Yet Another System that Wins
Quote: rdw4potusLOL. that's an odd number of events. How is that proof that things even out? We can all cherry pick results, and we can all make half-assed slanted arguments. The difference here is that you're the only person in this conversation who is actively trying to do those things.
The end. Nice talking to you guys as usual. 1 billion trials produces a 500 million unit win and you say that's proof the game breaks even. You do know anyone can run a million trial simulation of the D'Alembert on a 50/50 game and get the results. Popcan already did this on another thread. The game wins, period. I'm not the one B.S.ing anyone here. Don't believe me, run the simulation yourself. Give the results, I'm now very interested in what kinda B.S. your going to pull on this one. Amazing.
Quote: JyBrd0403The end. Nice talking to you guys as usual. 1 billion trials produces a 500 million unit win and you say that's proof the game breaks even. You do know anyone can run a million trial simulation of the D'Alembert on a 50/50 game and get the results. Popcan already did this on another thread. The game wins, period. I'm not the one B.S.ing anyone here. Don't believe me, run the simulation yourself. Give the results, I'm now very interested in what kinda B.S. your going to pull on this one. Amazing.
waaaaait, the trial produced a 50% rate of return and that doesnt set off a GIANT red flag for you? like, something must be wrong there? Really, 500 million profit in a billion spins. I'm glad this is the end. Go lose your money playing this way in the casino now.
Quote: JyBrd0403You do know anyone can run a million trial simulation of the D'Alembert on a 50/50 game and get the results. Popcan already did this on another thread. The game wins, period.
I don't get your point, this is like roulette 101. Even
the newest rookie knows you can beat a 50/50 game
with almost any progression. The problem is, no casino
offer a 50/50 game. So again, whats your point. Use
a d'Alem on bac, you have a tiny house edge and you'll
still lose.
Quote: JyBrd0403The end. Nice talking to you guys as usual. 1 billion trials produces a 500 million unit win and you say that's proof the game breaks even. You do know anyone can run a million trial simulation of the D'Alembert on a 50/50 game and get the results. Popcan already did this on another thread. The game wins, period. I'm not the one B.S.ing anyone here. Don't believe me, run the simulation yourself. Give the results, I'm now very interested in what kinda B.S. your going to pull on this one. Amazing.
I think the thing that you don't understand is that in a Casino you (the player) does not have unlimited money and even if you did, there is something called a table maximum. That little feature puts an end to ALL progressive betting systems. You can not recover from hitting the table max and losing. You can't go up one more unit at that point. System fails.
ZCore13
I tried to do some primitive math at work to figure out the flaw in Jy's modified system, but its just too much to do easily (at least for me)... But at least I see what he's getting at. I haven't seen any real mathematical reasons why he's wrong (while I'm sure he is)....
Quote: ewjones080I think a lot of you are missing the point of what Jy is saying. While I'm not defending him, I think there's still a shred of validity in what he's saying. Lets say a million trials doing d'lambert.. I "should" have 500,000 wins and 500,000 losses, producing 500,000 units ahead. There are two problems, one of which Jy hasn't addressed, which is what happens when early in your progression you get a bunch of wins at 1 unit... You can't go down.. This in effect raises where your bet will be when you hit your last bet by how many times you hit at min.. That's one thing that cuts your overall win.. Zeros increase even further.
I tried to do some primitive math at work to figure out the flaw in Jy's modified system, but its just too much to do easily (at least for me)... But at least I see what he's getting at. I haven't seen any real mathematical reasons why he's wrong (while I'm sure he is)....
One adjustment to address the multiple minimum win concern.....start at a unit count of greater than 1, that way you have room to move both up and down. for example: Start at 10 units, second bet will be either 9 or 11 depending on the outcome. If you are looking at an extremely large trial (like a million bets) the starting point would also have to be extremely large ( like 10,000 units).
Unfortunately, the Zero is still the death knell of a d'Alembert progression on roulette.
It occurs to me that a simulation of a 50/50 game would be incredibly easy to write. Here, I can throw most of it together in this window (hey, kids, count the bad habits):
#include <stdlib.h>
long trial(long session) {
long bet, total, i;
bet = 1;
total = 0;
for (i = 0; i < session; i++) {
if (rand() % 2) {
total += bet;
if (bet > 1) bet--;
} else {
total -= bet;
bet++;
}
}
return total;
}
I can't be arsed to do it, but I don't think a million hands would take long this way. (Of course, I'd have to think about overflow if I actually planned to run this, it being unlikely but not impossible that the progression should get into the tens of thousands, but screw it.)
Quote: thecesspitI could simulate this, but the OP wouldn't accept my results, as he doesn't accept 0.999... = 1, so no doubt he would suspect my programming to be flawed.
You'd think the bar would be set pretty low, since he accepted a sim showing a win of $500MM in 1B spins as valid. But you're still probably right.
Quote: rdw4potusYep, math is absolute. And 48.65/51.35 is not equal to 50/50. In roulette every bet you make has a 48.65% chance of winning. Your system attempts to overcome that by making additional bets with different dollar values. But they all also have a 48.65% chance of success. It's not hard to see how and why this will fail.
That is the only proof you need to guarantee that absolutely no betting system on roulette will work. That and the fact that each decision is independent from one another.
If you want to win at roulette, read American Roulette by Richard Marcus.
