Nobody will create a gambling site without any house edge.
They will lose money paying for site and support without any profit.
Topic: No House Edge? - page 5. (Read 4741 times)
Even with negative house edge is difficult to make profit,because you have to know to have a strategy
moneypot has a way to have negative edge,but even that way,it is really difficult to make profit
moneypot has a way to have negative edge,but even that way,it is really difficult to make profit
We are not talking about going all in or not, we started talking about if making a goal is better or not, and clearly is better since you can have a 99% chance of getting 50$ if thats your goal, you would need a lot of money? Yes. But that doesnt change the fact that you would have 99% chances
Yes we are talking about setting a goal but you are the one shifted it to a side topic , Remember this post below? you are the one started this side topic btw
What the hell are you talking about? Setting a goal does increase your chances of winning, you want a 0.01 profit? Bet 0.2 on 95% ta da your chances are 95% of winning your goal.
This has been discussed before and a 0% house edge will be the same, you would still lose whenever you start using a "strategy"
This has been discussed before and a 0% house edge will be the same, you would still lose whenever you start using a "strategy"
From the bolded part, if you put on some sense to it, you will know that if someone betting on the max win chance, it means that he is betting with max bet as well . I.e someone having 1 BTC will do a max bet at 98 % chance with 1 BTC and not 0.001 BTC at 98 %. It is logical though
P.S : I will leave you up here since you always twist you own words. Discussing it further will be useless with someone that always stick blindly to his opinion
We are not talking about going all in or not, we started talking about if making a goal is better or not, and clearly is better since you can have a 99% chance of getting 50$ if thats your goal, you would need a lot of money? Yes. But that doesnt change the fact that you would have 99% chances
Quoted it for you and this post support my argument to not go all-in as well . In the bolded part you can see dooglus statement that he put up calculation that if you go all in in the roulette you got a lower chance to win $50 with $8839 then if you split up your bet
Again, this post just supported my statement about it is better to not go all in
If you bet your whole bankroll on a single bet in a 1% house edge casino, your expected loss is 1% of your bankroll.
If instead you split your bankroll up into a series of bets, starting small and increasing on loss, you can reach the same goal with a slightly higher chance of success. You didn't change the house edge, you just expect to risk less, and so expect to lose less
For example, the guy earlier talking about how much you need to have a 99% chance of winning $50 in a casino.
Let's assume the house edge is (100/37)% like it is in roulette.
That means that the probability of winning a bet times the payout for that bet is 36/37.
"chance * payout = 36/37"
For example, when you bet on red, the payout is 2x. The chance of winning is 18/37 (there are 18 reds and 37 numbers). 2 times 18/37 is 36/37.
When you bet on a single number, the payout is 36x. The chance of winning is 1/37. 36 times 1/37 is 36/37.
And it's the same for all the possible bets.
So if we want a 99% chance of winning $50 from a single bet on roulette:
chance = 0.99
chance * payout = 36/37
payout = 36/37/0.99 = 0.9828x
So if we want a 99% chance of winning a single bet, the house will pay us back less than our stake if we win. You can't have a 99% chance of making a profit on a single bet when the house edge is bigger than 1%.
So we're going to have to split the bet into a two-step martingale sequence.
It turns out that you only need $8839 (*) to have a 99% chance of winning $50, and here's how:
We'll bet at 90% (we have to imagine they offer any percentage you like, like dice sites do).
The 90% chance bet has a payout of 36/37/0.9 = 1.081081x
First bet $616.67 at 90%.
If it hits, we get 1.081081 times 616.67 back.
1.081081 * 616.67 = 666.67 and we've made our $50 already, so stop.
If we lose that first bet, bet the remaining 8839 - 616.67 = $8222.33 at 90%.
If it hits, we get 1.081081 times 8222.33 back.
1.081081 * 8222.33 = 8889.00, and if we take off the 616.67 we lost on the first bet we get 8272.33, and we've made our $50 again.
So if we hit either the first or 2nd bet, we're $50 up, and if we lose both we're $8839 down.
