Some easy and nice properties of the theory of Russian Roulette

Tomohide Hashiba
Takahumi Inoue 
Yuta Nakagawa  
Akihiro Hyogu
Hiroshi Matsui
Ryohei Miyadera 
Kwansei Gakuin High School

1. Introduction

We have already made the theory of general Russian roulette, and submitted the item of the topic to MathSorce. See Miyadera [1]. Some part of the result was published in a well-known mathematics site Archimedes-lab. See Miyadera [2].

After we submitted the item, 5 high school students discovered new things. We are going to present these new facts.

We are going to study the Russian roulette game with p-players,n-chambers and m-bullets.  Throughout this article we denote by p the number of players, by n the number of chambers and by m the number of bullets.

Let me illustrate the Russian roulette game.  In the first round the first player takes up a gun and pulls the trigger to himself.  If he survives, then in the second round the second player does the same.  The game ends when one of the players gets killed.  If the p-th player survives the p-th round, then the first player does the same again.  
In our version of Russian roulette one does  not rotate the cylinder before he pulls the trigger.   We denote by F[p,n,m,v] the probability of death of the v-th player in this game.

If you make the list {F[p,n,m,v],m = 1,2,3,....,n, and n = 1,2,3,...} for fixed natural number p and v with v ≤ p, then you get a triangle of numbers. See Figure (2) in Example 1.  There exists a relation that is very similar to the relation found in Pascal's triangle, and we published this fact in Miyadera [1] and [2].

If you do not know how to use Mathematica, please skip the part printed in blue. Then you can read this paper as an ordinary mathematics paper.

2. Mathematical theory of general Russian roulette.

In the Russian roulette game we are going to study here, one does  not rotate the cylinder before he pulls the trigger.   We denote by F[p,n,m,v] the probability of death of the v-th player in this game, and by R[n,m,y] the probability of the game ends in the y-th round.

We define U[p, n, m, v] = \(∑\+\(z = 0\)\%\(t\ \  - 1\)\(\_\(n - v - pz\)\)\(C\_\(m - 1\)\)\), where t = [((n - m + p - v + 1)/p)].
Note that this number U[p, n, m, v] is an important number in this article.

Theorem 1.  F[p, n, m, v]= (∑_(z = 0)^(t   - 1))R[n, m, 
              v + pz] = U[p, n, m, 
                  v]/((Subscript[, n]))(Subscript[C, m]).

For a proof see Miyadera [1].

Mathematica calculation　1.  Mathematica function for general Russian roulette.  The following Mathematica function F[p,n,m,v] returns F[p,n,m,v];the probability of death of the v-th player in the Russian roulette  with n-chambers, m-bullets and p-players.

F[p_, n_, m_, v_] := 
        Block[{t, u}, t = IntegerPart[((1 + p - m + n - v)/p)]; 
          u = ((∑_(z = 0)^(t - 1))Binomial[n - v - p*z, m - 1])/
              Binomial[n, m]]

Theorem 2.   U[p, n, m, v]+U[p, n, m + 1, v]=U[p, n + 1, m + 1, v].

For a proof see Miyadera [1].

By Theorem 2 U[p,n,m,v]　has the property that is very similar to (Subscript[, n])(Subscript[C, m]).  Therefore F[p,n,m,v]=  U[p, n, m, 
            v]/((Subscript[, n]))(Subscript[C, m]) can form a Pascal's triangle-like pattern of fractions when we arrange them to make a triangle.  See the following Example 1.

Example 1.  The purpose of this example is to observe a Pascal's triangle-like pattern of fractions. The following is the probability of the 1st player's death when there are 2 players. In other words p = 2 and v = 1.

To do this we have to make a triangle.  Here we let p = 2 and v = 1, and make F[p,n,m,v] for any natural number n,m with n ≤ 7 and m ≤ n. Then we get a triangle(See Figure 1).  If we calculate F[2,n,m,1] for each n and m, then we have the following triangle of fractions (See Figure 2).　
    The pattern is quite obvious in Figure (2). For example look at 6th row.  F[2,6,2,1]= (9/(15 )) and F[2,6,3,1]= (13/20) are the second and third ones in the row.   F[2,7,3,1]=(22/35)= ((9 + 13)/(15 + 20)), which is the third one in the 7th row.  This reminds us of Pascal's triangle.  By Theorem 1 and 2 there exists the same kind of relation among F[p,n,m,v] ,F[p,n,m+1,v], F[p,n+1,m+1,v] for any natural number p,v,n and m with n ≥ m.

