Need help with some numbers

Hi Gents,

Looking for a little help in determining odds for certain events happening (or not happening) at the Craps table. Any and all positive comments would be greatly appreciated.

Here are the situations. It would also help if you could show the math too. I'm just not as good with this sort of thing.

What are the odds of a 6 and 8 not rolling for 5 shooters in a row?

What are the odds of a 5, 6, and 8 not rolling for 5 shooters in a row?

What are the odds of a 5, 6, 8, and 9 not rolling for 5 shooters in a row?

What are the odds of a 4, 5, 6, 8, and 9 not rolling for 5 shooters in a row?

What are the odds of a 4, 5, 6, 8, 9, and 10 not rolling for 5 shooters in a row?

I know it's a lot of situations. I just have an idea, but want to confirm a few things for our group.

Thanks again!

Quote: ROBROY01

Hi Gents,

Hi R
gentle
is my middle name

my true given initials = SGM

Quote: ROBROY01

Here are the situations. It would also help if you could show the math too. I'm just not as good with this sort of thing.

yes, but if you show what you have
some may point out what is correct and what is not correct.
we all can learn that way.

Quote: ROBROY01

What are the odds of a 6 and 8 not rolling for 5 shooters in a row?
What are the odds of a 5, 6, and 8 not rolling for 5 shooters in a row?
What are the odds of a 5, 6, 8, and 9 not rolling for 5 shooters in a row?
What are the odds of a 4, 5, 6, 8, and 9 not rolling for 5 shooters in a row?

per shooter = per 7out

Say X = the prob of an event NOT happening
then X*X*X*X*X = the prob of the event not happening over the very next 5 attempts, not 6 or more attempts
You want to know what X is for each type of event.

until I see what you have first here is what I had in a 1 million shooter simulation
group: probability
68: 0.200101
568: 0.113824
5689: 0.057305
5across-noPlace10: 0.025586

Quote: ROBROY01

What are the odds of a 4, 5, 6, 8, 9, and 10 not rolling for 5 shooters in a row?

How about this way
What are the odds of a 4, 5, 6, 8, 9, and 10 not rolling for 1 shooter?
A shooter must roll one of these #s to establish a point before a 7out can happen.

But me thinks you only are concerned about the rolls not counting the come out roll?
so now that is a totally different question from the many you first asked.
See?
Do you really want to know the number of place bets rolled per shooter per place bet group not counting come out rolls?
I think that is really your actual question, maybe

Quote: ROBROY01

I know it's a lot of situations.
I just have an idea, but want to confirm a few things for our group.

Oh.
Is your group in direct competition for fame and fortune as my group?
If yes,
hahaha


others may not be so demanding as I am
or you may just take my sim data and run with it as being close enough
SGMO

Hahaha.. Funny, Sally! Guess maybe I should've said "Ladies and Gents" just to be fair? I sometimes forget women are in the minority with Craps, but still play.

My "Group" is a bit different than yours, I think. They don't want fame and fortune - too much work to keep it! ;)

Wanting theoretical numbers and not sims, but thanks for it anyway!

OK. Here is what I was thinking.

Come Out wouldn't count. Bets to be placed after the Point is established.

For the 6/8 situation:

5 ways to make a 6, and 5 ways to make an 8. Total of 10 out of 36. That is our first fraction. 10/36

6 ways to 7-Out. As a fraction, that would be 6/36.

We multiply these fractions together and reduce down.

10/36 x 6/36 = 5/108 or 1/21.6

Then take 1/21.6 and multiply it by itself 5 times for 5 shooters.

1/21.6 x 1/21.6 x 1/21.6 x 1/21.6 x 1/21.6 = 1/4,701,849.845 or round up to 1/4,701,850

or 1 in 4,701,850 chance of the 6 or 8 not showing for 5 players in a row; or 99.9999787% chance that the 6 or 8 won't show for 5 players in a row.

Did I do my math correctly or did I screw up somewhere?

Thanks for your help!

Quote: ROBROY01

Hahaha.. Funny, Sally! Guess maybe I should've said "Ladies and Gents" just to be fair? I sometimes forget women are in the minority with Craps, but still play.

