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3 point molly meaning you make come bets until you have 3 points covered, including the passline bet. No odds bets made
Quote: WizardCan you remind me how the Three Point Molly works?
i dont know why but i call it 3 Point Betty.
What I do for this strategy:
Passline,max odds
2 Come bets, max odds.
stop and wait till one of my 3 #s hit then make another bet.
rinse/repeat
For 3 point molly, you will always have between $100 and $300 on the table. $100 on passline at all times, $0 or $100 on come, and $0, $100 or $200 on points. If there is $200 on points then there will be zero on come.Quote: ChumpChangeThe average bet would be $100 in the question above.
So the average total wager (total money on the table) per roll is between $100 and $300.
Incidentally, I don’t have the answer. I think I can calculate it but first I’m checking if someone already has
I think your question is how much, on average, do you have in action on any given roll rather than bets.
When the shooter is coming out (for the first time) obviously one bet.
When the shooter has had a point established then two or three bets. Initially there must be two bets, either because the point is made and the come goes to that point, or the come bet is coming out. Once the first come bet has established, then it will be between 1 and 3 bets depending on whether a 7 is on the Come out roll or a different point is established for either a Come or Pass bet.
(0-0) Shooter coming out: no points established to (0-0) = One bet - shooter rolls a 2 3 7 11 or 12 (essentially the shooter just rolls again).
(0-0) Shooter coming out: no points established to (1-0) = One bet - shooter rolls their point (shooter will have now made their point, so Come bets can be made).
(1-0) Shooter rolling: no come bets established to (0-0) = Two bets- shooter 7s out.
(1-0) Shooter rolling: no come bets established to (0-1) = Two bets - shooter makes their point.
(1-0) Shooter rolling: no come bets established to (1-0) = Two bets - shooter rolls a 2 3 11 or 12.
(1-0) Shooter rolling: no come bets established to (1-1) = Two bets - shooter rolls a different point.
(1-1) Shooter rolliing: one come bet established to (0-0) = Three bets - shooter 7s out.
(1-1) Shooter rolliing: one come bet established to (0-2) = Three bets - shooter makes their point.
(1-1) Shooter rolliing: one come bet established to (1-1) = Three bets - shooter makes a previous come point.
(1-1) Shooter rolliing: one come bet established to (1-1) = Three bets - shooter rolls 2 3 11 or 12.
(1-1) Shooter rolliing: one come bet established to (1-2) = Three bets - shooter makes a different point for the Come bet.
(1-2) Shooter rolliing: two come bets established to (0-0) = No new bets, three on table - shooter 7s out.
(1-2) Shooter rolliing: two come bets established to (0-2) = No new bets, three on table - shooter makes their point.
(1-2) Shooter rolliing: two come bets established to (1-1) = No new bets, three on table - shooter makes a Come point.
(1-2) Shooter rolliing: two come bets established to (1-2) = No new bets, three on table - shooter rolls 2 3 11 12 or non-point number.
(0-1) Shooter coming out: one come bet established to (0-0) = Two bets - shooter rolls a 7 on Come out.
(0-1) Shooter comingout: one come bet established to (0-1) = Two bets - shooter rolls a 2 3 11 or 12 on Come out.
(0-1) Shooter coming out: one come bet established to (1-0) = Two bets - shooter rolls the Come point.
(0-1) Shooter coming out: one come bet established to (1-1) = Two bets - shooter rolls a point different from the Come point.
(0-2) Shooter coming out: two come bets established to (0-0) = Three bets - shooter rolls a 7 on Come out.
(0-2) Shooter coming out: two come bets established to (0-2) = Three bets - shooter rolls a 2 3 11 or 12 on Come out.
(0-2) Shooter coming out: two come bets established to (1-1) = Three bets - shooter rolls one of the Come points.
(0-2) Shooter coming out: two come bets established to (1-2) = Three bets - shooter rolls a point different from either Come point.
Is anyone interested in solving this before I post the answer ?
Quote: Ace2I found the solution, which wasn’t that difficult after I thought it through.
Is anyone interested in solving this before I post the answer ?
I’d be interested in the answer for an always coming strategy with full odds at a 3/4/5 table.
That one is simple. Playing just the pass line, you have a resolution every 557/165 =~ 3.38 rolls. Always coming, every roll is a comeout roll, so your total wagers increase by a factor of 3.38.Quote: unJonI’d be interested in the answer for an always coming strategy with full odds at a 3/4/5 table.
For example, over a million rolls you will expect to have 1,000,000 * 165/557 = 296,230 bets resolved playing pass line only. Always coming, you’ll have 1 million resolutions. Always coming is effectively a “7 point molly” since you have a bet on the come even when all 6 points are covered
A “2 point molly” is also quite easy to calculate. 3 to 6 are a bit trickier
On average, a shooter takes 1.5 rolls to establish a point and then 6 rolls to seven out. If playing a 2-point molly, he’d always have 2 bets on the table (pass + come or point) except for come out rolls. So wagers increase by 6 / (1.5 + 6) = 80%.
For a 3-point molly, we need to calculate how often the shooter, after establishing a point, will roll (not necessarily make) at least one additional point before sevening out. If, for example, he establishes a point of 5 or 9, then there is a (24-4)/(30-4) = 10/13 chance he will roll at least one additional point during his turn. The probabilities for points 4&10 and 6&8 are 7/9 and 19/25 respectively. Therefore, that weighed probability is 1/4 * 7/9 + 1/3 * 10/13 + 5/12 * 19/25 = 449/585.
As previously shown, 6 of 7.5 bets will make it to a 2-point molly. Of those 6, 449/585 will make it to a 3-point molly. Therefore, ((449/585 + 1) * 6) / 7.5 = 4,136 / 2,925 =~ 141% increase.
This should say:Quote: Ace2On average, a shooter takes 1.5 rolls to establish a point and then 6 rolls to seven out.
On average, a shooter takes 1.5 rolls to establish a point and then 6 rolls to roll a seven.
However, once come bets have been made then any seven will make them lose, whether it be a seven out or a come out roll.
All place bets have higher house edge than a come bet. And you can’t put free odds on place bet. I’d never make a place betQuote: ChumpChangeI would just replace a winning Come bet with a place bet.
I may have just a dozen greens in my tray for odds bets on only the PL / Come 6 or 8, and take no odds on other numbers.
Could someone please do the math for a 4-point Molly ?Quote: Ace2If anyone is interested, playing a 3-point molly will increase your total wagers by an average of 4,136/2,925, or about 141%. So a 3-point molly is like playing 2.41 passline bets concurrently, in terms of total results. This is useful information to me since I believe that proper bet size & bet frequency are key to enjoying any game.
On average, a shooter takes 1.5 rolls to establish a point and then 6 rolls to seven out. If playing a 2-point molly, he’d always have 2 bets on the table (pass + come or point) except for come out rolls. So wagers increase by 6 / (1.5 + 6) = 80%.
For a 3-point molly, we need to calculate how often the shooter, after establishing a point, will roll (not necessarily make) at least one additional point before sevening out. If, for example, he establishes a point of 5 or 9, then there is a (24-4)/(30-4) = 10/13 chance he will roll at least one additional point during his turn. The probabilities for points 4&10 and 6&8 are 7/9 and 19/25 respectively. Therefore, that weighed probability is 1/4 * 7/9 + 1/3 * 10/13 + 5/12 * 19/25 = 449/585.
As previously shown, 6 of 7.5 bets will make it to a 2-point molly. Of those 6, 449/585 will make it to a 3-point molly. Therefore, ((449/585 + 1) * 6) / 7.5 = 4,136 / 2,925 =~ 141% increase.
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