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Galois and ory, h ai e tcasi the is craps at bet basic The enda olw.Tefis oli the is roll first The follows. as defined fanme eogn to belonging number a If f7o 1apas(a appears 11 or 7 If h eesot h euneo ol yteshooter the by rolls of the sequence called is The out. sevens she beginning roll, again. come-out process new the a is decision a following hc aetebto oe.Telte vn scalled is event latter The a loses. (or bettor repeated the is case which point the until made rolled be the to becomes continue number that appears, ,o 2apas(a appears 12 or 3, stepoaiiyo civn ada es slong as least at hand a achieving of probability the is h alo h itiuinof distribution the of tail the ad(.. h ubro ol)i admvariable random a by is denote rolls) will of decisions we The number losing the out. (i.e., and seven hand 7s the to winning prior contain can hand ee out seven hoe spritdt oltedc ni eor he until dice the roll to permitted is shooter A ,i hc aetebto is r7apas in appears, 7 or wins, bettor the case which in ), π j := i asee oe.Tefis roll first The money. even pays win A . t hoe’ hand shooter’s ( n 6 =P( := ) | − P 36 j L † := − u ocr eei with is here concern Our . rp number craps natural { L 7 4 | , j , ≥ 5 , oieta h shooter’s the that Notice . n 6 ) , n , ,tebto is f2, If wins. bettor the ), 8 L length asln bet pass-line , o example, For . 2 = 9 , 10 ,tebto loses. bettor the ), , } 3 ≥ point fteshooter’s the of , . . . , oeotroll come-out 1 , h dice The . 12 hc is which , . t (154) (1) (2) . as that of Ms. DeMauro. As can be easily verified matrix, which can be inferred from (1), is from (3), (6), or (8) below, t(154) 0.178882426 10−9; to state it in the way preferred≈ by the media,× 12 6 8 10 0 this amounts to one chance in 5.59 billion, approxi- 3270 0 6 1 mately. The 1 in 3.5 billion figure came from a simu- P := 4 0260 6 . (5) 36 lation that was not long enough. The 1 in 1.56 trillion 5 0 0256 154 figure came from (1 π7) , which is the right answer 0 0 0 036 to the wrong question.− The probability of sevening out in n 1 rolls or less is then just the probability that absorption− in state 7o occurs by the (n 1)th step of the Markov chain, Two methods starting in state co.− A marginal simplification results by considering the 4 by 4 principal submatrix Q of We know of two methods for evaluating the tail prob- (5) corresponding to the transient states. Thus, we abilities (2). The first is by recursion. As pointed out have in [3], t(1) = t(2) = 1 and 4 n−1 n−1 t(n)=1 (P )1 5 = (Q )1 . (6) − , ,j t(n) j=1 X n−2 = 1 πj t(n 1) + πj (1 πj π7) Clearly, (3) is not a closed-form expression, and we − ∈P − ∈P − − jX jX do not regard (6) as being in closed form either. Is n−1 l−2 there a closed-form expression for t(n)? + π (1 π π7) π t(n l) (3) j − j − j − j∈P l=2 X X Positivity of the eigenvalues for each n 3. Indeed, for the event that the shooter ≥ sevens out in no fewer than n rolls to occur, consider We begin by showing that the eigenvalues of Q are the result of the initial come-out roll. If a natural or positive. The determinant of a craps number occurs, then, beginning with the next roll, the shooter must seven out in no fewer than n 1 Q zI − − rolls. If a point number occurs, then there are two 12 36z 6 8 10 possibilities. Either the point is still unresolved after 1 −3 27 36z 0 0 = − . n 2 additional rolls, or it is made at roll l for some 36 4 0 26 36z 0 − − l 2, 3,...,n 1 and the shooter subsequently 5 0 025 36z sevens∈ { out in no− fewer} than n l rolls. − − The second method, first suggested, to the best of is unaltered by row operations. From the first row subtract 6/(27 36z) times the second row, 8/(26 our knowledge, by Peter A. Griffin in 1987 (unpub- − − lished) and rediscovered several times since, is based 36z) times the third row, and 10/(25 36z) times − on a Markov chain. The state space is the fourth row, cancelling the entries 6/36, 8/36, and 10/36 and making the (1,1) entry equal to 1/36 times
