An Essay on Bitcoin attacks (Phrased in terms of a Two-Phase Soccer Match)



                              By Shalosh B. Ekhad



Assume that  two Soccer teams, the Good Guys Team and the Bad Guys Team play\
     each other. We assume that

       the probability of the Bad Guys Team scoring the next goal is q,

and hence that the probability of the Good Guys Team  scoring the next goal \
    is 1-q.

We also assume that the outcome of each goal is INDEPENDENT of the outcomes \
     of the other goals.



We assume throughout this essay that q is strictly positive but less than 1/\
    2, so the Good Guys have an advantage.



                             There are two phases.



Phase I: The teams play until the Good Guys Team scores n goals. By this tim\
    e, the Bad Guys Team scored a certain number of goals

let's call it m. In the unlikely event that m is larger or equal to n, the g\
    ame ends immediately, and the Bad Guys Team is declared the winner.

                   Otherwise, the game proceeds to Phase II.



Phase II: Keep playing until the Bad Guys Team forces a tie, i.e. they both \
    have the same number of goals, and then

the Bad Guys Team is declared the winner, or the Good Guys Team is so much a\
    head (say by a Zillion goals), to make

it hopeless for the Bad Guys to catch up, in which case the Good Guys Team w\
    on.



Definition: Let, P[n](q),  that we will call the Rosenfeld polynomials (in ho\
    nor of Meni Rosenfeld, who wrote the great article

Analysis of hashrate-based double spending, ArXiv 1402.2009v1 [cs. CR], 9 Fe\
    b. 2014. )

               be the probability of the Bad Guys Team winning.



          Theorem 0 (Meni Rosenfeld, ArXiv 1402.2009v1, Eq. (1), p. 7)



                       /  n                            \
                       |-----                          |
                     n | \                            m|
P[n](q) = 1 - (1 - q)  |  )   binomial(n + m - 1, m) q |
                       | /                             |
                       |-----                          |
                       \m = 0                          /

          /  n                                  \
          |-----                                |
        n | \                                  m|
     + q  |  )   binomial(n + m - 1, m) (1 - q) |
          | /                                   |
          |-----                                |
          \m = 0                                /



Using the so-called Zeilberger algorithm, we find a second-order linear recu\
    rrence that enables the fast

computation of the first 10000, say, Rosenfeld polynomials, much faster than\
     using the definition.





Theorem 1: The Rosenfeld polynomials, P[n](q),

    satisfy the following linear recurrence with polynomial coefficients.

                  2              2
            (4 n q  - 4 n q - 6 q  - n + 6 q + 1) P[n - 1](q)
P[n](q) = - -------------------------------------------------
                                  n - 1

       2 q (q - 1) (2 n - 3) P[n - 2](q)
     + ---------------------------------
                     n - 1



                           with initial conditions:



                           P[0](q) = 1, P[1](q) = 2 q



                              and in Maple format



P[n](q) = -(4*n*q^2-4*n*q-6*q^2-n+6*q+1)/(n-1)*P[n-1](q)+2*q*(q-1)*(2*n-3)/(n-1
)*P[n-2](q)


P[0](q) = 1, P[1](q) = 2*q


In another interesting paper,  by Cyril Grunspan and Ricardo Perez-Marco, en\
    titled



                              Double Speed Races

                    arXiv: 1702.02867v2 [cs.CR] 17 Feb 2017



the authors stated and proved the following asymptotic formula for P[n](q) (\
    valid, of course, for 0<q<1/2)



                                              n
                                 (4 q (1 - q))
                              --------------------
                                1/2            1/2
                              Pi    (1 - 2 q) n



They used their representation formula in terms of the incomplete Beta funct\
    ion.



However, using the Birkhoff-Trijinski method, beautifully exposited and appl\
    ied in the article

"Resurrecting the Asymptotics of linear recurrences", by J.Wimp and D. Zeilb\
    erger, J. Math. Anal. Appl. 111 (1985), 162-177

available from http://sites.math.rutgers.edu/~zeilberg/mamarimY/Zeilberger_y\
    1985_p162.pdf ,

                      and implemented in the Maple package

           http://sites.math.rutgers.edu/~zeilberg/tokhniot/AsyRec.txt

(See http://sites.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/asy.html fo\
    r instructions)



 we can get a much more precise asymptotics, to order, 4,

    in the following theorem.





