Topic: Double-bet-on-lose method applied to bitcoin gambling - page 2. (Read 9219 times)
Anyone who doubts the futility of the martingale strategy should hop over to BitZino and play roulette. Using 17BTCs you'll be able to go 15 doublings deep (your account defaults to mBTCs). That's, what, almost $500 right now. You will get wiped out eventually. If you don't believe me, try it. Or, just go read the BitZino thread. 15 consecutive losses isn't that uncommon. Ok, so give yourself more room. Just to be safe allow yourself 20 doublings. Well, now you're at 524.288BTCs (almost $15k). That many losses in a row isn't unheard of, and that's an awful lot of money to wager when if you get 21 losses in a row you're completely wiped out. Hopefully you see where this is going. You might be able to make some money if you can go 25 deep, but, uh oh, that many BTCs don't even exist right now. What's worse is that you'll never even be able to go beyond 25 deep with BTCs. If you use mBTC you can go a little further and μBTC you can go further still, but your gains will be so small compared to how much money you'd need to put in to protect yourself against really unlucky, but possible, losing streaks that it's just silly to even try.
You double your bet every time you loose.
What people don't realize, is that doubling your bets is equivalent to an exponential increase. Here's a pic to help you grasp the exponential feeling:
http://www.regentsprep.org/Regents/math/ALGEBRA/AE7/fixpic2.gif
( I don't own copyrights on this image )
It's not sustainable. You run out of cash very quickly.
Additionally, if you are into statistics, you can calculate the expectation of this strategy:
https://en.wikipedia.org/wiki/Expected_value
The expectation value will be below 0.5, so the house will still have the advantage.
By the way, you will never find a casino game where the expectation value is over 0.5. The house would be loosing money over time in such a case.
Cheers!
What people don't realize, is that doubling your bets is equivalent to an exponential increase. Here's a pic to help you grasp the exponential feeling:
http://www.regentsprep.org/Regents/math/ALGEBRA/AE7/fixpic2.gif
( I don't own copyrights on this image )
It's not sustainable. You run out of cash very quickly.
Additionally, if you are into statistics, you can calculate the expectation of this strategy:
https://en.wikipedia.org/wiki/Expected_value
The expectation value will be below 0.5, so the house will still have the advantage.
By the way, you will never find a casino game where the expectation value is over 0.5. The house would be loosing money over time in such a case.
Cheers!
Disclaimer: This is a mathematical analysis of the probabilities of winning in bitcoin based gambling sites. Whereas there is never 100% success of winning, using this method, and with enough money* in your hand, you have high chances of winning. The more money you initially have the more chances you have of winning.
*Infinite money
Martingale Strategy is a good way to ensure you will loose everything.
Please read http://en.wikipedia.org/wiki/Martingale_%28betting_system%29 before applying this strategy.
Please read http://en.wikipedia.org/wiki/Martingale_%28betting_system%29 before applying this strategy.
This is called martingale strategy. In the end it's no better than a one-time bet.
Hi all, there is a method for gambling in the roulette that allows you to almost always win. The trick is that you need to have a lot of money to start with.
The idea is simple: bet to color (black or red), which has nearly 50% change of win and payout is 2x. Every time you lose you double your bet, and if you win, you start again with the original bet.
So, suppose you bet 1, lose, bet 2, lose, bet 4, win, you win 8 and spend 7, so you still win 1.
There is a mathematical formula for this which is not that hard to understand. Suppose you bet 'x' and lose n-1 times in a row, winning in the n-th time. What you win is x*2*2^(n-1) and what you spend is the sum of n bets, which gives x*(2^n - 1). The profit is then x*[2^n - (2^n - 1)] = x.
Losing 7 times in a row has a probability of 0.7%, and you need to spend 256 times you original bet. So, if you have enough money in your hand, unless you are extremely unlucky you will always win something.
Now, when applied to bitcoin gambling, let's name satoshidice, you have to consider also the transaction fee.
Going back to the previous example, with n-1 loses and winning in the n-th time, you have to pay n times the fee 'f', so the total profit is x-n*f which should be positive.
If you bet the minimum, BTC0.01, and the transaction fee is BTC0.005 you will always lose because you spend more in transaction fees that what you can win.
Making 'x' 10 times larger than the transaction fee will give us 0.1% change of having negative profit (if playing to 50% change of winning).
You can avoid loosing with transaction fees while also betting the minimum if you change the multiplication base by another number, like 3. This way, the amount of winnings increases a lot, but you also need to have more bitcoins to spend.
If we revise the previous example, now multiplying by 3 instead of 2 in a losing, what we win at the n-th roll is 2*3^(n-1), and the total spending, including transaction fee, is (3^n - 1)/2 + n*f.
Total profit is then x*(3^(n-1) + 1)/2 -n*f. Because the winnings increase exponentially with n, even betting the minimum you will never lose.
Disclaimer: This is a mathematical analysis of the probabilities of winning in bitcoin based gambling sites. Whereas there is never 100% success of winning, using this method, and with enough money in your hand, you have high chances of winning. The more money you initially have the more chances you have of winning.
If you think this post is helpful, or if you make lot of money with this method, please donate something
Discussions are welcome.
The idea is simple: bet to color (black or red), which has nearly 50% change of win and payout is 2x. Every time you lose you double your bet, and if you win, you start again with the original bet.
So, suppose you bet 1, lose, bet 2, lose, bet 4, win, you win 8 and spend 7, so you still win 1.
There is a mathematical formula for this which is not that hard to understand. Suppose you bet 'x' and lose n-1 times in a row, winning in the n-th time. What you win is x*2*2^(n-1) and what you spend is the sum of n bets, which gives x*(2^n - 1). The profit is then x*[2^n - (2^n - 1)] = x.
Losing 7 times in a row has a probability of 0.7%, and you need to spend 256 times you original bet. So, if you have enough money in your hand, unless you are extremely unlucky you will always win something.
Now, when applied to bitcoin gambling, let's name satoshidice, you have to consider also the transaction fee.
Going back to the previous example, with n-1 loses and winning in the n-th time, you have to pay n times the fee 'f', so the total profit is x-n*f which should be positive.
If you bet the minimum, BTC0.01, and the transaction fee is BTC0.005 you will always lose because you spend more in transaction fees that what you can win.
Making 'x' 10 times larger than the transaction fee will give us 0.1% change of having negative profit (if playing to 50% change of winning).
You can avoid loosing with transaction fees while also betting the minimum if you change the multiplication base by another number, like 3. This way, the amount of winnings increases a lot, but you also need to have more bitcoins to spend.
If we revise the previous example, now multiplying by 3 instead of 2 in a losing, what we win at the n-th roll is 2*3^(n-1), and the total spending, including transaction fee, is (3^n - 1)/2 + n*f.
Total profit is then x*(3^(n-1) + 1)/2 -n*f. Because the winnings increase exponentially with n, even betting the minimum you will never lose.
Disclaimer: This is a mathematical analysis of the probabilities of winning in bitcoin based gambling sites. Whereas there is never 100% success of winning, using this method, and with enough money in your hand, you have high chances of winning. The more money you initially have the more chances you have of winning.
If you think this post is helpful, or if you make lot of money with this method, please donate something
Discussions are welcome.
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