Craps Progression

Joined: Mar 29, 2011

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June 25th, 2018 at 1:07:38 PM permalink

Quote: Ace2
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.

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

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

Joined: Oct 2, 2017

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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’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.

It’s 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.....

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Steen

Joined: Apr 7, 2014

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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

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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’s all about making that GTA

Steen

Joined: Apr 7, 2014

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Posts: 126

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

Joined: Oct 2, 2017

Threads: 31
Posts: 2664

July 2nd, 2018 at 1:08:31 PM permalink

We’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 ?

Last edited by: Ace2 on Jul 2, 2018

It’s all about making that GTA

Steen

Joined: Apr 7, 2014

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Posts: 126

July 3rd, 2018 at 7:33:42 PM permalink

Quote: Ace2
We’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

Ace2

Joined: Oct 2, 2017

Threads: 31
Posts: 2664

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’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.

Last edited by: Ace2 on Jul 10, 2018

It’s all about making that GTA

mustangsally

Joined: Mar 29, 2011

Threads: 25
Posts: 2463

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

Joined: Mar 29, 2011

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Posts: 2463

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

Joined: Apr 7, 2014

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Posts: 126

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’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

mustangsally