General theory of Russian Roulette

Munetoshi Sakaguchi
Daisuke Minematsu
Satoshi Hashiba
Ryohei Miyadera 
Kwansei Gakuin

1. Introduction

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.  Russian roulette has a very interesting mathematical structure. See the figure (2) in Example 1.  There exists a relation that is very similar to the relation found in Pascal's triangle, but our relation is more elaborate.  See also Example 3 and 4.  The expectation of the first player's death in Russian rulette games with 2-players and 3-players are (Subscript[log, e])2 and (π/(3 (Sqrt[3])))respectively. It is very interesting to get very simple numbers such as (Subscript[log, e])2 and (π/(3 (Sqrt[3]))).  
   Let me illustrate the Russian rulette 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.  
   There are two versions of Russian roulette.  In the first version 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.
   In the second version one rotate the cylinder every time before he pulls the trigger.  We denote by F2[p,n,m,v] the probability of death of the v-th player in this game.
   Most of ideas about this thory of Russian rulette have been proposed by high school students and freshmen in a university, and a mathematician helped them to develope the ideas and make theories.
    In this article we only use mathematics that is understandable to freshmen in a university and perhaps high school students can understand most of our theory, but as you see later the theory is very rich in its contents.
    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 rulette 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.

Lemma 1.   
\(R[n, m, 
            y] = \((∏\+\(k = 0\)\%\(y - 2\)\(n - \((m + k)\)\)\/\(n - k\))\)×m\/\(n - \((y - 1)\)\)\ \ and\) 
R[n, m, y]  <  (((((n - m)/n)))^(y - 1)).

proof.  If the game is to end in the y-th round, the players have to survive from the first round to (y-1)-th round.  The probability to survive the first to (y-1)-th round are ((n - m)/n),((n - (m + 1))/(n - 1)),⋯ ,((n - (m + y - 2))/(n - (y - 2))) respectively and the probability of one of the players get killed in the y-th round is (m/(n - (y - 1))).
    Therefore  R[n,m,y]= \(\((∏\+\(k = 0\)\%\(y - 2\)\(n - \((m + k)\)\)\/\(n - k\))\)×m\/\(n - \((y - 1)\)\)\).  
    It is easy to see that R[n, m, y]  <  (((((n - m)/n)))^(y - 1)).

Remark.  There is another way to calculate R[n,m,y].  In our article we use these two ways to calculate R[n,m,y].

Lemma 2.  R[n, m, y]=\(\(\(\_\(n - y\)\)\(C\_\(m - 
                      1\)\)\/\(\_n\)\(C\_m\)\)\(.\)\)

proof.  We put m-bullets into n-chambers,  \(so\ we\ have\ \(\_n\)\(C\_m\)\ cases\ for\ \(\(this\)\(.\)\)\)  The game ends in the y-th round when there is a bullet in the y-th chamber and there are (m-1)-bullets in the following (n-y)-chambers.   \(There\ are\ \(\_\(n - y\)\)\(C\_\(m - 
                1\)\)\ cases\ for\ this . \ \ Therefore\ the\ probability\ is\ \(\(\(\_\(n - y\)\)\(C\_\(m - 
                      1\)\)\/\(\_n\)\(C\_m\)\)\(.\)\)\)

Lemma 3.  The v-\(th\ \ player\ can\ play\ [\(n - m + p - v + 1\)\/p]\ \ \ times\ at\ most\), where  [ ] is the Gauss function.

proof.  If there are m-bullets, then  the number of rounds is n-m+1 at most,　and hence the p-th player can play　q  =  [((n - m + 1)/p)]times at most.  Let b = (n-m+1)mod p,i.e.,　the remaining part of (n-m+1) when divided by p.  
      It is easy to see that the v-th player can play q+1 times at most if　v≤b,　and q times at most if v > b.
      Therefore the v-th player can play　q + [1 + ((b - v)/p)]=　[((p (q + 1) + b - v)/p)]=[((n - m + p - v + 1)/p)]times at most.
Note that -1 ≤ ((b - v)/p) < 1, since　0≤b<p and 1≤v≤p.

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]).

proof.  By Lemma 3  the  v-th player can play t= [((n - m + p - v + 1)/p)] times at most.  The v-th player plays the v-th, v+p-th, v+2p-th, ⋯v+(t-1)p-th round.  Therefore by Lemma 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]).

Mathemtica calculation　1.  Mathemtica function for general Russian roulette.  The following Mathemtaica 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].

proof.  U[p,n,m,v] = \(∑\+\(z = 0\)\%\(t\_1\ \  - 1\)\(\_\(n - v - pz\)\)\(C\_\(m - 1\)\)\),           ⋯(1)
where (Subscript[t, 1])  =  [((n - m + p - v + 1)/p)],                         ⋯(2)
U[p,n,m+1,v] = \(∑\+\(z = 0\)\%\(t\_2\ \  - 1\)\(\_\(n - v - pz\)\)\(C\_m\)\),                   ⋯(3)
where (Subscript[t, 2])  =  [((n - (m + 1) + p - v + 1)/p)]                        ⋯(4)
and 
U[p,n+1,m+1,v] = \(∑\+\(z = 0\)\%\(t\_3\ \  - 1\)\(\_\(n + 1 - v - pz\)\)\(C\_m\)\),                 ⋯(5)
where (Subscript[t, 3])  =  [((n + 1 - (m + 1) + p - v + 1)/p)]  =  
          (Subscript[t, 1]).
It is clear that
(Subscript[, n - v - pz])(Subscript[C, m - 
                  1]) + (Subscript[, n - v - 
                  pz])(Subscript[C, m]) = (Subscript[, n + 
                1 - v - pz])(Subscript[C, m + 1]) for z = 0, 1, ⋯, (Subscript[t, 2])    ⋯(6).
Here we used the fact that  (Subscript[t, 1]) ≥ (Subscript[t, 2]).
(a)  If (Subscript[t, 1]) = (Subscript[t, 2]), by (1), (3), (5) and (6)  we have
U[p,n,m,v]+U[p,n,m+1,v]=U[p,n+1,m+1,v]. 
(b)  If (Subscript[t, 1]) = (Subscript[t, 2]) + 1, then by (2) and (4) we have 
n-m+p-v+1= (Subscript[pt, 1]).  
From this we have
n - v - p ((Subscript[t, 1]) - 1) = m-1  
and
n + 1 - v - p ((Subscript[t, 3]) - 1) = m, 
which imply that
(Subscript[, n - v - 
                p ((Subscript[t, 1]) - 1)])(Subscript[C, m - 
                1]) = (Subscript[, m - 
                  1])(Subscript[C, m - 1])=1     ⋯(7)
and 
\(\(\(\ \)\(\(\_\(n + 1 - v - 
                p \((t\_3 - 
                      1)\)\)\)\(C\_m\) = \(\(\_m\)\(C\_m\) = 1\)\)\)\)       ⋯(8).
Note that (7) and (8) are the (Subscript[t, 1])-th terms of \(∑\+\(z = 0\)\%\(t\_1\ \  - 1\)\(\_\(n - v - pz\)\)\(C\_\(m - 1\)\)\) and \(∑\+\(z = 0\)\%\(t\_3\ \  - 1\)\(\_\(n + 1 - v - pz\)\)\(C\_m\)\) respectively.
By (1),(3),(5),(6),(7) and (8) we have
U[p,n,m,v]+U[p,n,m+1,v]=U[p,n+1,m+1,v].

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.

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}

Mathemtica calculation　2. Mathematica function to make the triangle. The following Mathemtaica 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]];

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

triangle[7, 2, 1]

The following calculation is for the case of v=7 and p = 2.

triangle[7, 7, 2]

3. Another version of Russian roulette.

Here we are going to study another version of Russian roulette.  In this game one rotate the cylinder each time before he pulls the trigger.   We denote by F2[p,n,m,v] the probability of death of the v-th player in this game and by R2[n,m,y] the probability of the game ends in the y-th round.

Lemma 4.  R2[n, m, y]=(((((n - m)/n)))^(y - 1)) (m/n)

We omit the proof,  
      because the calculation is almost trivial.

It is not difficult to predict the convergence of |R2[n,m,y]-R[n,m,y]|to 0, but the convergence is uniform with respect to m.  This uniform convergence is very important to prove the uniform convergence of |F2[p,n,m,v]-F[p,n,m,v]|to 0 later.

Lemma 5.  Let  K be a positive number.  we have (lim_(n → ∞))Max[|R2[n, m, y] - 
                  R[n, m, y]|;y<Min[K,n - m]and m≤n]=0.