S := co, p4-10, p5-9, p6-8, 7o 1, 2, 3, 4, 5 , (4) 6 8 10 { }≡{ } 12 36z 3 4 5 . (7) − − 27 36z − 26 36z − 25 36z whose five states represent the events that the shooter − − − is coming out, has established the point 4 or 10, has The determinant of Q zI, and therefore the char- established the point 5 or 9, has established the point acteristic polynomial q(−z) of Q is then just the prod- 6 or 8, and has sevened out. The one-step transition uct of the diagonal entries in the transformed matrix,
2 which is (7) multiplied by (27 36z)(26 36z)(25 appears three times in each root. Fortunately, α is 36z)/(36)4. Thus, − − − positive, as we see by writing ζ in polar form, that is, ζ = reiθ. We obtain q(z) = [(12 36z)(27 36z)(26 36z)(25 36z) − − − − 1 1 710369 18(26 36z)(25 36z) α =2√9829 cos cos− . − − − 3 − 9829√9829 32(27 36z)(25 36z) − − − 4 Q 50(27 36z)(26 36z)]/(36) . The four eigenvalues of can be expressed as − − − e1 := e(1, 1), We find that q(1), q(27/36), q(26/36), q(25/36), q(0) e2 := e(1, 1), alternate signs and therefore the eigenvalues are pos- − itive and interlaced between the diagonal entries (ig- e3 := e( 1, 1), − noring the entry 12/36). More precisely, denoting the e4 := e( 1, 1), eigenvalues by 1 >e1 >e2 >e3 >e4 > 0, we have − − where 27 26 25 1 >e1 > >e2 > >e3 > >e4 > 0. 5 u 349 + α 36 36 36 e(u, v) := + 8 72r 3 The matrix Q, which has the structure of an ar- v 698 α 3 rowhead matrix [5], is positive definite, although + − 2136u . not symmetric. This is easily seen by applying the 72s 3 − r349 + α same type of row operations to the symmetric part Next we need to find right eigenvectors correspond- A = 1 (Q + QT) of Q to show that the eigenvalues of 2 ing to the five eigenvalues of P . Fortunately, these A interlace its diagonal elements (except 12/36). But eigenvectors can be expressed in terms of the eigen- a symmetric matrix is positive definite if and only if values. Indeed, with r(x) defined to be the vector- all its eigenvalues are positive, and a non-symmetric valued function matrix is positive definite if and only if its symmetric part is positive definite, confirming that Q is positive 5+(1/5)x − 2 3 definite. 175 + (581/15)x (21/10)x + (1/30)x −275/2 (1199/40)−x + (8/5)x2 (1/40)x3 − − 1 A closed-form expression 0 The eigenvalues of Q are the four roots of the quartic we find that right eigenvectors corresponding to equation q(z)=0or eigenvalues 1,e1,e2,e3,e4 are T 4 3 2 (1, 1, 1, 1, 1) , r(36e1), r(36e2), r(36e3), r(36e4), 23328z 58320z + 51534z 18321z +1975 = 0, − − respectively. Letting R denote the matrix whose while P has an additional eigenvalue, 1, the spectral columns are these right eigenvectors and putting radius. We can use the quartic formula (or Mathe- L := R−1, the rows of which are left eigenvectors, matica) to find these roots. We notice that the com- we know by (6) and the spectral representation that plex number n−1 n−1 n−1 n−1 1/3 9829 t(n)=1 R diag(1,e1 ,e2 ,e3 ,e4 )L 1 5. α := ζ + , −{ } , ζ1/3 After much algebra (and with some help from where Mathematica), we obtain
n−1 n−1 n−1 n−1 ζ := 710369+ 18i√1373296647, t(n)= c1e1 + c2e2 + c3e3 + c4e4 , (8) −
