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tuttigym
tuttigym
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May 30th, 2010 at 9:16:49 AM permalink
Recently a second time female player descended on a craps table in Atlantic City. She threw a remarkable 153 times without a 7 out which occurred on her 154th toss. It was touted as a new world record for both consecutive tosses w/o a 7 out and in time of holding the dice at one turn in excess of 4 hours, I believe.

What are the world records for:

1. Consecutive successful passes?

2. Consecutive successful point conversions?

3. Consecutive Come Out natural winners (7 & 11)?

4. Consecutive Come Out craps?

5. Consecutive rolls w/o hitting a 6? (not point conversion, just the number)

6. Consecutive rolls w/o hitting an 8? (not point conversion, just the number)

Are such records kept??

tuttigym
pacomartin
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May 30th, 2010 at 12:48:36 PM permalink
I have never heard of records kept to this detail. Some of the older records were expressed as consecutive points made, and as hours holding the dice. I am not sure how many points were made on this record 154 rolls of the dice. The only anecdotal comment that I read was that she only rolled two or three 7's on the come out roll.

The Borgata says they only lost $185K on the table during the roll. It was full of small betters.
Wizard
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May 30th, 2010 at 6:55:37 PM permalink
I agree, I doubt anybody keeps track of that. The walkway between the California and Main Street Station has a craps "Hall of Fame" with plaques for players who hold the dice more than a defined period of time, around two hours I think.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
tuttigym
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May 31st, 2010 at 7:00:56 AM permalink
When I was very young and living in Calif., my dad would take us to LV about three times a year - the '50s. One time, at the old Sahara, a player created a huge amount of excitement. He made thirty (30) straight passes before the 7 out. The hotel placed the dice on a pink silk pillow in a glass case at the entrance to the casino in the lobby. The narrative stated that the shooter made (from memory a long time ago) $7,500 and the table paid out about $35,000 to the other players. Huge money back then.

It is hard to conceive that that might be the current world record after all this time. It would seem that with all the current technology and improvements in communications, such records or numbers would be available even with all the privacy concerns. Or perhaps not.

tuttigym
Nareed
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May 31st, 2010 at 7:24:34 AM permalink
Quote: tuttigym

It is hard to conceive that that might be the current world record after all this time. It would seem that with all the current technology and improvements in communications, such records or numbers would be available even with all the privacy concerns. Or perhaps not.



As far as I know there's no way to measure and account for every individual bet at a craps table. Sure the dealers keep track of individual bets, but not a list of all individual bets over time. Once they collect a bet or pay it off, that's it.

You could measure and track all the action in two different ways: 1) with a Rapid Craps table or a similar semi-electronic deal; 2) by using RFID enabled chips and recording the result of each roll in the same system tracking the chips. With that info, assuming you only track pass line bets, you could have accurate results for any number of rolls you wanted, or any time period you chose. Tracking all other bets would require more ifnormation input, such as which player bet which hardway, hop, etc.
Donald Trump is a fucking criminal
cclub79
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May 31st, 2010 at 7:45:08 AM permalink
Quote: Nareed

As far as I know there's no way to measure and account for every individual bet at a craps table. Sure the dealers keep track of individual bets, but not a list of all individual bets over time. Once they collect a bet or pay it off, that's it.

You could measure and track all the action in two different ways: 1) with a Rapid Craps table or a similar semi-electronic deal; 2) by using RFID enabled chips and recording the result of each roll in the same system tracking the chips. With that info, assuming you only track pass line bets, you could have accurate results for any number of rolls you wanted, or any time period you chose. Tracking all other bets would require more ifnormation input, such as which player bet which hardway, hop, etc.



Score one (1!) for RapidCraps...finally.
tuttigym
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May 31st, 2010 at 2:27:36 PM permalink
When that gal in AC did her 154 roll for a "world record," I am sure that there wasn't anybody standing on the side with a clicker counting the rolls. They went to the video tape, I am guessing for the exact number. It would seem that if there was an anomaly happening during a session, there would be a record of such just to be placed on the record such as consecutive Come Out natural winners, Come out craps, consecutive successful passes, as well as very long rolls.

Oh well just a thought.

Thanks all for jumping in.

tuttigym
rudeboyoi
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May 31st, 2010 at 2:33:26 PM permalink
i think it was four queens downtown las vegas where they had a clicker on the table that the boxman would hit after each roll a 7 wasnt rolled.
7winner
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May 31st, 2010 at 2:42:55 PM permalink
Quote: rudeboyoi

i think it was four queens downtown las vegas where they had a clicker on the table that the boxman would hit after each roll a 7 wasnt rolled.