{F[2,1,1,1]}       ⋯Figure(1)
                           {F[2,2,1,1],F[2,2,2,1]}
                      {F[2,3,1,1],F[2,3,2,1],F[2,3,3,1]}
                {F[2,4,1,1],F[2,4,2,1],F[2,4,3,1],F[2,4,4,1]}
          {F[2,5,1,1],F[2,5,2,1],F[2,5,3,1],F[2,5,4,1],F[2,5,5,1]}
     {F[2,6,1,1],F[2,6,2,1],F[2,6,31,1],F[2,6,4,1],F[2,6,5,1],F[2,6,6,1]}
{F[2,7,1,1],F[2,7,1,1],F[2,7,3,1],F[2,7,4,1],F[2,7,5,1],F[2,7,6,1],F[2,7,7,1]}

1   ⋯Figure(2)
                                          (1/2), (1/1)
                                      (2/3), (2/3), (1/1)
                                  (2/4), (4/6), (3/4), (1/1)
                             (3/5), (6/10), (7/10), (4/5), (1/1)
                          (3/6),(9/15),(13/20),(11/15),(5/6),(1/1)
                      (4/7),(12/21),(22/35),(24/35),(16/21),(6/7),1

Example 2.  The purpose of this example is to observe a new property of fractions. The following is the probability of the 1st player's death when there are 2 players. In other words p = 2 and v = 1. The pattern is almost obvious.

Mathematica calculation　2. Mathematica function to make the triangle. The following Mathematica function triangle[t,p,v] returns the triangle made by F[p,n,m,v](See Figure (3)).

{F[p,1,1,v]}       ⋯Figure(3)
                           {F[p,2,1,v],F[p,2,2,v]}
                      {F[p,3,1,v],F[p,3,2,v],F[p,3,3,v]}
                            ⋮          ⋮          ⋮
{F[p,t,1,v],F[p,t,2,v],F[p,t,3,v],F[p,t,4,v],⋯,F[p,t,t-1,v],F[p,t,t,v]}

Clear[triangle,   death];
triangle[t_, p_, v_] := 
          Block[{death}, 
            death[p2_, n2_, m2_, v2_] := 
              Block[{t2, u}, t2 = IntegerPart[((1 + p2 - m2 + n2 - v2)/p2)]; 
                u = ((∑_(z = 0)^(t2 - 1))Binomial[n2 - v2 - p2*z, 
                        m2 - 1]); 
                FractionBox[u, Binomial[n2, m2]] // DisplayForm]; 
            ColumnForm[Table[Table[death[p, n, m, v], {m, 1, n}], {n, 1, t}], 
              Center]];
triangle[5, 2, 1]

The following triangle is for the case of p = 2 and v = 1.

Remark.  We made the following triangle by using triangle[t,p,v] and colored the fraction. Note that we could not produce this triangle only the the function triangle[t,p,v].

(1/1) 
                                                         (1/2),(1/1) 
                                                     (2/3),(2/3),(1/1) 
                                                 (2/4),(4/6),(3/4),(1/1) 
                                            (3/5),(6/10),(7/10),(4/5),(1/1) 
                                      (3/6),(9/15),(13/20),(11/15),(5/6),(1/1) 
                                 (4/7),(12/21),(22/35),(24/35),(16/21),(6/7),(1/1) 
                           (4/8),(16/28),(34/56),(46/70),(40/56),(22/28),(7/8),(1/1) 
                     (5/9),(20/36),(50/84),(80/126),(86/126),(62/84),(29/36),(8/9),(1/1) 
          (5/10),(25/45),(70/120),(130/210),(166/252),(148/210),(91/120),(37/45),(9/10),(1/1) 
   \(\(6\/11\)\(,\)\(30\/55\)\(,\)\(95\/165\)\(,\)\(200\/330\)\(,\)\(296\/462\)\(,\)\(314\/462\)\(,\)\(239\/330\)\(,\)\(128\/165\)\(,\)\(46\/55\)\(,\)\(10\/11\)\(,\)\(1\/1\)\(\ \)\)

The pattern is clear.1+1+2+2+3+3+4+4+5=25  and  1+3+7+13+22+34= 80.  It is easy to find a similar pattern in other part of the triangle.

