Let's discuss the craps "five count"

[mustangsally]

The rest of my discussion on this topic here will be added to the thread over at Wizard of Vegas.
Wizard of Vegas - 5 count
Alan had asked the same question there about a year ago and Frank also replied there too.

I want a larger audience to see what I have come up with, from my math that actually matched SN Ethier's about the 5 count on my first attempt.
My husband, who is a math genius, already told me I did a great job on the math without doing much computer programming on it.

Plus, more that search the internet for information about the 5 count will find it there earlier than finding it here. Just a fact of computer life.

I expanded a table to include the 3,4,5,6 and 7 count
as well as a table showing the Alan 4 and run count (and 3 and 5 too)

here is a sample of the expected loss per shooter
using just the Place 6 and Place 8 for $6 each. yes, boring,
but it is exactly what Don Catlin used in his first simulation on the 5 count.

X-count	expected loss per shooter
3	-0.277265999
4	-0.238852933
5	-0.206000126
6	-0.177704165
7	-0.153297351

The Alan 4 and run count has a lower expected loss than the 5 count does.
Alan-X	expected loss per shooter
3	-0.160493827
4	-0.203389346
5	-0.240464487

I'm telling you Alan, my psychic powers tell me that you will lose the house edge per bet as those who wait for the 5-Count to bet. Watch and see.

Scobedini

No craps player keeps a record of all the specific number of bets with the amounts wagered to even attempt to arrive at the result to see with their own eyes how accurate that statement is.

Actually,
"you will lose the house edge per bet"
is not a 100% correct mathematical statement.

example
the house edge on the pass line bet that Alan makes is exactly -7/495

and one could make a trillion bets X a trillion bets on the pass line over their lifetime and not lose exactly -7/495 per bet made.

The correct concept is one approaches that value
and can get quite close to it as a percentage per bet and will only be on it exactly by shear coincidence after X bets made.

any assumption that one "will lose" the house edge times all the action made the more they play and bet
is part of the Gambler's fallacy and of the law of averages - that really is no law.
I hope Frank does not believe in that
The law of averages
Belief that a sample's average must equal its expected value.
That false belief says variance does not exist, ever.

still having fun!

Sally

Alan, you are famous to some in SoCal TV land.
I was at a party with lots of seniors and brought up your name.
the ladies there knew who you were.
sad part is most think you passed away years ago.
TV I guess can be so cruel

[FrankScoblete]

I agree with you Sally. I must have expressed myself incorrectly. I'll be interested to see what you come up with on WOV.

[Alan Mendelson]

Sally... if I owe alimony or child support to any of those ladies then yes, I did pass away years ago.

[FrankScoblete]

MustangSally:

Could you do the math on placing a Come or Pass Line bet when the 5-Count is finished? That Catlin didn't do and I would be interested in the results for a random roller. Next if you could do the results for someone with an SRR of 1:7 with the placing of the Come or Pass Line after the 5-Count.

---Frank

[Alan Mendelson]

Frank, don't get me started with Seven to Rolls Ratios. That means nothing. For example:

Shooter gets the dice and here are his rolls:
1. 12
2. 3
3. 6
4. 3
5. 2
6. 2
7. 12
8. 3
9. 3
10. 7 out

Would you call that better than an SRR of 1:7 ??

The proverbial SRR ratio is meaningless unless you are betting on the numbers that are hitting.

[mustangsally]

For example:

Shooter gets the dice and here are his rolls:
1. 12
2. 3
3. 6
4. 3
5. 2
6. 2
7. 12
8. 3
9. 3
10. 7 out

The proverbial SRR ratio is meaningless unless you are betting on the numbers that are hitting.

Alan, you only selected 1 of a possible
25,937,424,601 (Billion) 10 roll sequences that could happen. And one of the extreme rare ones BTW.


The SRR is meaningless by itself I do agree here.
Even Don Catlin pointed that out in a few of his articles.

We really need to know the roll probability distribution of the DI.
And that data does not appear any where on the internet as I did look for any large sample data to use.
We can not assume that because a DI rolls less 7s all the other numbers are equally distributed. Just intuition says this is not correct.
So, a DI could be simulated with a known distribution as I recall SmartCraps did this before. But simulation data without calculated data can also be misleading and just error filled as any other set of data.

I also thought a DI that Frank wants to model, actually rolls more 7s on the come out roll and less during the point round.
so we really have two SRR distributions to deal with.

My numbers (and in WinCraps) shows a constant ratio of 36/7 (5.14) instead of 36/6 for the 7
by a team of DIs could totally change the game of craps being currently offered.
They would have a +EV on every roll on ALL Lay bets, even paying a vig up front and on the 3 Hop 7 bets. Even the pass line becomes a nice bet.

I wonder why this has never been taken advantage of by a Team of DIs.

Baccarat has syndicates with hundreds of million of dollars backing them.
Sports betting has the same as does Poker and Blackjack.
Not Craps

Sally
time to post my data over at WoV

Alan, you are funny in a good way
You should be as famous (and rich) as Bob Barker for being on TV for so long

[Alan Mendelson]

Thank you for saying I am funny, Sally. I am really a standup comic. Unfortunately, I haven't had a gig in 40 years.

I started on TV on December 17, 1973. See http://vegascasinotalk.com/forum/showth...cember-17-1973

[mustangsally]

MustangSally:

Could you do the math on placing a Come or Pass Line bet when the 5-Count is finished? That Catlin didn't do and I would be interested in the results for a random roller.

He really did not have to do that. The value you wanted was how often or the probability of a shooter to achieve the 5-count.
Since x = the probability of that (I show this x and what Mr. Ethier calculated too)
simply x * ev
of the pass/come bet.
My college math professor did that to us too.

of course to compare it to Mr. Catlin's ev sim results you need to have the average action being equal to compare apples to apples.
$12 in action per roll with place bets is way different than one $12 pass line or come bet.
Just wanted to point that out. That would be more challenging.