The greatest craps players ever are women.
Frank Scoblete says the greatest shooter he ever saw was a woman
The current record holder for the longest hand, a woman and a grandma! Most may not even know her name.
I do not include myself as a great craps player, just a lifetime winner.

These pics below show many other great women craps players too.
I bet no one can name them all!








Now back to your question

Quote: ROBROY01

Come Out wouldn't count. Bets to be placed after the Point is established.

For the 6/8 situation:

How about this roll sequence 6,6,8,8
The first 6 is the come out roll. You do not want that counted.
The second 6 is a point 6 winner. (ugh from the don't side)
do you want that counted?
No no 6 or 8 place bets rolled?
or are you really wanting to only count place bet rolls and not point winners.

That would be reasonable too
Sally

Amazing pictures. Thanks for sharing.

I stand corrected! I'm just not used to seeing a whole lot of women shooters at the tables I play at. Yes, I already knew your facts and could maybe name about half the women in those photos. :)

If the Come Out rolled is a 6 or 8, then I would put out my bets on the 6 and 8 (because a Point is established. Doesn't have to be a 6 or 8, just any Point is fine.). If it hits, then great.

Really, what I am after is to confirm if those numbers and calculations I gave were accurate or not. Can you confirm?

If they're good, then I could probably figure out the rest of the other situations I gave.

Quote: ROBROY01

If the Come Out rolled is a 6 or 8, then I would put out my bets on the 6 and 8 (because a Point is established. Doesn't have to be a 6 or 8, just any Point is fine.). If it hits, then great.

well, that makes the math super super easy

Quote: ROBROY01

Really, what I am after is to confirm if those numbers and calculations I gave were accurate or not. Can you confirm?

my sim data for one shooter looks like this

group	sim prob	calculated	error
68	0.374602	0.375 (6/16)	-0.000398
568	0.299894	0.3 (6/20)	-0.000106
5689	0.249666	0.25 (6/24)	-0.000334
5across-no10	0.221676	0.222222222 (6/27)	-0.000546222
across	0.199584	0.2 (6/30)	-0.000416

calculated^5 = (for 6 or 8)
0.007415771
or 1 in 134.85
Now this is the probability of no 6 or 8 by 5 shooters in a row, the very next 5 shooters.

what about the next 10 shooters?
we could have 5 in row or more
0.03059006 or about 1 in 32.69 about 4 times easier and more often
this is about streak calculating. a bit more challenging that just the next 5 shooters

the average number of shooters required to see a run of 5 without rolling a 6 or 8 = 214.2
That comes from p=6/16
p^5 * (1+p+p^2+p^3+p^4)
# of attempts * the average length of each attempt
there are easier formulas to use BTW

So I think your method is not correct.
Just my opinion here

You only need to consider a 7 (that 7 is a 7out) before the other numbers you do want
you can also go the other way too for a number of 6 or 8s before a 7out
Sally

Um, anything simpler, like how I previously explained showing all of my math? I am no math expert and I am definitely not stupid either, but what you have is above my head. It's all sim computer answers, which again, I do not want. I need long-hand mathematics with explanations where the numbers are coming from. I'd like to be able to understand it in my head, WITHOUT using computers.

Like for example, how did you come up with .007415771? What is this number? How did you get ANY of these numbers? What is with the "error" on your chart? What's with the "^" symbols?

So, were you saying that there is a probability of 1 in 134.85 of there being no 6 or 8 for the next 5 shooters in a row? If so, I still don't understand how you came up with that.

You said something about easier formulas? That would be good. LOL

No, I never said anything about 10 shooters. Only interested in 5. LOL

I definitely believe in the "Keep It Simple Stupid" principle.

Perhaps another member can confirm your numbers (no offense)?