Proof.  By Lemma 1 we let a =
 R[n,m,y]=\(TraditionalForm\`∏\+\(k = 0\)\%\(y - 2\)\(n - \((m + k)\)\)\/\(n - k\)\)×(m/(n - (y - 1)))	
 =\(\(n - m\)\/n×\(n - \((m + 1)\)\)\/\(n - 1\)×\(n - \((m + 2)\)\)\/\(n - 2\)×⋯×\(n - \((m + y - 2)\)\)\/\(n - \((y - 2)\)\)×m\/\(n - \((y - 1)\)\)\).
We let b=(((((n - m)/n)))^(y - 1))×
            (m/(n - (y - 1))), 
c=\(\(\((\(n - \((m + y - 2)\)\)\/\(n - \((y - 2)\)\))\)\^\(y - 1\)\)\(×\)\(m\/n\)\(\ \ \)\)
and 
d = R2[n, m, 
            y] = (((((n - m)/n)))^(y - 1)) (m/n).
Then we have 
c<a<b  and  c<d<b.
|R2[n,m,y]-R[n,m,y]|
<b-c
= \(\((\(n - m\)\/n)\)\^\(y - 1\)×
          m\/\(n - \((y - 1)\)\) - \((\(n - \((m + y - 2)\)\)\/\(n - \((y - 2)\)\))\)\^\(y - 1\)×m\/n\)
= \(\((\((\(n - m\)\/n)\)\^\(y - 1\) - \((\(n - \((m + y - 2)\)\)\/\(n - \((y - 2)\)\))\)\^\(y - 1\))\)×m\/\(n - \((y - 1)\)\)\)
\(\(+\((\(n - \((m + y - 2)\)\)\/\(n - \((y - 2)\)\))\)\^\(y - 1\)\)×\((m\/\(n - \((y - 1)\)\) - m\/n)\)\)
≤ \(\((\((\(n - m\)\/n)\)\^\(y - 1\) - \((\(n - \((m + y - 2)\)\)\/\(n - \((y - 2)\)\))\)\^\(y - 1\))\) + \((m\/\(n - \((y - 1)\)\) - m\/n)\)\).
Note that in the last inequality we used the fact that 0 ≤(m/(n - (y - 1))), (((((n - (m + y - 2))/(n - (y - 2)))))^(y - 1))≤ 1.
Here we are going to use a simple formulae of inequality.  For any  numbers s and t with 0 ≤ s,t ≤ 1  (s^(y - 1))-(t^(y - 1))= (s-t)((s^(y - 2))+(s^(y - 3))(t^1)+(s^(y - 4))(t^2)+⋯+(t^(y - 2)))≤ (s-t)(y-1).
By this inequality the last term of the above inequality is equal to  or smaller than the first term of the following equation.
\((\(n - m\)\/n - \(n - \((m + y - 2)\)\)\/\(n - \((y - 2)\)\))\)×(y - 1)+((m/(n - (y - 1))) - (m/n))
= \(\(mn \((y\^2 - 2  y + 1)\) + m \((3  y\^2\  - y\^3 - 2  y)\)\)\/\(n\^3 + \(n\^2\) \((3\  - 2\ \ y)\) + n \((y\^2 + 2 - 3  y)\)\)\)
⟶ 0 as  n ⟶ ∞,
since 0 ≤ y ≤ K and m ≤ n.
The conclusion of this lemma is direct from this inequality.

Mathemtica calculation　3.  We can use Mathematica to prove some part of the above proof.  For example to we can prove the following equation.

\(\((\(n - m\)\/n - \(n - \((m + y - 2)\)\)\/\(n - \((y - 2)\)\))\)×\((y - 1)\) + \((m\/\(n - \((y - 1)\)\) - m\/n)\)\)
= \(\(mn \((y\^2 - 2  y + 1)\) + m \((3  y\^2\  - y\^3 - 2  y)\)\)\/\(n\^3 + \(n\^2\) \((3\  - 2\ \ y)\) + n \((y\^2 + 2 - 3  y)\)\)\)
by Mathematica.

Simplify[(((n - m)/n) - ((n - (m + y - 2))/(n - (y - 2))))×(y - 1) + ((m/(n - (y - 1))) - 
            (m/n))
 - ((m n ((y^2) - 2  y + 1) + m (3  (y^2)  - (y^3) - 2  y))/((n^3) + (n^2) (3  - 2  y) + n ((y^2) + 2 - 3  y)))]

Theorem 3.   
F2[p, n, m, v]=(({{\(\((\(n - m\)\/n)\)\^\(v - 1\)×
                  m\/n×1\/\(1 - \((\(n - m\)\/n)\)\^p\)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \((if\ 1 ≤ m < n)\)\)}, {\(1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \((if\ m\  = \ \(n\ and\ v\  = \ 1\))\)\)}, {\(0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \((if\ m\  = \ 
                      n\ and\ v\  > \ 1)\)\)}}))

proof.  First we assume that 1≤m<n.  The v-th player plays the v-th,v+p-th,v+2p-th,⋯ round.  Therefore F2[p,n,m,v]=(∑_(z = 0)^∞)R2[n, m, v + pz].  By Lemma 4 R2[n, m, y] = (((((n - m)/n)))^(y - 1)) (m/n).  
Therefore we have F2[p,n,m,v]
  =  (∑_(z = 0)^∞)(((((n - m)/n)))^(v + pz - 1))×
                (m/n)=\(\((\(n - m\)\/n)\)\^\(v - 1\)×
              m\/n×\(\(1\/\(1 - \((\(n - m\)\/n)\)\^p\)\)\(.\)\)\)
If m = n, then the game ends in the first round.  This implies that the first player will die, and other player will not die.  Therefore F2[p,n,m,v]= 1 or 0 when v = 1 or v > 1 respectively.

Mathemtica calculation　4.  Mathemtica function forgeneral Russian roulette  version 2 .  The following Mathemtaica function returns F2[p,n,m,v];the probability of death of the v-th player in the Russian roulette version 2 with n-chambers, m-bullets and p-players.  Note that in this version of the game the player rotates the cylinder each time before he pulls the trigger.

F2[p_, n_, m_, v_] := 
        If[n > m, (((((n - m)/n)))^(v - 1))×
            (m/n)×(1/(1 - (((((n - m)/n)))^p))), If[v == 1, 1, 0]]

Theorem 4.  Let  H be a positive number such that 0 < H < 1.   Here we fix  p and v .   Then we have 
\(lim\+\(n → ∞\)Max[\(\(|\)\(F2[p, n, m, v] - 
                F[p, n, m, v]\)\(|\)\); 
            m ≤ n\ and\ m\/n > H] = 0. \)

proof.  Let ϵ be an arbitrary small positive number.  we can choose a sufficiently large natural number  q such that q depends only on ϵ,H and
 (∑_(z = q + 1)^∞)(((1 - H))^(v + pz - 1)) < ϵ.
By Lemma 1 and the fact that (m/n) > H we have
R[n,m,v+pz]< (((((n - m)/n)))^(v + pz - 1))<(((1 - H))^(v + pz - 1))< ϵ.
, and hence 
(∑_(z = q + 1)^∞)R[n, m, v + pz] <  (∑_(z = q + 1)^∞)(((1 - H))^(v + pz - 1)) < ϵ.  
Therefore by Theorem 1 we have 
|F[p,n,m,v]- (∑_(z = 0)^q) R[n, m, v + pz]|< ϵ  ⋯(1). 
By Lemma 4 
|F2[p,n,m,v]-(∑_(z = 0)^q) R2[n, m, v + pz]|
= (∑_(z = q + 1)^∞) R2[n, m, v + pz]
= (∑_(z = q + 1)^∞)(((((n - m)/n)))^(v + pz - 1))×(m/n)
< (∑_(z = q + 1)^∞)(((1 - H))^(v + pz - 1))< ϵ  ⋯(2).
Let K = v+pq.  Then we can apply Lemma 5 to R2[n,m,y]-R[n,m,y] for y = v, v+p ,⋯, v+pq.
Therefore by (1), (2) we have
|F2[p,n,m,v]-F[p,n,m,v]|< 2ϵ for sufficiently large natural number n.
Since ϵ is an arbitrary small positive number, we can finish the proof.

Mathemtica calculation　5. Theorem 3 asserts that \(lim\+\(n → ∞\)Max[\(\(|\)\(F2[p, n, m, v] - 
                  F[p, n, m, v]\)\(|\)\); \ 
              m ≤ n] = 0\), but if you try to calculate Max[|F2[p,n,m,v]-F[p,n,m,v]|;m≤n]for n =1,2,⋯, the result reveals a strange fact.  It seems that the conergence depends on Mod[n,3].