There's a 10% chance of losing each bet, so a 1% chance of losing them both, so a 99% chance of winning one of them.
ie. we only need $8839 to have a 99% chance of winning $50 in a casino game with the same house edge as single-zero roulette.
Note that if we put all $8839 on a single bet to win $50, the payout would need to be (8839+50)/8839 = 1.00566x
and so the chance of winning would be (36/37)/((8839+50)/8839) = 96.75%
So again, we see that trying to win $50 starting with $8839 has a lower chance (96.75%) if you make a single bet and a higher chance 99% if you split it into two bets, and do a mini 2-step martingale progression. And that's because 90% of the time you only have to make the small first bet, and so you're betting less, and so on average you lose less.
(*) set p = 36.0/37/0.9 - 1, then the amount you need is 50*(2*p + 1)/(p*p) = $8838.888888
(The two bet sizes are really 616.66666 recurring and 8,222.22222 recurring)
Again, this post just supported my statement about it is better to not go all in
If you bet your whole bankroll on a single bet in a 1% house edge casino, your expected loss is 1% of your bankroll.
If instead you split your bankroll up into a series of bets, starting small and increasing on loss, you can reach the same goal with a slightly higher chance of success. You didn't change the house edge, you just expect to risk less, and so expect to lose less
For example, the guy earlier talking about how much you need to have a 99% chance of winning $50 in a casino.
Let's assume the house edge is (100/37)% like it is in roulette.
That means that the probability of winning a bet times the payout for that bet is 36/37.
"chance * payout = 36/37"
For example, when you bet on red, the payout is 2x. The chance of winning is 18/37 (there are 18 reds and 37 numbers). 2 times 18/37 is 36/37.
When you bet on a single number, the payout is 36x. The chance of winning is 1/37. 36 times 1/37 is 36/37.
And it's the same for all the possible bets.
So if we want a 99% chance of winning $50 from a single bet on roulette:
chance = 0.99
chance * payout = 36/37
payout = 36/37/0.99 = 0.9828x
So if we want a 99% chance of winning a single bet, the house will pay us back less than our stake if we win. You can't have a 99% chance of making a profit on a single bet when the house edge is bigger than 1%.
So we're going to have to split the bet into a two-step martingale sequence.
It turns out that you only need $8839 (*) to have a 99% chance of winning $50, and here's how:
We'll bet at 90% (we have to imagine they offer any percentage you like, like dice sites do).
The 90% chance bet has a payout of 36/37/0.9 = 1.081081x
First bet $616.67 at 90%.
If it hits, we get 1.081081 times 616.67 back.
1.081081 * 616.67 = 666.67 and we've made our $50 already, so stop.
If we lose that first bet, bet the remaining 8839 - 616.67 = $8222.33 at 90%.
If it hits, we get 1.081081 times 8222.33 back.
1.081081 * 8222.33 = 8889.00, and if we take off the 616.67 we lost on the first bet we get 8272.33, and we've made our $50 again.
So if we hit either the first or 2nd bet, we're $50 up, and if we lose both we're $8839 down.
There's a 10% chance of losing each bet, so a 1% chance of losing them both, so a 99% chance of winning one of them.
ie. we only need $8839 to have a 99% chance of winning $50 in a casino game with the same house edge as single-zero roulette.
Note that if we put all $8839 on a single bet to win $50, the payout would need to be (8839+50)/8839 = 1.00566x
and so the chance of winning would be (36/37)/((8839+50)/8839) = 96.75%
So again, we see that trying to win $50 starting with $8839 has a lower chance (96.75%) if you make a single bet and a higher chance 99% if you split it into two bets, and do a mini 2-step martingale progression. And that's because 90% of the time you only have to make the small first bet, and so you're betting less, and so on average you lose less.
(*) set p = 36.0/37/0.9 - 1, then the amount you need is 50*(2*p + 1)/(p*p) = $8838.888888
(The two bet sizes are really 616.66666 recurring and 8,222.22222 recurring)
Here it is:
If you bet your whole bankroll on a single bet in a 1% house edge casino, your expected loss is 1% of your bankroll.
