rockBTC
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★☆★Bitin.io★☆★


March 18, 2015, 12:21:35 AM 

Since its all from the faucet, i propose you put it on 50% chance and roll twice and then withdraw remember......... you only live once

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JaredStein
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Blocklancer  Freelance on the Blockchain


March 18, 2015, 12:35:30 AM 

Like what others have said, there is no strategy.

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LiQuidx


March 18, 2015, 07:19:34 AM 

Trying to revive this thread rather than start a new one.
Anyone care to share their "winning" dice formulas?
there is no winning formula on dices. Maybe leave while in profit? Exactly that.. It's just a matter of luck. Sure you can try to 'minimize' your risk but the longer you play the bigger the risk. So be lucky and quit while ahead.




ETFbitcoin
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Use SegWit and enjoy lower fees


March 18, 2015, 07:51:55 AM 

You want to win a lot only with faucet ? It's impossible since PD faucet isn't too high Maybe you have to try allin trick just like i do

Use SegWit and enjoy lower fees



rfisher1968


March 18, 2015, 01:16:44 PM 

Do a modified martingale.
Instead of betting 100, 200, 400, 800, .... until you win @ 2x.
bet 10, 10, 10, 10, 10, 20 ,20, 20, 20, 20, 40, 40, 40, 40, 40, .... until you win @ 10x.
You could guess a max lose streak of 100 before you win. So that makes 100 / 5 = 20, then (2^20)*5 = 5242880 which equals the minimum amount of Satoshi you need to bet using this technique with a starting bet of 1.




shanem
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March 18, 2015, 01:24:03 PM 

Losing streaks will destroy your bankroll. It is not uncommon to face losing streaks of more than 10 times in dice.




ranochigo
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March 18, 2015, 01:53:02 PM 

Do a modified martingale.
Instead of betting 100, 200, 400, 800, .... until you win @ 2x.
bet 10, 10, 10, 10, 10, 20 ,20, 20, 20, 20, 40, 40, 40, 40, 40, .... until you win @ 10x.
You could guess a max lose streak of 100 before you win. So that makes 100 / 5 = 20, then (2^20)*5 = 5242880 which equals the minimum amount of Satoshi you need to bet using this technique with a starting bet of 1.
Any martingale is the same as the original x2 martingale. This method isn't 100% safe as the varience is bound to get you after playing for sometime I tried various kinds of martingale but I still ended up losing BTC. No method is safe to be honest, its a matter of time before you lose if you are lucky due to the house edge.

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  BitBlender 
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rfisher1968


March 18, 2015, 02:31:42 PM 

Do a modified martingale.
Instead of betting 100, 200, 400, 800, .... until you win @ 2x.
bet 10, 10, 10, 10, 10, 20 ,20, 20, 20, 20, 40, 40, 40, 40, 40, .... until you win @ 10x.
You could guess a max lose streak of 100 before you win. So that makes 100 / 5 = 20, then (2^20)*5 = 5242880 which equals the minimum amount of Satoshi you need to bet using this technique with a starting bet of 1.
Any martingale is the same as the original x2 martingale. This method isn't 100% safe as the varience is bound to get you after playing for sometime I tried various kinds of martingale but I still ended up losing BTC. No method is safe to be honest, its a matter of time before you lose if you are lucky due to the house edge. Its not the same as a standard martingale. In my example; 1st 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+40) = 210 win 2nd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*2)) = 170 win 3rd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*3)) = 130 win 4th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*4)) = 90 win 5th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*5)) = 50 win I'm basically dividing the original martingale bet by 5 giving me 5 times a chance to win instead of 1. Now the chance of winning a 10x bet is much harder then a 2x times, but you only have 1 chance to win on 2x. By splitting the bet you have a range of potential gains as compared to fixed gain. 1st 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+80) = 370 win 2nd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*2)) = 290 win 3rd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*3)) = 210 win 4th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*4)) = 130 win 5th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*5)) = 50 win As you can see as the lose streak increases the win range increases also.




