i saw i video on youtube. a man explains (in german) how to win at roulette with no special skill

example:

1.round: bet 1â‚¬ red

---> u loose

2.round: bet 2â‚¬ red

---> u loose

3.round: bet 4â‚¬ red

---> u loose

4.round: bet 8â‚¬ red

---> u WIN

ok...before u lost: 1+2+4= 7â‚¬

at last game u won 8â‚¬

u got the lost money back and also won 1â‚¬!!!! ------>OWNED

if u dont belief me tried it out on this game:

http://www.miniclip-...games/3005.html

(btw its free)

and it really works

the guy in the vid said when u do this trick in real casino, they would !kick u out

but they dont in online game

im really thinking about "gambling" online with money

what do u guys think???

# WIN easily @ roulette

Started by krAzy :), Jan 06 2010 10:32 AM

9 replies to this topic

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#1
Posted 06 January 2010 - 10:32 AM

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#2
Posted 06 January 2010 - 10:39 AM

cant be true, and even that i didnt believe it at begin i tried that online game and did what you said and red came in 3third spin...

well i still won 1$ overall but imo you cant guess those even with some tactic

well gambling is fine in internet if you just play for fun with small amounts...

you can easily lose much money if you think too much of yourself

good luck if you start gambling

well i still won 1$ overall but imo you cant guess those even with some tactic

well gambling is fine in internet if you just play for fun with small amounts...

you can easily lose much money if you think too much of yourself

good luck if you start gambling

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#3
Posted 06 January 2010 - 02:04 PM

there is no guarantee you'll win on the 4th bet. every game you have a 50/50 chance of winning/loosing

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#4
Posted 06 January 2010 - 02:07 PM

there is no guarantee you'll win on the 4th bet. every game you have a 50/50 chance of winning/loosing

Not 50/50, the 0 (and 00 depending on the roulette) exist. No game gives you 50/50, it's always less then 50 else the casino wouldn't make money.

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#5
Posted 06 January 2010 - 03:12 PM

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#6
Posted 06 January 2010 - 05:30 PM

it's 18/38 ~ 47% in roulette if you bet even/odd or red/black. there is a green '0' ( which isn't counted as even, even though 0/2 is an integer. go figure )

the trick is to double your bet each round, and to quit as soona s you win ( or start over with a 1 $ bet )

there is no guarantee that you will win any round, but as long as you double your bet, when you win you will make more than what you have bet so far:

round 1 bet x = x*2^0

round 2 bet 2x = x*2^1

round 3 bet 2(2x)=4x = x*2^2

round 4 bet 2(4x)=8x = x*2^3

...

round n bet x*2^(n-1)

suppose you win on the kth round after losing rounds 1... k-1

then you have lost x*[2^0 + 2^1 + ... + 2^(k-2)] = x*[1 + 2 + 4 + ... + 2^(k-2)]

and you have won x*2^(k-1)

your profit is (win)-(lose), or:

x*2^(k-1) - x*[2^0 + 2^1 + ... + 2^(k-2)]

and we hope that this is > 0. lets check.

how the heck do we check?

well first lets set up the inequality we want, and see if it's true or not:

x*2^(k-1) - x*[2^0 + 2^1 + ... + 2^(k-2)] > 0 ??

now what?

ooh i know. let's divide everything by x*2^(k-2)

then we have:

2^(k-1)/2^(k-2)-[ (2^0)/2^(k-2) + (2^1)/2^(k-2) + (2^2)/2^(k-2) + ... + (2^(k-2))/2^(k-2) ] > 0/2^(k-2) > 0 ?

2 - [1/2^(k-2) + 1/2^(k-3) + ... + 1/2^3 + 1/2^2 + 1/2^1 + 1] > 0 ?

