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I’ll try the common progression where you double after a win, double again after two wins but stop there and return to the base bet. So max bet is 4 x base.

Can someone quantify how much my handle and variance (in terms of base bet) will increase assuming I always bet on the pass line, ignoring free odds bets.

Thanks

Ace

Why not use the wins for odds instead?

Low HE, and huge volatility. Best of both worlds.

Doesn’t make sense to ignore odds.

But to each his own.

Quote:klimate10If you want volatility, then why not go to the Cromwell, where they have the only 100x odds in LV, make a min pass bet ($10), and max odds that you can afford?

Low HE, and huge volatility. Best of both worlds.

Doesn’t make sense to ignore odds.

But to each his own.

I'll tell you why, nobody DOES that.

The ratio of people proclaiming their own intelligence (even in their on HEAD) by knowledge of this phenomenon divided by the number of people that actually DO 100x odds yields an extremely high number.

Some of the smartest math guys out there make the most expensive plays (like tipping too much) because of the positive reinforcement of the environment to placing the high house edge bets.

This conversation about "why don't you bet the odds" is sorta like saying "why don't you go to the strip club, get in free, don't tip anyone, then go home and have sex with your wife."

The reason is that the "free volatility" causes too much f'ing trouble! It's ANYTHING but free more than half the time.

Just go home and F your wife and forget the strippers.

Or in your case, bet as much as you can on the pass line with no odds and enjoy your comps.

Are you also recommending that the bettor bet as much pass line as possible with no odds?

a 3 step anti-MartyQuote:Ace2I’ll try the common progression where you double after a win, double again after two wins but stop there and return to the base bet. So max bet is 4 x base.

sounds fun and frustrating too

I get an average unit bet of 1.7 (just a tad higher actually)Quote:Ace2Can someone quantify how much my handle <snip>

there seems to be something missing here

Sally

Quote:mustangsally

there seems to be something missing here

Sally

Odds

I agree with that number.Quote:mustangsallyI get an average unit bet of 1.7 (just a tad higher actually)

After some introspection I found the formulaic solution.

The average wait for a single win is 2.03 bets, for two consecutive wins it’s 6.14 bets. You have to adjust these numbers for the bets immediately following a raise which don’t count toward the waiting time.

The adjustment is 1.5 for the single win (half the time progression ends there and half the time it goes to 4x) and 1 for the double win (next bet counts toward nothing).

So 1 / 3.53 x 1 + 1 / 7.14 x 3 =

70.3 % handle increase.

"Ace 2 you ! you dirty bird ! How could you ?"

I was working hard to share my WISDOM with the forum by having the most recent 5 posts !

BONUS POINTS for figuring out the obscure reference WITHOUT using google.....

If initializing script Then

cs1.betamount = 10 :

autohandle winning bets = "take bet and winnings" :

autohandle losing bets = "no bet"

EndIf

If passline wins Then

multiply by 2 on cs1.betamount

If cs1.betamount > 40 Then cs1.betamount = 10 EndIf

start new session(preserve checkstacks)

ElseIf passline loses Then

cs1.betamount = 10 :

start new session(preserve checkstacks)

EndIf

bet cs1.betamount on passline

Steen

The variance should be around .58 + .28 x 2^2 + .14 x 4^2 = 3.94.

Quote:Ace2Sounds high.

The variance should be around .58 + .28 x 2^2 + .14 x 4^2 = 3.94.

That's not how you compute variance.

Variance is the average of the squared distances from the mean.

You concluded that your handle would increase 70.3% (with which we all agree). Therefore, the mean for a $1 base bet is 1.703 and the variance is:

0.58 * (1.703-1)^2 +

0.28 * (1.703-2)^2 +

0.14 * (1.703-4)^2 =

----------------------

approx 1.05 (standard deviation of approx 1.025)

Since I used a $10 base bet, my mean was $17.03 and so....

0.58 * (17.03-10)^2 +

0.28 * (17.03-20)^2 +

0.14 * (17.03-40)^2 =

----------------------

approx 105 (standard deviation approx 10.25)

Steen

Does that sound like a standard deviation of 1.025 to you ?

Quote:Ace2We’re taking a bet with a standard deviation of basically 1 (very close to a 50/50 chance of going up/down 1 unit every bet) and replacing with a scenario that has the same chance of winning but now goes up/down 1 unit 58% of the time, 2 units 28% of the time, and 4 units 14% of the time.

Does that sound like a standard deviation of 1.025 to you ?

Yes.

It's not a 1 unit bet that goes up 1, 2, and 4 units.

It's a :

1 unit bet that goes up/down 1 unit 58% of the time

2 unit bet that goes up/down 2 units 28% of the time

4 unit bet that goes up/down 4 units 14% of the time

Now just do the simple math:

1 x 0.58 +

2 x 0.28 +

4 x 0.14 =

-----------

1.7 <--- this is the mean (the average from which variance and standard deviation are measured)

Steen

I suggest you run a basic simulation in order to grasp how this works. Try 100,000 bets and do some trials, 100 is probably enough. You will see that the result is 100,000 x 1.70 x -0.014 +/- (100,000 * 3.94)^.5 = -2,380 +/ 628.