I know at the Freemont Casino downtown they have an electronic display that the box keeps rolls of each shooters hand. Someone should keep track of records and have bonus payouts when someone breaks a record!
7 winner chicken dinner!
DJTeddyBear
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May 31st, 2010 at 5:19:33 PM permalink
A few months ago, Showboat A.C. put in a similar clicker. High roll of the week won some silly minor prize.

The problem is, the boxman frequently forgot to click, then would remember and estimate the number of rolls he missed. It was a joke.


Yeah, that roll at Borgata had to go to the video to get the accurate count.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Tiltpoul
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May 31st, 2010 at 5:40:44 PM permalink
For a long time at Harrah's St Louis, they featured a promotion called Monster Roll. Each month, the person with the highest roll would win some sort of cash prize, along with the other top ten finishers. Then, at the end of the year, the top rollers from the months and possibly overall (not a heavy craps player here so don't remember) came back to do a "Finals" for 100,000 dollars I believe.

Anyway, it was not uncommon toward the end of the month to see the numbers be in the high 90s to low 100s. Not sure what the record was for that promotion, but 154 seems like a low number to me to be the world record.
"One out of every four people are [morons]"- Kyle, South Park
Nareed
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May 31st, 2010 at 6:12:46 PM permalink
Quote: 7winner

I know at the Freemont Casino downtown they have an electronic display that the box keeps rolls of each shooters hand. Someone should keep track of records and have bonus payouts when someone breaks a record!



The Fremont runs a promo called the "Sharpshooter Roll-athon," for which they need to keep track of all shots made. See The Wizard's review for more info.
Donald Trump is a fucking criminal
tuttigym
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June 1st, 2010 at 6:35:35 AM permalink
Quote: Tiltpoul


Anyway, it was not uncommon toward the end of the month to see the numbers be in the high 90s to low 100s. Not sure what the record was for that promotion, but 154 seems like a low number to me to be the world record.



That 154 roll was touted as a "world record" not yet challenged as inaccurate. Apparently the documentation is irrefutable.

The Wizard has written and stated that rolling the dice 75+ times w/o a 7 out is a 47,619 to 1 shot. That is where his calculations end. Doubling that number at 154 would be astronomical I guess.

Is that "high 90s to low 100s" any given month or consecutive months, and if so, how many months???

Thanks for the links, I am learning some stuff here and have some questions down the road.

tuttigym
Wizard
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June 1st, 2010 at 6:44:26 AM permalink
154 rolls is the documented record (source). My Ask the Wizard #81 gives the probability of making it to 200 rolls.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
rudeboyoi
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June 1st, 2010 at 11:36:24 AM permalink
Quote: 7winner

I know at the Freemont Casino downtown they have an electronic display that the box keeps rolls of each shooters hand. Someone should keep track of records and have bonus payouts when someone breaks a record!



ahh yes. i knew it was some casino downtown. i got the first letter right at least :p
goatcabin
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June 1st, 2010 at 1:34:24 PM permalink
Quote: Wizard

154 rolls is the documented record (source). My Ask the Wizard #81 gives the probability of making it to 200 rolls.



It's important to distinguish between a number of rolls without a seven and a number of rolls without a seven-out. It's simple to figure the probability of rolls without a seven; it's just .833333^ n, where n is a number of rolls. I once rolled 56 times without a seven, which has a probability of .0000368, or odds of 27,172 to 1 against. It's also relatively easy to calculate the probability of different numbers of passline decisions without a seven-out, since the probability of a seven-out on any decision is 784/1980, taken from "The Perfect 1980". So, the probability of 60 decisions without a seven-out is (1 - (784/1980) )^ 60 = 7.31 * 10^ -14, or odds of 13,678,568,090,000 to 1 against. Since the average number of rolls in a decision is 3.375, that would come out to roughly 200 rolls (202.5).

However, the Wizard's calculation using "states" is the correct method; it's just a lot more laborious.
Cheers,
Alan Shank
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
lennyjacky
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September 30th, 2010 at 1:20:29 PM permalink
I have a similar memory but with exceptions. I recall those dice on a pink silk pillow under glass on top of a pillar about five feet high in the casino at the Desert Inn. They were, as I recall, rolled more than 70 times by some school teacher on vacation. Many people at that table that celebrated night won big but the roller went home with more glory than cash. And, yes, it was in the 1950s. Maybe we're both right or maybe memory going back that far isn't all it's cut out to be.

lennyjacky
guido111
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September 30th, 2010 at 1:46:37 PM permalink
Quote: goatcabin

Quote: Wizard

154 rolls is the documented record (source). My Ask the Wizard #81 gives the probability of making it to 200 rolls.