Theorem 2: The Rosenfeld polynomials, P[n](q), i.e. the probability of the B\
    ad Guys Team winning

              have the following asymptotic expansion to order, 4



                         /           2
                       n | 1     12 q  - 12 q - 1
P[n](q) = (4 q (1 - q))  |---- + -----------------
                         | 1/2              2  3/2
                         \n      8 (2 q - 1)  n

            4        3        2
       400 q  - 800 q  + 120 q  + 280 q + 1
     + ------------------------------------
                            4  5/2
               128 (2 q - 1)  n

                6         5        4         3         2
       5 (1344 q  - 4032 q  + 112 q  + 6496 q  - 3444 q  - 476 q + 1)
     + -------------------------------------------------------------- + 21 (
                                          6  7/2
                            1024 (2 q - 1)  n

           8          7          6           5           4          3
    20224 q  - 80896 q  - 88832 q  + 549632 q  - 514400 q  + 18368 q

                                                         \
              2                 /                 8  9/2 |   /              1/2
     + 92368 q  + 3536 q - 1)  /  (32768 (2 q - 1)  n   )|  /  ((1 - 2 q) Pi
                              /                          | /
                                                         /

           1
    ) + O(----)
            5
           n



                                                    1
               and in Maple format ,without the, O(----), we have
                                                     5
                                                    n



P[n](q) = 1/(1-2*q)/Pi^(1/2)*(4*q*(1-q))^n*(1/n^(1/2)+1/8*(12*q^2-12*q-1)/(2*q-\
1)^2/n^(3/2)+1/128*(400*q^4-800*q^3+120*q^2+280*q+1)/(2*q-1)^4/n^(5/2)+5/1024*(
1344*q^6-4032*q^5+112*q^4+6496*q^3-3444*q^2-476*q+1)/(2*q-1)^6/n^(7/2)+21/32768
*(20224*q^8-80896*q^7-88832*q^6+549632*q^5-514400*q^4+18368*q^3+92368*q^2+3536*
q-1)/(2*q-1)^8/n^(9/2))


Suppose that the Bad Guys Team is successful, how long should Phase II last \
    until it finishes?



              Meni Rosenfeld, in deriving Theorem 0, used the fact

(e.g. Feller's classic) that if a random walker starts at L, and at each ato\
    mic step moves one unit left with probability q

and  one unit to the right with probability 1-q, assuming q<1/2, the probabi\
    lity of making it to 0 is



                                    /  q  \L
                                    |-----|
                                    \1 - q/



  Also in Feller (but see the Georgiadis-Zeilberger paper for another proof)

is  the famous fact that the (conditional) expected duration, in case of suc\
    cess is

                                       L
                                    -------
                                    1 - 2 q



      By an analogous argument to the one used  by Meni Rosenfeld we have



                                Theorem 3: Let



                       /n - 1                                        \
                       |-----                               m        |
                     n | \    binomial(n + m - 1, m) (1 - q)  (n - m)|
          A[n](q) = q  |  )   ---------------------------------------|
                       | /                    1 - 2 q                |
                       |-----                                        |
                       \m = 0                                        /



then the (conditional) expectation of the duration of Phase II (in case of s\

    uccess to the Bad Guys Team), let's call it, E[n](q), is given by



                                         A[n](q)
                               E[n](q) = -------
                                         P[n](q)



Like with, P[n](q),  we can compute many terms of the sequence, A[n](q),

    and hence , E[n](q), using the second-order recurrence given

                                  by Theorem 4



Theorem 4:, A[n](q), satisfies the following second-order linear recurrence e\
    quation with polynomial coefficients:



