 255 or 510 and blow out your 200 + additional that would take you hours to rebuild.
Playing it correctly means you've already won.
techsavant
techsavant
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September 23rd, 2015 at 7:59:40 AM permalink
Totally get what your saying, that's why I put a stop in at 10 or so losses so a long losing streak can't wreck me. This is why I don't see the downfall of my system. In normal martingale, yes, a long losing streak will ruin you, but in this system, you break away and start over before any significant loss ocurs.

As far as rules of the dice game, I'm sorry, I just assumed everyone has played satoshi dice at one point or another.

It's a 1 - 100 roll. You can set your odds of winning anywhere from 1.01 to 9999 or something like that. If you set it to 9999 and choose "low" as your roll you need to roll a 00.00 to win. You win 9999 to 1 if you win that.

So if I set mine to 2.5x and I choose "low" I have to roll a 39.6 or lower to win. And when I win, I am paid out at 2.5x.

Hope that helps clear that up a little.

House edge is 1 percent BTW. So if you bet 2x odds. Then your roll has to be below 49.5 for a "low" roll or about 50.5 for a high roll. Which is also why my 39.6 number is not 40. Because of the house edge.

And I get what your saying about a bunch of tiny wins to make 200 and then losing it all in a few minutes, but the amount of time it takes to recoup isn't really important as the amount of chips i win compared to what i lose. If I lose 1000 chips, it's really not even a dent in what I can accumulate between those losses.

I will test this more obviously and maybe find out i was just on a hot streak, but it seemed pretty solid to me.
ThatDonGuy
ThatDonGuy
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September 23rd, 2015 at 9:44:00 AM permalink
Here's what I get:

Let p = the probability of a single bet winning = 0.396,
and q = the probability of a single bet losing = 1 - q = 0.604

Let "a win" be where you win a bet before losing 10 in a row, and "a loss" be where you lose 10 in a row

The probability of a win = 1 - p10 = about 615/619

The EV of a win =
p * 1.5 + qp * 2 + q2p * 3 + ... + q9p * 257
= p (1 + q + q2 + ... + q9) + p/2 * (1 + 2q + 4q2 + ... + 512 q9)
= p (1 + q + q2 + ... + q9) + p/2 * (1 + 2q + (2q)2 + ... + (2q)9)
= p (1 - q10) / (1 - q) + p/2 * (1 - (2q)10) / (1 - 2q)
= p (1 - q10) / p + p/2 * (1 - (2q)10) / (p + q - 2q)
= 1 - q10 + p (1 - 1024 q10) / (2p - 2q)
= about 6.340609

The probability of a loss = q10 = about 4/619

The EV of a loss = -1023


However, I don't know the Risk of Ruin formula where the bet pays something other than even money off the top of my head.

How many times have you tried this with a complete bankroll? I calculate an EV for each "result" (where you either win, or lose 10 in a row) of about -0.311, and you did say your bankroll was 6,500,000.

That's strange...I did a simulation on this, and while it does take quite a bit of time (something like 30 bankrolls per minute), I have yet to reach 13 million before reaching zero. (On the other hand, if the probability of winning is 0.404, I never lose.)
Romes
Romes
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September 23rd, 2015 at 10:55:50 AM permalink
Quote: techsavant

Totally get what your saying, that's why I put a stop in at 10 or so losses so a long losing streak can't wreck me. This is why I don't see the downfall of my system. In normal martingale, yes, a long losing streak will ruin you, but in this system, you break away and start over before any significant loss ocurs.


You don't need a super large streak. As I pointed out earlier in this thread you can simply lose your 10 in a row, which is 500 something, and that could happen in 10 minutes which wipes out hours and hours of winning. The fact that you cut it short has relatively no effect on EV.

Quote: techsavant

As far as rules of the dice game, I'm sorry, I just assumed everyone has played satoshi dice at one point or another.

It's a 1 - 100 roll. You can set your odds of winning anywhere from 1.01 to 9999 or something like that. If you set it to 9999 and choose "low" as your roll you need to roll a 00.00 to win. You win 9999 to 1 if you win that.

So if I set mine to 2.5x and I choose "low" I have to roll a 39.6 or lower to win. And when I win, I am paid out at 2.5x.


