I have tried some times and I can say I'm in loss with martingale, so it's not for me
its better to do all in on 50 % chance as doing martingale makes the house edge bigger when you roll bunch of times. i never use this strategy
Apparently this is not true. You can actually "lower" the house edge if you bet with a certain strategy other than 50% bet. There was a long discussing on previous pages.
It's not really lowering the house edge, per se, but rather exploiting the rounding of numbers. Some sites have already fixed that issue, though, and it relies on very specific sets of numbers to work even on those that haven't fixed it. With that said, it's less of a "beating the house" and more of an exploit than anything.
Oh I see. I thought it had something to do with the fact that you on average reach the goal earlier before wagering all of the 1 BTC and thus incurring less house edge.
That post I made with all the math I think has to be off. I spent a lot of time thinking about it and I don't see how raising bet amounts (without changing the % chance) could possibly affect the house edge. I just don't know where I went wrong with it...
I guess it's possible that it's correct. It just doesn't make logical sense.
Yes, it boggles my mind too. My explanation made the most sense in my head, but could be dead wrong.
I'll probably need to get in touch with Dooglus and see his opinion on this. He's pretty solid when it comes to figuring these things out and if he has time, should be able to see what's wrong.
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)