                  2              2
            (4 n q  - 4 n q - 6 q  - n + 6 q) A[n - 1](q)
A[n](q) = - ---------------------------------------------
                                n - 1

       2 (q - 1) (2 n - 3) q A[n - 2](q)
     + ---------------------------------
                     n - 2



                       subject to the initial conditions



                                                   2
                                q               2 q  (2 - q)
                   A[1](q) = -------, A[2](q) = ------------
                             1 - 2 q              1 - 2 q



                              and in Maple format



A[n](q) = -(4*n*q^2-4*n*q-6*q^2-n+6*q)/(n-1)*A[n-1](q)+2*(q-1)*(2*n-3)*q/(n-2)*
A[n-2](q)


                       subject to the initial conditions



A[1](q) = q/(1-2*q), A[2](q) = 2*q^2*(2-q)/(1-2*q)


Using AsyRec.txt again, we get an asymptotic formula for  . (A[n](q)),

    that we will not state, that implies an asymptotic formula

 for the (conditional) expected duration of Phase II (if the Bad Guys Team w\

    on) , E[n](q)





Theorem 5: The (conditional) expected duration of Phase II (in case the Bad \

    Guys Team made it), E[n](q)

               has the following asymptotic expansion to order, 4



                  /                           2
                  |               1        6 q  - 6 q - 1
E[n](q) = (1 - q) |1 - 2 q + ----------- + --------------
                  |          (2 q - 1) n            3  2
                  \                        (2 q - 1)  n

           4       3       2
       36 q  - 72 q  + 12 q  + 24 q + 1
     + --------------------------------
                         5  3
                (2 q - 1)  n

            6        5        4        3        2           \
       216 q  - 648 q  + 276 q  + 528 q  - 306 q  - 66 q - 1|   /          3
     + -----------------------------------------------------|  /  (1 - 2 q)
                                    7  4                    | /
                           (2 q - 1)  n                     /

          1
     + O(----)
           5
          n



                                                    1
               and in Maple format ,without the, O(----), we have
                                                     5
                                                    n



E[n](q) = (1-q)/(1-2*q)^3*(1-2*q+1/(2*q-1)/n+(6*q^2-6*q-1)/(2*q-1)^3/n^2+(36*q^
4-72*q^3+12*q^2+24*q+1)/(2*q-1)^5/n^3+(216*q^6-648*q^5+276*q^4+528*q^3-306*q^2-\
66*q-1)/(2*q-1)^7/n^4)


                         Note that the leading term is



                                     1 - q
                                   ----------
                                            2
                                   (1 - 2 q)



                              and in Maple format



(1-q)/(1-2*q)^2


and that it is INDEPENDENT of, n,

    the number of goals that the Good Guys Team scored in Phase II

it follows that while the probability of success of the Bad Guys Team strong\

    ly depends on, n, and decays exponentially

in the unlikely event that they do succeed, for large n, it takes roughly th\

                                       1 - q
    e same time in Phase II, namely, ----------
                                              2
                                     (1 - 2 q)



                                  1 - q
             Here is a plot of, ----------, for q from 0.1 to 0.45
                                         2
                                (1 - 2 q)




  +                                                                          H 
  +                                                                          H 
50+                                                                         H  
  +                                                                         H  
  +                                                                         H  
  +                                                                        HH  
40+                                                                        H   
  +                                                                       HH   
  +                                                                       H    
30+                                                                      H     
  +                                                                     HH     
  +                                                                    HH      
  +                                                                   HH       
20+                                                                  HH        
  +                                                                HHH         
  +                                                              HHH           
  +                                                           HHHH             
10+                                                       HHHHH                
  +                                               HHHHHHHHH                    
  +                              HHHHHHHHHHHHHHHHHH                            
  ********************************+-+-+--+-+-+-+-+-+-+--+-+-+-+-+-+-+--+-+-+-+ 
  0.1      0.15       0.2       0.25        0.3       0.35       0.4       0.45


For example, if q=0.45, the probability of the Bad Guys Team succeeding, fol\
    lowed by the expected duration of Phase II (if they do succeed) is