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]];
F2[p_, n_, m_, v_] := (((((n - m)/n)))^(v - 1))×
            (m/n)×(1/(1 - (((((n - m)/n)))^p)));

p = 3; v = 1; 
      dat = Table[
          N[Max[Table[Abs[F[p, n, m, v] - F2[p, n, m, v]], {m, 1, n - 1}]], 
            10], {n, 100, 110}]

p = 3; v = 1; 
      dat = Table[
          N[Max[Table[Abs[F[p, n, m, v] - F2[p, n, m, v]], {m, 1, n - 1}]], 
            10], {n, 200, 210}]

4. Monotonicity of the probability of death.

If you are going to play a Russian rulette game with a gun of fixed number of chambers, is it better for you to get more bullets?  This is the problem of the monotonicity of F[p,n,m,v]and F2[p,n,m,v] as to the variable m.  It seems obvious that with a fixed number of chambers the probability of death of the first player is an increasing function of the number of bullets m, but as to the other players the situation is far from obvious.
    First we are going to study the first version of the Russian roulette game.

Theorem 5.  If we fix n, then F[p,n,m,1]   is an increasing function of m,i.e., F[p,n,m,1]≤F[p,n,m+1,1] for any  m = 1,2,⋯n-1.

Proof.  (a).  First we are going to show that
 F[p,n,1,1] ≤ F[p,n ,2,1].
 We are going to calculate F[p,n,2,1] by Theorem 1.
 \(Let\ t\_1\  = \ \(\([\(n - 2 + p - 1 + 1\)\/p]\) = \(\([\(n + p - 2\)\/p]\)\(.\)\(\ \)\)\)\)
 Let q be the remaining part of n+p-2 when it is divided by p.
 Then 
 0≤q<p 
 and
 n + p - 2  =  (Subscript[t, 1]) p+q ⋯(3)
 By (3) we have
 n - 1 - ((Subscript[t, 1]) - 1) p = q + 1  ⋯(4)
 and
 (Subscript[t, 1]) p - n + 1  =  
        p - q - 1 ⋯ (5).
 Therefore by Theorem 1 and (4) we have
 F[p,n,2,1]
= (((Subscript[, n - 1])(Subscript[C, 1])+(Subscript[, n - 1 - p])(Subscript[C, 1])+  ⋯  +(Subscript[,   n - 
                  1 - ((Subscript[t, 1]) - 1) p])(Subscript[C, 1]))/((Subscript[, n])(Subscript[C, 2]) ))
= (((n - 1) + (n - 1 - 
              p)  +  ⋯  +   (n - 
              1 - ((Subscript[t, 1]) - 1) p))/((Subscript[, n])(Subscript[C, 2]) ))
= \(\(\((n - 1)\) + \((n - 1 - \((t\_1 - 1)\) p)\)\)\/2×t\_1×2\/\(n \((n - 1)\)\)\)
= \(\(\((n - 1)\) + \((q + 1)\)\)\/2×t\_1×2\/\(n \((n - 1)\)\)\)
=  (((n + q) (Subscript[t, 1]))/(n (n - 1))) ⋯(6).
We are going to calculate F[p,n,1,1] by Theorem 1.
\(Let\ t\_2\  = \ \(\([\(n - 1 + p - 1 + 1\)\/p]\) = \(\([\(n + p - 1\)\/p]\)\(.\)\)\)\)
By (3) n + p - 1 = (Subscript[t, 1]) p + (q + 1),and hence we have
(Subscript[t, 2])=(({{(Subscript[t, 1]) + 1, (if p  =  q + 1)}, {(Subscript[t, 1]), (if p  >  q  + 1)}}))⋯(7).
By Theorem 1 
F[p,n,1,1]
= (((Subscript[, n - 1])(Subscript[C, 0])+(Subscript[, n - 1 - p])(Subscript[C, 0])+  ⋯  +(Subscript[,   n - 
                  1 - ((Subscript[t, 2]) - 1) p])(Subscript[C, 0]))/((Subscript[, n])(Subscript[C, 1]) ))
= ((Subscript[t, 2])/n) ⋯(8).
By (6) and (8)
F[p, n, 2, 1] - 
          F[p, n, 1, 
            1] = (((Subscript[nt, 1]) + (Subscript[qt, 1]) - (Subscript[nt, 2]) + 
                (Subscript[t, 2]))/(n (n - 1))) ⋯(9).
We are going to show that (Subscript[nt, 1]) + (Subscript[qt, 1]) - (Subscript[nt, 2]) + (Subscript[t, 2]) ≥ 0.
(a-1) First we assume that (Subscript[t, 1]) = (Subscript[t, 2]).
Then we have
(Subscript[nt, 1]) + (Subscript[qt, 1]) - (Subscript[nt, 2]) + (Subscript[t, 2]) = (Subscript[qt, 1]) + (Subscript[t, 2]) ≥ 0. 
By (9) we have F[p,n,2,1]-F[p,n,1,1]≥ 0.
(a-2) Secondly we assume that (Subscript[t, 2]) = (Subscript[t, 1]) + 1⋯(10). 
Then by (7) we have p = q +1 ⋯(11).
By using (10),(11) and (5) we have 
(Subscript[nt, 1]) + (Subscript[qt, 1]) - (Subscript[nt, 2]) + (Subscript[t, 2]) 
= (Subscript[nt, 1]) + (Subscript[qt, 1]) - n ((Subscript[t, 1]) + 1) + ((Subscript[t, 1]) + 1)
= (q+1)(Subscript[t, 1])-n+1
= p(Subscript[t, 1])-n+1
= p-q-1 = 0,
and hence by (9) we have F[p,n,2,1]≥F[p,n,1,1].
(b).  Now we are going to show that F[p,n,m,1] ≤ F[p,n,m+1,1] for any  m ≥2.  Here we use  mathematical induction as to the variable n　and the result of (a).  We assume that F[p,n-1,k,1] ≤ F[p,n-1,k+1,1] ⋯(12)
for any  k = 2,⋯,n-1.
Then by (12) and the result of (a) we have
F[p,n-1,m-1,1]≤ F[p, n - 1, m, 1]≤ F[p,n-1,m+1,1]⋯(13)
for any m = 2,⋯,n-2.
By Theorem 1 and (13) we have
((U[p, n - 1, m - 1, 1])/((Subscript[, n - 1])(Subscript[C, m - 
                  1])))
≤ ((U[p, n - 1, m, 1])/((Subscript[, n - 1])(Subscript[C, m]) ))
≤((U[p, n - 1, m + 1, 1])/((Subscript[, n - 1])(Subscript[C, m + 1]))).
, and hence by Theorem  2 we have 
F[p,n+1,m,1]

= ((U[p, n - 1, m - 1, 1] + U[p, n - 1, m, 1])/((Subscript[, n - 1])(Subscript[C, m - 
                    1]) + (Subscript[, n - 
                    1])(Subscript[C, m]) ))

≤((U[p, n - 1, m, 1] + U[p, n - 1, m + 1, 1])/((Subscript[, n - 
                      1])(Subscript[C, m])  + (Subscript[, n - 
                      1])(Subscript[C, m + 1])))

=F[p,n+1,m+1,1].

As you see in the above proof, to prove Theorem 5 is not so easy to prove although the result seems to be obvious.

Theorem  6.  If we fix n, then F[p,n,m,p] is a decreasing function of m,i.e., F[p,n,m,p]≥F[p,n,m+1,p] for any  m = 1,2,⋯n-1.