Another point I meant to make earlier. You're not getting paid 2.5 to 1, you're getting paid 2.5 FOR 1, which means you bet 10, you get 15 in profit, essentially a 3-2 payout in blackjack. Be careful with the verbiage, 2.5 to 1 would mean you bet 10, and you WIN 25 for a TOTAL of 35.

Next, I've never heard of or even seen this game. I assume you mean the roll is 00.00 to 99.99? Also, when you set your odds to "2.5 FOR 1" you need a 39.6 or lower, which puts the house at 60.4, a 1.53 favorite. Thus, you're giving up .03, or 3% when you do this. And yes that extra .4 does greatly affect the game. There are blackjack games where that extra .4% would triple the house edge.

Quote: techsavant

House edge is 1 percent BTW. So if you bet 2x odds. Then your roll has to be below 49.5 for a "low" roll or about 50.5 for a high roll. Which is also why my 39.6 number is not 40. Because of the house edge.


The house edge is only 1% if you're playing EVEN MONEY (2 FOR 1 odds, not 2x odds).

Quote: techsavant

And I get what your saying about a bunch of tiny wins to make 200 and then losing it all in a few minutes, but the amount of time it takes to recoup isn't really important as the amount of chips i win compared to what i lose. If I lose 1000 chips, it's really not even a dent in what I can accumulate between those losses.

I will test this more obviously and maybe find out i was just on a hot streak, but it seemed pretty solid to me.


You realize your entire testimony involves your supposed 5k throws, with no numbers/stats recorded, and you saying "it seemed pretty solid to me" right? I can tell you for a fact if you won you had a "hot streak" and if you continue to do this you will run in to streaks that will inevitably destroy your bankroll down to the EV of the game.

Simple EV Explanation... EV = NumEvents*AvgBet*HouseEdge. If you're playing the 1% house edge you referred to above, then let's say you bet $10 for 5000 spins... your EV is:

EV = (5000)(10)(-.01) = -$500

This doesn't include standard deviations, which from basic high school mathematics is the "plus or minus" number due to the natural variance of the game and the low amount of trials. To hit your EV near exactly you'd need to play a million throws or more. 5,000 is a drop in the bucket and means you SHOULD lose $500, but since it's the short run you might be up some, you might be down some. Assuming a pure coin flip (50/50) has a SD of 1, to be 99.9% confident you could say that I'd expect you to lose $500, plus or minus $1500... and this is also using a SD of 1, which is not the case in your game. So you see how here you could in fact be up $1,000 even though you're playing a negative edge losing game? This is called variance. The more trials you get, the more you'll be closer to your true EV. How about an example?

After 1,000,000 throws of $10 each, your EV = (1,000,000)(10)(-.01) = -$100,000
SD = Sqrt(NumEvents)*OriginalSD = Sqrt(1,000,000)*10 = $10,000.... 3xSD gives you 99.9% confidence, thus 3xSD = $30,000.

*Note I used a "50/50" SD of 1, which your game has more variance.

So what does this mean? After 1,000,000 throws you're expected to be DOWN $100,000.. plus or minus $30,000 to be 99.9% confident. This means the BEST you can do 99.9% mathematically speaking after 1,000,000 throws is to be DOWN $70,000. At no point did you change the house edge. All you're doing by doubling your bet is upping your average bet and the total amount you've wagered on the house advantage game.

I hope you can see why you absolutely can not beat a game just by doubling your bet, in any fashion. It doesn't matter if you "reset" after a 10 streak, a 5 streak, a 15 streak, etc. It doesn't matter if you triple your wager every time you lose, etc. All that matters is how much you wager and what the house edge is. You've experienced short term variance on the positive side of the bell curve. Do not attribute this to some "working system." Many have tried before you, and learned the hard way. I myself nearly a decade ago though I could make some money now and then for fun at roulette. I'd wait until 4 of the same thing landed (hi/low, odd/even, red/black, etc) and then I'd Marty on the opposite. Worked great short term until I hit another eventual streak that took my money from me. All of this is an attempt to help you see past the very short term "success" you've had and realize the truth. MARTINGALE, in any form, DOES NOT WORK.
Playing it correctly means you've already won.
OnceDear
OnceDear
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September 23rd, 2015 at 11:18:28 AM permalink
Quote: techsavant

Hi, I was wondering if I could get some insight into this system I was using. It's a variation on Martingale I guess....