However, the Wizard's calculation using "states" is the correct method; it's just a lot more laborious.
Cheers,
Alan Shank



For those of you that really LOVE working out math problems, there is another formula that can be used that is different from the Wizard's but arrives at the same answers... and it is not recursive or a matrix.

From the article:
"Our aim here is to give an explicit closed-form expression for them, showing that the distribution of L is a linear combination (not a convex combination) of four geometric distributions"

see: http://arxiv.org/abs/0906.1545v2 (for the article)
S. N. Ethier, Fred M. Hoppe

To read the complete article and see the formulas:
you can download a PDF free from the site.
or:http://arxiv.org/pdf/0906.1545v2
(to download just the pdf)

I re-created The Wizard's 200 roll table in excel using their formula in less than 3 minutes.
focd
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September 30th, 2010 at 3:50:42 PM permalink
Quote: tuttigym

3. Consecutive Come Out natural winners (7 & 11)?


I don't think this is a record, but I just "rolled" 7 consecutive 7's on a come out roll yesterday (card craps). This was ridiculous as I was betting on the don't pass side the whole time. So I lost 7 in a row. Then I got upset and moved it back to the pass side. Then I 7ed out later on. This is ridiculous.
goatcabin
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September 30th, 2010 at 4:49:46 PM permalink
Quote: focd

I don't think this is a record, but I just "rolled" 7 consecutive 7's on a come out roll yesterday (card craps). This was ridiculous as I was betting on the don't pass side the whole time. So I lost 7 in a row. Then I got upset and moved it back to the pass side. Then I 7ed out later on. This is ridiculous.



Odds against seven consecutive 7s are almost 280,000 to 1, if that makes you feel any better. Of course, once you switched, your chances of sevening out on the pass line were not affected. Perhaps it's a good thing that you lost the "dice". >:-)

Where was this unfortunate series of events?
Cheers,
Alan Shank
Woodland, CA
Cheers, Alan Shank "How's that for a squabble, Pugh?" Peter Boyle as Mister Moon in "Yellowbeard"
focd
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September 30th, 2010 at 5:55:02 PM permalink
Quote: goatcabin

Odds against seven consecutive 7s are almost 280,000 to 1, if that makes you feel any better. Of course, once you switched, your chances of sevening out on the pass line were not affected. Perhaps it's a good thing that you lost the "dice". >:-)


Seriously, I wanted to pass the dice in the middle, but it maybe be perceived rude. It wasn't that I was losing money. When I see patterns I don't like, I just want to quit or pass the dice. It happened in California.
Ayecarumba
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January 19th, 2012 at 11:59:13 AM permalink
Quote: Wizard

154 rolls is the documented record (source). My Ask the Wizard #81 gives the probability of making it to 200 rolls.



Has the "Ask the Wizard" archive been re-indexed? The link to ATW #81 is intact, but the content of the response is different. In this case, the description of the 200 roll probability is missing.
Simplicity is the ultimate sophistication - Leonardo da Vinci
TIMSPEED
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January 19th, 2012 at 12:59:10 PM permalink
I think it was in 09 or 10, but I rolled consecutively 104 rolls at Siena Casino in Reno, NV. (84 of which were box numbers, 20 horns)
Yes I was counting and had a backup counter aswell. (I counted the total rolls, he counted box numbers hit) I did this in just under 2½ hours (like 2:20 I think).
The table wasn't that full and no one made more than $10k.
Gambling calls to me...like this ~> http://www.youtube.com/watch?v=4Nap37mNSmQ
pacomartin
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January 19th, 2012 at 4:36:11 PM permalink
Quote: Ayecarumba

Quote: Wizard

154 rolls is the documented record (source). My Ask the Wizard #81 gives the probability of making it to 200 rolls.



Has the "Ask the Wizard" archive been re-indexed? The link to ATW #81 is intact, but the content of the response is different. In this case, the description of the 200 roll probability is missing.



Craps introduction mentions the calculation as a in 5.6 billion. (precisely it is 1: 5,590,264,072) Do you want the answer or how to do the calculation? There is a link to the problem in MathProblems.

You can have my spreadsheet using Markov transition matrix if you like. It uses the matrix multiplication function in Excel.