Sorry, Sally. I know you are trying to help. I would just rather use long-hand math - again, so I can understand it all in my poor brain. :)

P.S. I am talking about place bets on the 6 AND 8 after a Point is established, ANY Point - not playing the dark side. Maybe using a progression after a 7-out. Once the 6 or 8 hits, I drop back to basic place betting 1 unit again. I am simply wanting to know what are the odds of a 6 or 8 not hitting for the next 5 shooters in a row NOT 10. Same goes for my other scenarios too. Does this make better sense? :) Just need something simple and not overthinking things too much. Okey Dokey, cow pokey? :) By the way, I really thought those photos of your were pretty cool "vintage Vegas". Sweet!

Quote: ROBROY01

Like for example, how did you come up with .007415771?
What is this number? How did you get ANY of these numbers? What is with the "error" on your chart? What's with the "^" symbols?

The ^ = *to the power of* which is the same as a*a*a*a*a (a # multiplied by itself 5 times)

I placed the calculated values in the table next to the simulated values to compare how close the simulation values can be.

example
a 7out before any 6 or 8
you know 6/36 for the 7
and 10/36 for the 6 and 8
so
6/36 / 6/36 + 10/36 = (this is known as the craps principle)
6/36 / 16/36 =

6/36 * 36/16 =
6/16 = 0.375
or simply
6 / 6+10

that is for no 6 or 8 for one shooter (a 7out)
for 5 in a row (the very next 5) = 0.375^5 or
0.375*0.375*0.375*0.375*0.375 = 0.007415771484375

Quote: ROBROY01

So, were you saying that there is a probability of 1 in 134.85 of there being no 6 or 8 for the next 5 shooters in a row? If so, I still don't understand how you came up with that.

yes.

the others follow the same path
a 7 before a 5,6 or 8
6 ways for the 7
14 ways for the 5,6,8
that makes 6/6+14 = 6/20 = 3/10 or 0.30

for 5 in a row (the very next 5 shooters)
0.30^5 = 0.00243 = 1 in 411.52 (1/0.00243) (411 = # of groups of 5 shooters)

you should be able now do the rest

you can also go the other way too if that is ever needed
6 or 8 before a 7?
10 / 10+6 = 10/16 = 5/8 = 0.625
5, 6s or 8s before a 7 = 0.625^5 = 0.0953674

Sally

Mustangsally




I also believe women can be great craps players, but not for the same reason you do.

Now in the case of Pat Demauro who has the longest roll we know of at 154, i have met
her and i cant remember for sure, but i think that was her second time at a craps
table. That would hardly make her the worlds best player, nor would the fact that
Jean Arthur played craps with John Wayne make much of a difference.

I think women can be better than men because they tend to have smaller softer hands
and if they learn to roll the dice correctly, they are very very hard to beat by men.

Franks talks about the ARM as being the best player he has seen. I surely have not seen
her, but i understand she had a very soft toss. Frank has seen alot of good players and
if he says she was the best, i am inclined to believe him. You indicate your a lifetime
winner at craps, that would place you on a level not reached by 99% craps players.

I hope to see you play someday.


Dicesetter

Quote: dicesitter

I also believe women can be great craps players, but not for the same reason you do.

Now in the case of Pat Demauro who has the longest roll we know of at 154, i have met
her and i cant remember for sure, but i think that was her second time at a craps
table. That would hardly make her the worlds best player, nor would the fact that
Jean Arthur played craps with John Wayne make much of a difference.

I think women can be better than men because they tend to have smaller softer hands
and if they learn to roll the dice correctly, they are very very hard to beat by men.

Franks talks about the ARM as being the best player he has seen. I surely have not seen
her, but i understand she had a very soft toss. Frank has seen alot of good players and
if he says she was the best, i am inclined to believe him. You indicate your a lifetime
winner at craps, that would place you on a level not reached by 99% craps players.

I hope to see you play someday.
Dicesetter

Do you honestly believe that one person can be a better craps player than another?

How about roulette, then? Is there a world's best roulette player?

Soft toss, hard toss, dice off the table every other roll, it makes no difference. A 154 roll turn at the table is just luck, nothing more, nothing less.

Terribletom


There is no question one person can be better at craps that another, even further
there is no question a person can have a much better throw than another depending on
hand size etc. the size of your hand is vital in terms of how you can grip the dice
and therefore how they release from your hand.