ETFbitcoin
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Use SegWit and enjoy lower fees


March 18, 2015, 02:52:34 PM 

Do a modified martingale.
Instead of betting 100, 200, 400, 800, .... until you win @ 2x.
bet 10, 10, 10, 10, 10, 20 ,20, 20, 20, 20, 40, 40, 40, 40, 40, .... until you win @ 10x.
You could guess a max lose streak of 100 before you win. So that makes 100 / 5 = 20, then (2^20)*5 = 5242880 which equals the minimum amount of Satoshi you need to bet using this technique with a starting bet of 1.
Any martingale is the same as the original x2 martingale. This method isn't 100% safe as the varience is bound to get you after playing for sometime I tried various kinds of martingale but I still ended up losing BTC. No method is safe to be honest, its a matter of time before you lose if you are lucky due to the house edge. Its not the same as a standard martingale. In my example; 1st 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+40) = 210 win 2nd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*2)) = 170 win 3rd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*3)) = 130 win 4th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*4)) = 90 win 5th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*5)) = 50 win I'm basically dividing the original martingale bet by 5 giving me 5 times a chance to win instead of 1. Now the chance of winning a 10x bet is much harder then a 2x times, but you only have 1 chance to win on 2x. By splitting the bet you have a range of potential gains as compared to fixed gain. 1st 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+80) = 370 win 2nd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*2)) = 290 win 3rd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*3)) = 210 win 4th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*4)) = 130 win 5th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*5)) = 50 win As you can see as the lose streak increases the win range increases also. Any martingale tricks will lead you to lose But, i'm agree that trick is less risky than classic martingale Just do it for short terms

Use SegWit and enjoy lower fees



Micon
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FPV Drone Pilot


March 18, 2015, 03:01:57 PM 

There is major strategy in some parts of gambling. Not in 99% payout games with posted 1% holds. Dude, get serious. Any martingale variation is a sure loser in the long run, the house always wins
Martingale is a longterm losing strategy for those that can't wrap their heads around independent probabilities. It's amazing how many times it's been explained yet almost daily it's presented as a possible winning strategy vs. unbeatable (unless you cheat) dice.




mrhollis1
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September 11, 2015, 10:24:27 AM 

I can win every time. If anyone is willing to pay me for the secret we can set up escrow




plost24


September 11, 2015, 10:29:53 AM 

Its not the same as a standard martingale. In my example;
1st 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+40) = 210 win 2nd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*2)) = 170 win 3rd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*3)) = 130 win 4th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*4)) = 90 win 5th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*5)) = 50 win
I'm basically dividing the original martingale bet by 5 giving me 5 times a chance to win instead of 1. Now the chance of winning a 10x bet is much harder then a 2x times, but you only have 1 chance to win on 2x. By splitting the bet you have a range of potential gains as compared to fixed gain.
1st 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+80) = 370 win 2nd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*2)) = 290 win 3rd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*3)) = 210 win 4th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*4)) = 130 win 5th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*5)) = 50 win
As you can see as the lose streak increases the win range increases also.
i was trying this strategie in a long time but it lead me to a lose :/ it have a nice chance to win but it's not a safe one

For rent 1.4 Bitcoin for 11 months starting Feb 1 2017



mrhollis1
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September 11, 2015, 10:46:41 AM 

Its not the same as a standard martingale. In my example;
1st 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+40) = 210 win 2nd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*2)) = 170 win 3rd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*3)) = 130 win 4th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*4)) = 90 win 5th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*5)) = 50 win
I'm basically dividing the original martingale bet by 5 giving me 5 times a chance to win instead of 1. Now the chance of winning a 10x bet is much harder then a 2x times, but you only have 1 chance to win on 2x. By splitting the bet you have a range of potential gains as compared to fixed gain.
1st 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+80) = 370 win 2nd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*2)) = 290 win 3rd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*3)) = 210 win 4th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*4)) = 130 win 5th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*5)) = 50 win
As you can see as the lose streak increases the win range increases also.
i was trying this strategie in a long time but it lead me to a lose :/ it have a nice chance to win but it's not a safe one Hint: What are the odds off flipping heads on a coin 20 times in a row? If you can use those odds to your advantage you cannot lose