Probability is 1 in 8.81 you will 7-out on 17th roll or higher.
Probability is 1 in 92 you will 7-out on 33rd roll or higher.
Probability is 1 in 10,686 you will 7-out on 65th roll or higher.

I am trying to calculate A^153 where A is a matrix, but Excel does not have that function. So I have to calculate A, A^2, A^4,A^32,A^64,A^128,A^(128+32), etc.
If you use a more sophisticated software (like MatLab or hundreds of other programs) you can calculate it directly. But everyone has EXCEL. You need to read up on matrix multiplication if you didn't learn it, or you forgot it.
Ayecarumba
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January 19th, 2012 at 6:51:16 PM permalink
Quote: pacomartin

Quote: Ayecarumba

Quote: Wizard

154 rolls is the documented record (source). My Ask the Wizard #81 gives the probability of making it to 200 rolls.



Has the "Ask the Wizard" archive been re-indexed? The link to ATW #81 is intact, but the content of the response is different. In this case, the description of the 200 roll probability is missing.



Craps introduction mentions the calculation as a in 5.6 billion. (precisely it is 1: 5,590,264,072) Do you want the answer or how to do the calculation? There is a link to the problem in MathProblems.

You can have my spreadsheet using Markov transition matrix if you like. It uses the matrix multiplication function in Excel.

Probability is 1 in 8.81 you will 7-out on 17th roll or higher.
Probability is 1 in 92 you will 7-out on 33rd roll or higher.
Probability is 1 in 10,686 you will 7-out on 65th roll or higher.

I am trying to calculate A^153 where A is a matrix, but Excel does not have that function. So I have to calculate A, A^2, A^4,A^32,A^64,A^128,A^(128+32), etc.
If you use a more sophisticated software (like MatLab or hundreds of other programs) you can calculate it directly. But everyone has EXCEL. You need to read up on matrix multiplication if you didn't learn it, or you forgot it.



Thanks Paco. My original inquiry was the odds of rolling 117 times without throwing a seven (which is apparently the "Sharpshooter Roll-A-Thon" record at the Fremont). I'll check it out on Excel.

Edit: my search came up with the link to Ask the Wizard, but it does not display the craps question (if it is there). Could just be my browser.
Simplicity is the ultimate sophistication - Leonardo da Vinci
pacomartin
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January 20th, 2012 at 11:22:02 AM permalink
Quote: Ayecarumba

Thanks Paco. My original inquiry was the odds of rolling 117 times without throwing a seven (which is apparently the "Sharpshooter Roll-A-Thon" record at the Fremont). I'll check it out on Excel.



The question of not getting a 7 is a very simple calculation. No matrices are required. You need a matrix multiplication or a equivalently a set of recursive formulas to do the harder problem where you are permitted to roll a 7 on "come-out" rolls.

Probability of rolling n time without getting a 7 is P=(36/30)^n
n=117 1:1,837,408,781
n=123 1:5,486,473,222

Probability of rolling a 7 out on roll 154 or higher 1:5,590,264,072
buzzpaff
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January 20th, 2012 at 12:53:37 PM permalink
" The question of not getting a 7 is a very simple calculation." Not to the other 99% of us. No way !
charliepatrick
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January 20th, 2012 at 2:11:00 PM permalink
Many years ago I vaguely remember seeing a hall of fame list at a casino once, but don't think it went over 2 hours. However with a table of small betters, presumably there would be more players - if there weren't I'm sure it would become full fairly soon - and hence the dealers would be paying out more bets, possibly resulting in less rolls per hour. Nevertheless 4 hours sounds a very long time!
Wizard
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January 20th, 2012 at 3:59:35 PM permalink
Quote: Ayecarumba

Has the "Ask the Wizard" archive been re-indexed? The link to ATW #81 is intact, but the content of the response is different. In this case, the description of the 200 roll probability is missing.



When the site was revamped in December not everything went perfectly. I'll try to find the missing question.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
riverbed
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June 9th, 2012 at 8:33:32 AM permalink
I cannot find the chart that I used to access on this site that shows the probability for 'length of shooter', and it showed 1 to 200 rolls. I would love to see that chart again.
riverbed
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June 13th, 2012 at 7:19:48 PM permalink
I am eagerly awaiting a reference to that chart.
7craps
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ChumpChange
June 13th, 2012 at 8:15:10 PM permalink
Quote: riverbed

I am eagerly awaiting a reference to that chart.

Looks like it even disappeared from the Craps Q&A page.
That has to be a popular page to misplace.