Next if you could do the results for someone with an SRR of 1:7 with the placing of the Come or Pass Line after the 5-Count.

---Frank

This can be done the same way after re-calculating the probability distribution using the probability of 1/7 instead of 1/6 for the 7. the challenge is what are the other 10 numbers probabilities?

Once that is in hand,
the transition matrix probabilities can then easily be changed to arrive at correct values
(like the probability of a shooter to achieve the 5-count and the proportion of rolls to bet on).

But that requires some more effort and a known distribution for the roll probabilities.
I might look into that later on this week as I know how to do it, it might be a fun challenge.

You should really start to learn how to do this math stuff yourself.
Many just need basic math skills (adding etc to start) and most do really know matrix algebra.
It is many times on your grocery receipt. We just use spreadsheets to help with the many calculations required.
A time saver!

Two years ago I was useless when it came to math. Totally. Thick headed too.
If I can learn how to do one math problem any one can.

Sally

added:
here are some photos that might help explain what I mean by a roll probability distribution.
we all should know there are 36 possible outcomes for the roll of 2 fair dice. 2d6
die#1,die#2
example:
1,1 for a 2
1,2 or 2,1 for a 3
(we really do not know which one rolled when the color of each die is the same)
and so on.
1,2,3,4,5,6,5,4,3,2,1 are the ways the sums could roll out of 36
{2,3,4,5,6,7,8,9,10,11,12}
here is what it looks like in WinCraps and what a 1/7 could look like if all the other sums increased equally.
first the normal distribution


now for a 1/7.
But this shows every way still being equal and that could never be as a person rolls less 7s, they must roll less of something else and more of others.

for 1/7 as a percentage

hope this helps out

this is why it is important to know how the less # of 7s came to be.
It really does change the whole probability roll distribution and any values arrived from that distribution

[mustangsally]

MustangSally:

Could you do the math on placing a Come or Pass Line bet when the 5-Count is finished? That Catlin didn't do and I would be interested in the results for a random roller. Next if you could do the results for someone with an SRR of 1:7 with the placing of the Come or Pass Line after the 5-Count.

---Frank

Frank, an update.
I downloaded Smart Craps and the very large user file
I will take some time this week (at night) to look it over and read. I am really interested in how Smart Craps takes your Pro test data and comes up with a SRR to use in its simulations.
My first look thru the program I did not see how to extract the roll probability distribution so I can enter that in Excel worksheets.

Maybe you could find out or you may know how this is done.
I hope it can be found using Smart Craps. Would make things even easier and more fun.
BTW, I do not mind at all sharing my work and worksheets.
I mentioned in my last post at WoV I have converted my work in Excel to use recursion (where no macros or script has to run).
Everyone can do this easily.
It is really a very simple method to calculate seemingly difficult "state" type problems.
I would just like to be recognized as having contributed to any further advancement of any gained knowledge about the 5 count or any other concept.
Like the Alan 4 Count and Run - that was fun for me.

I still give a lot of my credit to Don Catlin and my Uncle who both have helped me out during my learning journey.
It is all about the small things in life that make the largest differences.

Sally

[FrankScoblete]

Sally, when the time comes I'll send you Dan Pronovost's and Jerry "Stickman's" email addresses. He invented it and probably can help you more than anyone.

[mustangsally]

Sally, when the time comes I'll send you Dan Pronovost's and Jerry "Stickman's" email addresses. He invented it and probably can help you more than anyone.

Nice suggestion.
I was thinking of contacting Smart Craps too.

After reading more and using Smart Craps it looks like the info I want is in the program. It calculates it based off the hardway set and the Pro Test results and goes from there.
I do not see a roll probability distribution.

The Wizard if Odds shows some info on this and it relates to the hardway set and the shooter skill. WinCraps does the same.
BoneTracker and Dice Tool (Wong) both Excel workbooks, have this and makes it very clear how it is calculated and arrived at.
This was nice to find. They show dates from 2004 and 2008
It really should be important to any DI to know what their roll distribution is, not just SRR.

Passing the Pro tests in Smart Craps is an opinion of control at the 1% level.
I would really love (hate) to see 100 million blind squirrels shoot the dice and have 1 in 100 pass these tests. ( 1 million of these critters) I would have a proven skill just as they would. How would I know the difference?

You should know your roll probability distribution so any person using math, say an 11 year old, can use it for advantage calculations, expectation results and just the feeling of having a nice day.
Sally

[Alan Mendelson]

Frank Scoblete just did another article about the 5-Count and this paragraph made a big impression on me:

"You can’t do anything about the shooters who make it past the 5-Count and then seven-out right away. That’s the price you pay. There are no hedge bets to help you out either as they come with far higher house edges. And, yes, there are some nights when many players will seven-out right after the 5-Count. An awful fact of craps life."

It was written in response to a question about the 5-count which came from a new craps player who was having bad luck with the shooter sevening-out once the 5-count gave the all-clear to bet.

I still like my idea about the "reverse 5-count" which means bet with the shooter for the first four or so rolls, and then take your bets down or mark them off for the arrival of the inevitable 7. After all, just how many monster rolls a day are you seeing?

I'm starting to think that in craps it's not the number of bets you win, but limiting the number of bets your lose.

Link to Frank's new article: http://scoblete.casinocitytimes.com/...wagering-63349

[RoeIncarnate]

A bit like the stock market too. If you are a trader the key is to limit the losses.

[Alan Mendelson]

A bit like the stock market too. If you are a trader the key is to limit the losses.

Uh oh. This is a can of worms whenever you talk to the math guys about positive expectation games.