Mathematically 3point Betty and continuous Come is the same?
April 28th, 2012 at 6:58:48 AM
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From this thread:
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continuous passline/come w/max odds should be self explanatory.
3point betty: passline + 2 come bets w/max odds.
-after hitting 3 points, i do a 3rd come bet to have 4 #s.
-after 4 points, add a 4th come bet for a max of 5 #s.
-and after 5 points, continuous come.
mathematically both strategies are the same?
and should yield the same profit/loss over the long run???
LINK
continuous passline/come w/max odds should be self explanatory.
3point betty: passline + 2 come bets w/max odds.
-after hitting 3 points, i do a 3rd come bet to have 4 #s.
-after 4 points, add a 4th come bet for a max of 5 #s.
-and after 5 points, continuous come.
mathematically both strategies are the same?
and should yield the same profit/loss over the long run???
Craps is paradise (Pair of dice).
Lets hear it for the SpeedCount Mathletes :)
April 28th, 2012 at 8:06:45 AM
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Not under default conditions. The more points you establish and take max odds for, the greater will be your amount of "odds off on the comeout". That increases your ratio of line bet to odds action and slightly increases your realized long-term house edge. But not by much. If you always work your odds this is a non-issue, but almost nobody works odds.
The real effect is that making more bets means you expose more bets to the house edge.
The real effect is that making more bets means you expose more bets to the house edge.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
April 28th, 2012 at 8:12:23 AM
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No and No. because of what ME just posted.Quote: 100xOdds
mathematically both strategies are the same?
and should yield the same profit/loss over the long run???
From my come bet simulation results: average # of come points
per 100 rolls:
having max 1 come point avg: 16 per 100 to 105 rolls
having max 2 come point avg: 29
having max 3 come point avg: 38, stdev:4.1, min:23, max:60
having max 4 come point avg: 44
having max 5 come point avg: 47
having max 6 come point avg: 48, stdev:6.3, min:26, max:85
Now the classic Ponzer method(3point Molly etc), 1 pass+2comepoints max
averages 29 CP per 100rolls (add for the pass)
one can calculate the ev and sd for this
same with a come bet on every roll that averages 48 CP per/100.
Your ev and sd for method two will be lower compare to constant come bets.
You have to calculate the expected number of extra comepoints compared to just the Ponzer method.
A finite Markov chain can be used to describe the state of the system and the expected number of visits to each state.
So can a simulation.
I have this programmed into WinCraps already but do not have the time right now to run the needed sim.
Back to the Future
winsome johnny (not Win some johnny)
May 6th, 2012 at 6:13:38 AM
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Q: should yield the same profit/loss over the long run???
A: "The real effect is that making more bets means you expose more bets to the house edge."
lets say I do 100 3point Bettys:
100 x (1pass + 2 come)= 300 #'s
Thats not the same as continuous pass/come for 300 #s??
In both instances i have 300 #s exposed to the house edge.
A: "The real effect is that making more bets means you expose more bets to the house edge."
lets say I do 100 3point Bettys:
100 x (1pass + 2 come)= 300 #'s
Thats not the same as continuous pass/come for 300 #s??
In both instances i have 300 #s exposed to the house edge.
Craps is paradise (Pair of dice).
Lets hear it for the SpeedCount Mathletes :)
May 7th, 2012 at 9:18:34 PM
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True, but in the latter case you're playing for a shorter amount of time. That wasn't your original question. Over the same period of time, making more bets exposes more of your money to the EV.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