                               for various n are

n1=, 10, Prob=, 0.657928175156784470315025329598, Expected Duration=,

    15.7998735892050790240681753309

n1=, 30, Prob=, 0.439334368905251216167096171316, Expected Duration=,

    24.2364193397083477708747026795

n1=, 50, Prob=, 0.317304397874197589395304019275, Expected Duration=,

    28.7518018461666223264484589303

n1=, 70, Prob=, 0.236383131350518027134957810133, Expected Duration=,

    31.8203651506003540535042521518

n1=, 90, Prob=, 0.179224062416643702150405300463, Expected Duration=,

    34.1132916612751723024497804904

n1=, 110, Prob=, 0.137469491687045789029645365173, Expected Duration=,

    35.9209553967353878736923684366

n1=, 130, Prob=, 0.106321006739133517940020491602, Expected Duration=,

    37.3969939216570850484493306692

n1=, 150, Prob=, 0.0827480032540806332334106290346, Expected Duration=,

    38.6328557673279370884832113294

n1=, 170, Prob=, 0.0647201558345036148781848847372, Expected Duration=,

    39.6874163827172628615175737290

n1=, 190, Prob=, 0.0508225993247899387517701338396, Expected Duration=,

    40.6007815537453098845822002819

n1=, 210, Prob=, 0.0400415507146447022013807232206, Expected Duration=,

    41.4014643955490283623658282651

n1=, 230, Prob=, 0.0316355938665163891115549279474, Expected Duration=,

    42.1104288449561236100309878941

n1=, 250, Prob=, 0.0250540060603111219702843930753, Expected Duration=,

    42.7435127590482451667505757918

n1=, 270, Prob=, 0.0198827167721962129022620901218, Expected Duration=,

    43.3129530043146865685132714826

n1=, 290, Prob=, 0.0158073714818615613646818147440, Expected Duration=,

    43.8283845445301647537310252923

n1=, 310, Prob=, 0.0125874334065934349185114995862, Expected Duration=,

    44.2975171233707153126822904898

n1=, 330, Prob=, 0.0100376500296295838119347875278, Expected Duration=,

    44.7266066931359509612134684037

n1=, 350, Prob=, 0.00801456884166863808192764491188, Expected Duration=,

    45.1207918940499396589679405596

n1=, 370, Prob=, 0.00640659272843284758379118602052, Expected Duration=,

    45.4843393135857439707647424126

n1=, 390, Prob=, 0.00512656291802691721067853475718, Expected Duration=,

    45.8208255794355697598517231525

and for q=0.49, the probability of the Bad Guys Team succeeding, followed by\
     the expected duration of Phase II (if they do succeed) is