I have a copy of it on my computer.
Here is a link to it
I will remove it once the Wizard gets the info back online.
No Charge :)

The question and table (Probability States in Craps Recursive Table 1 to 200) is about half way down

"According to the Las Vegas Advisor the record for the longest time a single shooter held the dice in craps is held by Stanley Fujitake, who once held the dice for three hours and six minutes at a downtown casino before sevening out.
(1) What are the odds that Mr. Fujitake could have accomplished his feat, assuming typical craps speed?
(2) What are the odds that this happened in Vegas since 1950?
- Veggie Boy"

Ask the Wizard: Craps

FYI:
riverbed, just go out and roll a 160 roll hand in Las Vegas.
That will get everyone interested in that table again!

added:
Length of a shooters hand
my table using SN Ethier's closed form formula
can be found at his website
Stewart Ethier, Professor

rollsor more1 inor lessrollsrelative1 in
30.888888888888881.111.1111111111%20.1111111119.0
40.772119341563781.322.7880658436%30.1167695478.6
50.667352537722901.533.2647462277%40.1047668049.5
60.576128908829951.742.3871091170%50.09122362911.0
70.497210870421172.050.2789129579%60.07891803812.7
80.429044106625212.357.0955893375%70.06816676414.7
90.370191348541172.762.9808651459%80.05885275817.0
100.319390698651603.168.0609301348%90.0508006519.7
110.275546561987293.672.4453438013%100.04384413722.8
120.237710425961294.276.2289574039%110.03783613626.4
130.205061925293064.979.4938074707%120.03264850130.6
140.176891903460855.782.3108096539%130.02817002235.5
150.152587568839846.684.7412431160%140.02430433541.1
160.131619560348467.686.8380439652%150.02096800847.7
170.113530703351438.888.6469296649%160.01808885755.3
180.0979262489642310.290.2073751036%170.01560445464.1
190.0844654102266111.891.5534589773%180.01346083974.3
200.0728540292655713.792.7145970734%190.01161138186.1
210.0628382289224915.993.7161771078%200.010015899.8
220.0541989199951018.594.5801080005%210.008639309115.7
230.0467470511745221.495.3252948825%220.007451869134.2
240.0403195029904024.895.9680497010%230.006427548155.6
250.0347755397068228.896.5224460293%240.005543963180.4
rollsor more1 inor lessrollsrelative1 in
260.0299937442632333.397.0006255737%250.004781795209.1
270.0258693711590338.797.4130628841%260.004124373242.5
280.0223120607722744.897.7687939228%270.00355731281.1
290.0192438661117752.098.0756133888%280.003068195325.9
300.0165975495492560.298.3402450451%290.002646317377.9
310.0143151127776969.998.5684887222%300.002282437438.1
320.0123465281949781.098.7653471805%310.001968585508.0
330.0106486442115293.998.9351355788%320.001697884589.0
340.00918424070887108.999.0815759291%330.001464404682.9
350.00792121410618126.299.2078785894%340.001263027791.7
360.00683187428823146.499.3168125712%350.00108934918.0
370.00589233806802169.799.4107661932%360.0009395361,064.4
380.00508200594980196.899.4917994050%370.0008103321,234.1
390.00438311076688228.199.5616889233%380.0006988951,430.8
400.00378032833215264.599.6219671668%390.0006027821,659.0
410.00326044158919306.799.6739558411%400.0005198871,923.5
420.00281205091828355.699.7187949082%410.0004483912,230.2
430.00242532425864412.399.7574675741%420.0003867272,585.8
440.00209178157754478.199.7908218422%430.0003335432,998.1
450.00180410896744554.399.8195891033%440.0002876733,476.2
460.00155599830003642.799.8444001700%450.0002481114,030.5
470.00134200892494745.299.8657991075%460.0002139894,673.1
480.00115744838330864.099.8842551617%470.0001845615,418.3
490.000998269522471,001.799.9001730478%480.0001591796,282.2
500.000860981757421,161.599.9139018243%490.0001372887,284.0
rollsor more1 inor lessrollsrelative1 in
510.000742574533821,346.799.9257425466%500.0001184078,445.4
520.000640451315591,561.499.9359548684%510.0001021239,792.1
530.000552372649601,810.499.9447627350%528.80787E-0511,353.5
540.000476407059782,099.099.9523592940%537.59656E-0513,163.9
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570.000305644193263,271.899.9694355807%564.87366E-0520,518.5
580.000263610138403,793.599.9736389862%574.20341E-0523,790.2
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rollsor more1 inor lessrollsrelative1 in
760.0000183818189354,401.699.9981618181%752.93108E-06341,170.8
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rollsor more1 inor lessrollsrelative1 in
1010.000000455025442,197,679.399.9999544975%1007.25563E-0813,782,392.5
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rollsor more1 inor lessrollsrelative1 in
