(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.
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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.
very well wordedQuote: 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.
"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."
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)
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)
You need a large enough sample to ensure the result you get isn't just luck.Quote: 7craps
why is a larger simulation more important than a shorter one?...
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.
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.
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.
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?
I call heads.
Lose three in a row (-3) (Stop)
Win 20 out of 30 (+10) (Stop because I lost three in a row)
Win 6 out of 10 (+4) (Stop because I lost three in a row)
Lose three in a row to start (-3)
Lose 3 out of 6 (-3)
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.