Proof.  (a) First we are going to show that 
F[p,n,1,p]≥F[p,n,2,p].
We are going to calculate F[p,n,2,p] using Theorem 1.
Let (Subscript[t, 1])  =  [((n - 2 + p - p + 1)/p)]=[((n - 1)/p)].
Let q be the remaining part of n-1 when it is divided by p.
Then 0≤q<p  and  n-1 = (Subscript[t, 1]) p+q    ⋯         (14).
By (14)  we have n - 
          p - ((Subscript[t, 1]) - 1) 
            p = n - (Subscript[t, 1]) p  =  q+1  ⋯                    (15).
Therefore by Theorem 1 and (2) we have 
F[p,n,2,p] 
= (((Subscript[, n - p])(Subscript[C, 1])+(Subscript[, n - 2  p])(Subscript[C, 1])+  ⋯  +(Subscript[,   n - 
                  p - ((Subscript[t, 1]) - 1) p])(Subscript[C, 1]))/((Subscript[, n])(Subscript[C, 2]) ))
= (((n - p) + (n - 2  p) + ⋯ + (n - 
              p - ((Subscript[t, 1]) - 1) p))/((Subscript[, n])(Subscript[C, 2]) ))
= \(\(\((n - p)\) + \((n - p - \((t\_1 - 1)\) p)\)\)\/2×t\_1×2\/\(n \((n - 1)\)\)\)
= \(\(\((n - p)\) + \((q + 1)\)\)\/2×t\_1×2\/\(n \((n - 1)\)\)\)
= (((n - p + q + 1) 
              (Subscript[t, 1]))/(n (n - 1))) ⋯(16).
Let (Subscript[t, 2])  =  [((n - 1 + p - p + 1)/p)] = [(n/p)]⋯(17)
By (14) n = (Subscript[t, 1]) p + (q + 1)
, and hence we have
(Subscript[t, 2])=(({{(Subscript[t, 1]) + 1, (if p  =  q + 1)}, {(Subscript[t, 1]), (if p  >  q  + 1)}}))⋯(18)
By Theorem 1 
F[p, n, 1, p] = ((Subscript[t, 2])/n) ⋯(19).
By (16) and (19)
F[p, n, 1, p] - F[p, n, 2, p]=(((n - 1) (Subscript[t, 2]) - (n - p + q + 1) 
                (Subscript[t, 1]))/(n (n - 1)))⋯(20).
We are going to show that (n - 1) (Subscript[t, 2]) - (n - p + q + 1) (Subscript[t, 1])≥ 0.
(a-1)  First we suppose that (Subscript[t, 1]) = (Subscript[t, 2])⋯  (21).  
Then  by (18) we have q+1<p , and  hence
p-q ≥2  ⋯  (22).
By (21) and (22) we have 
(n - 1) (Subscript[t, 2]) - (n - p + q + 1) (Subscript[t, 1])
= (p - q) (Subscript[t, 1]) - 2  (Subscript[t, 1]) ≥ 0⋯(23).
Therefore by (20) and (23) we have
F[p,n,1,p]-F[p,n,2,p] ≥ 0.
(a-2)  Secondly we suppose that  (Subscript[t, 2]) = (Subscript[t, 1]) + 1 ⋯(24),
then by (18) we have p = q +1⋯(25).
By (25) and (24) we have 
(n - 1) (Subscript[t, 2]) - (n - p + q + 1) (Subscript[t, 1])
= (n - 1) (Subscript[t, 2]) - (Subscript[nt, 1])
= (n - 1) ((Subscript[t, 1]) + 1) - (Subscript[nt, 1])
= n - (Subscript[t, 1]) - 1
= n - (Subscript[t, 2]) > 0 ⋯(26).
By (17)   n  >  (Subscript[t, 2]), and hence by (20) and (26) 
 we have F[p,n,1,p]≥F[p,n,2,p].
(b) Now we are going to show that F[p,n,m,p]≥F[p,n,m+1,p] for any  m≥2.  Here we use  mathematical induction as to the variable n and the result of (a).
 We assume that F[p,n-1,k,p]≥ F[p,n-1,k+1,p]for any  k ≥2 ⋯(27).
 Then by (27) and the result of (a) we have
 F[p,n-1,m-1,p] ≥ F[p,n-1,m,p] ≥ F[p,n-1,m+1,p]⋯(28)
for any m ≥2.
By Theorem 1 and (28) we have
U[p, n - 1, m - 1, 
              p]/((Subscript[, n - 1]))(Subscript[C, m - 
                    1])
≥((U[p, n - 1, m, p])/((Subscript[, n - 1])(Subscript[C, m]) ))
≥ ((U[p,n-1,m+1,p])/((Subscript[, n-1])(Subscript[C, m+1]))).
, and hence by Theorem  2 we have F[p,n+1,m,p]=((U[p, n - 1, m - 1, p] + U[p, n - 1, m, p])/((Subscript[, n - 1])(Subscript[C, m - 1])+(Subscript[, n - 1])(Subscript[C, m]) ))
≥((U[p, n - 1, m, p] + U[p, n - 1, m + 1, p])/((Subscript[, n - 
                      1])(Subscript[C, m])  + (Subscript[, n - 
                      1])(Subscript[C, m + 1])))
=F[p,n+1,m+1,p].

Only the probabilities of the first and the last player have monotonicity.  For example in a game with 3 players the probabilities of death for the second player is not monotone.  See the next example.

Example 2.  A counter example of non-monotonicity of F[p, n, m, v] 
              as to the variable m when v≠1,p.

F[3, 4, 1, 2]= \(\(\_\(4 - 2\)\)\(C\_\(1 - 1\)\)\/\(\_4\)\(C\_1\)\) = (1/4), F[3,4,2,2] =  \(\(\_\(4 - 2\)\)\(C\_\(2 - 1\)\)\/\(\_4\)\(C\_2\)\)= (1/3)and  F[3,4,3,2] = \(\(\_\(4 - 2\)\)\(C\_\(3 - 1\)\)\/\(\_4\)\(C\_3\)\) = (1/4).  Here we have F[3,4,1,2] < F[3,4,2,2] > F[3,4,3,2].  In this case the probability of death for the third player in not monotone as to m.

Mathemtica calculation　6.  It is interesting to see the monotonicity and non-monotonicity of F[p,n,m,v].  Here is the case of 7 players and 25 chambers.  We are going to plot graphs of {F[7,25,m,v];m = 1,2,3,⋯25} for v = 1,2,3,⋯7.

<< Graphics`MultipleListPlot`

n = 25;
p = 7;
Do[V[v] = Table[F[p, n, m, v], {m, 1, n}], {v, 1, p}]

MultipleListPlot[V[1], V[2], V[3], V[4], V[5], V[6], V[7], 
      PlotStyle -> {RGBColor[1, 0, 0], RGBColor[0, 0, 1], 
          RGBColor[0, 1, 0], RGBColor[1, 0, 0], RGBColor[0, 1, 0], 
          RGBColor[0, 0, 1], RGBColor[1, 0, 0]}, PlotJoined -> True, 
      Frame -> True, PlotRange -> {0, 1}]

By theorem 5 F[p,n,m,1] is an increasing function of m.  Therefore it would be natural to ask if　(m/n)≤(t/s)  implies that　F[p,n,m,1]≤F[p,s,t,1].　The following example shows that this is not the case.

Example 3.  A counter example of non-monotonicity of F[2, n, m, 1] as to (m/n).

By Theorem  3 and Theorem 4 for any fixed integer k
(lim_(n → ∞))F2[2, 2  n, n + k, 1]
= (lim_(n → ∞))F[2, 2  n, n + k, 1]
= \(lim\+\(n → ∞\)\((\(n - k\)\/\(2  n\)×1\/\(1 - \((\(2  n - \((n - k)\)\)\/\(2  n\))\)\^2\))\)\)
= \(lim\+\(n → ∞\)\((\ \(n - k\)\/\(2  n\)×1\/\(1 - \((\(n + k\)\/\(2  n\))\)\^2\))\)\)
= (1/2)×(1/(1 - ((((1/2)))^2)))
Therefore we have 
(lim_(n → ∞))F[2,2n,n-k,1] =  (2/3).
Since F[2,6,3,1] = (13/20)=0.65,  
F[2,2n,n-k,1] > F[2,6,3,1]  for sufficiently large n.
In particular 
F[2,32,15,1]=(184101389/282861360), which is approximately 0.650854,
so we have
F[2,32,15,1] > F[2,6,3,1] although (15/32)<(3/6).

Mathemtica calculation　7.  We can present an counterexample to the monotonicity of F[p,n,m,v] as to (m/n).  Here we present F[2,6,3,1], F[2,32,15,1] and F[2,128,60,1].

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]]

F[2, 6, 3, 1]

F[2, 32, 15, 1]

% // N

F[2, 128, 60, 1] // N

F2[2, 128, 60, 1] // N

Next we are going to study the second version of the game.  As to the monotonicity this version of Russian roulette is easier to study.

Lemma 6.  Let g[x]= ((x^t)/((∑_(k = 0)^s) (x^k))) , where t and s are non-negative integers.  
(a) y is decreasing on (0,1) if t=0.
(b)y is increasing on (0,1)if   2t≥s, 
and y is not monotone function on (0,1) if 0< 2t<s.