About 68 % of the trials will be within 1 SD (628) and about 95% within 2 SD (1,256). This matches the expectation of a normal distribution.

If you do the same exercise assuming a varíance of 1.02, you will quickly see that about 1 in 20 trials are 4 or more standard deviations from expectations when a 4 SD occurence should be very rare...only about 1 in every 15,000 triais.

As stated previously, you don’t even have to do any math to know that 1.02 is not a reasonable number. 1.02 implies that this progression doesn’t really increase variance, and that’s way off base.

agreeQuote:Ace2The variance should be around .58 + .28 x 2^2 + .14 x 4^2 = 3.94.

and so does my simulations in WinCraps

that gets back to what exactly was Steen calculating?

Sally

added:

*****

thought I would share my math for the mean bet and the variance

I used a Markov chain solution (did anyone doubt that?)

very simple 4x4 and raised to a high power to get the steady state probabilities (actually at 32 it is there. I went up to 256 for the photo shoot)

mean bet (1 unit) = 1.703877595

variance = 3.951467176

standard deviation = 1.987829765

I guess many call this the Paroli betting system after he who first wrote about it.

also know as the reverse or anti Martingale

for Ace requested simulation of 100,000 bets

I only ran it 1000 times (and it was close to the calculated values)

calculated

ev=-2409.52

sd=628.607

I think you are correct when thinking about an independent eventQuote:SteenThat's not how you compute variance.

Variance is the average of the squared distances from the mean.

clearly a bet system of 1,2,4,1 on a win and 1 on a loss has to be dependent event.

I can not explain it well, others have and I will search for that.

I failed this stuff way back in math class.

(at least I was NOT by myself but in the majority, iirc)

I never 'got it' in school

Sally

Quote:Ace2You are incorrect Steen.

I suggest you run a basic simulation in order to grasp how this works. Try 100,000 bets and do some trials, 100 is probably enough. You will see that the result is 100,000 x 1.70 x -0.014 +/- (100,000 * 3.94)^.5 = -2,380 +/ 628.

I have no problem "grasping" these concepts. I've run the simulations and posted the code. I suggest you do the same.

Quote:If you do the same exercise assuming a varíance of 1.02, you will quickly see that about 1 in 20 trials are 4 or more standard deviations from expectations when a 4 SD occurence should be very rare...only about 1 in every 15,000 triais.

As stated previously, you don’t even have to do any math to know that 1.02 is not a reasonable number. 1.02 implies that this progression doesn’t really increase variance, and that’s way off base.

You''re being imprecise with your terminology and confounding your data.

I never said the variance was 1.02.

1.02 is an entirely reasonable approximation of the STANDARD DEVIATION of the BET HANDLE.

Your original question asked, "Can someone quantify how much my handle and variance (in terms of base bet) will increase assuming I always bet on the pass line, ignoring free odds bets?"

Do you see that word "handle" that you used? Well, it means something. If you run the simulations which you suggest, you'll discover that in regards to BET HANDLE you'll get approximately:

Mean 1.7025

Variance 1.0479

Standard deviation 1.0237

I specifically responded to your request as stated.

Now, if you want to change your question to "What's the average outcome (amount won or lost)?" then you'll get the numbers you've been posting which my simulation shows are approximately:

Mean -0.024

Variance 3.9457

Standard deviation 1.986

Steen

I had wondered about the use of handle in the OP question early onQuote:SteenYour original question asked, "Can someone quantify how much my handle and variance (in terms of base bet) will increase assuming I always bet on the pass line, ignoring free odds bets?"

Do you see that word "handle" that you used? Well, it means something. <snip>

I specifically responded to your request as stated.

Now, if you want to change your question to "What's the average outcome (amount won or lost)?" then you'll get the numbers you've been posting which my simulation shows are approximately:

Mean -0.024

Variance 3.9457

Standard deviation 1.986

Steen

I even mentioned something was missing.

and it looks like all has been found

pays to understand what it written (I miss that lots)

great work!

Sally

added for Original Post

with a max bet of 4 and after a win either return to 1,2 or stay at 4

looks fun to try!

data

data | return to 1 | return to 2 | stay at 4 |
---|---|---|---|

mean bet >> | 1.703877595 | 1.983014897 | 1.978887869 |

variance >> | 3.951467176 | 5.12193288 | 5.394539333 |

standard deviation | 1.987829765 | 2.26316877 | 2.322614762 |

bet handle | . | . | . |

variance >> | 1.048268318 | 1.189584799 | 1.478542136 |

standard deviation | 1.023849754 | 1.090680888 | 1.21595318 |