Example 3.  The following triangle is the same as the one in Example 2. Here we are going to introduce a new fact.

(1/1) 
                                                         (1/2),(1/1) 
                                                     (2/3),(2/3),(1/1) 
                                                 (2/4),(4/6),(3/4),(1/1) 
                                           (3/5),(6/10),(7/10),(4/5),(1/1) 
                                     (3/6),(9/15),(13/20),(11/15),(5/6),(1/1) 
                                (4/7),(12/21),(22/35),(24/35),(16/21),(6/7),(1/1) 
                          (4/8),(16/28),(34/56),(46/70),(40/56),(22/28),(7/8),(1/1) 
                   (5/9),(20/36),(50/84),(80/126),(86/126),(62/84),(29/36),(8/9),(1/1) 
       (5/10),(25/45),(70/120),(130/210),(166/252),(148/210),(91/120),(37/45),(9/10),(1/1) 
 (6/11),(30/55),(95/165),(200/330),(296/462),(314/462),(239/330),(128/165),(46/55),(10/11),(1/1)

Let (Subscript[a, n]) and (Subscript[b, n]) be the sum of the numerators and denominators in the n-th row respectively. For example (Subscript[a, 7]) = 4+12+22+24+16+6+1=85 and (Subscript[b, 7])= 7+21+35+35+21+7+1=127.
Then we can make the following table.

The most interesting thing is that the values of (Subscript[a, n])/(Subscript[b, n]) = (2/3)when n is even. We are not going to prove this mathematically here.

Mathematica function for (Subscript[a, n]) and (Subscript[b, n]) are given by a[n] and b[n].

U[p_, n_, m_, v_] := 
          Block[{t, u}, t = IntegerPart[((1 + p - m + n - v)/p)]; 
            u = ((∑_(z = 0)^(t - 1))Binomial[n - v - p*z, 
                    m - 1])];
a[n_]:=(∑_(k = 1)^n) U[2, n, k, 1];b[n_] := (∑_(k = 1)^n) Binomial[n, k];

a[7]

b[7]

({{(Subscript[a, n]), (Subscript[b, n]), (Subscript[a, n])/(Subscript[b, n])}, {1, 1, 1}, {2, 3, (2/3)}, {5, 7, (5/7)}, {10, 15, (2/3)}, {21, 31, (21/31)}, {42, 63, (2/3)}, {85, 127, (85/127)}, {170, 255, (2/3)}, {341, 511, (341/511)}, {682, 1023, (2/3)}, {1365, 2047, (1365/2047)}})

Mathematica calculation for the above table. Note that we produced the table, and after that we colored some fractions.

Clear[aa, a, b];
U[p_, n_, m_, v_] := 
          Block[{t, u}, t = IntegerPart[((1 + p - m + n - v)/p)]; 
            u = ((∑_(z = 0)^(t - 1))Binomial[n - v - p*z, 
                    m - 1])];
a[n_]:=(∑_(k = 1)^n) U[2, n, k, 1];b[n_] := (∑_(k = 1)^n) Binomial[n, k];
aa = 
        Table[{a[n], b[n], a[n]/b[n]}, {n, 1, 
            11}];
\(FrameBox[
        GridBox[Join[{{\*""(Subscript[a, n])"", \*""(Subscript[b, n])"", \*""(Subscript[a, n])/(Subscript[b, n])""}}, aa], RowLines -> True, 
          ColumnLines -> True]] // DisplayForm\)

Miyadera [1] General theory of Russian Roulette
http://library.wolfram.com/infocenter/MathSource/5710/ - 23k - 2005-08-05

R. Miyadera, D. Minematsu, S. Hashiba [2],
Mathematical Theory of Magic Fruits,
-Interesting patterns of fractions- ,
Archimedes' Laboratory,
http://www.archimedes-lab.org/fraction_patt/Magic_fruits.html