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dicesitter
dicesitter
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March 15th, 2015 at 3:12:53 PM permalink
some one may break the 154 but it wont be any time soon, and I don't think it will be anyone
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
mustangsally
mustangsally
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March 15th, 2015 at 3:26:22 PM permalink
Quote: dicesitter

some one may break the 154 but it wont be any time soon,

why not 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?

Quote: dicesitter

and I don't think it will be anyone in the AP or DI community, it will have to be some one completely random.

the 154 roller was 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
Quote: dicesitter

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

maybe just getting to 50 is all shooters of today want

maybe it be 50 years until another gets past Stanley

that is such a cool name too!

Sally
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GWAE
GWAE
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March 15th, 2015 at 4:54:21 PM permalink
I wonder how much money the right side bettors made? I can imagine playing continuous come bets and getting crushed on a 4 hour roll. That would be so sad.
Expect the worst and you will never be disappointed. I AM NOT PART OF GWAE RADIO SHOW
dicesitter
dicesitter
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March 16th, 2015 at 7:35:14 AM permalink
ON the 154 roll, the table lost $280,000.... just think of that, there was no one at the table
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
ThatDonGuy
ThatDonGuy
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March 16th, 2015 at 9:56:57 AM permalink
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.
Sabretom2
Sabretom2
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March 16th, 2015 at 11:09:48 AM permalink
Common factor; both times, it happened in May.

ill be waiting for the math.
mustangsally
mustangsally
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March 16th, 2015 at 12:04:28 PM permalink
Quote: ThatDonGuy

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.

i think in his paper (he did this with another)
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
I Heart Vi Hart
ThatDonGuy
ThatDonGuy
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March 16th, 2015 at 1:43:12 PM permalink
Some quick analysis shows that the probability of eventually sevening out before N rolls is 1 - A x BN, 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 0.862473752N.

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.
mustangsally
mustangsally
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March 16th, 2015 at 1:58:07 PM permalink
Quote: ThatDonGuy

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)
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 Heart Vi Hart
dicesitter
dicesitter
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March 17th, 2015 at 12:28:24 PM permalink
Sally



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

what do you think

dicesetter

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