Primedice Strategy A.K.A Ultimate Puzzle

[rockBTC]

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

[1713408305]

1713408305

[1713408305]

1713408305

[JaredStein]

Like what others have said, there is no strategy. 

[LiQuidx]

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.

[rfisher1968]

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]

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

[ranochigo]

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.

[rfisher1968]

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.

[Micon]

There's no strategy in Gambling. Hope you know      

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 long-term 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]

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

[plost24]

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

[mrhollis1]

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]

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

[mrhollis1]

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]

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]

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