Here is something that I found on the CHANCE wiki:
"A new record in craps"
Holy Craps! How a Gambling Grandma Broke the Record
Time.com, 29 May 2009 by Claire Suddath
On May 23, a New Jersey woman named Patricia Demauro set a new world record for the longest turn at craps without "sevening out" by rolling the dice 154 consecutive times. Unfortunately, the article misstates the probability calculation needed to compute the odds of her feat:It sounds like a homework problem out of a high school math book: What is the probability of rolling a pair of dice 154 times continuously at a craps table, without throwing a seven? The answer is roughly 1 in 1.56 trillion...
It is true that the chance of going 154 (or more) consecutive rolls without rolling a 7 is
(56)154, which is indeed about 1 in 1.56 trillion. Since presumably Patricia failed on the last roll, it would be more accurate to find her chance of rolling more than 153 times. The more serious issue, however, is that "sevening out" is a more complicated event, and refers to rolling a seven after the player has established a "point".
To describe this more fully, we need to review the rules of craps. The player rolling the dice is called the "shooter". The basic bet is called the "Pass" bet, and although many side bets that can be placed during the course of action, it is the Pass bet that governs the play. The shooter begins her turn with an initial roll of the dice. If this is a 2, 3, or 12, the Pass bet loses; if this is a 7 or 11 the pass bet wins; but in either case, the shooter maintains possession of the dice and rolls again. This continues until a 4, 5, 6, 8, 9, or 10 appears, which establishes the shooter's "point". Once a point is established, the game enters a second phase, with the shooter now rolling repeatedly until she either reproduces the point, in which case the Pass bet wins, or else she rolls a seven, in which case the Pass bet loses. The latter option is called "sevening-out", and this is the event upon which the shooter must surrender the dice. In the former case, where the shooter reproduces her point before rolling a 7, she maintains possession of the dice and the whole process starts over.
The initial sequence of rolls prior to establishing a point are called "come-out rolls". It is important to recognize that any number of come-out rolls may produce 7's without ending the shooter's turn with the dice. The flaw in the Time.com analysis was the assumption that any 7 would end the turn. The more complicated process of sevening out can be modeled using an absorbing Markov chain, with state space {0,1,2,3,4} defined by
| 0: |
come out rolls |
|---|
| 1: |
point is 4 or 10 |
|---|
| 2: |
point is 5 or 9 |
|---|
| 3: |
point is 6 or 8 |
|---|
| 4: |
sevened out |
|---|
| n |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
| nth roll |
_ |
2 |
7 |
11 |
5 |
8 |
12 |
4 |
5 |
7 |
4 |
10 |
7 |
| state of chain |
0 |
0 |
0 |
0 |
2 |
2 |
2 |
2 |
0 |
0 |
1 |
1 |
4 |
The probability transition matrix P for the Markov chain is given below:
|
0 |
1 |
2 |
3 |
4 |
|---|
| 0 |
12/36 |
6/36 |
8/36 |
10/36 |
0 |
|---|
| 1 |
3/36 |
27/36 |
0 |
0 |
6/36 |
|---|
| 2 |
4/36 |
0 |
26/36 |
0 |
6/36 |
|---|
| 3 |
5/36 |
0 |
0 |
25/36 |
6/36 |
|---|
| 4 |
0 |
0 |
0 |
0 |
1 |
|---|
To analyze the chain, we use standard absorbing chain theory, following notation from Grinstead and Snell's Introduction to Probability text. (Concise notes from a recent Dartmouth course are here). The leading 4x4 submatrix corresponding to the transient states is denoted
Q
. The probabilities of still being in particular transient states after 153 rolls is given by entries in the vector(1,0,0,0)Q153
, and the sum of these probabilities is the chance of not having sevened out after 153 rolls. It is easy to implement this computation iteratively, giving1.788824×10?10
, which approximately 1 in 5.59 billion. This certainly is a small probability, but much larger than the 1 in 1.56 trillion from Time.com.You can find a discussion of this problem in Crunching the numbers on a craps record from Carl Bialik's Wall Street Journal column, "The Numbers Guy". He explains why the sevening-out event is complicated, and reports on various attempts to solve the problem. In the final update, he reports that Keith Crank of the American Statistical Association used a Markov chain analysis (which presumably corresponds to what we did above) to find a probability of 1 in 5.6 billion. Earlier in the article he cites simulation results from Professor Michael Shackleford, referencing Shackleford's Wizard of Odds website. Apart from the simulation results, Shackleford presents a recursive computation scheme that is equivalent to the Markov chain. You can also read there interesting historical notes on craps records.
Returning to the Time.com piece, we read "The average number of dice rolls before sevening out? Eight." The true value is about 8.5, but it doesn't follow from their original analysis, which asked only how long it takes to roll a seven (the expected number of rolls is 6). Using the Markov chain model, the expected time to absorption is the sum of the row zero entries in the fundamental matrix
N=(I?Q)?1
. Computing this gives an expected 8.5 rolls to seven-out.Time also interviewed Professor Thomas Cover of Stanford, who pointed out that while the probability of the event in question is small, one needs to remember that there are many people playing craps at any time, all of whom are in principle contending for the record! Shackleford estimates their are about 50 million craps turns per year in the US, giving about a 1% chance that a feat like Demauro's would occur in a given year.
DISCUSSION QUESTION:
How do you think Shackleford arrived at the 50 million estimate? Can you use the earlier post on Guesstimating to come up with a figure?
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Originally Posted by Oneoffthecount
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