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Ace2
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May 5th, 2020 at 11:23:01 AM permalink
If you like to play a 3 point molly, flat betting $100 on pass/come, what is your average total wager per roll?

3 point molly meaning you make come bets until you have 3 points covered, including the passline bet. No odds bets made
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Wizard
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May 5th, 2020 at 1:41:54 PM permalink
Can you remind me how the Three Point Molly works?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
100xOdds
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May 5th, 2020 at 2:45:20 PM permalink
Quote: Wizard

Can 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
Craps is paradise (Pair of dice). Lets hear it for the SpeedCount Mathletes :)
ChumpChange
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May 5th, 2020 at 6:37:09 PM permalink
My turns are either 2-3 rolls or 20 to 30 rolls. Can't really give you an average that means anything. The average bet would be $100 in the question above.
Ace2
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May 5th, 2020 at 6:57:01 PM permalink
Quote: ChumpChange

The average bet would be $100 in the question above.

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.

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
Last edited by: Ace2 on May 5, 2020
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charliepatrick
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May 5th, 2020 at 7:45:24 PM permalink
I haven't worked out the answer but I sometimes play that way, keeping placing bets until I've got three different numbers going for me. During that time some bets may win or lose (or the shooter 7's out before I get to three numbers). So I guess it's one of those state diagrams.

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.
Ace2
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May 7th, 2020 at 4:55:50 PM permalink
I found the solution, which wasn’t that difficult after I thought it through.

Is anyone interested in solving this before I post the answer ?
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unJon
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May 7th, 2020 at 6:04:44 PM permalink
Quote: Ace2

I 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.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Ace2
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May 7th, 2020 at 6:43:57 PM permalink
Quote: unJon

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.

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
Last edited by: Ace2 on May 7, 2020
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Ace2
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May 13th, 2020 at 10:23:13 AM permalink
If 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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Ace2
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June 7th, 2020 at 10:55:19 AM permalink
Quote: Ace2

On average, a shooter takes 1.5 rolls to establish a point and then 6 rolls to seven out.

This should say:

On average, a shooter takes 1.5 rolls to establish a point and then 6 rolls to roll a seven.
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ChumpChange
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June 7th, 2020 at 11:09:41 AM permalink
Wouldn't that be 1.5 rolls to establish a point, 6 rolls of anything but a 7, then 1 roll of a 7-out, for 8.5 rolls per shooter?
Ace2
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June 7th, 2020 at 12:21:07 PM permalink
Average roll length for a shooter is 1671/196 =~ 8.53

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.
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ChumpChange
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June 7th, 2020 at 12:56:54 PM permalink
I would just replace a winning Come bet with a place bet.
Ace2
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June 7th, 2020 at 1:07:42 PM permalink
Quote: ChumpChange

I would just replace a winning Come bet with a place bet.

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 bet
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ChumpChange
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June 7th, 2020 at 1:19:22 PM permalink
If I have $6 on the Come and $25 on Odds, on the 6 or 8, I'd win $36. So I could put $36 on the PB 6 or 8 after the Come 6 or 8 wins. After each win on the PB 6 or 8 I could press by $6. I could either not make another Come bet until a place bet hits, or put one up right away. If it comes up an existing PB, I can move the PB to the opposite number or take it down and use it for odds. I may have to put bet the line a little so I can match the odds progression I'm making from the PB.

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.
Ace2
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November 16th, 2021 at 9:26:43 PM permalink
Quote: Ace2

If 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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Could someone please do the math for a 4-point Molly ?
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