                                 various n are

n1=, 1000, Prob=, 0.371105318863023324849343861245, Expected Duration=,

    623.452416132876347794878663715

n1=, 2000, Prob=, 0.205886196670458273986688397908, Expected Duration=,

    768.007655813596489951113501508

n1=, 3000, Prob=, 0.121313527058695533732377320370, Expected Duration=,

    852.089159958663419879399071125

n1=, 4000, Prob=, 0.0736184550505632011681068524280, Expected Duration=,

    909.378435721442497353895384011

n1=, 5000, Prob=, 0.0454840646491492841228477544989, Expected Duration=,

    951.662612193057511706841035078

n1=, 6000, Prob=, 0.0284471816423606747174779329692, Expected Duration=,

    984.464859797074226407826441401

n1=, 7000, Prob=, 0.0179510458573885184494416098902, Expected Duration=,

    1010.80442953511358828141608494

n1=, 8000, Prob=, 0.0114050942836705973527808825251, Expected Duration=,

    1032.50162383745055273806966109

n1=, 9000, Prob=, 0.00728532104673207479883674402498, Expected Duration=,

    1050.73178882869473828765050772

n1=, 10000, Prob=, 0.00467411866022033570793431482146, Expected Duration=,

    1066.29356620363739899172000780

n1=, 11000, Prob=, 0.00300973013205387710307073082754, Expected Duration=,

    1079.75157350150280692688813163

n1=, 12000, Prob=, 0.00194395145803715930340345782521, Expected Duration=,

    1091.51786955442921463271795443

n1=, 13000, Prob=, 0.00125887046137457750950859003260, Expected Duration=,

    1101.90113072251327196603617860

n1=, 14000, Prob=, 0.000817074316699566219112146529272, Expected Duration=,

    1111.13770195175581423285886491

n1=, 15000, Prob=, 0.000531377683140489852349494159652, Expected Duration=,

    1119.41194626313084072060895889

n1=, 16000, Prob=, 0.000346182256071088985813488811407, Expected Duration=,

    1126.87000523861939572768511935

n1=, 17000, Prob=, 0.000225882021858711273474542303094, Expected Duration=,

    1133.62935641977148433928838971

n1=, 18000, Prob=, 0.000147592052875913960129123899786, Expected Duration=,

    1139.78560700076808910148721785

n1=, 19000, Prob=, 0.0000965580559263843407589941722994, Expected Duration=,

    1145.41742182739518351090132151

n1=, 20000, Prob=, 0.0000632421634970668798247481160022, Expected Duration=,

    1150.59016255337079815474838184



But what about the variance of the random variable (conditional) "duration o\
    f Phase II". We have the following theorem





                Theorem 6:,Let, B[n](q), be defined as follows.



             /n - 1                                                           \
             |-----                               a               2           |
           n | \    binomial(n + a - 1, a) (1 - q)  n (2 n q + 4 q  - n - 4 q)|
B[n](q) = q  |  )   ----------------------------------------------------------|
             | /                                     3                        |
             |-----                         (2 q - 1)                         |
             \a = 0                                                           /



Then the second moment of the (conditional) random variable "duration of Pha\

    se II (in case of the Bad Guys Team winning)", let,s call it, M2[n](q), is



                                          B[n](q)
                               M2[n](q) = -------
                                          P[n](q)



Like with, P[n](q),  and , A[n](q), we can compute many terms of the sequence ,

    B[n](q)

and hence , M2[n](q),

    using the second-order recurrence given by the next theorem.



Theorem 7:, B[n](q), satisfies the following second-order linear recurrence e\
    quation with polynomial coefficients:



                 2  5         6       2  4          5       6      2  3
B[n](q) = - (32 n  q  + 32 n q  - 64 n  q  - 176 n q  - 48 q  + 8 n  q

              4        5       2  2         3        4      2            2
     + 216 n q  + 192 q  + 36 n  q  + 20 n q  - 192 q  - 8 n  q - 126 n q

           3      2                 2                                 /
     - 42 q  - 3 n  + 31 n q + 138 q  + 3 n - 54 q + 6) B[n - 1](q)  /  (
                                                                    /

                  3      4        2       3              2
    (n - 1) (8 n q  + 8 q  - 8 n q  - 24 q  - 4 n q + 8 q  + 3 n + 15 q - 6))

     + 2 (q - 1) (2 n - 3)

          3      4        2       3
    (8 n q  + 8 q  - 8 n q  - 16 q  - 4 n q + 3 n + 11 q - 3) q B[n - 2](q)

       /
      /  ((n - 2)
     /

          3      4        2       3              2
    (8 n q  + 8 q  - 8 n q  - 24 q  - 4 n q + 8 q  + 3 n + 15 q - 6))



                       subject to the initial conditions



                     2                            2     3       2
               q (4 q  - 2 q - 1)              2 q  (4 q  - 10 q  + q + 3)
     B[1](q) = ------------------, B[2](q) = - ---------------------------
                            3                                   3
                   (2 q - 1)                           (2 q - 1)