Proof.  It is easier to study the function  y  = h[
            x] = (((∑_(k = 0)^s) (x^k))/(x^t)).
(a) If t=0, it is clear that y is increasing on (0,1).
Therefore g is decreasing on (0,1).
(b) Let t>0. First we assume that s ≤ t, then it is easy to see that y is decreasing on (0,1).
Therefore g is increasing on (0,1). 
Next we assume that s > t, then we have
y = (1/(x^t))+(1/(x^(t - 1)))+⋯+(1/x)+1+x+⋯+(x^(s - t)),
and hence we have 
\(\(\(\ \)\(y\^′ = \(-\((t\/x\^\(t + 1\) + \(t - 1\)\/x\^t + \(t - 2\)\/x\^\(t - 1\) + ⋯  2\/x\^3 + 1\/x\^2)\)\)\)\)\)+(1+2x+⋯+(s - t) (x^(s - t - 1))).
It is clear that   - (k/(x^(k + 1))) + (kx^(k - 1)) < 
        0  on (0, 1)      ⋯(29),
(lim_(x → 1))(-(k/(x^(k + 1))) + 
              (kx^(k - 1)))=0      ⋯(30).
and
\(\(\(\ \)\(lim\+\(x → 0\)y\^′ = \(-∞\)\)\)\)      ⋯(31).
If t ≥ s - t, by (29)  (y^′) < 0.  Therefore y is decreasing, and hence g is increasing.
If  t<s-t , then by (30)　\(\(\(\ \)\(\(lim\+\(x → 1\)y\^′\)\(>\)\(0\)\(\ \)\)\)\), and hence by (31)　(y^′)can be positive for some x ϵ (0,1) and negative for some x ϵ (0,1).  Therefore y is not monotone, and hence g is not monotone on (0,1).

Theorem 7.   
(a) When v =1, F2[p,n,m,v]is an increasing function of (m/n),i.e.,F2[p,n,m,v]≥F2[p,(n^′),(m^′),v]if (m/n)≥((m^′)/(n^′)).
(b) When 2v≥p+1, F2[p,n,m,v]is a decreasing function of (m/n),i.e.,F2[p,n,m,v]≤F2[p,(n^′),(m^′),v]if (m/n)≥((m^′)/(n^′)).
(c) When 2v<p+1, 
the function F2[p,n,m,v]with a fixed n is not a monotone function of (m/n).

Proof.  By Theorem 3 
F2[p, n, m, p] = (((((n - m)/n)))^(v - 1))×
          (m/n)×(1/(1 - (((((n - m)/n)))^p))), 
and hence we are going to study the function\(\(\(\ \)\(f[
          x]\  = \(x \((1 - x)\)\^\(v - 1\)\)\/\(1 - \((1 - x)\)\^p\)\ on\ \(\((0, 1)\)\(.\)\)\)\)\)
By exchanging 1-x with x and we get a simpler function  g[x] =\(\(\((1 - x)\) x\^\(v - 1\)\)\/\(1 - x\^p\) = \(\(x\^\(v - 1\)\/\(∑\+\(k = 0\)\%\(p - 1\)x\^k\)\)\(\ \)\(.\)\(\ \ \)\)\)
(a)   If v =1, then by Lemma 6   g is decreasing, and hence f is increasing.
(b)   If 2v≥p+1, then 2(v-1)≥ p-1 and by Lemma 6  g is increasing, and hence  f  is decreasing.
(c)    If 2v<p+1, then 2(v-1)< p-1 by Lemma 6  g is not monotone, and hence  f  is not monotone.

If you are going to play a Russian rulette game with a fixed number of bullets,　is it better for you to use a gun with larger number of chambers?  
    This is the problem of the monotonicity of F[p,n,m,v]and F2[p,n,m,v] as to the variable n.

{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}

Remark .   In Figure (2) you see that{(1/1),(2/3),(4/6),(6/10),(9/15),⋯} is decreasing and  {1, (1/2), (2/3), (2/4), (3/5), ⋯} in not monotone .   We are going to study these facts in the next theorem.

Theorem 8.  F[p,n,m+1,1]is decreasing function of  n (≥ m) when we fix a natural number m.　　F[p,n,1,1] is not a monotone function of n.

Proof.  By Theorem 5 F[p,n,m,1]≤ F[p,n,m+1,1], and hence by Theorem 1 we have
((U[p, n, m, 1])/((Subscript[, n])(Subscript[C, m]) ))≤  ((U[p, n, m + 1, 1])/((Subscript[, n])(Subscript[C, m + 1]) )),
which implies that
((U[p, n, m, 1] + U[p, n, m + 1, 1])/((Subscript[, n])(Subscript[C, m])+(Subscript[, n])(Subscript[C, m + 1]) ))≤((U[p, n, m + 1, 1])/((Subscript[, n])(Subscript[C, m + 1]) )).
By the above inequality and Theorem 1,2 we have 
F[p,n+1,m+1,1] ≤ F[p,n,m+1,1].
F[p, n, 1, 1]=(t/n) with  t =[((n + p - 
                1)/p)] = [((n - 1)/p)] + 
            1,
and hence F[p,n,1,1] is not a monotone function of n.  For example if we let p = 5 and calculate F[5,n,1,1] for n = 1, 2, 3, ⋯,8, then we have {1, (1/2), (1/3), (1/4), (1/5), (1/3), (2/7), (1/4)}.  This is clearly not monotone.

Mathemtica calculation　8.  We can calculate F[p,n,1,1] for n = 1,2,3,⋯ to see that this is not a monotone function of the variable n.

p = 5; Table[IntegerPart[((n + p - 1)/p)]/n, {n, 1, 60}]

5.  The Limit of the expectation of death of the v-th player in a Russian roulette game with p players.

If you have guns with varieties of numbers of chambers and bullets, which gun should you use?  
    Suppose that you have a gun with 10 chambers and a gun with 100 chambers, but you do not know the number of bullets.  Which is better for you?  This is a problem of average.  We have to compare the average of probability of death of each gun.  
    We consider E[p,n,v]= ((F[p, n, 1, v] + F[p, n, 2, v] + ⋯ + 
              F[p, n, n, v])/n) and E2[p,n,v]=((F2[p, n, 1, v] + F2[p, n, 2, v] + ⋯ + 
              F2[p, n, n, v])/n).  
    E[p,n,v]and E2[p,n,v]are the expectations of the death of the v-th player in the first and second version of Russian rulette games respectively when we know the number  n  of the chamber and do not know the number of bullets.  Here we suppose that the probability is the same for each number of bullets.  We found out that (lim_(n → ∞))E[p,n,v]and (lim_(n → ∞))E2[p,n,v] have very interesting properties.
    (lim_(n → ∞))E[p, n, v]and (lim_(n → ∞))E2[p, n, v]have the same value, because by Theorem 4 |F2[p,n,m,v]-F[p,n,m,v]|→0 uniformly for m≤ n.

Theorem 10.   (lim_(n → ∞))E[p, n, v]=(lim_(n → ∞))E2[p, n, v]=\(∫\_0\%1\(\( 
                      x \((1 - x)\)\^\(v - 1\)\)\/\(1 - \((1 - x)\)\^p\)\) dx = ∑\+\(k = 0\)\%∞\(\( 
                    1\/\(\((v + kp)\) \((v + kp + 1)\)\)\)\(.\)\)\)

proof.   By Theorem 3　

(lim_(n → ∞))E2[p, n, v]

= (lim_(n → ∞))((F2[p, n, 1, v] + F2[p, n, 2, v]+⋯+F2[p, n, n, v])/n)
\(\(=\)\(lim\+\(n → ∞\)\ \(∑\+\(m = 1\)\%n\( m\/n\) \(\((1 - m\/n)\)\^\(v - 
                      1\)\) \(1\/\(1 - \((1 - m\/n)\)\^p\)\) 1\/n\)\)\)
\(\(=\)\(lim\+\(ε → 0\)∫\_\(0 + ε\)\%1\(\( 
                      x \((1 - x)\)\^\(v - 1\)\)\/\(1 - \((1 - x)\)\^p\)\) \(\(dx\)\(.\)\)\)\)
If we exchange x with 1-x, then
\(lim\+\(ε → 0\)∫\_\(0 + ε\)\%1\(\( 
                    x \((1 - x)\)\^\(v - 1\)\)\/\(1 - \((1 - x)\)\^p\)\) dx\)
\(\(=\)\(lim\+\(ε → 0\)∫\_0\%\(1 - ε\)\(\(\((1 - x)\) 
                      x\^\(v - 1\)\)\/\(1 - 
                      x\^p\)\) \(\(dx\)\(.\)\)\)\)
Since 1 + (x^p) + (x^(2  p)) + (x^(3  p)) + ⋯=(1/(1 - (x^p))),
\(lim\+\(ε → 0\)∫\_0\%\(1 - ε\)\(\(\((1 - x)\) x\^\(v - 1\)\)\/\(1 - 
                  x\^p\)\) dx\)
\(\(=\)\(lim\+\(ε → 0\)∫\_0\%\(1 - ε\)\((1 - x)\) \(x\^\(v - 1\)\) \((1 + 
                    x\^p + x\^\(2  p\) + 
                    x\^\(3  p\) + ⋯)\) dx\)\)

\(\(=\)\(lim\+\(ε → 0\)\(∑\+\(k = 0\)\%∞∫\_0\%\(1 - ε\)\((1 - 
                      x)\) \(x\^\(v - 1 + kp\)\) dx\)\)\)

=(∑_(k = 0)^∞)(∫_0^1)(1 - 
                    x) (x^(v - 1 + kp)) dx
=(∑_(k = 0)^∞) 
                  (1/((v + kp) (v + kp + 1))).