Thanks for Reading :)



Tech,

What did you set as the house edge?
With no house edge, you can estimate the probability of reaching a goal before busting out, and vice versa. from

IN ABSENSE of House Edge or Player Edge
Probability of hitting goal = (Starting Bankroll)/(Starting Bankroll + Target Profit)

Probability of going bust = 1-probability of hitting goal

So, probability of doubling money before losing everything is 50%, subject to certain constraints*
No edge, no bets placed that could take you over target profit, no topping up bankroll.

And no bankroll management or progressive system that you can come up with can improve on that.
NONE!
Psalm 25:16 Turn to me and be gracious to me, for I am lonely and afflicted. Proverbs 18:2 A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
ThatDonGuy
ThatDonGuy
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September 23rd, 2015 at 11:20:40 AM permalink
Question: what site are you playing on? (Is there more than one site that plays Satoshi Dice?)

It's possible that someone is setting you up for some sort of con - you start off winning, then you increase your bankroll and suddenly you discover that you have lost everything.

Then again, it's also possible that you found a glitch in some software, but you won't know until it's too late when the glitch is fixed...or maybe somebody wants you to think there's a glitch (similar to when the winning card's corner is turned up in 3-Card Monte; you bet everything...only to discover that, somehow, the card with the bent corner is now one of the losing ones).
mustangsally
mustangsally
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September 24th, 2015 at 8:45:40 AM permalink
Quote: ThatDonGuy

However, I don't know the Risk of Ruin formula where the bet pays something other than even money off the top of my head.

there is not one and the closest attempt would be a Markov chain and have fun finding a computer ready and willing to do it for you.

Quote: ThatDonGuy

How many times have you tried this with a complete bankroll? I calculate an EV for each "result" (where you either win, or lose 10 in a row) of about -0.311, and you did say your bankroll was 6,500,000.

your ev result for win/lose should be about -.27 as the avg bet = abouts 27
that right there shows the HE of 1%


this thread is just a advertisment for bitcoin-online gambling
all-a-scam-to-take-your-cash, imo

Quote: ThatDonGuy

That's strange...I did a simulation on this,

a BIG waste of time and energy imo
Quote: ThatDonGuy

and while it does take quite a bit of time (something like 30 bankrolls per minute), I have yet to reach 13 million before reaching zero.

probability is about 0 to double a 6.5 million bankroll where the prob of winning is less than 40% and the average win is only 6.340608937
in other words
the average win when one wins (finally) is only 6.340608937
the average win when one wins (finally) is only 6.340608937
the average win when one wins (finally) is only 6.340608937
the average win when one wins (finally) is only 6.340608937
the average win when one wins (finally) is only 6.340608937

*******************************************
try starting with a bankroll of 1023 and try to double that
ok
well it takes on average 161.3409706 wins in a row
to win that amount, the OP claims it only takes a few rounds to win that, proof of a scam thread,
0.993537964^162=0.349851361
UGLY!! 35% (as expected)
so doubling a 6.5 million bank with the OP super-betting-system (false)

is way way less than that the
1st crash trying to double 1023 units
*************************************************************
advertise the scam bitcoin gambling websites, my opinion

free spinach rules!

remember this
"All the while I kept wracking my brain and telling myself,
there has to be some catch, there has to be something that can happen to ruin me in this system, it seems to easy, too flawless.

After doubling my 6.5m bankroll,
I was just bored with it and didn't even want to continue because there are no exciting moments.
You just sit there watching yourself wreck the house."

"After doubling my 6.5m bankroll"
thank you for sharing that pure comedy
hahahahaha
I Heart Vi Hart
ThatDonGuy
ThatDonGuy
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September 24th, 2015 at 11:04:23 AM permalink
Quote: mustangsally

there is not one and the closest attempt would be a Markov chain and have fun finding a computer ready and willing to do it for you.


How hard can it be? It's only 2.6 million equations in 2.6 million unknowns ("don't you mean 1.3 million?" No, as you have to allow for half-chip values, since winning a bet of 1 increases your total by 1.5), so that's only about 6.8 trillion cells you have to calculate 2.6 million times each; that's about 17.7 quintillion calculations. With a 5 GHz processor, and let's say 10 cycles per operation (which is actually quite fast), that's only about 410 days of nonstop processing...
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