Bitcoin Forum

Me and my mate are in an argument right now.

I think that 2x (49.5%) win chance with 100% bet increase on loss is the best (16x losses in a row and I'm bust). Is there a better way? The house always win in the end and with those margins I should last the longest, yes? I've made profit from it so far, those 16 losses in a row rarely come.

Yeah, what your referring to right there is the martingale system http://en.wikipedia.org/wiki/Martingale_(betting_system)  (http://en.wikipedia.org/wiki/Martingale_(betting_system))and no in the short term there is no guarantee that the house wins. There is no "best" method despite what people will tell you, it's simply about luck and variance in probability. Of course as you approach an infinite number of bets you approach that 0.5% or whatever house edge they have, but a) you'll never reach that number of bets and b) there is always a crap ton of variance anyways.

And despite what you think 16 losses in a row sounds good (1/65536) but you must remember that the past has no influence on the future and hence don't think just because you've lost 15 times in a row you won't lose a 16th.

If you are scaling your bets so that you win 1 unit per incarnation then the larger the odds, the longer you'll play (so a 1% martingale will last longer then a 49.5% which will last longer then a 90%).

I've been running this system for 3 days now with 20 satoshi as basebet and a bankroll at 0.01 to begin with. I got my streak of losses, but the scond try has been working out so far. Here are my stats (I used facuet propably 1000 times+ and managed to get to 50k+ satoshi which is why there are so many bets and not too much profit).

https://i.imgur.com/36fyFEK.png

But can elabroate why 1% will last longer?

The way that they calculate the house edge.  For a 1% house edge site they take the payout that you should get and multiply it by .99.  This makes the payout on shorter odd bets exponentially worse.

Let's look at the max chance to win on JD.  98.99%.  At 0% edge, you'll win 9899 times out of 10000 and lose 101 times out of 10000.  So in order to win 1 unit, you'll need to risk 98.99 units.  Over 10,000 spins you'll (theoretically) break exactly even, winning 1 unit 9899 times and losing 98.99 units 101 times.

But when you multiple that already small payout and multiply it by .99, the odds decrease exponentially.  So 98.99% to win on a 1% house edge site only pays out 1.00010102.  So in order to win 1 unit, you'll need to risk 9899.  Over 10,000 spins, you'll win 1 unit 9899 times, but the 101 times that you lose, you'll lose 9899 units.  For a net loss of -989900 units.

Let's look at 60% to win.  1% house edge makes the payout 1.65.  To win 1 unit, you'll need to risk 1.53846153 units.  Let's spin 10,000 times again.  6000 wins (1 each time), 4000 losses (-1.53846153 each time).  Net loss of 153.84612 units.

50% to win = 1.98x payout odds.  To win 1 unit, you'll need to risk 1.02040816 units.  10,000 spins, 5000 wins, 5000 losses, net loss of 102.0408 units.

25% to win = 3.96x payout.  To win 1 unit, you'll need to risk 0.33783783 units.  10,000 spins, 2500 wins, 7500 losses, net loss of 33.783725.

2% to win = 49.5x payout.  To win 1 unit, you'll need to risk 0.02061855 units.  10,000 spins, 200 wins, 9800 losses, net loss of 2.06179.

The longer the odds, the less the 1% reduction of them hurts.

Cheers for the reply :) Forgot to thank you earlier!

Also you can start with a lower chance like 47%.