Next we are going to show that   (lim_(n → ∞))E[p, n, v]=(lim_(n → ∞))E2[p, n, v].

For any positive number  ϵ> 
  0 there exists a positive number  H such that  0 < H < 1 and We consider E[p,n,v] = ((F[p, n, 1, v] + F[p, n, 2, v] + ⋯ + 
            F[p, n, n, v])/n) and E2[p,n,v]  = ((F2[p, n, 1, v] + 
                F2[p, n, 2, v] + ⋯ + F2[p, n, n, v])/n).

Lemma 7.  Before we proceed to calculate expectations for some examples,we need some formulas of the infinite series in Theorem 5.  We present these formulas without proof, because they are famous formuals.

\(Let\ L[p, 
              v] = ∑\+\(k = 0\)\%∞ 1\/\(\((v + kp)\) \((v + kp + 1)\)\)\), then L[2, 1]= (Subscript[log, e])2, L[3, 1] =  (π/(3 (Sqrt[3]))),L[3, 2]=(1/2) (Subscript[log, e])3-(((Sqrt[3]) π)/18),L[3, 3]=1-(((Sqrt[3]) π)/18)-(1/2) (Subscript[log, e])3,L[4, 1]=(π/8)+(1/4) (Subscript[log, e])2,L[4, 2]=(π/8)-(1/4) (Subscript[log, e])2,L[4, 3]=(3/4) (Subscript[log, e])2-(π/8) and L[4, 4]=1-(π/8)-(3/4) (Subscript[log, e])2.

Mathemtica calculation　8.  We can calculate \(∑\+\(k = 0\)\%∞ 1\/\(\((v + kp)\) \((v + kp + 1)\)\)\) for v,p = 1,2,3,⋯

L[p_, 
          v_] := (∑_(k = 0)^∞) (1/((v + k p) (v + k p + 1)));

L[2, 1]

L[3, 1]

L[4, 1]

Example 4.   We are going to calculate the expectation of the first player's death in a Russian roulette with 2 players.

The answer is L[2, 1]=(Subscript[log, e])2.

Example 5.   We are going to calculate the expectation of death in a Russian roulette with 3 players.

The expectations for the first, the second and the third players are L[3, 
              1] =  (π/(3 (Sqrt[3]))),L[3, 2]=(1/2) (Subscript[log, e])3-(((Sqrt[3]) π)/18) and  L[3, 3]=1-(((Sqrt[3]) π)/18)-(1/2) (Subscript[log, e])3.  In Example 4 we get (Subscript[log, e])2, but this time π .   This is amazing!

Example 6.   We are going to calculate the expectation of death in a Russian roulette with 4 players.

The expectations for the first, the second, the third and the forth players are L[4, 1]=(π/8)+(1/4) (Subscript[log, e])2,L[4, 2]=(π/8)-(1/4) (Subscript[log, e])2,L[4, 3]=(3/4) (Subscript[log, e])2-(π/8) and L[4, 4]=1-(π/8)-(3/4) (Subscript[log, e])2.

6.  Appendix.  Further studies.

In section 4 we studied the monotonicity of F[p,n,m,v]and F2[p,n,m,v].  We could determine when F2[p,n,m,v] becomes a monotone function of m.  As to F[p,n,m,v] we could prove that F[p,n,m,v]  is monotone when v = 1 or p, and made a counter example when p = 3, n = 4 and v = 2.  Perhaps we still have many things to prove about the monotonicity of F[p,n,m,v] for v = 2,3,⋯p-2.  
    As to F[p,n,m,p-1] we have the following lemma.

Lemma 8.  Let p>2 and q=Mod[n,p],i.e.,the remainning part of n divided by p. 
(a) If p>q+2 or p = q+1,then  F[p,n,1,p-1]≥F[p,n,2,p-1] 
(b) If p = q+2, then F[p,n,1,p-1]<F[p,n,2,p-1]

Proof.  We are going to calculate F(p,n,2,p-1).
Let (Subscript[t, 1]) = \(\([\(n - 2 + p - \((p - 1)\) + 1\)\/p]\) = \(\([n\/p]\)\(.\)\)\)
Then there exists integer  q  such that　 0≤q<p and  n =  (Subscript[t, 1]) p+q ⋯(32).
By (32)
n - p + 
          1 - ((Subscript[t, 1]) - 1) p = n - (Subscript[t, 1]) p + 
            1  =  q+1  ⋯(33).
Therefore by Theorem 1 and (33) we have 
F[p,n,2,p-1] 
=(((n - p + 1) + (n - p + 1 - 
                  p) + ⋯ + (n - p + 
                  1 - ((Subscript[t, 1]) - 1) p))/((Subscript[, n])(Subscript[C, 2]) ))
=(((n - p + 1) + (q + 1))/2)×
            (Subscript[t, 1])×(2/(n (n - 1)))
=(((n - p + q + 2) 
                  (Subscript[t, 1]))/(n (n - 1))) ⋯(34).
\(Let\ t\_2\  = \ \(\([\(n - 1 + p - \((p - 1)\) + 1\)\/p]\) = \(\([\(n + 1\)\/p]\)\(.\)\)\)\)
By (33)  n + 1 = (Subscript[t, 1]) p + (q + 1), 
and hence we have
(Subscript[t, 2])=(({{(Subscript[t, 1]) + 1, (if p  =  q + 1)}, {(Subscript[t, 1]), (if p  >  q  + 1)}}))⋯(35)
By Theorem 1 
F[p, n, 1, p - 1] = ((Subscript[t, 2])/n)⋯(36).
By (34) and (36)
\(F[p, n, 1, p - 1] - 
            F[p, n, 2, 
              p - 1] = \(\((n - 1)\) t\_2 - \((n - p + q + 2)\) t\_1\)\/\(n \((n - 1)\)\)\)    ⋯(37).
First We suppose that  q+1<p⋯(38). 
By (38) we have p-q ≥2  ⋯(39).
By (35) and (38)
we have (Subscript[t, 1]) = (Subscript[t, 2]) , 
and  hence  
(n - 1) (Subscript[t, 2]) - (n - p + q + 2) (Subscript[t, 1])
=(p - q) (Subscript[t, 1]) - 3  (Subscript[t, 1])⋯(40)
We are going to study the case of (a).
(a-1) If p>q+2, then by (37)and(40) 
F[p,n,1,p-1]≥F[p,n,2,p-1].
(a-2)  Next we suppose that  p = q+1⋯(41).
By (35) we have  (Subscript[t, 1]) + 1 = (Subscript[t, 2])   ⋯(42).
Therefore by (41) and (42) we have
(n - 1) (Subscript[t, 2]) - (n - p + q + 2) (Subscript[t, 1])
=(n - 1) ((Subscript[t, 1]) + 1) - (n + 1) (Subscript[t, 1])
=(Subscript[nt, 1]) - (Subscript[t, 1]) + n - 1 - (Subscript[nt, 1]) - (Subscript[t, 1])
=n - 2  (Subscript[t, 1]) - 1   ⋯(43).
By (32) and the fact that p ≥ 3 we have
n = (Subscript[pt, 1]) + q ≥ 2  (Subscript[t, 1]) + 1. 
Therefore by (37) and (43) we have
F[p,n,1,p-1]≥F[p,n,2,p-1] .
(b) If p=q+2, then by (37) and (40) F[p,n,1,p-1]<F[p,n,2,p-1].

Example 7.   Interesting properties of a Russian roulette game with 2 players.