If instead you split your bankroll up into a series of bets, starting small and increasing on loss, you can reach the same goal with a slightly higher chance of success. You didn't change the house edge, you just expect to risk less, and so expect to lose less
For example, the guy earlier talking about how much you need to have a 99% chance of winning $50 in a casino.
Let's assume the house edge is (100/37)% like it is in roulette.
That means that the probability of winning a bet times the payout for that bet is 36/37.
"chance * payout = 36/37"
For example, when you bet on red, the payout is 2x. The chance of winning is 18/37 (there are 18 reds and 37 numbers). 2 times 18/37 is 36/37.
When you bet on a single number, the payout is 36x. The chance of winning is 1/37. 36 times 1/37 is 36/37.
And it's the same for all the possible bets.
So if we want a 99% chance of winning $50 from a single bet on roulette:
chance = 0.99
chance * payout = 36/37
payout = 36/37/0.99 = 0.9828x
So if we want a 99% chance of winning a single bet, the house will pay us back less than our stake if we win. You can't have a 99% chance of making a profit on a single bet when the house edge is bigger than 1%.
So we're going to have to split the bet into a two-step martingale sequence.
It turns out that you only need $8839 (*) to have a 99% chance of winning $50, and here's how:
We'll bet at 90% (we have to imagine they offer any percentage you like, like dice sites do).
The 90% chance bet has a payout of 36/37/0.9 = 1.081081x
First bet $616.67 at 90%.
If it hits, we get 1.081081 times 616.67 back.
1.081081 * 616.67 = 666.67 and we've made our $50 already, so stop.
If we lose that first bet, bet the remaining 8839 - 616.67 = $8222.33 at 90%.
If it hits, we get 1.081081 times 8222.33 back.
1.081081 * 8222.33 = 8889.00, and if we take off the 616.67 we lost on the first bet we get 8272.33, and we've made our $50 again.
So if we hit either the first or 2nd bet, we're $50 up, and if we lose both we're $8839 down.
There's a 10% chance of losing each bet, so a 1% chance of losing them both, so a 99% chance of winning one of them.
ie. we only need $8839 to have a 99% chance of winning $50 in a casino game with the same house edge as single-zero roulette.
Note that if we put all $8839 on a single bet to win $50, the payout would need to be (8839+50)/8839 = 1.00566x
and so the chance of winning would be (36/37)/((8839+50)/8839) = 96.75%
So again, we see that trying to win $50 starting with $8839 has a lower chance (96.75%) if you make a single bet and a higher chance (99%) if you split it into two bets, and do a mini 2-step martingale progression. And that's because 90% of the time you only have to make the small first bet, and so you're betting less, and so on average you lose less.
(*) set p = 36.0/37/0.9 - 1, then the amount you need is 50*(2*p + 1)/(p*p) = $8838.888888
(The two bet sizes are really 616.66666 recurring and 8,222.22222 recurring)
I dont know how to quote from other posts but this was made by dooglus here :
https://bitcointalk.org/index.php?topic=610339.780
If you bet your whole bankroll on a single bet in a 1% house edge casino, your expected loss is 1% of your bankroll.
If instead you split your bankroll up into a series of bets, starting small and increasing on loss, you can reach the same goal with a slightly higher chance of success. You didn't change the house edge, you just expect to risk less, and so expect to lose less
For example, the guy earlier talking about how much you need to have a 99% chance of winning $50 in a casino.
Let's assume the house edge is (100/37)% like it is in roulette.
That means that the probability of winning a bet times the payout for that bet is 36/37.
"chance * payout = 36/37"
For example, when you bet on red, the payout is 2x. The chance of winning is 18/37 (there are 18 reds and 37 numbers). 2 times 18/37 is 36/37.
When you bet on a single number, the payout is 36x. The chance of winning is 1/37. 36 times 1/37 is 36/37.
And it's the same for all the possible bets.
So if we want a 99% chance of winning $50 from a single bet on roulette:
chance = 0.99
chance * payout = 36/37
payout = 36/37/0.99 = 0.9828x
So if we want a 99% chance of winning a single bet, the house will pay us back less than our stake if we win. You can't have a 99% chance of making a profit on a single bet when the house edge is bigger than 1%.
So we're going to have to split the bet into a two-step martingale sequence.