1260.0000000112637588,780,411.199.9999988736%1251.79607E-09556,772,080.0
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1490.000000000374832,667,852,146.399.9999999625%1485.97693E-1116,730,997,178.0
1500.000000000323283,093,256,045.499.9999999677%1495.15494E-1119,398,856,069.4
rollsor more1 inor lessrollsrelative1 in
1510.000000000278823,586,492,968.099.9999999721%1504.446E-1122,492,132,184.8
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1530.000000000207414,821,456,026.899.9999999793%1523.30721E-1130,236,966,550.2
1540.000000000178885,590,264,071.899.9999999821%1532.85237E-1135,058,517,488.9
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1560.000000000133067,515,200,124.799.9999999867%1552.12177E-1147,130,474,563.3
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1590.0000000000853711,713,937,159.499.9999999915%1581.36124E-1173,462,191,132.4
1600.0000000000736313,581,789,749.499.9999999926%1591.17404E-1185,176,071,932.7
1610.0000000000635015,747,481,848.999.9999999936%1601.01258E-1198,757,735,373.5
1620.0000000000547718,258,505,628.399.9999999945%1618.73324E-12114,505,088,285.8
1630.0000000000472421,169,926,149.299.9999999953%1627.5322E-12132,763,387,399.6
1640.0000000000407424,545,588,904.699.9999999959%1636.49625E-12153,935,010,249.7
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1670.0000000000261438,259,192,196.899.9999999974%1664.16778E-12239,936,048,341.5
1680.0000000000225444,359,833,703.099.9999999977%1673.59457E-12278,197,462,851.4
1690.0000000000194451,433,256,510.999.9999999981%1683.1003E-12322,549,659,972.8
1700.0000000000167759,634,576,022.799.9999999983%1692.67386E-12373,991,000,446.0
1710.0000000000144669,143,641,656.399.9999999986%1702.30616E-12433,622,147,830.8
1720.0000000000124780,168,980,822.099.9999999988%1711.98896E-12502,774,169,954.8
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rollsor more1 inor lessrollsrelative1 in
1760.00000000000690144,884,796,662.599.9999999993%1751.10056E-12908,624,962,649.1
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1780.00000000000513194,774,026,407.499.9999999995%1778.18678E-131,221,480,777,697.5
1790.00000000000443225,831,830,861.899.9999999996%1787.05991E-131,416,449,010,023.7
1800.00000000000382261,841,975,396.199.9999999996%1796.09068E-131,641,851,851,028.3
1810.00000000000329303,594,138,247.599.9999999997%1805.25135E-131,904,270,455,547.8
1820.00000000000284352,003,916,250.799.9999999997%1814.53082E-132,207,105,918,829.0
1830.00000000000245408,132,903,260.599.9999999998%1823.90687E-132,559,590,581,057.4
1840.00000000000211473,211,970,190.799.9999999998%1833.36953E-132,967,775,701,726.9
1850.00000000000182548,668,257,185.099.9999999998%1842.90656E-133,440,488,638,174.6
1860.00000000000157636,156,469,839.799.9999999998%1852.50577E-133,990,783,896,650.9
1870.00000000000136737,595,165,784.299.9999999999%1862.1616E-134,626,193,762,065.2
1880.00000000000117855,208,827,358.799.9999999999%1871.86517E-135,361,428,127,822.0
1890.00000000000101991,576,643,014.799.9999999999%1881.6076E-136,220,441,474,268.6
1900.000000000000871,149,689,067,182.799.9999999999%1891.38778E-137,205,759,403,792.8
1910.000000000000751,333,013,398,924.699.9999999999%1901.19571E-138,363,230,505,794.8
1920.000000000000651,545,569,817,469.899.9999999999%1911.0314E-139,695,585,850,098.0
1930.000000000000561,792,019,542,039.899.9999999999%1928.90399E-1411,230,921,764,016.2
1940.000000000000482,077,767,049,248.8100.0000000000%1937.67164E-1413,035,020,629,147.6
1950.000000000000422,409,078,589,639.8100.0000000000%1946.61693E-1415,112,750,427,417.8
1960.000000000000362,793,219,602,341.5100.0000000000%1955.70655E-1417,523,733,958,640.1
1970.000000000000313,238,614,041,259.3100.0000000000%1964.92939E-1420,286,484,807,975.2
1980.000000000000273,755,029,106,716.0100.0000000000%1974.24105E-1423,579,055,640,683.2
1990.000000000000234,353,789,433,581.8100.0000000000%1983.66374E-1427,294,543,196,184.8
2000.000000000000205,048,025,432,896.4100.0000000000%1993.16414E-1431,604,207,911,371.9
2010.000000000000175,852,961,232,947.0100.0000000000%2002.72005E-1436,764,078,590,779.6
winsome johnny (not Win some johnny)
riverbed
riverbed
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Joined: Jan 14, 2010
June 14th, 2012 at 12:36:19 PM permalink
Thank you so much! I've been looking for this chart for a couple of weeks. And I certainly do intend to go out and roll a 160 roll hand as soon as possible.

