EV for a D'Alembert
Quote: JyBrd0403Ahh, thank you ThatDonGuy, you just showed that the D'Alembert wins on a 50/50 game. Now feel free to take that back and state with 1000% certainty that there is no Betting System that Works on an HE or 50/50 game. The crap you have to go through to get to the obvious, huh.
Why would I say that? In fact, didn't I (and a few others) say that Martingale also always wins on a 50/50 game if you don't have to worry about things like time and bankroll? The same applies to D'Alembert.
I think I have been able to calculate the EV for an infinite D'Alembert under the given stop conditions.
It appears that there are (2n)! / (n! (n+1)!) ways to have n wins and n losses such that the first result is a loss and losses > wins until the (2n)th bet.
If p is the probability of winning a particular bet, then the stop conditions are:
(a) the first bet is a win (probability p, result +1)
(b) the first two results are LW (probability (1-p) p, result +2)
(c) the first two results are LL, then n Ls and Ws that do not result in the number of Ws = the number of Ls (including the first 2) at any point, then WW (probability (2n)! / (n! (n+1)!) * pn+2 * (1-p)n+2, result (n+2))
For a 50/50 game, this appears to diverge
For double-zero roulette, it appears to converge at around 5.210526
For a 25% game, it appears to converge at 0.625
Remember the key word - infinite. A finite D'Alembert has expected value zero for 50/50 and negative for a less than 50/50 game.
Quote: ThatDonGuyWhy would I say that? In fact, didn't I (and a few others) say that Martingale also always wins on a 50/50 game if you don't have to worry about things like time and bankroll? The same applies to D'Alembert.
For a 50/50 game, this appears to diverge
For double-zero roulette, it appears to converge at around 5.210526
For a 25% game, it appears to converge at 0.625
Remember the key word - infinite. A finite D'Alembert has expected value zero for 50/50 and negative for a less than 50/50 game.
You're getting close to understanding it Don. If we play until it "diverges" that would be the finite. You just can't determine when that would be, stinking probabilities :)
Now the funny thing is Flat betting would break even when it "diverges" and the Marty would, and maybe this is what is giving the problems, the Marty would also just break even when it "diverges". So you've got a finite timeline, and a winning system right there in front of you.
So, like I said, if you can say a 50/50 flat bettor will break even, then you can say a D'Alembert player will win. Simple as that. Otherwise, you have to say that a Flat bettor would lose and never break even, and that's just silly.
Quote: JyBrd0403You're getting close to understanding it Don. If we play until it "diverges" that would be the finite.
If the result is some finite value, then it converges. Diverges means that it approaches infinity as the number of bets increases.
Quote: JyBrd0403So, like I said, if you can say a 50/50 flat bettor will break even, then you can say a D'Alembert player will win. Simple as that.
Eventually. Just one problem with that; both you and the casino need to be immortal. Otherwise, you're not taking into account the fact that you can end on a losing streak long enough to cancel out the expected wins. I understand this just fine, thank you very much.
Quote: ThatDonGuyIf the result is some finite value, then it converges. Diverges means that it approaches infinity as the number of bets increases.
Eventually. Just one problem with that; both you and the casino need to be immortal. Otherwise, you're not taking into account the fact that you can end on a losing streak long enough to cancel out the expected wins. I understand this just fine, thank you very much.
Good, then you UNDERSTAND that EVENTUALLY, flat bettors break even, marty players break even, and D'Alembert Players win.
Also, you're now going back on the math you did. The game doesn't end on a losing streak, so you can't end on a losing streak. The second point is that the D'Alembert doesn't have to go all the way back to even to win. You can be betting 100 units and still be positive on your bankroll. So, maybe, it would be easier to understand if you stop when you have won money, as opposed to breaking even on wins and losses. So, if you're point was flat betting you go up +100 units and then - 100 units that you're even on wins/losses, and the D'alembert would be betting 100 units even though the wins/losses are even. I still don't think that would matter, but you can just play until the bankroll is positive that way you don't have to break even, or go back to +100 to stop the progression. I don't know why that would matter since the longer you play, according to you guys, the higher you go up. Eventually, you'll be able to go right past +100. But, if that's giving you trouble, this stopping on a losing streak, you can just stop when you're bankroll is positive.
But, the whole point, which you seemed to agree with is that the D'Alembert eventually wins, flat betting and marty break even.
Quote: JyBrd0403Quote: ThatDonGuyIf the result is some finite value, then it converges. Diverges means that it approaches infinity as the number of bets increases.
Eventually. Just one problem with that; both you and the casino need to be immortal. Otherwise, you're not taking into account the fact that you can end on a losing streak long enough to cancel out the expected wins. I understand this just fine, thank you very much.
Good, then you UNDERSTAND that EVENTUALLY, flat bettors break even, marty players break even, and D'Alembert Players win.
Just as EVENTUALLY, six monkeys in front of six typewriters can write the complete works of Shakespeare.