                              and in Maple format



B[n](q) = -(32*n^2*q^5+32*n*q^6-64*n^2*q^4-176*n*q^5-48*q^6+8*n^2*q^3+216*n*q^4
+192*q^5+36*n^2*q^2+20*n*q^3-192*q^4-8*n^2*q-126*n*q^2-42*q^3-3*n^2+31*n*q+138*
q^2+3*n-54*q+6)/(n-1)/(8*n*q^3+8*q^4-8*n*q^2-24*q^3-4*n*q+8*q^2+3*n+15*q-6)*B[n
-1](q)+2*(q-1)*(2*n-3)*(8*n*q^3+8*q^4-8*n*q^2-16*q^3-4*n*q+3*n+11*q-3)*q/(n-2)/
(8*n*q^3+8*q^4-8*n*q^2-24*q^3-4*n*q+8*q^2+3*n+15*q-6)*B[n-2](q)


                       subject to the initial conditions



B[1](q) = q*(4*q^2-2*q-1)/(2*q-1)^3, B[2](q) = -2*q^2*(4*q^3-10*q^2+q+3)/(2*q-1
)^3


Using AsyRec.txt yet one more time, we get an asymptotic formula for  .

    (B[n](q)), that we will not state, and hence for M2[n](q)

      and hence for the (conditional) variance, let us call it , Var[n](q)



Theorem 8: The (conditional) variance of the duration of Phase II (in case t\

    he Bad Guys Team made it), Var[n](q)

               has the following asymptotic expansion to order, 4



                        2                          2
            (q - 1) (4 q  - 3 q - 2)   (q - 1) (4 q  + 2 q - 7)
Var[n](q) = ------------------------ - ------------------------
                            4                         6
                   (2 q - 1)                 (2 q - 1)  n

                    4       3       2
       (q - 1) (24 q  - 12 q  - 58 q  + 29 q + 18)
     - -------------------------------------------
                               8  2
                      (2 q - 1)  n

                     6        5        4        3        2
       (q - 1) (144 q  - 216 q  - 348 q  + 444 q  + 328 q  - 312 q - 41)
     - ----------------------------------------------------------------- -
                                         10  3
                                (2 q - 1)   n

                  8         7         6         5         4         3         2
    (q - 1) (864 q  - 2160 q  - 1704 q  + 4956 q  + 4632 q  - 9972 q  + 1586 q

                      /           12  4       1
     + 1711 q + 88)  /  ((2 q - 1)   n ) + O(----)
                    /                          5
                                              n



                                                    1
               and in Maple format ,without the, O(----), we have
                                                     5
                                                    n



Var[n](q) = (q-1)*(4*q^2-3*q-2)/(2*q-1)^4-(q-1)*(4*q^2+2*q-7)/(2*q-1)^6/n-(q-1)
*(24*q^4-12*q^3-58*q^2+29*q+18)/(2*q-1)^8/n^2-(q-1)*(144*q^6-216*q^5-348*q^4+
444*q^3+328*q^2-312*q-41)/(2*q-1)^10/n^3-(q-1)*(864*q^8-2160*q^7-1704*q^6+4956*
q^5+4632*q^4-9972*q^3+1586*q^2+1711*q+88)/(2*q-1)^12/n^4


Once again, like the expectation the leading term of the variance does not d\
    epend on n

                                            2
                                (q - 1) (4 q  - 3 q - 2)
                        and is, ------------------------
                                                4
                                       (2 q - 1)

and hence the standard deviation is, for large n, tends to a constant (that \
    only depends on q)



                         /            2           \1/2
                         |(q - 1) (4 q  - 3 q - 2)|
                         |------------------------|
                         |                4       |
                         \       (2 q - 1)        /



                              and in Maple format



((q-1)*(4*q^2-3*q-2)/(2*q-1)^4)^(1/2)




note that there is no concentration about the mean. Not only that, the stand\
    ard deviation is larger than the expectation



     the duration of Phase II, in case of success, is very unpredictable.