It turns out that you only need $8839 (*) to have a 99% chance of winning $50, and here's how:
We'll bet at 90% (we have to imagine they offer any percentage you like, like dice sites do).
The 90% chance bet has a payout of 36/37/0.9 = 1.081081x
First bet $616.67 at 90%.
If it hits, we get 1.081081 times 616.67 back.
1.081081 * 616.67 = 666.67 and we've made our $50 already, so stop.
If we lose that first bet, bet the remaining 8839 - 616.67 = $8222.33 at 90%.
If it hits, we get 1.081081 times 8222.33 back.
1.081081 * 8222.33 = 8889.00, and if we take off the 616.67 we lost on the first bet we get 8272.33, and we've made our $50 again.
So if we hit either the first or 2nd bet, we're $50 up, and if we lose both we're $8839 down.
There's a 10% chance of losing each bet, so a 1% chance of losing them both, so a 99% chance of winning one of them.
ie. we only need $8839 to have a 99% chance of winning $50 in a casino game with the same house edge as single-zero roulette.
Note that if we put all $8839 on a single bet to win $50, the payout would need to be (8839+50)/8839 = 1.00566x
and so the chance of winning would be (36/37)/((8839+50)/8839) = 96.75%
So again, we see that trying to win $50 starting with $8839 has a lower chance (96.75%) if you make a single bet and a higher chance (99%) if you split it into two bets, and do a mini 2-step martingale progression. And that's because 90% of the time you only have to make the small first bet, and so you're betting less, and so on average you lose less.
(*) set p = 36.0/37/0.9 - 1, then the amount you need is 50*(2*p + 1)/(p*p) = $8838.888888
(The two bet sizes are really 616.66666 recurring and 8,222.22222 recurring)
I dont know how to quote from other posts but this was made by dooglus here :
https://bitcointalk.org/index.php?topic=610339.780
Of course it depends on luck, but 99% is better than 50% isnt it.
99% is no better than 50 %. The logic will be based on your bet amount. Assuming you are betting with 99 % , you will the bet with your maximum bankroll to ensure greater profit for it rather than a spare change. If your bad luck struck up, you will lose your entire balance with a single click
On the other hand if you are betting with 50 % assuming you wont be betting the whole bankroll for it which is much more safer if you "know when to stop" especially after hitting some bad luck stroke
P.S : it should be 98 % AFAIK in most dice sites and not 99 %
the house always must have an edge to operate a gambling casino or other operation, if people play long enough they lose ,
True, operating a gambling sites with no edge at all is pointless as people are making a gambling sites to make some profit as well
You have no idea what you are talking about, someone already calculated, i think dooglus that if you wanted a profit of 50$ and you had 5000$ (im not sure about the exact numbers) you had a 99% chance of doing it.
Someone calculated what? who? post the quote here
You are missleading your own post, You claimed that 98 % chance with max bet is better than splitting it up to a few bet and please dont put up dooglus name in here if you cant quote his post saying that it is better to risk 98 % rather than splitting it up
Dug up this old post just for you conversation between Light and Dooglus (found it faster than I expected):
I already made a couple of posts on why martingale sucks - but I didn't actually address OP.
I haven't done the math on the martingale method per se - but I did dig up an old conversation me a dooglus had about losing 'less' by betting less (ie. rather than all in betting, splitting your bets into n portions where you aim to get to x as a value then stop).
And some quick maths (plus graphing)
I haven't given it a lot of thought beyond this - doog explained it quite elegantly.
I haven't done the math on the martingale method per se - but I did dig up an old conversation me a dooglus had about losing 'less' by betting less (ie. rather than all in betting, splitting your bets into n portions where you aim to get to x as a value then stop).
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 theinvisibility 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.
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
And some quick maths (plus graphing)
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.
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.
I haven't given it a lot of thought beyond this - doog explained it quite elegantly.
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.
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.
Of course it depends on luck, but 99% is better than 50% isnt it.
