EV for a D'Alembert

Quote: JyBrd0403

Yes, but do you know of a formula that calculates the EV on the D'Alembert?

Yes: house edge = 2/38.

Quote: Wizard

Yes: house edge = 2/38.

I'll take that as a no. I don't get it, I've asked around and can't find any formula for it, weird.

Quote: JyBrd0403

Quote: 24Bingo

By that strategy, technically, the EV will be positive, but your expected time at the table will be infinite.

The time at the table would be finite, under those rules. You'd reach equal number of wins and losses well before infinity.

Two problems with that statement.

First, "well before infinity" is pretty much meaningless, given that infinity is very, very, very, veryveryvery, very, did I mention very, very large. "You'd reach equal number of wins and losses" may take years.

Second, you would also reach a loss limit well before infinity.

For limited plays, I can show that EV is zero. For example, if you stop after three plays, there are 8 possible results:
WWW: +3
WWL: +1
WLW: +2 (+1, -1, +2)
WLL: -2 (+1, -1, -2)
LWW: +2 (-1,+2, +1)
LWL: 0 (-1, +2, -1)
LLW: 0 (-1, -2, +3)
LLL: -6 (-1, -2, -3)
In a 50/50 game, each is equally likely, so the EV = 1/8 x (the sum of the 8 values) = 0.

I am trying to see if I can work out an EV given the conditions (win the first game or eventually reach wins = losses) over certain loss limits (as well as one with infinite bankroll and time), but there are quite a few states to consider. For example, if you lose your first 2 bets, you are at 2 more losses than wins and bankroll -3, but a win and loss (in either order) results in still having 2 more losses than wins but bankroll -2.

Quote: ThatDonGuy

Quote: JyBrd0403

Quote: 24Bingo

By that strategy, technically, the EV will be positive, but your expected time at the table will be infinite.

The time at the table would be finite, under those rules. You'd reach equal number of wins and losses well before infinity.

Two problems with that statement.

First, "well before infinity" is pretty much meaningless, given that infinity is very, very, very, veryveryvery, very, did I mention very, very large. "You'd reach equal number of wins and losses" may take years.

Second, you would also reach a loss limit well before infinity.

For limited plays, I can show that EV is zero. For example, if you stop after three plays, there are 8 possible results:
WWW: +3
WWL: +1
WLW: +2 (+1, -1, +2)
WLL: -2 (+1, -1, -2)
LWW: +2 (-1,+2, +1)
LWL: 0 (-1, +2, -1)
LLW: 0 (-1, -2, +3)
LLL: -6 (-1, -2, -3)
In a 50/50 game, each is equally likely, so the EV = 1/8 x (the sum of the 8 values) = 0.

I am trying to see if I can work out an EV given the conditions (win the first game or eventually reach wins = losses) over certain loss limits (as well as one with infinite bankroll and time), but there are quite a few states to consider. For example, if you lose your first 2 bets, you are at 2 more losses than wins and bankroll -3, but a win and loss (in either order) results in still having 2 more losses than wins but bankroll -2.

Let me try to explain, what I'm trying to show, that may help. My only point, is that the D'alembert wins, if it takes years for it to win it take years, but I hope you also realize that after years of waiting, the payoff will be HUGE. You win 1 unit per win so after a million trials, you'd win 500,000 units. Point being, you'd end up making the same amount of money, whether it takes years to complete 1 sequence or you complete 100 sequences daily.

As for bankroll, that's always a question, even on a +HE game you have to worry about the bankroll. The thing with the D'Alembert is that you gain 1 unit for every win, so you're adding to your bankroll all the time. So, with the D'alembert the only thing that really hurts is sharp downturns, you need a bunch of trials or a sharp upturn to make up for the losses. In reality, though, if you know you will eventually win, you can always go back and put more money together if you lose your bankroll while playing. Just keep adding to the bankroll, if you know you will eventually win and turn a profit.

As you can probably tell, I try to do most of this in my head, so I just try to see the big picture (like you'll have equal wins/losses before reaching Infinity). Doing all the technical stuff is just not as easy for me.

Quote: JyBrd0403

Quote: 24Bingo

By that strategy, technically, the EV will be positive, but your expected time at the table will be infinite.

The time at the table would be finite, under those rules. You'd reach equal number of wins and losses well before infinity.

In one sense, that's true. In a 50/50 game (although NOT one with an edge), you have probability 1 of ultimately coming to that point. And, of course, hand #googolplex is just as long "before infinity" as the fourth.

However, just like the return of the St. Petersburg game, that doesn't mean the expected number of hands you'll need is finite.

First, the easy part: half of all progressions will be over in one hand, and another quarter will be over in two. If we were to keep a running total, it would be at 1 hand, but that doesn't really matter for reasons you'll soon see.

It makes sense to look at the remaining quarter of progressions as pairs of trials, with the first pair fixed at LL. So starting at -1, with each step bringing us up 1 (.25), down 1 (.25) or standing still, what's the expected number of steps it'll take us to get to 0? Let's call it x (so EV = (2x+2)/4 + 1).

To "answer" the question, we'll do a bit of recursion. If you drop down, the expected number of steps it'll take you to get back to where you were for the first time will be the same as the expected number of steps it'll take you to then get back up to the top. Likewise, if you stay where you are, the expected number of steps will increase by 1. So let x be the expected number of steps.

x = .25*1 + .5*(x+1) + .25*(x+x)
x = x + .75

In short, there can be no finite x.

I'm sure I could come up with a less hackish demonstration that x blows up, but it's 1 AM and I've been drinking.

In fact, it occurs to me that your expected winnings, if you could follow the strategy you describe, would be not only positive, but infinite. However, you can't follow such a strategy, because no matter how trusting your credit card company is, sooner or later you'll keel over, and whether it takes a hundred years, or a googol, or even 10^10^10^10^10^1.1 years, that's still finite, and any finite limit of time or funds is -EV.

I don't think you understand how math works, OP.

Quote: RS

I don't think you understand how math works, OP.

What are you questioning?

Quote: JyBrd0403

What are you questioning?

Your sanity.

I guess the real question is what game are you trying to beat and how? Perhaps someone will then show you how or why its not possible. Not that it would help you anyways but do you ave a game that's a true 50 /50?

I just don't know why people believe they are smarter then EVERYONE. Years and years of very intelligent people have been analyzing table games and systems.
Nothing works unless you have something more then just some betting system. IE card counting

Quote: JyBrd0403

What are you questioning?

In a perfect world, the D'alem works perfectly. In a
real casino the pit smiles while they take your money.
Maybe they'll comp you a pack of smokes for
your trouble.

Quote: EvenBob

In a perfect world, the D'alem works perfectly. In a
real casino the pit smiles while they take your money.
Maybe they'll comp you a pack of smokes for
your trouble.

In a perfect world no one would be looking for easy money betting schemes.

BOB did you see the new Episode of Heel on wheels yet? I started it but feel asleep .