plost24


September 11, 2015, 10:52:37 AM 

Its not the same as a standard martingale. In my example;
1st 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+40) = 210 win 2nd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*2)) = 170 win 3rd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*3)) = 130 win 4th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*4)) = 90 win 5th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*5)) = 50 win
I'm basically dividing the original martingale bet by 5 giving me 5 times a chance to win instead of 1. Now the chance of winning a 10x bet is much harder then a 2x times, but you only have 1 chance to win on 2x. By splitting the bet you have a range of potential gains as compared to fixed gain.
1st 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+80) = 370 win 2nd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*2)) = 290 win 3rd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*3)) = 210 win 4th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*4)) = 130 win 5th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*5)) = 50 win
As you can see as the lose streak increases the win range increases also.
i was trying this strategie in a long time but it lead me to a lose :/ it have a nice chance to win but it's not a safe one Hint: What are the odds off flipping heads on a coin 20 times in a row? If you can use those odds to your advantage you cannot lose trust me you can lose with any strategie if there is a safe strategie all dice site will close :p after losing all the bankroll

For rent 1.4 Bitcoin for 11 months starting Feb 1 2017



mrhollis1
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September 11, 2015, 10:57:19 AM 

Its not the same as a standard martingale. In my example;
1st 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+40) = 210 win 2nd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*2)) = 170 win 3rd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*3)) = 130 win 4th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*4)) = 90 win 5th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*5)) = 50 win
I'm basically dividing the original martingale bet by 5 giving me 5 times a chance to win instead of 1. Now the chance of winning a 10x bet is much harder then a 2x times, but you only have 1 chance to win on 2x. By splitting the bet you have a range of potential gains as compared to fixed gain.
1st 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+80) = 370 win 2nd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*2)) = 290 win 3rd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*3)) = 210 win 4th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*4)) = 130 win 5th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*5)) = 50 win
As you can see as the lose streak increases the win range increases also.
i was trying this strategie in a long time but it lead me to a lose :/ it have a nice chance to win but it's not a safe one Hint: What are the odds off flipping heads on a coin 20 times in a row? If you can use those odds to your advantage you cannot lose trust me you can lose with any strategie if there is a safe strategie all dice site will close :p after losing all the bankroll Or ban you




mrhollis1
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September 11, 2015, 11:14:35 AM 

Here.... 95.3*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2*2= about one Bitcoin. You guys can figure out the rest




plost24


September 11, 2015, 11:25:19 AM 

Its not the same as a standard martingale. In my example;
1st 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+40) = 210 win 2nd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*2)) = 170 win 3rd 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*3)) = 130 win 4th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*4)) = 90 win 5th 40 satoshi bet wins the (40*10)  ((10*5)+(20*5)+(40*5)) = 50 win
I'm basically dividing the original martingale bet by 5 giving me 5 times a chance to win instead of 1. Now the chance of winning a 10x bet is much harder then a 2x times, but you only have 1 chance to win on 2x. By splitting the bet you have a range of potential gains as compared to fixed gain.
1st 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+80) = 370 win 2nd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*2)) = 290 win 3rd 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*3)) = 210 win 4th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*4)) = 130 win 5th 80 satoshi bet wins the (80*10)  ((10*5)+(20*5)+(40*5)+(80*5)) = 50 win
As you can see as the lose streak increases the win range increases also.
i was trying this strategie in a long time but it lead me to a lose :/ it have a nice chance to win but it's not a safe one Hint: What are the odds off flipping heads on a coin 20 times in a row? If you can use those odds to your advantage you cannot lose trust me you can lose with any strategie if there is a safe strategie all dice site will close :p after losing all the bankroll Or ban you you said that from your experience because there is the third option the site turn to scam and run with all balance :p

For rent 1.4 Bitcoin for 11 months starting Feb 1 2017