2 - [1 + 1/2 + 1/4 + 1/8 + ... + 1/2^(k-2)] > 0?

well, the sum in the line above is a geometric series with ratio - 1/2 ( every term is 1/2 the term before it ). as k approaches infinity, a bit of calculus II ( actually analysis, but nvm ) tells us that the limit of this series is 1/(1-r) = 1/(1-1/2) = 2

which would mean our winnings equation becomes

2 - 2 = 0

bummer. if it takes you infinite number of plays to win, your going to end up doing no more than break even.

the good news is that you won't ever be able to play infinite times; you're going to win after some less-than-infinite number of plays.

now some actual analysis that i won't confuse you with tells us that the geometric series that approaches 2 approaches it from the lower side of 2, so for a non-infinite k, the series will sum to something less than 2. that's really good news, because then for some arbitrarily small, positive number e ( for epsilon =p ), we have

2 - ( 2 - e ) = e > 0

i know you are probably not following this anymore, dear reader, but it means that you will get positive winnings. unless you play an infinite number of times before you win lol.

this strategy works for any game with a 1-1 payout.

the problem with using this strategy is that you have to keep doubling your bet! At some point, you either won't be able to cover your bet, or the house won't take your bet because it's too large. say you start with $100. you don't want to waste your time trying to win $1 at a time, so you bet $10. lets say you lose. now you have $90. well, you've got this great strategy, so double your bet ( $20 ) and play again. and lose. dang. you now have $70, and a great strategy to win, so double your bet again. $40. crap. you lose. now you are down to $30 - if you lose, you can't double your bet! after only 3 losses in a row, you got pwned! statistically, you're going to see 5, 6, 7 losses in a row sometimes on a fair table on the 1-1 bets. that means you have to walk in and bet ( and stand to win ) small change in order to make sure you have enough of a reserve to keep doubling on a losing streak. this in turn means you aren't going to stand to win very much $$ compared to what you need to have in your pocket.

all in all though, it's a sound strategy that WILL win, even if it won't win much, but to be sure you're going to come out ahead, you need a big bankroll, and you need to bet small.

if you were going to play with real money, i would keep my bets at like 1% of your bankroll. even at 1% you will bankrupt yourself if you lose 6 times in a row - you won't be able to double your bet a 7th time. the chances of losing 6 times in a row? 1/64 for coin tosses - a little higher for a roulette wheel with a '0' and higher still for one with a '00' - not entirely unlikely.

it's 18/38 ~ 47% in roulette if you bet even/odd or red/black. there is a green '0' ( which isn't counted as even, even though 0/2 is an integer. go figure )

the trick is to double your bet each round, and to quit as soona s you win ( or start over with a 1 $ bet )

there is no guarantee that you will win any round, but as long as you double your bet, when you win you will make more than what you have bet so far:

round 1 bet x = x*2^0

round 2 bet 2x = x*2^1

round 3 bet 2(2x)=4x = x*2^2

round 4 bet 2(4x)=8x = x*2^3

...

round n bet x*2^(n-1)

suppose you win on the kth round after losing rounds 1... k-1

then you have lost x*[2^0 + 2^1 + ... + 2^(k-2)] = x*[1 + 2 + 4 + ... + 2^(k-2)]

and you have won x*2^(k-1)

your profit is (win)-(lose), or:

x*2^(k-1) - x*[2^0 + 2^1 + ... + 2^(k-2)]

and we hope that this is > 0. lets check.

how the heck do we check?

well first lets set up the inequality we want, and see if it's true or not:

x*2^(k-1) - x*[2^0 + 2^1 + ... + 2^(k-2)] > 0 ??

now what?

ooh i know. let's divide everything by x*2^(k-2)

then we have:

2^(k-1)/2^(k-2)-[ (2^0)/2^(k-2) + (2^1)/2^(k-2) + (2^2)/2^(k-2) + ... + (2^(k-2))/2^(k-2) ] > 0/2^(k-2) > 0 ?

2 - [1/2^(k-2) + 1/2^(k-3) + ... + 1/2^3 + 1/2^2 + 1/2^1 + 1] > 0 ?

