MichaelBluejay
MichaelBluejay
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September 7th, 2019 at 12:09:18 PM permalink
Jordan1, here's my understanding of your system. Is this right?

(1) Bet $5 each on Pass / Don't Pass.

(2) Lay Odds on the Pass point ($310 on 4/10, $210 on 5/9, $102 on 6/8).

(3) If no point has hit yet, then bet $5 each on Come/Don't Come until 3 #'s established (1 Pass point + 2 Come points).

(4) Lay Odds on each new Come point set, using the same schedule as for the Pass point.

(5) If a # hits, take down the remaining Odds bets. Don't make any new Come/Don't Come bets.

(6) Repeat the whole sequence on each new Come-out roll.

I think that's what you described, but even if so, what happens when Come Point is hit on a Come-Out Roll? Do we take odds down on the other Come Point if there is one? Do we lay odds on the new Pass Point that has just been set? Do we try to establish 3 points during this Come-Out round, or wait for the next come-out round?

BTW, here's how you could figure the average loss for a Pass or Don't Pass on billion come-out rolls:

1B x $5 wager x 1.41%

It's that simple. For betting Pass / Don't Pass together, double the loss, since you're doubling your bet. The Odds bets have no house edge, so we ignore them in the calculations. Yes, the Odds bets dilute the house edge %, but they don't dilute the overall loss in dollars.
Last edited by: MichaelBluejay on Sep 8, 2019
MichaelBluejay
MichaelBluejay
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Thanks for this post from:
7craps
September 8th, 2019 at 11:53:14 AM permalink
Okay, I took my best guess at the system rules (see questions noted above), and finished the simulation. Here it is.

Jordan1, you asked for a billion rolls. That many would choke some web browsers, so I did 2 million come-outs, which is nearly 7 million rolls total. That returns a player loss, and trust me, it won't get better by increasing from 7 million rolls to a billion.
7craps
7craps
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Thanks for this post from:
MichaelBluejay
September 8th, 2019 at 7:59:25 PM permalink
Quote: MichaelBluejay

Okay, I took my best guess at the system rules (see questions noted above), and finished the simulation. Here it is.

Jordan1, you asked for a billion rolls. That many would choke some web browsers, so I did 2 million come-outs, which is nearly 7 million rolls total.

very well worded
"the system hasn't even put a dent in the house edge: It's still the expected ~-1.39% for the regular bets, and ~0% on the Odds bets."
A++

I just looked and extended that sim to 100 million rounds (about 337 million rolls)
and got about the same results

photo below (I changed a few things in the code on my machine only)

winsome johnny (not Win some johnny)
Boz
Boz
Joined: Sep 22, 2011
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September 8th, 2019 at 8:23:39 PM permalink
Solved @https://easy.vegas/
7craps
7craps
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September 9th, 2019 at 7:37:53 AM permalink
why is a larger simulation more important than a shorter one?
some would think one way and others another
to some, a simulation of any size is proof their system works

the latest sim is for only 1000 rounds (a large sim to many)

and some sessions easily do show a profit and that easily leads to one believing that their system is a winner to whomever plays it correctly. (a false belief)

photo
winsome johnny (not Win some johnny)
charliepatrick
charliepatrick
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September 9th, 2019 at 7:52:17 AM permalink
Quote: 7craps

why is a larger simulation more important than a shorter one?...

You need a large enough sample to ensure the result you get isn't just luck.

For instance I could try suggesting that betting Red always wins at roulette. If I only ran a simulation of 1000 spins (for simplicity the single zero causes a loss) the average #reds is 486.5 (np) but the sd is 15.8 (SQRT(npq)) - there's a reasonable chance of being ahead by betting red. However if you ran simulations of 1m then the average is 486k but the sd is 499.8, so there's not a ghostly chance of any simulation coming out ahead.

Very similar logic applies for Craps due to it's either win 1 or lose 1 property.
JORDAN1
JORDAN1
Joined: Sep 1, 2019
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September 10th, 2019 at 7:00:17 PM permalink
Hello. I first want to thank you for all of your mathematical evaluations for my "system", but there is one thing that it seems you might have figured incorrectly, and it might be my fault for not being clear. I am "laying odds" not "placing odds". Laying odds is betting a 7 will show before the pass or come number that has been established is made. "Placing odds is betting a number will be hit before the dreaded 7 shows. If I was placing odds, and established three numbers, the hateful 7 knocks down all my bets, which is pretty bad. My thinking is that I always have the 7 working for me, which is the most common number to be rolled on a pair of dice. So when I get one two or three "don't"s set (with the laying of odds) a seven will be always be my hero! If at any time one of my numbers is rolled, by any particular shooter, I take all the "laying odds bets" down and wait for another shooter (or if I am the only one at the table, when I seven out) and start the process over.