Just the other day a lady on my right rolled 63 times, which is only one in 7949 shooters, fewer than I thought. The probability really jumps along as the roll lengthens. I used to chart the intervals of the sevens, and I found that process quite instructive in understanding the game. I don't think that very many players that I have played with realize that half the shooters roll 8.5 times or less, and this has profound implications for your betting and money management strategies. I notice from the chart that the proportion of any roll number to the previous one is .8562 or so, whereas the intervals between sevens were at a proportion to the next interval of .83333, a 6:5 ratio. I guess the difference is the seven-out/come-out factor.

Another question: I notice that the median roll in 8.5, yet the probability of 2 in the table (I assume this might be expressed as 'one chance in two') is right at 7. Why the difference?
7craps
7craps
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Joined: Jan 23, 2010
June 14th, 2012 at 1:14:53 PM permalink
Quote: riverbed

Thank you so much! I've been looking for this chart for a couple of weeks. And I certainly do intend to go out and roll a 160 roll hand as soon as possible.

Just the other day a lady on my right rolled 63 times, which is only one in 7949 shooters, fewer than I thought. The probability really jumps along as the roll lengthens. I used to chart the intervals of the sevens, and I found that process quite instructive in understanding the game. I don't think that very many players that I have played with realize that half the shooters roll 8.5 times or less, and this has profound implications for your betting and money management strategies. I notice from the chart that the proportion of any roll number to the previous one is .8562 or so, whereas the intervals between sevens were at a proportion to the next interval of .83333, a 6:5 ratio. I guess the difference is the seven-out/come-out factor.

Nice observations.
The ratio for the 7 is always the same.
For the shooter 7out, there is a difference in the first few iterations but as you have shown once past that, the ratio converges very quickly.
A transition matrix also shows this.
Quote: riverbed

Another question: I notice that the median roll in 8.5, yet the probability of 2 in the table (I assume this might be expressed as 'one chance in two') is right at 7. Why the difference?

No, the median is not 8.5 rolls
8.5 is the average number of rolls per shooter (8.525510204 or 1671/196 to be exact)

6 is the median. (look at my above table or the photo below)
about ~50% (50.2789129579%) of all shooters will 7out by the 6th roll so the other half will get past the 6th roll.
length of a shooter's hand - cumulative probability
winsome johnny (not Win some johnny)
7craps
7craps
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Joined: Jan 23, 2010
June 14th, 2012 at 2:44:31 PM permalink
Quote: pacomartin

You can have my spreadsheet using Markov transition matrix if you like. It uses the matrix multiplication function in Excel.

I am trying to calculate A^153 where A is a matrix, but Excel does not have that function. So I have to calculate A, A^2, A^4,A^32,A^64,A^128,A^(128+32), etc.
If you use a more sophisticated software (like MatLab or hundreds of other programs) you can calculate it directly. But everyone has EXCEL. You need to read up on matrix multiplication if you didn't learn it, or you forgot it.

The matrix algebra program does not have to be sophisticated.
One needs to use something else than Excel sledgehammer methods.

I am not a Matrix Algebra expert by any means, some may think I am, unlike ChrystalMath
A good lesson here
Markov Systems

FREE online program
JavaScript. I have yet to see this one fail.
Matrix Algebra Tool
winsome johnny (not Win some johnny)
guido111
guido111
  • Threads: 10
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Joined: Sep 16, 2010
June 14th, 2012 at 3:19:16 PM permalink
Quote: pacomartin


Craps introduction mentions the calculation as a in 5.6 billion. (precisely it is 1: 5,590,264,072) Do you want the answer or how to do the calculation? There is a link to the problem in MathProblems.