2 - [1 + 1/2 + 1/4 + 1/8 + ... + 1/2^(k-2)] > 0?

well, the sum in the line above is a geometric series with ratio - 1/2 ( every term is 1/2 the term before it ). as k approaches infinity, a bit of calculus II ( actually analysis, but nvm ) tells us that the limit of this series is 1/(1-r) = 1/(1-1/2) = 2

which would mean our winnings equation becomes

2 - 2 = 0

bummer. if it takes you infinite number of plays to win, your going to end up doing no more than break even.

the good news is that you won't ever be able to play infinite times; you're going to win after some less-than-infinite number of plays.

now some actual analysis that i won't confuse you with tells us that the geometric series that approaches 2 approaches it from the lower side of 2, so for a non-infinite k, the series will sum to something less than 2. that's really good news, because then for some arbitrarily small, positive number e ( for epsilon =p ), we have

2 - ( 2 - e ) = e > 0

i know you are probably not following this anymore, dear reader, but it means that you will get positive winnings. unless you play an infinite number of times before you win lol.

this strategy works for any game with a 1-1 payout.

the problem with using this strategy is that you have to keep doubling your bet! At some point, you either won't be able to cover your bet, or the house won't take your bet because it's too large. say you start with $100. you don't want to waste your time trying to win $1 at a time, so you bet $10. lets say you lose. now you have $90. well, you've got this great strategy, so double your bet ( $20 ) and play again. and lose. dang. you now have $70, and a great strategy to win, so double your bet again. $40. crap. you lose. now you are down to $30 - if you lose, you can't double your bet! after only 3 losses in a row, you got pwned! statistically, you're going to see 5, 6, 7 losses in a row sometimes on a fair table on the 1-1 bets. that means you have to walk in and bet ( and stand to win ) small change in order to make sure you have enough of a reserve to keep doubling on a losing streak. this in turn means you aren't going to stand to win very much $$ compared to what you need to have in your pocket.

all in all though, it's a sound strategy that WILL win, even if it won't win much, but to be sure you're going to come out ahead, you need a big bankroll, and you need to bet small.

if you were going to play with real money, i would keep my bets at like 1% of your bankroll. even at 1% you will bankrupt yourself if you lose 6 times in a row - you won't be able to double your bet a 7th time. the chances of losing 6 times in a row? 1/64 for coin tosses - a little higher for a roulette wheel with a '0' and higher still for one with a '00' - not entirely unlikely.

the trick is to double your bet each round, and to quit as soona s you win ( or start over with a 1 $ bet )

there is no guarantee that you will win any round, but as long as you double your bet, when you win you will make more than what you have bet so far:

round 1 bet x = x*2^0

round 2 bet 2x = x*2^1

round 3 bet 2(2x)=4x = x*2^2

round 4 bet 2(4x)=8x = x*2^3

...

round n bet x*2^(n-1)

suppose you win on the kth round after losing rounds 1... k-1

then you have lost x*[2^0 + 2^1 + ... + 2^(k-2)] = x*[1 + 2 + 4 + ... + 2^(k-2)]

and you have won x*2^(k-1)

your profit is (win)-(lose), or:

x*2^(k-1) - x*[2^0 + 2^1 + ... + 2^(k-2)]

and we hope that this is > 0. lets check.

how the heck do we check?

well first lets set up the inequality we want, and see if it's true or not:

x*2^(k-1) - x*[2^0 + 2^1 + ... + 2^(k-2)] > 0 ??

now what?

ooh i know. let's divide everything by x*2^(k-2)

then we have:

2^(k-1)/2^(k-2)-[ (2^0)/2^(k-2) + (2^1)/2^(k-2) + (2^2)/2^(k-2) + ... + (2^(k-2))/2^(k-2) ] > 0/2^(k-2) > 0 ?

2 - [1/2^(k-2) + 1/2^(k-3) + ... + 1/2^3 + 1/2^2 + 1/2^1 + 1] > 0 ?

