What are odds of going on 154 roll run?

While playing craps at the Borgata Hotel Casino & Spa on Saturday night, Patricia Demauro of Denville set a new record for the longest craps roll by hanging on for four hours and 18 minutes. Borgata officials said that bested the previous mark of three hours and six minutes, set nearly 20 years ago at a Las Vegas casino.

Demauro bought into a game for $100 at about 8:13 p.m. and quickly amassed a large crowd of cheering supporters. When she eventually "sevened out" around 12:31 a.m., after 154 rolls of the dice, she was greeted by Borgata with a champagne toast.

I was just going to post and ask this!

what did he cash out with? thats so crazy

It was a she and rumor has it about 10k.

I really want to know the odds. Anyone?

"According to Michael Shackleford -- a statistician dubbed the Wizard of Odds and an adjunct professor at the University of Nevada, Las Vegas -- DeMauro had only one chance in about 3.5 billion to roll 154 times."

A freak set of unbalanced dice? Or loaded dice somehow slipped in?

I wonder who started counting and when...

I'm sure after about an hour or so they did a video replay to count it up.

You mean they have cameras in casinos now?

Hmm...makes sense.

Can some of the more math-minded among us come up with real-life ANALOGIES to help someone understand the magnitude of this roll?

Something along the lines of:

"Someone rolling 154 times in craps without sevening out is like...."
-getting dealt pocket aces in texas holdem ____ times in a row.
-hitting your number ___ times in a row on a double-zero roulette wheel.
-an average skilled player winning __ ____-player poker touraments in a row.
Or some other easily-understandable analogy(?)

Also correct me if I'm wrong but wouldn't one need to know how many POINTS this woman established (and how many come out 7s and 11s she rolled) during the course of her 154 throws to get a true calculation of the odds?

Simply stated, her ROLL lasted for 154 rolls total. She (likely) didn't establish a point on her first roll and then dodge a seven 152 times. Similarly, she (likely) didn't avoid throwing sevens or elevens during her come-out rolls (which would be included in the 154 total, but during which she was never at risk of her roll ending on that throw). And lastly, she (likely) didn't make her point on every first try, and thus half of her rolls were come-out-rolls (during which she was neve at risk of her roll ending on that throw).

I suppose one could assume an 'average' number of points established and 'come out 7s and 11' and work from there.

Right?

It is roughly the same chance as a player being dealt AA five times during one orbit of a 9 player table.


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100->10k on a 3.5billion to 1 shot
I would be pissed!!!!

It is easy to solve this problem exactly, and you just need a program that can multiply 4x4 matrices. Excel can do this. I found the exact probability of rolling at least 154 times before missing a point to be about 1 in 5.6 billion. The wizardofodds website is reporting the number 1 in 5.3 billion (not 3.5 billion as Mizzles has posted), but this is based on a monte carlo simulation rather than an exact calculation, and he admits that he did not run enough samples for accuracy at 154 rolls, and he is extrapolating his error. This isn't the first time that the wizard has relied on simulation for something that can be computed exactly with relative ease, and then reported more digits than could be justified by the size of the simulation. This problem cries out for a state space solution as shown below.

I heard the number 1 in 1.56 trillion in the media, but this is simply (30/36)^154 which is the probability of no sevens in 154 rolls. This is not sufficient since it is OK to have sevens on the come out roll. It is also OK to LOSE on the come out roll (with 2,3, or 12) according to the article at wizardofodds.com, as the dice don't get passed in this case. If you wish to know the probability of at least 154 rolls before losing a point or losing the come out bet, I can compute this by only changing a single number in my matrix, and this gives about 1 in 260 billion.