You can have my spreadsheet using Markov transition matrix if you like. It uses the matrix multiplication function in Excel.

I am trying to calculate A^153 where A is a matrix, but Excel does not have that function. So I have to calculate A, A^2, A^4,A^32,A^64,A^128,A^(128+32), etc.
If you use a more sophisticated software (like MatLab or hundreds of other programs) you can calculate it directly. But everyone has EXCEL. You need to read up on matrix multiplication if you didn't learn it, or you forgot it.

I thought I posted this here before, maybe at another forum.
One can figure this in Excel with a simple closed form formula (no matrix multiplication needed)

Quite easy actually.
closed form formula using eigenvectors and eigenvalues (remember math class with matrix algebra???)

The explicit closed-form expression for the length of a craps shooter hand is:

=(_c1*(e1_^(Roll))+(_c2*(e2_^(Roll))+(_c3*(e3_)^(Roll))+(_c4*(e4_)^(Roll))))

values
c1 1.211844812464510000
c2 -0.006375542263784770
c3 -0.004042671248651500
c4 -0.201426598952082000

e1 0.862473751659322000
e2 0.741708271459795000
e3 0.709206775794379000
e4 0.186611201086502000

From the pdf:
A world record in Atlantic City
and the length of the shooter痴 hand at craps
S. N. Ethier and Fred M. Hoppe
length of the shooter痴 hand at craps free pdf

yes, this is sick math.
These guys had nothing better to do one afternoon!

Enjoy!
mustangsally
mustangsally
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June 14th, 2012 at 5:31:14 PM permalink
removed
silly
Sally
I Heart Vi Hart
riverbed
riverbed
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Joined: Jan 14, 2010
June 14th, 2012 at 6:45:33 PM permalink
Many thanks, 7craps. As a back line better, predominantly, this huge difference between the median (6) and the mean (8.5) is the crux of the game for me. I run a progression on the back line, and I'm standing there trying desperately to avoid those long rolls. Or at least recognize the hot shooter who is going to get my whole stake onto the back line before I switch to the front line. I'm constantly experimenting w/ switching over earlier in these long rolls so that my losses on individual shooters (those that 7out right after I have switched to the front line) are smaller (but more frequent) and my wins on the long rolls are bigger. Your chart illustrates so well where that 'ideal' switching-over point might be according to how hot/cold the situation seems to be.

Again, many thanks.
7craps
7craps
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Joined: Jan 23, 2010
June 14th, 2012 at 7:29:35 PM permalink
Quote: riverbed

Many thanks. But what accounts for the median and the average rolls per shooter being different?

The distribution. The probabilities of each possible outcome.

It is not normal where the mean (average), median and mode are about the same.

here is a photo of the relative probabilities
One can see the mode is equal to 3.
We can not see the median unless we look at the cumulative graph.
And the average is nowhere near a peak.

In most geometric type distributions, the median is less than the mean.
crank out the old stats book for more
winsome johnny (not Win some johnny)
riverbed
riverbed
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June 15th, 2012 at 10:52:58 AM permalink
Yes, I get this now. The few long rolls skew the mean 40% above the median. Fully 60& of shooters have shorter rolls than average (the mean). This is a crucial description of the game. Your strategy had better accommodate all these short rolls! Your chart is highly instructive.
Ace2
Ace2
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November 13th, 2019 at 10:48:48 AM permalink
A decent estimate for this probability, expressed as 1 in x, is 1.2^((n-1) * 0.802). So the estimate of lasting 154 rolls is 1.2^122.7 = 1 in 5.2 billion. Accurate within 7%

On average there is a comeout roll every 5.06 rolls, so 4.06/5.06 = 0.802 rolls aren稚 comeout rolls. For each of these rolls there痴 a 5 in 6, or 1 in 1.2, chance a seven won稚 be rolled.
It痴 all about making that GTA
ChumpChange
ChumpChange 
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November 13th, 2019 at 1:57:10 PM permalink
So am I supposed to add one PB to the table every 6 rolls? #Damn7outs
Ace2
Ace2
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Joined: Oct 2, 2017
November 14th, 2019 at 6:12:16 AM permalink
For more accuracy (within 1%), use:

3.131 * 1.159456^(n-10)

3.131 being the value for 10, 1.159456 being the sustained rate for each iteration after 10.

So the value for 154 is 3.131 * 1.159456^144 = 5.601 x 10^9
It痴 all about making that GTA
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