2 - [1 + 1/2 + 1/4 + 1/8 + ... + 1/2^(k-2)] > 0?

well, the sum in the line above is a geometric series with ratio - 1/2 ( every term is 1/2 the term before it ). as k approaches infinity, a bit of calculus II ( actually analysis, but nvm ) tells us that the limit of this series is 1/(1-r) = 1/(1-1/2) = 2

which would mean our winnings equation becomes

2 - 2 = 0

bummer. if it takes you infinite number of plays to win, your going to end up doing no more than break even.

the good news is that you won't ever be able to play infinite times; you're going to win after some less-than-infinite number of plays.

now some actual analysis that i won't confuse you with tells us that the geometric series that approaches 2 approaches it from the lower side of 2, so for a non-infinite k, the series will sum to something less than 2. that's really good news, because then for some arbitrarily small, positive number e ( for epsilon =p ), we have

2 - ( 2 - e ) = e > 0

i know you are probably not following this anymore, dear reader, but it means that you will get positive winnings. unless you play an infinite number of times before you win lol.

this strategy works for any game with a 1-1 payout.

the problem with using this strategy is that you have to keep doubling your bet! At some point, you either won't be able to cover your bet, or the house won't take your bet because it's too large. say you start with $100. you don't want to waste your time trying to win $1 at a time, so you bet $10. lets say you lose. now you have $90. well, you've got this great strategy, so double your bet ( $20 ) and play again. and lose. dang. you now have $70, and a great strategy to win, so double your bet again. $40. crap. you lose. now you are down to $30 - if you lose, you can't double your bet! after only 3 losses in a row, you got pwned! statistically, you're going to see 5, 6, 7 losses in a row sometimes on a fair table on the 1-1 bets. that means you have to walk in and bet ( and stand to win ) small change in order to make sure you have enough of a reserve to keep doubling on a losing streak. this in turn means you aren't going to stand to win very much $$ compared to what you need to have in your pocket.

all in all though, it's a sound strategy that WILL win, even if it won't win much, but to be sure you're going to come out ahead, you need a big bankroll, and you need to bet small.

if you were going to play with real money, i would keep my bets at like 1% of your bankroll. even at 1% you will bankrupt yourself if you lose 6 times in a row - you won't be able to double your bet a 7th time. the chances of losing 6 times in a row? 1/64 for coin tosses - a little higher for a roulette wheel with a '0' and higher still for one with a '00' - not entirely unlikely.

it's 18/38 ~ 47% in roulette if you bet even/odd or red/black. there is a green '0' ( which isn't counted as even, even though 0/2 is an integer. go figure )

the trick is to double your bet each round, and to quit as soona s you win ( or start over with a 1 $ bet )

there is no guarantee that you will win any round, but as long as you double your bet, when you win you will make more than what you have bet so far:

round 1 bet x = x*2^0

round 2 bet 2x = x*2^1

round 3 bet 2(2x)=4x = x*2^2

round 4 bet 2(4x)=8x = x*2^3

...

round n bet x*2^(n-1)

suppose you win on the kth round after losing rounds 1... k-1

then you have lost x*[2^0 + 2^1 + ... + 2^(k-2)] = x*[1 + 2 + 4 + ... + 2^(k-2)]

and you have won x*2^(k-1)

your profit is (win)-(lose), or:

x*2^(k-1) - x*[2^0 + 2^1 + ... + 2^(k-2)]

and we hope that this is > 0. lets check.

how the heck do we check?

well first lets set up the inequality we want, and see if it's true or not:

x*2^(k-1) - x*[2^0 + 2^1 + ... + 2^(k-2)] > 0 ??

now what?

ooh i know. let's divide everything by x*2^(k-2)

then we have:

2^(k-1)/2^(k-2)-[ (2^0)/2^(k-2) + (2^1)/2^(k-2) + (2^2)/2^(k-2) + ... + (2^(k-2))/2^(k-2) ] > 0/2^(k-2) > 0 ?

2 - [1/2^(k-2) + 1/2^(k-3) + ... + 1/2^3 + 1/2^2 + 1/2^1 + 1] > 0 ?

