longtimelancer
Joined: Apr 12, 2014
• Posts: 54
January 20th, 2015 at 7:37:20 PM permalink
Last April I won over \$2000 starting with \$200 by playing the Don't with full odds (345) and playing 1 or 2 Don't Come with full odds. I am going back to Vegas this April and I wanted to try the same strategy and was wondering what my chances are.

Starting with \$200 on a \$10 table and playing the Don't Pass with full 345 odds and playing one Don't Come also with full odds, what are my chances that:

I will lose \$200 without ever being up.
I will win \$100.
I will win \$200.
I will win \$1000.
I will win \$2000.
I will win \$10,000.

Thanks!

P.S.: Mustang Sally was a great help last year with all of his calculations. I hope he is still here!
rainman
Joined: Mar 28, 2012
• Posts: 1462
January 20th, 2015 at 7:43:26 PM permalink
I can't help with the math, however I can help with MSally. My first pointer is I wouldn't call her a he. :)
goatcabin
Joined: Feb 13, 2010
• Posts: 664
January 21st, 2015 at 9:55:41 AM permalink
Quote: longtimelancer

Last April I won over \$2000 starting with \$200 by playing the Don't with full odds (345) and playing 1 or 2 Don't Come with full odds. I am going back to Vegas this April and I wanted to try the same strategy and was wondering what my chances are.

Starting with \$200 on a \$10 table and playing the Don't Pass with full 345 odds and playing one Don't Come also with full odds, what are my chances that:

I will lose \$200 without ever being up.

What about losing the \$200 after being up a bit?

Quote: longtimelancer

I will win \$100.
I will win \$200.
I will win \$1000.
I will win \$2000.
I will win \$10,000.

Your question is sort of meaningless without a stop condition for time/number of bets. Or, do you mean that you will either lose the \$200 or reach a goal, no matter how long it takes, and you are asking really four questions, i.e.

lose \$200 or win \$100, no matter how long it takes
lose \$200 or win \$200, " " " " " "
etc. etc.

\$10 DP and 3, 4, 5x odds is a lot of risk for a \$200 BR, even without the DC bet(s).
For just the \$10 DP with full odds, I ran a sim with \$200 BR, \$200 win goal.
Bust rate was 53.1%
win goal reached 46.9%

Trying to win \$1000:
Bust rate was 84.5%
win goal reached 15.5%

Trying to win \$2000
Bust rate was 91.6%
win goal reached 7.7%
max bets of 1600 reached .7% (my program requires a maximum number of bets, but I set it real high)

For a more realistic scenario, tell me how long you are willing to play before quitting, if you have neither busted nor reached your win goal.
Cheers,
Alan Shank
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
January 21st, 2015 at 10:37:10 AM permalink
Quote: longtimelancer

Starting with \$200 on a \$10 table and playing the Don't Pass with full 345 odds and playing one Don't Come also with full odds, what are my chances that:
<snip> <snip>

the probabilities to hit a win goal given no time restraints will be abouts the same, adding DC just shortens the time to ruin/success

a simple way to estimate without any simulations/calculations is (for most simple bets)

(start bank/target bankroll) - (2* HE)

(it gets close most times and really who knows the difference in a session between 46.9% or 47.2%)

example:
turn \$200 into \$400

200/400= 0.50
HE = 1.40% * 2 = 2.8%
50% - 2.8% =

have fun!
Sally
I Heart Vi Hart
Romes
Joined: Jul 22, 2014
• Posts: 5496
January 21st, 2015 at 10:37:19 AM permalink
Quote: longtimelancer

Last April I won over \$2000 starting with \$200 by playing the Don't with full odds (345) and playing 1 or 2 Don't Come with full odds. I am going back to Vegas this April and I wanted to try the same strategy and was wondering what my chances are.

Starting with \$200 on a \$10 table and playing the Don't Pass with full 345 odds and playing one Don't Come also with full odds, what are my chances that:

I will lose \$200 without ever being up.
I will win \$100.
I will win \$200.
I will win \$1000.
I will win \$2000.
I will win \$10,000.

Thanks!

P.S.: Mustang Sally was a great help last year with all of his calculations. I hope he is still here!