It is necessary to establish target win goals and proper bankroll size that you are willing to part with. Say a win goal of 200 units, bankroll of 1000 units. Start your progression at some set point that allows you to adjust both up and down to avoid the multiple minimum concern, say 10 units. Establish a positive stop point as your progression reduces, say 3 units.
Finally, you need to apply it to a bet that truly approximates 50/50 (or at least has a house edge of less than 2%).
The strength of the d'Alembert progression seems to lie in the intermediate run (20 - 500 bets). In the extreme short run, anything can and will happen (read: streaks). In the long run, the law of large numbers comes into play and will almost assuredly crush the progression.
If applied intelligently to the correct bet, with the appropriate stops in place, it is very reasonable to expect a success rate of around 90% or better, which can be extremely profitable. It won't make you independently wealthy, nor will it break the casino, but it can provide a rather steady flow of considerable income.
Quote: AlembertThe d'Alembert progression can be a very powerful and profitable tool if used effectively.
It is necessary to establish target win goals and proper bankroll size that you are willing to part with. Say a win goal of 200 units, bankroll of 1000 units. Start your progression at some set point that allows you to adjust both up and down to avoid the multiple minimum concern, say 10 units. Establish a positive stop point as your progression reduces, say 3 units.
Finally, you need to apply it to a bet that truly approximates 50/50 (or at least has a house edge of less than 2%).
The strength of the d'Alembert progression seems to lie in the intermediate run (20 - 500 bets). In the extreme short run, anything can and will happen (read: streaks). In the long run, the law of large numbers comes into play and will almost assuredly crush the progression.
If applied intelligently to the correct bet, with the appropriate stops in place, it is very reasonable to expect a success rate of around 90% or better, which can be extremely profitable. It won't make you independently wealthy, nor will it break the casino, but it can provide a rather steady flow of considerable income.
I think I will try this strategy. When I play I always bet on black. However every time I play roulette my evil twin from an alternate universe always shows up. He always only matches my bet (edit: same strategy) on red. He just tries to piss me off. Will we both be successful 90% of the time?
Quote: supermaxhdI think I will try this strategy. When I play I always bet on black. However every time I play roulette my evil twin from an alternate universe always shows up. He always only matches my bet on red. He just tries to piss me off. Will we both be successful 90% of the time?
Vs. roulette with a double Zero, NO. You will both lose. Furthermore, if his bets always match your bets, then NO he would not enjoy the same success rate as you.
However, if you each take opposite sides of a 50/50 wager and run independent progressions, then yes I would expect both could anticipate a success rate around 90%, AND you could both expect to end up winning during the same session(s).
Opposite bettors on a 50/50 proposition can both expect to be successful IF they are running independent progressions. Given the concept of the Evil Twin that always just matches bets, then the evil twin's progression is not independent and will not be moving in the appropriate directions to take advantage of the progression. Given the evil twin is matching your bets, then his loss would equal your profit or vice versa.
Example: Coin Flip: 10 trials H H H T T H T T T H
Head bettor progression: 10 win, 9 win, 8 win, 7 lose, 8 lose, 9 win, 8 lose, 9 lose, 10 lose, 11 win.
Result: Win 47 units, lose 42 units, net 5 units profit.
Tail bettor progression: 10 lose, 11 lose, 12 lose, 13 win, 12 win, 11 lose, 12 win, 11 win, 10 win, 9 lose.
Result: Win 58 units, lose 53 units, net 5 units profit.
Notice, only very occasionally do the bets ever actually match value.
Also, if you can not find one, and you are saying you can use this system on a game that is close to 50/50 and overcome the small house edge, please tell me about the game that has no table maximum, so that when a bad streak does come, the table max doesn't ruin your progression.
Thanks,
ZCore13
Quote: 24BingoSo somehow the progression stops working if someone else is opposite-betting you with a similar progression? How does that work?
It depends on who that other person is. Particularly if he is Statman oe EvenBob !
Quote: AlembertNotice, only very occasionally do the bets ever actually match value.
And that's the problem.
I see your system is a little different from mine, in that you start well above one unit, so that you can absorb quite a few wins in excess of losses. That's actually good, because under the assumption that your base bet, we'll call it x, is high enough that you never have to bet one unit (or even if you start betting the opposite in a similar positive progression), the math gets much simpler:
If your wins outstrip your losses, you will be ahead by (wins + (2*base bet - wins + losses - 1)*(wins-losses)/2).
If your losses outstrip your wins, you will be ahead if ((2*base bet + losses - wins + 1)*(losses-wins)/2) is less than your total number of losses, by the difference. You'll be behind by the difference if it's greater.
(And of course, if you have an equal number of wins and losses, that's what you'll be up by.)
Add it up and you'll find you break even.
Quote: Zcore13This is quite funny. But assuming it's true, please tell me about a wager in the casino that you can use this system on that is 50/50?
Also, if you can not find one, and you are saying you can use this system on a game that is close to 50/50 and overcome the small house edge, please tell me about the game that has no table maximum, so that when a bad streak does come, the table max doesn't ruin your progression.
Thanks,
ZCore13
Moving from theoretical to practical application, table Maximum is almost irrelevant. If you are running off a finite bankroll that approximates 2 standard deviations your bets will never approach any tables maximum. The highest bet I have ever had to place (on my way to a losing session) was 350.
I have tinkered with playing against Pai Gow Poker, Bacarat, Craps