99% is no better than 50 %. The logic will be based on your bet amount. Assuming you are betting with 99 % , you will the bet with your maximum bankroll to ensure greater profit for it rather than a spare change. If your bad luck struck up, you will lose your entire balance with a single click
On the other hand if you are betting with 50 % assuming you wont be betting the whole bankroll for it which is much more safer if you "know when to stop" especially after hitting some bad luck stroke
P.S : it should be 98 % AFAIK in most dice sites and not 99 %
the house always must have an edge to operate a gambling casino or other operation, if people play long enough they lose ,
True, operating a gambling sites with no edge at all is pointless as people are making a gambling sites to make some profit as well
You have no idea what you are talking about, someone already calculated, i think dooglus that if you wanted a profit of 50$ and you had 5000$ (im not sure about the exact numbers) you had a 99% chance of doing it.
Of course it depends on luck, but 99% is better than 50% isnt it.
99% is no better than 50 %. The logic will be based on your bet amount. Assuming you are betting with 99 % , you will the bet with your maximum bankroll to ensure greater profit for it rather than a spare change. If your bad luck struck up, you will lose your entire balance with a single click
On the other hand if you are betting with 50 % assuming you wont be betting the whole bankroll for it which is much more safer if you "know when to stop" especially after hitting some bad luck stroke
P.S : it should be 98 % AFAIK in most dice sites and not 99 %
the house always must have an edge to operate a gambling casino or other operation, if people play long enough they lose ,
True, operating a gambling sites with no edge at all is pointless as people are making a gambling sites to make some profit as well
the house always must have an edge to operate a gambling casino or other operation, if people play long enough they lose ,
Gambling always will be about luck. Someone will win someone loses - it will always be like that 
0% or 1% House Edge doesn't change that much.
0% or 1% House Edge doesn't change that much.
i don't believe that greediness has anything to do with a casino's profits if they are at 0% house edge. The house will win some and they will lose some, they will lose as much as they win in the long run. Someone might be greedy and win, someone might be greedy and they lose. Someone might not be greedy and win. Someone might not be greedy and lose. in the end, they will win as much as they lose.
The thing is that as a casino owner, the fluctuations might be unmanageable and the risk of these fluctuations will be enough to scare them away from offering 0%.
The thing is that as a casino owner, the fluctuations might be unmanageable and the risk of these fluctuations will be enough to scare them away from offering 0%.
no house edge is risky statisticly looking but based on people greed casino would still profit 
Statistically looking is not risky, is neutral. If you were to bet flat bets (say 0.01) without increasing the bet or the odds, statistically you would end up getting 0 profit, no loss no win. But if you use, lets say, martingale you would end up loosing anyways, statistically you would survive a little bit longer on average compared to a 1% house edge but the difference is not too significative
no house edge is risky statisticly looking but based on people greed casino would still profit
Statistically looking is not risky, is neutral. If you were to bet flat bets (say 0.01) without increasing the bet or the odds, statistically you would end up getting 0 profit, no loss no win. But if you use, lets say, martingale you would end up loosing anyways, statistically you would survive a little bit longer on average compared to a 1% house edge but the difference is not too significative
i always find these types of threads funny. Mainly because this situation will always be a hypothetical. There will be no casino owner that will ever have a house edge of 0% or lower. NEVER.
They might have it as a promotion for an extremely short amount of time, and very restricted limits on amount of play, but they will never have it long lasting.
That is, if they want to stay in business.
If there are any casinos that boast 0 or negative edges for the house, i'd be wary and would do a lot of due diligence on their trustworthiness. It would scream scam to me at first glance.
They might have it as a promotion for an extremely short amount of time, and very restricted limits on amount of play, but they will never have it long lasting.
That is, if they want to stay in business.
If there are any casinos that boast 0 or negative edges for the house, i'd be wary and would do a lot of due diligence on their trustworthiness. It would scream scam to me at first glance.
no house edge is risky statisticly looking but based on people greed casino would still profit
I remember negative house edge being discussed in another thread. So, I'm not going to talk about this.
No house edge you said? Well, I'm pretty sure both parties will be happy then. The operator can earn from donation or maybe advertisement from his or her site. The user will gamble peacefully as there is no house edge. Well... I'm like others.. There is no peace if you started gambling.