According to the Wizards of Odds website a "single" laying or placing of odds bets the formula is (3/220)/(1+X). X is the amount of multiples the lay bet is compared to the original line bet. For example of 100 times odds. $5 don't pass, $500 odds = (3/220)/(1+100) this comes to a very small house advantage of .000135. My bets are not at 100 times odds, but let's say I have the funds to make such bets. I was just wondering if anything improves by setting multiple don'ts instead of just the one don't and adjusting to any loss by taking down odds bets that I have not lost. Probably not. Let's say that I try to establish even more lay numbers than 3. What changes on getting possibly 4, 5 or an amazing 6 numbers, from any particular shooter (or whatever one would call the roller's time throwing dice on a come out roll) and lay 100 times odds against all of them. Does that hurt or help my case or does it all just stay at the .000135? Or would I be doing a bunch more harm due to having 26 ways to lose and 6 ways to win if I did manage to establish all six numbers, even though if a seven shows I win 6 bets at once?

Dunno. Up to three works for me. I know gambling is a fool's gambit. I have been working on ways to win for years, and it is fun to think about, but yes I understand the house always has the edge.

Thanks again for everyone's input.
MichaelBluejay
MichaelBluejay
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September 10th, 2019 at 8:41:11 PM permalink
Hi Jordan. First, please know that I do understand craps. My How to Play Craps article has been on my site for nearly 20 years, and in it I explain the difference between taking odds after Pass/Come bets vs. laying them after Don't Pass/Don't Come bets. My knowledge is also how I knew what the exact results of the simulation would be even before I programmed and ran it.

So, rest assured that my simulation does indeed lay odds, not take them. You can see that in the play-by-play. (Not that it matters. The house take % is the same whether laying or taking.)

As you saw from the Wizard's site, making big odds bets dilutes the house edge, but doesn't erase it. Think of it this way: The average of zero plus any set of negative numbers is always negative. For example, the average of 0, -5, -3, -8, -9, -7, -2 is negative. The odds bet has no house edge (0), and every other bet in craps does have a house edge (negative), so whatever combination of bets you make, however and whenever you make them and how much you bet, your expectation will always be negative. Whatever modifications you think you might make to your system, it's always gonna be the average of zero plus negative numbers, which is why it can't ever win.

Rather than looking at the combined house edge, split the bets up. Your expectation on Pass / Don't Pass / Don't Come together is -1.39%, and your expectation on the Odds bets is 0%. And that's exactly what the simulation shows, as expected.
ChumpChange
ChumpChange
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September 10th, 2019 at 9:31:31 PM permalink
Every Passline point number loses on average. Odds bets can win or lose, and maybe make up for the losing average of Passline point numbers. Probably need triple odds minimum.
JORDAN1
JORDAN1
Joined: Sep 1, 2019
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September 11th, 2019 at 5:38:00 PM permalink
I saw your computer simulation, and it is close to what I do, but still there is one variant that, to me seems to be part of the program, which probably doesn't matter. If a shooter makes any of my don't numbers, I stop betting against that shooter, take all of the odds on all the numbers I have established don't bets with, and wait until another shooter turn arrives. The shooter that knocked down one of my don't numbers might roll for 30 minutes or more, but I don't dare make another bet. Your simulation seems to think that I going to bet on the same shooter if he makes a point, and try to start betting don't numbers again once he made it. Again it probably doesn't matter though. I guess the $5 I lose every 36 rolls or so is what cost me so much in the long run. If odds bets have no house advantage, then in a million rolls the eventual result would be a tie, but $5 times 1,000,000/36 is where the problem lies with my system.

Another question would be just a single coin toss bet. No house advantage, just a 50/50 bet. Is there a way to come out ahead on such a game? I am not playing for a million flips in a row, but something to the effect of sessions. Everyone hates sessions, but don't they deserve some merit, especially in a game of coin toss? What would happen if you played a coin toss game with four simple rules.

1 -Whenever you lose three coin tosses in a row, stop a session. (Could be a hundred flips occur but you only stop when you lose three in a row)

2 - If you lose three in a row starting out a session, stop the session.

3- Lose a total of three tosses (like this -1, 0, -1, -2, -1, -2 ,-3) stop the session.

4 - If you get down to -2 and get the losses back to 0 (even) stop the session.

Is there any way that one could turn this into a winning game?

Example:
I call heads.

First session:
Lose three in a row (-3) (Stop)
Second session:
Win 20 out of 30 (+10) (Stop because I lost three in a row)
Third session
Win 6 out of 10 (+4) (Stop because I lost three in a row)
Forth session:
Lose three in a row to start (-3)
Fifth Session:
Lose 3 out of 6 (-3)
Sixth Session
Lose the first two flips (-2) but manage to get back to even (0). Stop the session.

These sessions would add up to 14+/-9 for a net of +5.

The reason behind my question if the house odds can get down to .0004, isn't the game as close to a coin toss as possible? So if it is, then couldn't sessions be used to possibly obtain a possible (ever so slightly) advantage? That simulator you could be used to figure this (with max lay odds on don'ts). Probably wishful thinking, but it is thinking....

Thanks again for all of your help. Sorry to be so persistent.

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