Without running the actual numbers and just looking at the facts, I'd say you got extremely lucky last time. Playing full 3-4-5 odds with a \$10 DP/DC... If you take 2 DC points then your bankroll is essentially on the table each shooter. If you're allowed to wager double 3-4-5 from the don't (to get 3-4-5 odds) then that's even more! Let's take an "generic" roll... come out 6, you have \$10 DP, \$60 odds (to win \$50... 5x). Then the next point is a 9, and you have your \$10 DC, \$60 odds (to win 4x). At this point you have 70% of your 'bankroll' on the table. While yes it might be a bit harder to re-throw a 6, then a 9, before the 7... If this happens, then you just lost a massive portion of your 'bankroll' on one shooter. Notice you originally said 1 or 2 DC bets. I'm rounding down to 1 for my example, so I assume you aren't actually taking max 3-4-5 odds, or the casino doesn't let you actually win max 3-4-5 odds (which most do, because that's what's mathematically right and what they're advertising).

Odds of 7: 6/36 = 1/6 = .16667 -- Resulting in total win of \$110, which is 55% of your starting bankroll.
Odds of 6, then 7: (5/36) * (6/36) = .13889 * .16667 = .02315 -- Resulting in total loss of \$20, which is 10% of your starting bankroll.
Odds of 9, then 7: (4/36) * (6/36) = .11111 * .16667 = .01852 -- Resulting in total loss of \$10, which is 5% of your starting bankroll.
Odds of 6, then 9: (5/36) * (4/36) = .13889 * .11111 = .01543 -- Resulting in total loss of \$140, which is 70% of your starting bankroll.
(note 9 then 6 is the same math)

So of all the rolls the shooter could throw that could effect you (after you've gotten your DP and 1 DC established):
16.7% of the time you'll win both.
2.3% of the time you lose the 6, win the 9.
1.9% of the time you lose the 9, win the 6.
3% of the time you lose both.

Again, this is all assuming you establish both points (when in reality your DP and DC bets are subjective to a disadvantage come out roll).
Playing it correctly means you've already won.
mustangsally
Joined: Mar 29, 2011
• Posts: 2463
January 21st, 2015 at 11:08:57 AM permalink
Quote: goatcabin

\$10 DP and 3, 4, 5x odds is a lot of risk for a \$200 BR, even without the DC bet(s).

but at the same time gives one of the highest chances of hitting the bankroll target
without single bet Bold Play, betting all \$200 on one game/bet that is

BTW, Thanks for sharing the WCPro Risk of Ruin (by Steen) Don't Pass version

I do most of my simple RoR and target hitting calculations
in Excel
using a transition matrix and sometimes a recursive table.
I love Markov Chains and I do have a few programs that do them very quickly
but most can see and grasp the process when seen in a spreadsheet, in my opinion

Sally

snow snow snow
I Heart Vi Hart
goatcabin
Joined: Feb 13, 2010
• Posts: 664
January 21st, 2015 at 12:51:39 PM permalink
Quote: mustangsally

the probabilities to hit a win goal given no time restraints will be abouts the same, adding DC just shortens the time to ruin/success

a simple way to estimate without any simulations/calculations is (for most simple bets)

(start bank/target bankroll) - (2* HE)

(it gets close most times and really who knows the difference in a session between 46.9% or 47.2%)

example:
turn \$200 into \$400

200/400= 0.50
HE = 1.40% * 2 = 2.8%
50% - 2.8% =

have fun!
Sally

Something is fishy about this. The probability of reaching a win goal is mostly dependent on the variance. Clearly, the probability of doubling a \$200 bankroll with a \$10 DP bet (no odds) is nowhere near .472; my sim gives .243 for no odds, .42 for single odds, .457 for double odds, .472 for 3, 4, 5x. That formula does not take into account the odds bet.
Cheers,
Alan Shank
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
longtimelancer
Joined: Apr 12, 2014
• Posts: 54
January 21st, 2015 at 3:34:42 PM permalink
But the edge for full odds DP is .281% which would make your example 49.438%. Correct?
goatcabin
Joined: Feb 13, 2010
• Posts: 664
January 21st, 2015 at 5:34:26 PM permalink
Quote: longtimelancer

But the edge for full odds DP is .281% which would make your example 49.438%. Correct?

Edge for DP, 3, 4, 5x is 0.37%, but my sims do not show that high a probability of winning \$200.
It doesn't seem to me that this formula works, although for formula without the HE part does, i.e. for "fair" bets.
Cheers,
Alan Shank
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
longtimelancer
Joined: Apr 12, 2014