3 where the coefficients are defined in terms of the e2 0.741708271459795977, ≈ eigenvalues and the function e3 0.709206775794379015, ≈ f(w,x,y,z) := ( 25+36w)[4835 5580(x + y + z) e4 0.186611201086502979, − − ≈ + 6480(xy + xz + yz) 7776xyz] − and the coefficients in (8) are /[38880(w x)(w y)(w z)] − − − c1 1.211844812464518572, as follows: ≈ c2 0.006375542263784777, ≈− c1 := f(e1,e2,e3,e4), c3 0.004042671248651503, ≈− c2 := f(e2,e3,e4,e1), c4 0.201426598952082292. ≈− c3 := f(e3,e4,e1,e2), These numbers will give very accurate results over a c4 := f(e4,e1,e2,e3). wide range of values of n. The result (11) shows that the leading term in (8) Of course, (8) is our closed-form expression. may be adequate for large n; it can be shown that Incidentally, the fact that t(1) = t(2) = 1 implies n−1 −m that 1
In particular, we have the inequality S0 := co, p2-12, p3-11, p4-10, p5-9, p6-8, 7o { } n−1 1, 2, 3, 4, 5, 6, 7 , t(n) n−1 e1 0.862473751659322030, t0(n) := P(L0 n)=1 (P0 )1 7. ≈ ≥ − , 4 There is an interesting distinction between this [3] Ethier, S. N. A Bayesian analysis of the shooter’s game and regular craps. The nonunit eigenvalues of hand at craps. In: S. N. Ethier and W. R. Ead- P0 are the roots of the sextic equation ington (eds.) Optimal Play: Mathematical Stud- ies of Games and Gambling, pp. 311–322. Insti- 6 5 4 15116544z 59206464z + 93137040z tute for the Study of Gambling and Commercial − 3 2 73915740z + 30008394z 5305446z + 172975 Gaming, University of Nevada, Reno, 2007. − − =0, [4] Grosjean, J. Exhibit CAA: Beyond Counting— Exploiting Casino Games from Blackjack to and the corresponding Galois group is, according to Video Poker. South Side Advantage Press, Las Maple, the symmetric group S6. This means that our Vegas, 2009. sextic is not solvable by radicals. Thus, it appears that there is no closed-form expression for t0(n). [5] Horn, R. A. and Johnson, C. R. Topics in Ma- Nevertheless, the analogue of (8) holds (with six trix Analysis. Cambridge University Press, Cam- terms). All nonunit eigenvalues belong to (0, 1) and bridge, 1991. all coefficients except the leading one are negative. Thus, the analogues of (10) and (11) hold as well. [6] Paik, E. Denville woman recalls setting the craps record in AC. Newark Star-Ledger, 27 May 2009. Also, the distribution of L0 is a linear combination of six geometric distributions. These results are left http://www.nj.com/news/local/index.ssf/2009/ as exercises for the interested reader. 05/pat demauro remembers only one.html −10 Finally, t0(154) 0.296360068 10 , which is ≈ × [7] Peterson, B. A new record in craps. Chance to say that a hand of length 154 or more is only about News 49 (2009). http://chance.dartmouth.edu/ one-sixth as likely as at regular craps (one chance in chancewiki/index.php/Chance News 49 33.7 billion, approximately). [8] Scarne, J. and Rawson, C. Scarne on Dice. The Military Service Publishing Co., Harrisburg, PA, Acknowledgment 1945. We thank Roger Horn for pointing out the interlacing [9] Scoblete, F. The Virgin Kiss and Other Adven- property of the eigenvalues of Q. tures. Research Services Unlimited, Daphne, AL, 2007. [10] Shackleford, M. Ask the Wizard! No. 81. 1 June References 2009. http://wizardofodds.com/askthewizard/ askcolumns/askthewizard81.html [1] Akane, K. The man with the golden arm, Parts I and II. Around Hawaii, 1 May 2008 and 1 June [11] Suddath, C. Holy craps! How a gambling 2008. http://www.aroundhawaii.com/lifestyle/ grandma broke the record. Time, 29 May 2009. travel/2008-05-the-man-with-the-golden-arm- http://www.time.com/time/nation/article/ part-i.html and http://www.aroundhawaii.com/ 0,8599,1901663,00.html lifestyle/travel/2008-06-the-man-with-the- golden-arm-part-ii.html [2] Bialik, C. Crunching the numbers on a craps record. The Numbers Guy, Wall Street Journal blog. 28 May 2009. http://blogs.wsj.com/ numbersguy/crunching-the-numbers-on-a- craps-record-703/ 5