Quote: JyBrd0403Also, you're now going back on the math you did. The game doesn't end on a losing streak, so you can't end on a losing streak.
It can if you drop dead during one, or the casino shuts down during one.
Quote: JyBrd0403But, the whole point, which you seemed to agree with is that the D'Alembert eventually wins, flat betting and marty break even.
Wrong. Marty "eventually wins" just as much as D'Alembert.
And the phrase "D'Alembert eventually wins" is meaningless when you realize that "eventually" includes "millions of years from now."
Quote: JyBrd0403Good, then you UNDERSTAND that EVENTUALLY, flat bettors break even, marty players break even, and D'Alembert Players win.
Where do you get that marty players break even? Eventually, with infinite time and bankroll, they win, too. One unit per win, with no regard to losses, just like D'Alembert players. In fact, that's what they're ahead by after every single win, unlike D'Alembert players, who only get there by resetting their progression. In fact, also unlike D'Alembert players, the expected time until their progression ends is finite. In their case, it's only the expected nadir of their progressions that's infinite.
Quote: ThatDonGuyEventually. Just one problem with that; both you and the casino need to be immortal. Otherwise, you're not taking into account the fact that you can end on a losing streak long enough to cancel out the expected wins. I understand this just fine, thank you very much.
Don't forget that the game has to have no edge, or the probability of an eternal progression converges to a nonzero value.
Quote: 24BingoWhere do you get that marty players break even? Eventually, with infinite time and bankroll, they win, too. One unit per win, with no regard to losses, just like D'Alembert players. In fact, that's what they're ahead by after every single win, unlike D'Alembert players, who only get there by resetting their progression. In fact, also unlike D'Alembert players, the expected time until their progression ends is finite. In their case, it's only the expected nadir of their progressions that's infinite.
Right, I'm trying to explain that the D'Alembert will produce a profitable game. By this, I mean that eventually, you will be playing with the houses money. All the losses you sustain will be covered by the money you won, and you will just keep adding to your profit. How is this possible? Because the profits will eventually outpace the loss limits. You may be betting 5000 units, but still playing on the houses money after 100 million trials. My point is that not only will you be winning at that point, but you have the benefit of being assured that you will come back to a 1 unit bet eventually. This is what is apparently being disputed, and maybe I didn't explain it well enough, but the point is that the D'Alembert profits will outpace the loss limits.
One of my ways of showing this is with the Marty. The marty will always break even. The wins will equal the losses. So, while you win 100 million dollars, a losing streak hits and you lose 100 million dollars. Then you have to pony up a 200 million dollars from you're own money and make a 200 million dollar bet. You never are in the Black you always have to pony up your own money. I can show you quickly that the wins and losses are equal for the marty. You lose 4 in a row once every 30 trials, that's 1-2-4-8 = 15 unit loss. In 30 trials you get 15 wins. 15 losses = 15 wins. This holds true for all cases. 10 in a row 18 in a row 30 in a row etc. You break even, and you are never in the black, you always have to pony up you're own money to cover. So your bankroll goes up +100 million then down -100 million.
The D'Alembert Bankroll should be in the Black at some point, going up let's say +100 million down let's say to +85 million, staying in the black. The reason this is better than the Marty is obvious. Losing 18 in a row in a million trials doesn't hurt the D'Alembert at all! The losses in a row don't push the game down like it does for a marty. If you aren't in the Black after a million trials, you know what the big worry is, you would now lose 19 in a row. The D'Alembert should easily outpace the loss limits on a 50/50 game.
Here's the problem, the math these guys are giving are saying that the losses somehow magically keep up with the wins the D'Alembert produces. I haven't a clue how they think that's possible. The only thing I can see is they are basing the math on streaks ( losses in a row). The losses in a row, can grow the more trials you have, but there's no reason why the total losses would grow any higher than the losses in a row. I mean if the 50/50 game is on the positive side of wins and losses you can play a million trials and only go down -18 units. Again, my only point is that the D'Alembert wins should easily outpace the losses eventually, because the losses in a row are nothing to the D'Alembert. And, that means that the player is winning, or always in the BLACK, always playing on the houses money eventually. I can see no other possibility. This what will eventually happen with the D'Alembert, the Marty will always break even, and the Player will always be playing with his own money.
To try to clarify, the total losses should reach a limit over the losses in a row. Let's say you're at -5000 at some point. That -5000 is going to be the limit for Billions of trials. It was the worst run possible for the 50/50 game, and nothing will push that further down, because the losses in a row are only at 40 in billions of trials. So, there should be a definitive loss limit, that the D'alembert should eventually, easily, overcome.
That's how the casino's can make money, Buzz. There's a LIMIT on how much a player can win on a 50/50 or 49% or 47.5% game. Also, a LIMIT on how much the Casino can lose on a 51%, 52.5% game. It's not Infinity.
Easy Babs, finger off that trigger. Lots of stores on that show !
