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Ace2
Ace2
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June 23rd, 2018 at 4:35:23 PM permalink
I知 normally a flat better but on my upcoming trip to Vegas I知 going to try a progression to make things more interesting (volatile).

I値l 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
It痴 all about making that GTA
BleedingChipsSlowly
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DeMango
June 23rd, 2018 at 11:37:54 PM permalink
I知 not a math wiz, but let痴 start with a pass line bet win probability of 0.49293 which has a house edge of 1.414%. You plan to 斗et it ride twice and keep a 4x win. Probability you win is 0.49293^3 ~= 0.11977 by 4 for a ~0.479088 return. Sum that with the expected loss of -0.50707 for a house edge of ~2.7982%. I could be wrong. Can稚 help you with variance or handle, but perhaps that痴 a start.

Why not use the wins for odds instead?
Last edited by: BleedingChipsSlowly on Jun 24, 2018
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klimate10
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June 24th, 2018 at 9:06:59 PM permalink
If 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稚 make sense to ignore odds.

But to each his own.
Ahigh
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DeMango
June 25th, 2018 at 8:20:12 AM permalink
Quote: klimate10

If 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稚 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.
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klimate10
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June 25th, 2018 at 8:32:37 AM permalink
Are you saying that it痴 better to make high HE bets because you know smart people who make high HE bets?

Are you also recommending that the bettor bet as much pass line as possible with no odds?
mustangsally
mustangsally
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June 25th, 2018 at 1:07:38 PM permalink
Quote: Ace2

I値l 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.

a 3 step anti-Marty
sounds fun and frustrating too

Quote: Ace2

Can someone quantify how much my handle <snip>

I get an average unit bet of 1.7 (just a tad higher actually)

there seems to be something missing here
Sally
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DeMango
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June 26th, 2018 at 12:35:51 AM permalink
Quote: mustangsally


there seems to be something missing here
Sally


Odds
When a rock is thrown into a pack of dogs, the one that yells the loudest is the one who got hit.
Ace2
Ace2
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June 27th, 2018 at 6:28:49 PM permalink
Quote: mustangsally

I get an average unit bet of 1.7 (just a tad higher actually)

I agree with that number.

After some introspection I found the formulaic solution.

The average wait for a single win is 2.03 bets, for two consecutive wins it痴 6.14 bets. You have to adjust these numbers for the bets immediately following a raise which don稚 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.
It痴 all about making that GTA
JohnnyQ
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June 27th, 2018 at 6:34:56 PM permalink
AceVentura 2 the Sequel:

"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.....
There's emptiness behind their eyes There's dust in all their hearts They just want to steal us all and take us all apart
Steen
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June 30th, 2018 at 5:12:50 PM permalink
The average bet from simulation of 100,000 trials is 1.70249 times the base bet. This agrees with everyone else but I don't think I saw any answers to your variance question. I get a variance of 104.7918 and standard deviation of 10.2368 starting with a base bet of $10. The average number of rolls per decision is 3.38.



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
Last edited by: Steen on Jun 30, 2018
Ace2
Ace2
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July 1st, 2018 at 11:26:49 AM permalink
Sounds high.

The variance should be around .58 + .28 x 2^2 + .14 x 4^2 = 3.94.
It痴 all about making that GTA
Steen
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July 2nd, 2018 at 2:45:08 AM permalink
Quote: Ace2

Sounds 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
Last edited by: Steen on Jul 2, 2018
Ace2
Ace2
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July 2nd, 2018 at 1:08:31 PM permalink
We池e 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 ?
Last edited by: Ace2 on Jul 2, 2018
It痴 all about making that GTA
Steen
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July 3rd, 2018 at 7:33:42 PM permalink
Quote: Ace2

We池e 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
Ace2
Ace2
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July 9th, 2018 at 6:09:44 PM permalink
You 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.

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稚 even have to do any math to know that 1.02 is not a reasonable number. 1.02 implies that this progression doesn稚 really increase variance, and that痴 way off base.
Last edited by: Ace2 on Jul 10, 2018
It痴 all about making that GTA
mustangsally
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July 9th, 2018 at 8:57:47 PM permalink
Quote: Ace2

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

agree
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
Last edited by: mustangsally on Jul 10, 2018
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mustangsally
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July 10th, 2018 at 9:33:15 AM permalink
Quote: Steen

That's not how you compute variance.

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

I think you are correct when thinking about an independent event
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
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Steen
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July 10th, 2018 at 6:41:46 PM permalink
Quote: Ace2

You 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稚 even have to do any math to know that 1.02 is not a reasonable number. 1.02 implies that this progression doesn稚 really increase variance, and that痴 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
mustangsally
mustangsally
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July 10th, 2018 at 7:50:35 PM permalink
Quote: Steen

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. <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 had wondered about the use of handle in the OP question early on
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
datareturn to 1return to 2stay at 4
mean bet >>1.7038775951.9830148971.978887869
variance >>3.9514671765.121932885.394539333
standard deviation1.9878297652.263168772.322614762
bet handle...
variance >>1.0482683181.1895847991.478542136
standard deviation1.0238497541.0906808881.21595318
Last edited by: mustangsally on Jul 10, 2018
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RogerKint
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July 10th, 2018 at 7:55:34 PM permalink
All I want to know is when MustangSally is going to share her place to lose progression with the class ;)
100% risk of ruin

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