{1}       ⋯Figure(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}       ⋯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}

{1}        ⋯Figure(3)
                                          {(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}

Theorem 11. F[2,2n,3,1]+F[2,2n+2,3,1]=2F[2,2n+1,3,1].

Proof.       F[2,2n,3,1] = (((Subscript[, 2  n - 1])(Subscript[C, 2])+(Subscript[, 2  n - 3])(Subscript[C, 2])+(Subscript[, 2  n - 5])(Subscript[C, 2])+⋯)/((Subscript[, 2  n])(Subscript[C, 3])))
=(\(TraditionalForm\`∑\+\(k = 1\)\%\(n - 1\)\(\((2  n - 2  k + 1)\) \((2  n - 2  k)\)\)\/2\))/(((2  n(2  n - 1) (2  n - 2))/6))
= 3(∑_(k = 1)^(n - 1))(2  (n^2)-n(4k-1)+k(2k-1))/(n(2  n - 1) (2  n - 2))
= 3(∑_(k = 1)^(n - 1))((2  (n^2) + n) - (4  n + 1) k + 
            2  (k^2))/(n(2  n - 1) (2  n - 2))
=3((2  (n^2) + n) (n - 1) - (4  n + 
              1)×((n (n - 1))/2) + 
        2×((n (n - 1) (2  n - 1))/6))/(n(2  n - 1) (2  n - 2))
= ((4 n + 1)/(8 n - 4)).  Here if we put n+1 into n, we have
F[2,2n+2,3,1] = ((4 n + 5)/(8 n + 4)).
F[2,2n+1,3,1]= (((Subscript[, 2  n])(Subscript[C, 2])+(Subscript[, 2  n - 2])(Subscript[C, 2])+(Subscript[, 2  n - 4])(Subscript[C, 2])+⋯)/((Subscript[, 2  n + 1])(Subscript[C, 3])))
= 3 (∑_(k = 1)^n)(2  n - 2  k + 
                2) (2  n - 2  k + 1)/((2  n + 1) 2  
                  n (2  n - 1))
= 3 (∑_(k = 1)^n)((4  (n^2) + 6  n + 2) - (8  n + 6) 
                  k + 4  (k^2))/((2  n + 1) (2  n - 1) 2  
                n)
= 3 ((4  (n^2) + 6  n + 2) 
                n - (8  n + 6)×((n (n + 1))/2) + 
              4×((n (n + 1) (2  n + 1))/6))/((2  n + 
                  1) (2  n - 1) 2  n)
= ((4 (n^2) + 3  n - 1)/(8 (n^2) - 2)). 
Therefore we have F[2,2n,3,1]+F[2,2n+2,3,1]= 2F[2,2n+1,3,1].

Mathemtica calculation　10.  We can use Theorem 11 by the calculation of Mathematica.  Mathematica can calculate very fast even when we use (∑_(k = 1)^n).  We can see that Mathematica can be very effective in doing this kind of calculation.  Of course in mathematical result mathematical proof is essential, but you can compare the calculation of Mathematica and your own calculation when you make mathematical proof.

First we show that F[2,2n,3,1] = (((Subscript[, 2  n - 1])(Subscript[C, 2])+(Subscript[, 2  n - 3])(Subscript[C, 2])+(Subscript[, 2  n - 5])(Subscript[C, 2])+⋯)/((Subscript[, 2  n])(Subscript[C, 3])))=(\(TraditionalForm\`∑\+\(k = 1\)\%\(n - 1\)\(\((2  n - 2  k + 1)\) \((2  n - 2  k)\)\)\/2\))/(((2  n(2  n - 1) (2  n - 2))/6))=((1 + 4 n)/(-4 + 8 n))

((∑_(k = 1)^(n - 1))(((2  n - 2  k + 1) (2  n - 2  k))/2))/(((2  n (2  n - 1) (2  n - 2))/6))//Simplify

Secondly we show that F[2,2n+2,3,1]= ((5 + 4 n)/(4 + 8 n)) by substituting n with n+1 in F[2,2n,3,1].

f[n_] := ((1 + 4 n)/(-4 + 8 n)); 
    f[n + 1] // Simplify

Thirdly we show that F[2, 2  n + 1, 3, 1]=(((Subscript[, 2  n])(Subscript[C, 2])+(Subscript[, 2  n - 2])(Subscript[C, 2])+(Subscript[, 2  n - 4])(Subscript[C, 2])+⋯)/((Subscript[, 2  n + 1])(Subscript[C, 3])))=3 (∑_(k = 1)^n)(2  n - 2  k + 
                  2) (2  n - 2  k + 1)/((2  n + 1) 2  
                    n (2  n - 1))= \(\(\((1 + n)\)\ \((\(-1\) + 4\ n)\)\)\/\(2\ \((\(-1\) + 2\ n)\)\ \((1 + 2\ n)\)\)\).

3 (∑_(k = 1)^n)(2  n - 2  k + 
              2) (2  n - 2  k + 1)/((2  n + 1) 2  
                n (2  n - 1))

We prove F[2,2n,3,1]+F[2,2n+2,3,1]= 2F[2,2n+1,3,1] by calculation of Mathemtaica.

((1 + 4 n)/(-4 + 8 n)) + ((5 + 4 n)/(4 + 8 n)) - ((2 (1 + n) (-1 + 4 n))/(2 (-1 + 2 n) (1 + 2 n))) // 
      Simplify

Theorem 12. F[2,2n+1,2,1]+F[2,2n+1,4,1]=2F[2,2n+1,3,1].

Proof.  F[2,2n+1,2,1] = (((Subscript[, 2  n])(Subscript[C, 1])+(Subscript[, 2  n - 2])(Subscript[C, 1])+(Subscript[, 2  n - 4])(Subscript[C, 1])+⋯)/((Subscript[, 2  n + 1])(Subscript[C, 2]) )) =((2  n+(2  n - 2)+⋯+2)/((Subscript[, 2  n + 1])(Subscript[C, 2]) )) =(2(∑_(k = 1)^n)(2  n - 2  k + 2))/((2n+1)2n) = ((n + 1)/(2  n + 1)).
F[2,2n+1,4,1]= (((Subscript[, 2  n])(Subscript[C, 3])+(Subscript[, 2  n - 2])(Subscript[C, 3])+(Subscript[, 2  n - 4])(Subscript[C, 3])+⋯)/((Subscript[, 2  n + 1])(Subscript[C, 4]) )) = (4(∑_(k = 1)^(n - 1))(2  n - 2  k + 2) (2  n - 2  k + 
              1)(2n-2k))/((2n+1)2n(2n-1)(2n-2)).
(∑_(k = 1)^(n - 1))(2  n - 2  k + 2) (2  n - 2  k + 
              1) (2  n - 2  k)
=(24  (n^2) + 24  n + 
                4) ((1/2) n (n - 1)) + (8 (n^3) + 12 (n^2) + 
                4 n) (n - 1)
=-2 (n^2) + 2 (n^4).
Therefore F[2,2n+1,4,1]= \(TraditionalForm\`\(4 \((\(-2\)\ n\^2 + 2\ n\^4)\)\)\/\(\((2  n + 1)\) 2 \( n(2  n - 1)\) \((2  n - 2)\)\)\)
= ((2 n (1 + n))/(4 (n^2) - 1)).
By the previous theorem we have
F[2,2n+1,3,1]
= ((4 (n^2) + 3  n - 1)/(8 (n^2) - 2)).
F[2,2n+1,4,1]+F[2,2n+1,2,1]=((2 n (1 + n))/(4 (n^2) - 1))+((n + 1)/(2  n + 1))= \(\(\((4  n - 1)\) \((2  n + 1)\) \((n + 1)\)\)\/\(\((4\ n\^2 - 1)\) \((2  n + 1)\)\)\)=(((4  n - 1) (n + 1))/(4 (n^2) - 1))=2F[2,2n+1,3,1].
Therefore we have F[2,2n,3,1]+F[2,2n+2,3,1]= 2F[2,2n+1,3,1].

Mathemtica calculation　11.  We can prove Theorem 12 by the calculation of Mathematica. In this calculation Mathemtaica is very powerful.

First we show that F[2,2n+1,2,1] = (((Subscript[, 2  n])(Subscript[C, 1])+(Subscript[, 2  n - 2])(Subscript[C, 1])+(Subscript[, 2  n - 4])(Subscript[C, 1])+⋯)/((Subscript[, 2  n + 1])(Subscript[C, 2]) )) =((2  n+(2  n - 2)+⋯+2)/((Subscript[, 2  n + 1])(Subscript[C, 2]) )) =(2(∑_(k = 1)^n)(2  n - 2  k + 2))/((2n+1)2n) = ((n + 1)/(2  n + 1)).