Let's conclude with a table of the limiting expected duration and standard d\
    eviations for q from 0.1 to 0.49



q = 0.1, av = 1.40625000000000000000000000000,

    sd = 2.22841206075088365825756968210

q = 0.11, av = 1.46285338593030900723208415516,

    sd = 2.34220849625700825009235050388

q = 0.12, av = 1.52354570637119113573407202216,

    sd = 2.46436220641351562289957731115

q = 0.13, av = 1.58875091307523739956172388605,

    sd = 2.59576097112618104567354696816

q = 0.14, av = 1.65895061728395061728395061728,

    sd = 2.73741465191665259661010169621

q = 0.15, av = 1.73469387755102040816326530612,

    sd = 2.89047611042534106885922219027

q = 0.16, av = 1.81660899653979238754325259516,

    sd = 3.05626644750175904054481973468

q = 0.17, av = 1.90541781450872359963269054178,

    sd = 3.23630561712396331742050984592

q = 0.18, av = 2.00195312500000000000000000000,

    sd = 3.43234976843289211497582498370

q = 0.19, av = 2.10718002081165452653485952133,

    sd = 3.64643706566880468785631500613

q = 0.20, av = 2.22222222222222222222222222222,

    sd = 3.88094426590510681029356525132

q = 0.21, av = 2.34839476813317479191438763377,

    sd = 4.13865704948845604122579689458

q = 0.22, av = 2.48724489795918367346938775510,

    sd = 4.42285807159232683476423349658

q = 0.23, av = 2.64060356652949245541838134431,

    sd = 4.73743804201631700979175223811

q = 0.24, av = 2.81065088757396449704142011834,

    sd = 5.08703700105368631946596987065

q = 0.25, av = 3.00000000000000000000000000000,

    sd = 5.47722557505166113456969782801

q = 0.26, av = 3.21180555555555555555555555556,

    sd = 5.91473971783915024181441740093

q = 0.27, av = 3.44990548204158790170132325142,

    sd = 6.40778781177300587832740978155

q = 0.28, av = 3.71900826446280991735537190083,

    sd = 6.96645685201344644145587999407

q = 0.29, av = 4.02494331065759637188208616780,

    sd = 7.60325609472428198879995243262

q = 0.30, av = 4.37500000000000000000000000000,

    sd = 8.33385415039164217313826018951

q = 0.31, av = 4.77839335180055401662049861496,

    sd = 9.17809256202894183307075346312

q = 0.32, av = 5.24691358024691358024691358025,

    sd = 10.1614013365475275666853109533

q = 0.33, av = 5.79584775086505190311418685121,

    sd = 11.3168098835873478609863928726

q = 0.34, av = 6.44531250000000000000000000000,

    sd = 12.6878583885650792677999380829

q = 0.35, av = 7.22222222222222222222222222222,

    sd = 14.3329026636418660486517568539

q = 0.36, av = 8.16326530612244897959183673469,

    sd = 16.3316318561163727543592474929

q = 0.37, av = 9.31952662721893491124260355030,

    sd = 18.7952039038360041900904215676

q = 0.38, av = 10.7638888888888888888888888889,

    sd = 21.8824943070657608271110221661

q = 0.39, av = 12.6033057851239669421487603306,

    sd = 25.8270743725277890486608751919

q = 0.40, av = 15.0000000000000000000000000000,

    sd = 30.9838667696593350814341231983

q = 0.41, av = 18.2098765432098765432098765432,

    sd = 37.9137993408771377729242081036

q = 0.42, av = 22.6562500000000000000000000000,

    sd = 47.5464924606694831217372179671

q = 0.43, av = 29.0816326530612244897959183673,

    sd = 61.5156342821984659274560603921

q = 0.44, av = 38.8888888888888888888888888889,

    sd = 82.9137585568564812122950131502

q = 0.45, av = 55.0000000000000000000000000000,

    sd = 118.194754536739065897591254936

q = 0.46, av = 84.3750000000000000000000000000,

    sd = 182.762012259112533552269863816

q = 0.47, av = 147.222222222222222222222222222,

    sd = 321.430060745128945617905398072

q = 0.48, av = 325.000000000000000000000000000,

    sd = 715.227236617845811186548937294

q = 0.49, av = 1275.00000000000000000000000000,

    sd = 2828.31221756014765478174834134



              ----------------------------------------------------



               This concludes this essay. Thanks for reading it.