A Unique and Stable $\hbox {Se}{\mathcal {C}}\hbox {ure}$ Reversion Protocol Improving Efficiency: A Computational Bayesian Approach for Empirical Analysis

Abstract

In Bayesian mechanism we demonstrate the unicity and the stability of secure reversion protocols in which risk-averse players have no incentive to cheat or to deviate from the meditator’s recommendation and that can greatly improve their equilibrium expected payoffs as compared to those generated through correlation device. The main idea of this work is to show the ease with which we can compute the equilibrium expected payoffs for players. Furthermore, we emphasize those results through a numerical simulation for which the method is able to be used for empirical analysis.

Appendices

Appendix 1

Data showed in Table 3 below allow us to define the value of $c_{4}$ and $c_{3}$ for the collective choice problem with missing data in Table 1. Scenarii are implemented and computed in the following way in $t=1$: when the project is undertaken and player 1¹³ pays for it his expected payoff is computed as the hydroelectric power station income minus contribution to the MRC minus government charge minus financial costs: $23.64-1-2.31-13=7.33$ for type 1.0 in which we remove running costs: $23.64-1-2.31-13-3.70=3.63$ for type 1.1 and player 2’s expected payoff is computed as contribution to the MRC plus government charge: $1+2.31=3.31$. When the project is undertaken and player 2 pays for it his expected payoff is computed as contribution to the MRC plus government charge minus financial costs: $1+2.31-13=-9.69$ and player 1’s expected payoff is computed as the hydroelectric power station income minus contribution to the MRC minus government charge: $23.64-1-2.31=20.33$ for type 1.0 in which we remove running costs: $23.64-1-2.31-3.70=16.63$ for type 1.1.

Appendix 2

We describe constraint ICC for player 1 by:

$$\begin{aligned} 7.33\alpha _{4}^{0}+20.33\alpha _{3}^{0}+0\alpha _{0}^{0}\ge & {} 7.33\alpha _{4}^{1}+20.33\alpha _{3}^{1}+0\alpha _{0}^{1} \end{aligned}$$

(5.1)

$$\begin{aligned} 3.63\alpha _{4}^{0}+16.63\alpha _{3}^{0}+0\alpha _{0}^{0}\le & {} 3.63\alpha _{4}^{1}+16.63\alpha _{3}^{1}+0\alpha _{0}^{1} \end{aligned}$$

(5.2)

where $\alpha _{4}^{0}+\alpha _{3}^{0}+\alpha _{0}^{0}=1$, $\alpha _{4}^{1}+\alpha _{3}^{1}+\alpha _{0}^{1}=1$, for all $\alpha _{c}^{\theta ^{\prime },\theta }\ge 0$. The first (or the second) inequality states that player 1 should not want to assert that he is of type 1.1 (or of type 1.0) if his true type is 1.0 (or type 1.1). The incentive-feasible set is determined according to the vectors estimated by the mechanism $\alpha$:

$$\begin{aligned} {\widehat{u}}_{1.0}= & {} 7.33\alpha _{4}^{0}+20.33\alpha _{3}^{0}+0\alpha _{0}^{0} \nonumber \\ {\widehat{u}}_{1.1}= & {} 3.63\alpha _{4}^{1}+16.63\alpha _{3}^{1}+0\alpha _{0}^{1}\nonumber \\ {\widehat{u}}_{2}= & {} 0.5\left( 3.31\alpha _{4}^{0}-9.69\alpha _{3}^{0}+0\alpha _{0}^{0}\right) \nonumber \\&\cdots +0.5\left( 3.31\alpha _{4}^{1}-9.69\alpha _{3}^{1}+0\alpha _{0}^{1}\right) \end{aligned}$$

(5.3)

in which $\alpha$ verifies constraint ICC. We obtain the extremals vectors

$$\begin{aligned} ( 0,0,0)\; \hbox { for }\; \alpha _{0}^{0}= & {} \alpha _{0}^{1}=1 \\ \left( 20.33,16.63,-9.69\right) \; \hbox { for }\; \alpha _{3}^{0}= & {} \alpha _{3}^{1}=1 \\ \left( 7.33,3.63,3.31\right) \; \hbox { for }\; \alpha _{4}^{0}= & {} \alpha _{4}^{1}=1. \end{aligned}$$

This system makes it possible to implement the optimals values $\alpha _{c}^{\theta ^{\prime },\theta ^{*}}$in computing the following Program $\mathcal {P}$ under Karush–Kuhn–Tucker conditions:

Values $\alpha _{c}^{\theta ^{\prime },\theta ^{*}}$ enable us to determine optimals payoffs of players $( {\widehat{u}}_{1.0}^{*},{\widehat{u}}_{1.1}^{*}, {\widehat{u}}_{2}^{*})$. For instance, we obtain $\left( 7.33,3.63,3.31\right)$ for optimals parameters $\alpha _{3}^{{0}^{*}}=0$, $\alpha _{4}^{{0}^{*}}=1$, $\alpha _{0}^{{0}^{*}}=0$, $\alpha _{0}^{{1}^{*}}=0$, $\alpha _{3}^{{1}^{*}}=0$ and $\alpha _{4}^{{1}^{*}}=1$. The socially optimal parameters show that, if player 1 claims to be type 1.0 or type 1.1, then the project is produced for sure and player 1 pays the entire cost. So, whatever player 1 claims to be, the socially optimal recommendation is unfair for him. The following solutions where $lm(\ge 0)$ stands for Lagrange Multipliers sum up the socially optimal parameters for each period:

• $solution_{t=1}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=0.,lm_{2}=0.,lm_{3}=17.07392202,lm_{4}=17.07392202,lm_{5}=17.07392202,lm_{6}=0.,lm_{7}=18.38826420,lm_{8}=17.07392202,lm_{9}=0.,lm_{10}=2.955711285\}$

• ${ solution}_{t=2}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=0.,lm_{2}=.3154343799,lm_{3}=16.38891142,lm_{4}=18.76350143,lm_{5}=16.38891142,lm_{6}=0.,lm_{7}=23.24722190,lm_{8}=18.76350143,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=3}:= \{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=0.,lm_{2}=.4050800176,lm_{3}=16.51395161,lm_{4}=19.67560115,lm_{5}=16.51395161,lm_{6}=0.,lm_{7}=25.20001491,lm_{8}=19.67560115,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=4}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=0.,lm_{2}=0.,lm_{3}=18.61037293,lm_{4}=18.61037293,lm_{5}=18.61037293,lm_{6}=0.,lm_{7}=20.71836573,lm_{8}=18.61037293,lm_{9}=0.,lm_{10}=6.404638421\}$

• $solution_{t=5}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=5.824607474,lm_{2}=6.404123440,lm_{3}=38.25933537,lm_{4}=0.,lm_{5}=38.25933537,lm_{6}=0.,lm_{7}=29.04375823,lm_{8}=0.,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=6}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=0.,lm_{2}=.6657277165,lm_{3}=16.77841004,lm_{4}=22.52684904,lm_{5}=16.77841004,lm_{6}=0.,lm_{7}=30.96350112,lm_{8}=22.52684904,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=7}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=0.,lm_{2}=0.,lm_{3}=20.17839129,lm_{4}=20.17839129,lm_{5}=20.17839129,lm_{6}=0.,lm_{7}=23.11594871,lm_{8}=20.17839129,lm_{9}=0.,lm_{10}=9.769144811\}$

• $solution_{t=8}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=0.,lm_{2}=.8344800191,lm_{3}=16.90251329,lm_{4}=24.57669196,lm_{5}=16.90251329,lm_{6}=0.,lm_{7}=34.76862056,lm_{8}=24.57669196,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=9}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=0.,lm_{2}=.9212819021,lm_{3}=16.87841706,lm_{4}=25.61185625,lm_{5}=16.87841706,lm_{6}=0.,lm_{7}=36.72589516,lm_{8}=25.61185625,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=10}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=0.,lm_{2}=0.,lm_{3}=21.78411673,lm_{4}=21.78411673,lm_{5}=21.78411673,lm_{6}=0.,lm_{7}=25.57688502,lm_{8}=21.78411673,lm_{9}=0.,lm_{10}=13.07364131\}$

• $solution_{t=11}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=0.,lm_{2}=0.,lm_{3}=22.32737401,lm_{4}=22.32737401,lm_{5}=22.32737401,lm_{6}=0.,lm_{7}=26.41159012,lm_{8}=22.32737401,lm_{9}=0.,lm_{10}=14.16663346\}$