(2(∑_(k = 1)^n)(2  n - 2  k + 2))/((2  n + 1) 2  n)//Simplify

Secondly we show that F[2,2n+1,4,1]= (((Subscript[, 2  n])(Subscript[C, 3])+(Subscript[, 2  n - 2])(Subscript[C, 3])+(Subscript[, 2  n - 4])(Subscript[C, 3])+⋯)/((Subscript[, 2  n + 1])(Subscript[C, 4]) )) = (4(∑_(k = 1)^(n - 1))(2  n - 2  k + 2) (2  n - 2  k + 
              1)(2n-2k))/((2n+1)2n(2n-1)(2n-2))= ((2 n (1 + n))/(4 (n^2) - 1))

(4(∑_(k = 1)^(n - 1))(2  n - 2  k + 2)(2  n - 2  k + 1)(2  n - 2  k))/((2  n + 1) 2  n (2  n - 1) (2  n - 2))//Simplify

Thirdly By the previous theorem we have
F[2,2n+1,3,1]
= ((4 (n^2) + 3  n - 1)/(8 (n^2) - 2)).

We prove F[2,2n,3,1]+F[2,2n+2,3,1]= 2F[2,2n+1,3,1]by Mathematica.

((1 + n)/(1 + 2 n)) + ((2 n (1 + n))/(4 (n^2) - 1)) - 
        2 (( 4 (n^2) + 3  n - 1)/(8 (n^2) - 2)) // Simplify

Theorem 13. F[2,2n,2,1]+F[2,2n+2,4,1]=2F[2,2n+1,3,1].

Proof.  F[2,2n,2,1]=　 (((Subscript[, 2  n - 1])(Subscript[C, 1])+(Subscript[, 2  n - 3])(Subscript[C, 1])+(Subscript[, 2  n - 5])(Subscript[C, 1])+⋯)/((Subscript[, 2  n])(Subscript[C, 2])))
= 2(∑_(k = 1)^n)(2  n - 2  k + 1)/(2  
                  n (2  n - 1))
=(∑_(k = 1)^n)(2  n - 2  k + 
                1)/(n (2  n - 1)).
(∑_(k = 1)^n)(2  n - 2  k + 1) = 
        2  (n^2) - 2 ((1/2) n (n + 1)) + n
=(n^2).
Therefore 
F[2,2n,2,1]=((n^2)/(n (2  n - 1))) = (n/(2  n - 
                  1)).
By the previous theorem 
F[2,2n+1,3,1]= ((4 (n^2) + 3  n - 
              1)/(2 (2  n - 1) (2  n + 1)))
F[2, 2  n + 2, 4, 1]=(((Subscript[, 2  n + 1])(Subscript[C, 3])+(Subscript[, 2  n - 1])(Subscript[C, 3])+(Subscript[, 2  n - 3])(Subscript[C, 3])+⋯)/((Subscript[, 2  n + 2])(Subscript[C, 4])))
=(4(∑_(k = 1)^n)((2  n - 2  k + 
                        3) (2  n - 2  k + 2) (2  n - 2  k + 
                        1)))/((2  n + 2) (2  n + 1) 2  n (2  n - 1))
(∑_(k = 1)^n)((2  n - 2  k + 3) (2  n - 2  k + 
                2) (2  n - 2  k + 1))
=(∑_(k = 1)^n)(-8 (k^3) + (24  n + 
                    24) (k^2)  + k (-22  - 48  n - 24  (n^2)) + 
              6 + 22 n + 24 (n^2) + 8 (n^3))
=-8 ((1/4) (n^2) (((n + 1))^2)) + (24  n + 
                24) ((1/6) n (1 + n) (1 + 
                    2 n)) + ((1/2) n (1 + n)) (-22  - 
                48  n - 24  (n^2)) + 
          n (6 + 22 n + 24 (n^2) + 8 (n^3))
= n (2 (n^2) + 2 n - 1) (n + 1).
\(F[2, 2  n + 2, 4, 
          1] = \(4  n \((2\ n\^2 + 2\ n - 1)\) \((n + 1)\)\)\/\(\((2  n + 2)\) \((2  n + 1)\) 2  n \((2  n - 1)\)\)\)
=((2 (n^2) + 2 n - 
              1)/((2  n + 1) (2  n - 1))).
F[2,2n,2,1]+F[2,2n+2,4,1]-2F[2,2n+1,3,1]
=((2 (n^2) + 2 n - 1)/((2  n + 1) (2  n - 1))) + 
        (n/(2  n - 1)) - 
        2 (( 4 (n^2) + 3  n - 1)/(2 (2  n - 1) (2  n + 1)))
=\(\(\((2\ n\^2 + 2\ n - 1)\) + n \((2  n + 1)\) - \((4\ n\^2 + 3  n - 1)\)\)\/\(\((2  n - 1)\) \((2  n + 1)\)\) = 0. \)

Mathemtica calculation　12.  We can prove Theorem 13 by the calculation of Mathematica. In this calculation Mathemtaica is very powerful.

First we show that F[2,2n,2,1]=　 (((Subscript[, 2  n - 1])(Subscript[C, 1])+(Subscript[, 2  n - 3])(Subscript[C, 1])+(Subscript[, 2  n - 5])(Subscript[C, 1])+⋯)/((Subscript[, 2  n])(Subscript[C, 2])))
= 2(∑_(k = 1)^n)(2  n - 2  k + 1)/(2  
                  n (2  n - 1))
=(n/(2  n - 1)).

2(∑_(k = 1)^n)(2  n - 2  k + 1)/(2  
                    n (2  n - 1))//Simplify

(n/(-1 + 2 n))

Secondly by the previous theorem we have
F[2,2n+1,3,1]= ((4 (n^2) + 3  n - 
              1)/(2 (2  n - 1) (2  n + 1))).

Thirdly we prove that F[2, 2  n + 2, 4, 1]=(((Subscript[, 2  n + 1])(Subscript[C, 3])+(Subscript[, 2  n - 1])(Subscript[C, 3])+(Subscript[, 2  n - 3])(Subscript[C, 3])+⋯)/((Subscript[, 2  n + 2])(Subscript[C, 4])))
=(4(∑_(k = 1)^n)((2  n - 2  k + 
                        3) (2  n - 2  k + 2) (2  n - 2  k + 
                        1)))/((2  n + 2) (2  n + 1) 2  n (2  n - 1))
=((2 (n^2) + 2 n - 
              1)/((2  n + 1) (2  n - 1))).

(4(∑_(k = 1)^n)((2  n - 2  k + 3) (2  n - 
                      2  k + 2) (2  n - 2  k + 1)))/((2  n + 2) (2  n + 1) 2  n (2  n - 1))//Simplify

We prove F[2,2n,2,1]+F[2,2n+2,4,1]=2F[2,2n+1,3,1]by Mathematica.

(n/(2  n - 1)) + ((2 (n^2) + 2 n - 1)/((2  n + 1) (2  n - 1))) - 2 (( 4 (n^2) + 3  n - 1)/(2 (2  n - 1) (2  n + 1))) // 
      Simplify

Remark.  Theorem 11, 12 and 13 are suggested by Kohsuke Morimoto and Naoya Urakawa when they were high school students in Kwansei Gakuin.

Example 8.  
    Let p = 3 and v = 1.  Theorem 3 asserts that \(lim\+\(n → ∞\)Max[\(\(|\)\(F2[p, n, m, v] - 
                F[p, n, m, v]\)\(|\)\); \ 
            m ≤ n] = 0\).  If we calculate Max[|F2[p,n,m,v]-F[p,n,m,v]|; m≤n] for n =100, 101, 102, ⋯,110, then we have the following results.  If you look the data carefully, then you find that small number appear every third times.

Mathemtica calculation　13.  To see how Max[|F2[p,n,m,v]-F[p,n,m,v]|; m≤n] converges to 0 is interesting from the view point of calculation.

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]];
F2[p_, n_, m_, v_] := 
        If[n > m, (((((n - m)/n)))^(v - 1))×
            (m/n)×(1/(1 - (((((n - m)/n)))^p))), If[v == 1, 1, 0]]

p = 3; v = 1; 
    Table[Max[
        Table[Abs[F2[p, n, m, v] - F[p, n, m, v]] // N, {m, 1, n}]], {n, 100, 
        110}]