In terms of the 154 roll by Pat, there is no question that was luck, but to assume that a person
could not in fact develope a toss where they could do better than others under the right conditions
is a stretch. But what you believe is fine with me, i would rather everybody feel that it was
impossible to have some influence on the table.

Dicesetter

No question? That seems a stretch.

What exactly constitutes a "better throw"?

I'd also be interested in reading more about "the right conditions", if you're inclined to expound upon the point.

I do operate under the premise that craps is a game of chance and not a game of skill.

TerribleTom



You have the right to your opinion about craps and i sure am not
going to try to change it.

I will only say this about condition.,...... Tiger woods is a great golfer and as
great as he is he would be the first to tell you not every course fits hit game.

craps is no different, everybody that has ever tried to get some degree
of influence with a repeatable roll is better on one type of craps table
surface as compared to another. Some tables have little bounce, some lots of
bounce, so are 10 feet long some are 16 feet long. NO throw no matter how
well perfected works the same on all surfaces.

I am very good on 12 foot tables with little bounce, i am terrible on tables
such as Silverton which is a 14' table with a good deal of bounce.

I also believe winning at craps needs a good deal of luck, i only say i get luckier
far more often on a table that fits my particular toss.


dicesetter

OK. I think i got most of that, except for one thing. How were you able to take .00741577 and convert that to 1/134.85? I think I am still missing something...

Thanks for your patience, Sally. I appreciate the help! :)

Quote: ROBROY01

How were you able to take .00741577 and convert that to 1/134.85?

1 / .00741577 = 134.8477636 = average number of attempts per 5 shooters

If I only knew the 1 in 134.85 and I wanted the probability
1 / 134.85 = 0.007415647

same method used as the probability of rolling a 6 with one 6-sided die
we know that = 1 in 6
so the probability = 1 / 6
or better known as 16.67%

have fun
Sally

Got it now. I needed to read that as "1 divided by 134.85". I sometimes forget that cmputer keyboards don't really have a standard symbol for division! LOL Sorry about that.

Is it also correct that I could take that .007415647 and convert it to a success percentage as:

100 - .007415647 = 99.9925844%

or is that totally off?

Quote: ROBROY01

Is it also correct that I could take that .007415647 and convert it to a success percentage as:

100 - .007415647 = 99.9925844%

actually it would only be (1 - .007415647)*100
100 - .7415647 = 99.9925844% looks better

and yes, as long as all possible outcomes are accounted for that will work just fine.
that would be the chance of at least 1 (1 or more)
6 or 8 in the very next 5 shooters

we know we accounted for the chances of NO (0) 6s and 8s before a 7out so that leaves all the rest of the possible outcomes, namely, 1 or more

Sally

So Sally, just to clarify, the chance that 2 shooters in a row won't hit a 6 or 8, is:

.375 * .375 = 0.140625

1/.140625 = 6.8376068 or once every 7 (rounded up) shooters?

or

85.9375 % chance of that not happening?

Quote: ROBROY01

So Sally,

yes

Quote: ROBROY01

just to clarify, the chance that 2 shooters in a row won't hit a 6 or 8, is:

.375 * .375 = 0.140625

yes. the very next 2 shooters

Quote: ROBROY01

1/.140625 = 6.8376068

that = 7.11 (look at the number of my posts)

Quote: ROBROY01

or once every 7 (rounded up) shooters?

so you rounded down but
no.

not once every 7 shooters on average...

but once every 7 groups of 2 shooters on average
FYI

Quote: ROBROY01

or 85.9375 % chance of that not happening?

yes, yes and yes!
over the very next 2 shooters
Sally

Las Vegas
here we come!

Thanks for the clarification, Miss Sally! I see my errors now. I think I was typing one thing and thinking another. That error was a "duh" moment! LOL Everything else should be no problem now.

Take care and thanks again for all of your help! :)

Now ya did it Sally... that Promo pix of Gilligan's Island. You just KNOW thats gonna bring up the Ginger vs. Mary-Ann all over again. Jeez....