Question of the Day - 14 March 2021

It makes me sick every time I hear the Borgata craps story, and the Stanford odds are quoted. I would have thought that you were above that, or at least will include a clarification about the game of craps. 

Yes, Patricia Demauro had an epic roll, but probably included many sevens. Just that they were all come out sevens. Stanford’s odds doesn’t consider that. 

The Borgata time is printed on the felt of the table where it occurred.  One time a young woman next to me asked the stickman about breaking the record to which he replied he would simply be impressed if she held on to the dice for the 18 minutes.

Dave is right, lots of sevens, just none after the point was set. for more than four hours.

I ever played craps I held the dice for 10 minutes at an empty craps table. Half that time was asking the stick man how to play. I made a point (5+1) on the come out roll and several throws latter threw a 3+3.  I got nervous thinking I was about to lose and the stick man said I could either collect my winnings (a couple of dollars) or bet the negative.  What's the negative?  He said bet that I would throw a losing number and told me what those were.  So I bet behind the line and rolled snake eyes.  The stick man was surprised. The look on his face told me he didn't expect a win so I grabbed my money and left the table.

Since then I studied everything about the game. Books don't tell you everything!

The probabililty of 154 consecutive throws (starting with a comeout) without missing a point is about 1 in 6,481,662,846.

Let P(N,K) be the probability of reaching 154 after N rolls, where K = 0 if it is a comeout or K = the current point
P(154,K) = 1 for all K
For all N < 154:
On a comeout roll, of the 36 possibilities, 12 don't establish a point, 3 are a 4, 4 are a 5, 5 are a 6, 5 are an 8, 4 are a 9, and 3 are a 10:
P(N,0) = 1/3 P(N+1,0) + 1/12 P(N+1,4) + 1/9 P(N+1,5) + 5/36 P(N+1,6) + 5/36 P(N+1,8) + 1/9 P(N+1,9) + 1/12 P(N+1,10)
If there is already a point, either you make the point, the point remains the same, or you miss the point (and the run ends unsuccessfully):
P(N,4) = 1/12 P(N+1,0) + 3/4 P(N+1,4)
P(N,5) = 1/9 P(N+1,0) + 13/18 P(N+1,5)
P(N,6) = 5/36 P(N+1,0) + 25/36 P(N+1,6)
P(N,8) = 5/36 P(N+1,0) + 25/36 P(N+1,8)
P(N,9) = 1/9 P(N+1,0) + 13/18 P(N+1,9)
P(N,10) = 1/12 P(N+1,0) + 3/4 P(N+1,10)

Work backwards from N = 153 to N = 0; assuming the first roll is a comeout, the answer is P(0,0).

This was the subject of a research paper, which can be found online at https://arxiv.org/pdf/0906.1545.pdf.  The first paragraph reads as follows:

It was widely reported in the media that, on 23 May 2009, at the Borgata Hotel Casino & Spa in Atlantic City, Patricia DeMauro, playing craps for only the second time, rolled the dice for four hours and 18 minutes, finally sevening out at the 154th roll. Initial estimates of the probability of this event ranged from one chance in 3.5 billion [6] to one chance in 1.56 trillion [11]. Subsequent computations agreed on one chance in 5.6 (or 5.59) billion [2, 7, 10].

I am a veteran craps player, and nothing holds up a dice game more than a penny-ante player who thinks he's going to hit the "lottery" by playing all of the props (in the middle of the table).  Those are by far the worst bets anyone can make in a craps game, and the dealers have to always slow down the game in order to pay the nut jobs that bet them the few times they hit.  In the case of those players that make several prop bets, the dealers have to consult one another about the calculated payout.  Nothing frustrates "pure" players more than someone who can't calculate the odds and make dumb bets.

So epic rolls that last an hour or more usually contain lots of "down" time when the dealers have to pay off prop bets, not to mention the 7's, 11's, and craps rolled on come-outs.  Don't think that the shooter is rolling the dice all of the time that is reported.

Does the Craps hall of fame include rolls from across the US or just from the properties owned by the Call. I have walked over the tunnel a few times and stopped to look.  I first learned to play Craps at the Tez.  I was and still am a avid BJ player. But made my way to the tables once and played just the come-out roll and odds for a long time before I nice pit boss said to me you do not know this game do you. The tables were empty and he spent some time teaching me about the come bet and buying points.  I still love the game and his time helping me made all the difference.  I think his name was Tim and as of a year or so ago he was still there in the BJ pit Silver hair

It's been bugging me for hours, so I busted open Excel to figure out just how wrong Stanford was.

30 / 36 = .833 333 333...
That's the odds of not rolling a seven.

( 30 / 36 ) ^ 154 = 0.000 000 000 000 639 864 637
That's the odds of not rolling a seven 154 times.

1 / ( ( 30 / 36 ) ^ 154 ) = 1,562,830,547,008.8
That's the reciprocal. That's where they got the 1 in 1.56 trillion figure.

A mathematician with a slight clue about craps should have come up with this:
1 / ( ( 30 / 36 ) ^ 153 ) * ( 6 / 63 ) = 124,034,170,397.5
That's 1 in 124 billion chances of exactly 153 rolls without a seven, followed by a seven.

The valid point has been made that it should be the number of rolls, not the amount of time, that decides if someone gets in a craps hall of fame. Paying off prop bets in the middle of the table is only 1 way delays happen; there's also no action when a guy disputes the amount he was paid on a placed or laid number, or someone buys in or cashes out, or a die bounces off the table, etc. Also, by the way (and you guys will all throw tomatoes at me for pointing this out), the shooter doesn't actually hold the dice for 3 or 4 hours; as soon as he/she throws the dice, he/she is no longer holding them. So the term that should be used is "having possession of the dice". What I'm curious about, but have never heard anything on it, is whether Mr. Fujitake or Ms. Demauro utilized "dice controlling", the dice-throwing method espoused by Frank Scoblete, invented and developed by an East Coast craps master named "The Captain". If neither did this, then their feats were even more impressive.

If something is possible then it will eventually occur, no matter how small the chance. Just need enough trials. Any craps turn over 30 minutes is impressive to me!

"Subsequent computations agreed on one chance in 5.6 (or 5.59) billion"

This is the probability of 153 consecutive rolls (starting with a comeout roll) without missing a point. Specifically, it is about 1 in 5,590,264,072. I could show you an exact fraction, but the denominator in lowest terms has 237 digits.

No, That Don Guy, 1 in 5.59 billion is correct.  It is the probability that the seven out occurs at the 154th roll or later.  You can confirm this number in the arXiv paper I cited, in three of the references in that paper, in the published version of that paper in The Mathematical Intelligencer (2010), in the book, The Doctrine of Chances (2010), and at the Wizard of Odds web site.  In short, your derivation contains a minor error.

For the scenario of 153 non-losing rolls followed by Roll 154 being a losing roll, couldn't that loser Roll 154 be your rolling craps (2, 3 or 12) on a come out roll? I would think so. The analysis comments from a couple contributors only mentioned Roll 154 losing due to a "seven out", which would be the majority of occurrences but not all of them. Does the calculation result number that was stated by more than one person only reflect Roll 154 losing via a "seven out", or does it include rolling craps on the come out?