For the original problem, the solution is as follows:

Explanation

Define 4 states:
State 0 is the come out roll.
State 1 is rolling for a point of 4 or 10, both of which are 6:3 underdogs
State 2 is rolling for a point of 5 or 9, both of which are 6:4 underdogs
State 3 is rolling for a point of 6 or 8, both of which are 6:5 underdogs

The 4x4 matrix A above completely characterizes the game of craps, as far as the pass line bets are concerned. The numbers in the matrix are state transition probabilities, with the 1/36 factored out for convenience. The number in (ROW, COLUMN) is the probability that we transition TO state ROW, FROM state COLUMN. So for example, numbers in the first column give the probabilities when we are on the come out roll, on which we have a 12/36 probability of rolling a 7/11 or 2/3/12, either of which would bring us to another come out roll. There is a 6/36 probability that we roll a 4 or a 10 which would put us in state 1. There is an 8/36 probability that we roll a 5 or a 9 which would put us in state 2. And there is a 10/36 probability that we roll a 6 or an 8 which would put is in state 3. The columns for the other states each give 2 non-zero probabilities; one for making the point and returning to the come out state 0, and the probability for any number besides the point or 7, which leave us in the same state. Note that we do not require a state for rolling a 7 which would terminate the player's rolls.

Now the column vector [1 0 0 0] represents the state before the first come out roll, where we are in state 0 with probability 1. When the transition matrix multiplies this vector, the elements of the resultant vector indicate the probabilities of being in each state after the roll. For each roll this matrix is multiplied by the resultant vector from the last roll, until after multiplying it 153 times, the resultant vector will give the probabilities of being in each of the 4 states after 153 rolls, each of which satisfy the conditions of the problem. Note that these will not sum to 1 since the cases where we seven out are not represented. We simply need to sum these 4 elements to give our desired probability.

EDIT: Points 8 and 10 were swapped in the state definition and explanation.
EDIT: Factor of (1/36) is part of matrix which gets raised to power of 153.

I contacted the wizard, and he has rewritten his article using a modified version of my method which eliminates the matrices, and he gets my results. He says that this is the longest answer to any "ask the wizard" question.

FWIW, it is pretty important that we compute the number exactly. I mean, I'd really hate to find out that a 5.3 billion to one shot was really 5.6 billion to one.



Sherman

Wouldn't that be the same probability as that of the Cubs winning the World Series?

Low blow. 2 month ban minimum imo.

Sherman

So who cares about a 1 in 5.6 billion chance. How many chances (count the number of 7 outs since each time they are ending a streak) have there been in Vegas casinos since the first casino opened? Where I come from the probability of winning the lottery is 1 in 14 million give or take, yet in approximately half the draws there is a winner.

good point, cboevey.
1 in 5.6 billion chance it will happen on my next throw, and it happens, I am impressed. It is remarkable.
It's not like it's a 1 in 5.6 billion chance to happen at some casino somewhere in the world at some year during our lifetime. Remarkable for that person who does it, but not quite as remarkable that it happened.

If we assume 100 active craps tables across North America playing 24x7 (seems vaguely reasonable). 100 rolls per hour and the average time between seven-out as 8.5 rolls, means 11.7 chances per hour per table of going on a monster roll (*).

Thats 11.7*24*100*365 = 10,249,200 chances per year of hitting a long roll.

Chances of a 1 in 5.6 billion shot coming up = 1 in 546. It's a 1 in 500 year event, by my approximations. This is heavily approximated and with more rolls per hour, more tables in action this number drops.

(*) I think... I've not really taken into account long and short rolls in there, so not convinced that 11.7 new sequences per hour is correct, but it'll do.

firebet ftw

The previous comment by BruceZ about how to do the calculation was completely correct in the setup and was explained very well, but somehow the answer is slightly off. The result of the calculation is 1 : 5,590,264,072 (5.6 billion).
What is interesting is how much higher these odds are than the previous record set 20 years ago of 119 rolls of the Dice at California casino in Las Vegas. The odds of 119 rolls of the dice are 1 : 31,516,527 (31.5 million).
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Excel does not let you do a matrix to a power, so you have to do a bunch of multiplications of matrices to evaluate the solution.

What do you mean? I said 1 in 5.6 billion in the first post in this thread. That's the only number I ever reported. I got the exact number 5,590,264,072 on my first run of the spreadsheet, and that number now appears in the solution to the Wizard's problem 204 http://mathproblems.info/group11.html. Many other numbers were reported by others in the media.


Right, but a single cut and paste of an MMULT function into 153*4 cells will allow you to multiply the matrix by a 4-vector 153 times.