No house edge you said? Well, I'm pretty sure both parties will be happy then. The operator can earn from donation or maybe advertisement from his or her site. The user will gamble peacefully as there is no house edge. Well... I'm like others.. There is no peace if you started gambling.
Well what if I say I have a determined goal? Like I reach 0.1 with 0.05? Would it change then knowing I will stop if I reach 0.1?
Setting a goal will not be helping you in reaching profit since at the end it will depends on your luck either but atleast it will help you in not to lose more. In order to set up a goal, you cant set a goal with just " only stop at X profit" but you must set up a goal to stop at X loss too.
Nevertheless, A goal doesnt stop a gambler from gambling to keep on gaining more and more. Mark my word for it. You can check on this thread for more proof of my words https://bitcointalksearch.org/topic/my-journey-can-i-win-gambling-1052406
Basically the OP of the site set a goal of only betting for 0.01 BTC and stop after he get a profit of at least 10% of it but guess what, He cant keep up with his words and thus keep on doubling each of his bankroll
What the hell are you talking about? Setting a goal does increase your chances of winning, you want a 0.01 profit? Bet 0.2 on 95% ta da your chances are 95% of winning your goal.
This has been discussed before and a 0% house edge will be the same, you would still lose whenever you start using a "strategy"
Playing with higher winning chance is also not a guarantee that a player will always get profit.
Do you remember someone who lost 7000btc in 96% winning chance?
And yes, it depends on "LUCK" as what arallmuss said.
no shit, sherlock. So your point is that playing with 99% doesnt matter because you can have bad luck :^)
The guy who lost 7000 btc in 96% didnt just lose it in 1 bet, he was betting for a long time from i think 2000 btc so yea...
Of course it depends on luck, but 99% is better than 50% isnt it.
Well what if I say I have a determined goal? Like I reach 0.1 with 0.05? Would it change then knowing I will stop if I reach 0.1?
Setting a goal will not be helping you in reaching profit since at the end it will depends on your luck either but atleast it will help you in not to lose more. In order to set up a goal, you cant set a goal with just " only stop at X profit" but you must set up a goal to stop at X loss too.
Nevertheless, A goal doesnt stop a gambler from gambling to keep on gaining more and more. Mark my word for it. You can check on this thread for more proof of my words https://bitcointalksearch.org/topic/my-journey-can-i-win-gambling-1052406
Basically the OP of the site set a goal of only betting for 0.01 BTC and stop after he get a profit of at least 10% of it but guess what, He cant keep up with his words and thus keep on doubling each of his bankroll
What the hell are you talking about? Setting a goal does increase your chances of winning, you want a 0.01 profit? Bet 0.2 on 95% ta da your chances are 95% of winning your goal.
This has been discussed before and a 0% house edge will be the same, you would still lose whenever you start using a "strategy"
Playing with higher winning chance is also not a guarantee that a player will always get profit.
Do you remember someone who lost 7000btc in 96% winning chance?
And yes, it depends on "LUCK" as what arallmuss said.
Well what if I say I have a determined goal? Like I reach 0.1 with 0.05? Would it change then knowing I will stop if I reach 0.1?
Setting a goal will not be helping you in reaching profit since at the end it will depends on your luck either but atleast it will help you in not to lose more. In order to set up a goal, you cant set a goal with just " only stop at X profit" but you must set up a goal to stop at X loss too.
Nevertheless, A goal doesnt stop a gambler from gambling to keep on gaining more and more. Mark my word for it. You can check on this thread for more proof of my words https://bitcointalksearch.org/topic/my-journey-can-i-win-gambling-1052406
Basically the OP of the site set a goal of only betting for 0.01 BTC and stop after he get a profit of at least 10% of it but guess what, He cant keep up with his words and thus keep on doubling each of his bankroll
What the hell are you talking about? Setting a goal does increase your chances of winning, you want a 0.01 profit? Bet 0.2 on 95% ta da your chances are 95% of winning your goal.
This has been discussed before and a 0% house edge will be the same, you would still lose whenever you start using a "strategy"
You will only have high chance of winning in your strategy to bet at 95% but still that 5% have a chance to kill your balance.
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