2 - [1 + 1/2 + 1/4 + 1/8 + ... + 1/2^(k-2)] > 0?

well, the sum in the line above is a geometric series with ratio - 1/2 ( every term is 1/2 the term before it ). as k approaches infinity, a bit of calculus II ( actually analysis, but nvm ) tells us that the limit of this series is 1/(1-r) = 1/(1-1/2) = 2

which would mean our winnings equation becomes

2 - 2 = 0

bummer. if it takes you infinite number of plays to win, your going to end up doing no more than break even.

the good news is that you won't ever be able to play infinite times; you're going to win after some less-than-infinite number of plays.

now some actual analysis that i won't confuse you with tells us that the geometric series that approaches 2 approaches it from the lower side of 2, so for a non-infinite k, the series will sum to something less than 2. that's really good news, because then for some arbitrarily small, positive number e ( for epsilon =p ), we have

2 - ( 2 - e ) = e > 0

i know you are probably not following this anymore, dear reader, but it means that you will get positive winnings. unless you play an infinite number of times before you win lol.

this strategy works for any game with a 1-1 payout.

the problem with using this strategy is that you have to keep doubling your bet! At some point, you either won't be able to cover your bet, or the house won't take your bet because it's too large. say you start with $100. you don't want to waste your time trying to win $1 at a time, so you bet $10. lets say you lose. now you have $90. well, you've got this great strategy, so double your bet ( $20 ) and play again. and lose. dang. you now have $70, and a great strategy to win, so double your bet again. $40. crap. you lose. now you are down to $30 - if you lose, you can't double your bet! after only 3 losses in a row, you got pwned! statistically, you're going to see 5, 6, 7 losses in a row sometimes on a fair table on the 1-1 bets. that means you have to walk in and bet ( and stand to win ) small change in order to make sure you have enough of a reserve to keep doubling on a losing streak. this in turn means you aren't going to stand to win very much $$ compared to what you need to have in your pocket.

all in all though, it's a sound strategy that WILL win, even if it won't win much, but to be sure you're going to come out ahead, you need a big bankroll, and you need to bet small.

if you were going to play with real money, i would keep my bets at like 1% of your bankroll. even at 1% you will bankrupt yourself if you lose 6 times in a row - you won't be able to double your bet a 7th time. the chances of losing 6 times in a row? 1/64 for coin tosses - a little higher for a roulette wheel with a '0' and higher still for one with a '00' - not entirely unlikely.

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#7
Posted 06 January 2010 - 06:37 PM

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#8
Posted 07 January 2010 - 12:04 PM

omg connection . . . . ur a math-genius

didnt read all of it....some parts i understood...other i didnt

and....the "trick" i wrote isnt much about 50/50 chance.....its just 100% of Win....

but u need big bankroll (as connection wrote) and the max bet has to be high enough....

i just figured out: betting 1â‚¬ at beginning is crap...

start with 2â‚¬

with the example i made (start with 1â‚¬) u just win 1â‚¬

if u start with 2â‚¬ its better : u win 2â‚¬ and its the same thing

example:

2â‚¬ red

-->loose

4â‚¬ red

--->loose

8â‚¬ red

--->win: lost 6â‚¬ and won 8â‚¬=WIN of 2â‚¬

only thing is: amount of betting gets faster .. bigger

didnt read all of it....some parts i understood...other i didnt

and....the "trick" i wrote isnt much about 50/50 chance.....its just 100% of Win....

but u need big bankroll (as connection wrote) and the max bet has to be high enough....

i just figured out: betting 1â‚¬ at beginning is crap...

start with 2â‚¬

with the example i made (start with 1â‚¬) u just win 1â‚¬

if u start with 2â‚¬ its better : u win 2â‚¬ and its the same thing

example:

2â‚¬ red

-->loose

4â‚¬ red

--->loose

8â‚¬ red

--->win: lost 6â‚¬ and won 8â‚¬=WIN of 2â‚¬

only thing is: amount of betting gets faster .. bigger

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#9
Posted 07 January 2010 - 01:19 PM

lol sorry i couldn't help myself. maybe later i'll prove that you win exactly the amount of your first bet =D

yeah the biggest problem is like you say:

doubling your bet every round adds up quick! so you need to start with a SMALL amount compared to the total of what you have.

yeah the biggest problem is like you say:

only thing is: amount of betting gets faster .. bigger

doubling your bet every round adds up quick! so you need to start with a SMALL amount compared to the total of what you have.

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#10
Posted 07 January 2010 - 02:25 PM

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