## Poll

5 votes (21.73%) | |||

18 votes (78.26%) |

**23 members have voted**

in the AP or DI community, it will have to be some one completely random.

Now I know folks who have had 50's 60's 70's 80 and 90's, but those are damn hard to come by, and they

are light years away from 154.

But I hope it is beaten.. that would be great

dicesetter

why not soon?Quote:dicesittersome one may break the 154 but it wont be any time soon,

it took 20 years to smash the record

should come quicker this time around, more craps tables in the world

more hands per day

maybe the poor quality of the dice the casinos now use?

the 154 roller was randomQuote:dicesitterand I don't think it will be anyone in the AP or DI community, it will have to be some one completely random.

would love to see a few rolls from the video, if there really is a video

The one at the California Casino (The Cal), he was a pro

there are still people downtown Vegas, some the old timers, that remember Stanley

that IS so cool

maybe just getting to 50 is all shooters of today wantQuote:dicesitterNow I know folks who have had 50's 60's 70's 80 and 90's, but those are damn hard to come by, and they

are light years away from 154.

But I hope it is beaten.. that would be great

dicesetter

maybe it be 50 years until another gets past Stanley

that is such a cool name too!

Sally

that happened to be a big better...

Now I cant say for sure this is correct, but I was told from a person that should know that

the casino actually came out ahead that day, all the winners put it back on the table

looking for the next 154 roll.

Sally indicated it could be beat again soon and she is right, but then again it may never be beaten

The tables are harder to beat, the dice vary more than they used to so I have a hard time seeing a

dice controller beating that. For a dice controller, on every roll you are closer to a 7 than is a random

player.... we all dream of a roll like that, but when you get to 50,51,52,53,54,55,56 you begin to understand

what your trying to do.

In the world of baseball this would be like hitting 130 home runs in a year...possible, maybe, likely....not so much

dicesetter

Quote:mustangsally

SN Ethier went and found, using eigenvalues and eigenvectors, a closed form formula even that is super super SUPER accurate from 2 to 200 rolls

That's how I tried to solve it as well, but in order to get the eigenvalues, you have to solve an order-5 polynomial, and I don't know of an "exact" way of doing it. It has been proven that there's no "universal formula" like there are for quadratics, cubics, and quartics; probably the easiest way is to stumble across a root, and then solve the quartic that results when you divide the original polynomial by (x minus the root).

For those of you playing at home, I get this:

`[S(N,C)] [1/3 1/12 1/9 5/36 0] [1]`

[S(N,4)] [1/6 3/4 0 0 0] [0]

[S(N,5)] = [2/9 0 13/18 0 0]^N [0]

[S(N,6)] [5/18 0 0 25/36 0] [0]

[S(N,E)] [0 1/6 1/6 1/6 1] [0]

where the vector on the left is the probability of being in each of the five possible states (Comeout, Point 4, Point 5, Point 6, End (i.e. seven out)) after N rolls (or N or fewer rolls for the end state), and the probability of "surviving" N consecutive rolls is 1 - S(N,E).

However, the first step is to find the five eigenvalues for the 5x5 matrix, which are the values of X for which, if you subtract X from the five elements in the main diagonal, the determinant of the resulting matrix is zero.

ill be waiting for the math.

i think in his paper (he did this with another)Quote:ThatDonGuythe first step is to find the five eigenvalues for the 5x5 matrix, which are the values of X for which, if you subtract X from the five elements in the main diagonal, the determinant of the resulting matrix is zero.

he used a 5x5 matrix

have you read that paper?

EthierHoppe_WorldRecordCraps.pdf

"Is there a closed-form expression, simple enough to

be used by a journalist the next time the record is

broken?"

he also has this that i never really looked at

But we would like something still simpler, usable

on a handheld calculator. We use the approximation

¯t(n) := ¯c1(¯e1)n^−1, (9)

where

¯c1 := 1.211844813 and ¯e1 := 0.862473752,

which are the nine-decimal-place upper bounds.

Then ... should suffice

gosh

a table from 2 to 200 rolls is so easy to make

once made, it will last forever (made of gold)

and one can just look at it

at anytime

Sally

^{N}, where A and B are constants.

Doing some "plug in some numbers and see what you get" analysis with Excel, I get an approximate value of the probability of N consecutive rolls without sevening out = 1.211844707 x 0.862473752

^{N}.

However, for N = 154 rolls, this = 1 in 6,481,663,063 instead of the 1 in 5,590,264,072 that pacomartin got.

Even if I use the 1.211844813 that Hoppe used, I get 1 in 6,481,662,449.

6,481,679,532.80 is getting past 154 rolls (making roll 155)Quote:ThatDonGuyHowever, for N = 154 rolls, this = 1 in 6,481,663,063 instead of the 1 in 5,590,264,072 that pacomartin got.

Even if I use the 1.211844813 that Hoppe used, I get 1 in 6,481,662,449.

the record was to get past 153

5,590,264,220

5,590,277,389.56 = the 1 in chance of making the 154th roll

and NOT the chance to 7 out on the 154th roll

that = 1 in 40,648,588,884.24

so looks to be close enough for an approximation

you have to say yes

Sally

I would make a wager with you that 154 roll is not broken in the next 20 years.

what do you think

dicesetter