• $solution_{t=12}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=7.822186137,lm_{2}=8.995737398,lm_{3}=45.74989209,lm_{4}=0.,lm_{5}=45.74989209,lm_{6}=0.,lm_{7}=42.50937791,lm_{8}=0.,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=13}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=8.137486524,lm_{2}=9.394636595,lm_{3}=46.85394532,lm_{4}=0.,lm_{5}=46.85394532,lm_{6}=0.,lm_{7}=44.44475180,lm_{8}=0.,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=14}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=8.460603532,lm_{2}=9.801173778,lm_{3}=47.96697240,lm_{4}=0.,lm_{5}=47.96697240,lm_{6}=0.,lm_{7}=46.38464534,lm_{8}=0.,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=15}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=8.791703132,lm_{2}=10.21557488,lm_{3}=49.08916955,lm_{4}=0.,lm_{5}=49.08916955,lm_{6}=0.,lm_{7}=48.33000958,lm_{8}=0.,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=16}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=9.130884797,lm_{2}=10.63795945,lm_{3}=50.22073169,lm_{4}=0.,lm_{5}=50.22073169,lm_{6}=0.,lm_{7}=50.28093682,lm_{8}=0.,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=17}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=9.478274557,lm_{2}=11.06848364,lm_{3}=51.36185207,lm_{4}=0.,lm_{5}=51.36185207,lm_{6}=0.,lm_{7}=52.23774964,lm_{8}=0.,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=18}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=9.834014017,lm_{2}=11.50733406,lm_{3}=52.51265068,lm_{4}=0.,lm_{5}=52.51265068,lm_{6}=0.,lm_{7}=54.20120745,lm_{8}=0.,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=19}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=10.19818986,lm_{2}=11.95460558,lm_{3}=53.67324465,lm_{4}=0.,lm_{5}=53.67324465,lm_{6}=0.,lm_{7}=56.17114170,lm_{8}=0.,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=20}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=10.57097504,lm_{2}=12.41051214,lm_{3}=54.84388939,lm_{4}=0.,lm_{5}=54.84388939,lm_{6}=0.,lm_{7}=58.14823710,lm_{8}=0.,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=21}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=10.95246281,lm_{2}=12.87516306,lm_{3}=56.02462704,lm_{4}=0.,lm_{5}=56.02462704,lm_{6}=0.,lm_{7}=60.13259717,lm_{8}=0.,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=22}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=2.761237326,lm_{2}=0.,lm_{3}=57.21570909,lm_{4}=0.,lm_{5}=57.21570909,lm_{6}=0.,lm_{7}=.1514510394,lm_{8}=0.,lm9=0.,lm_{10}=61.97326520\}$

• $solution_{t=23}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=lm_{1},lm_{2}=lm_{2},lm_{3}=29.20858690+10.51454000*lm_{1}-6.814540000*lm_{2},lm_{4}=29.20858690-10.51454000*lm_{1}+6.814540000*lm_{2},lm_{5}=29.20858690+10.51454000*lm_{1}-6.814540000*lm_{2}, lm_{6}=0.,lm_{7}=36.96428096-13.*lm_{1}+13.*lm_{2},lm_{8}=29.20858690-10.51454000*lm_{1}+6.814540000*lm_{2},lm_{9}=0.,lm_{10}=27.16038250+13.*lm_{1}-13.*lm_{2}\}$

• $solution_{t=24}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=12.15060496,lm_{2}=14.32330390,lm_{3}=59.62926965,lm_{4}=0.,lm_{5}=59.62926965,lm_{6}=0.,lm_{7}=66.18377182,lm_{8}=0.,lm_{9}=0.,lm_{10}=0.\}$

• $solution_{t=25}:=\{\alpha _{0}^{0,*}=0.,\alpha _{4}^{0,*}=1.,\alpha _{3}^{0,*}=0.,\alpha _{0}^{1,*}=0.,\alpha _{4}^{1,*}=1.,\alpha _{3}^{1,*}=0.,lm_{1}=-2.256275035+lm_{2},lm_{2}=lm_{2},lm_{3}=6.001052068+3.700000000 *lm_{2},lm_{4}=54.85112135-3.700000000*lm_{2},lm_{5}=6.001052068+3.700000000*lm_{2},lm_{6}=0.,lm_{7}=68.14973803,lm_{8}=54.85112135-3.700000000*lm_{2},lm_{9}=0.,lm_{10}=0.\}$

Appendix 3

Assume a string of y symbols $a_{1},\ldots ,a_{y}$ with circular permutation of each symbols showed by the matrix

$$\begin{aligned} \begin{pmatrix} a_{1} &{} \cdots &{} a_{y} \\ \vdots &{} \ddots &{} \vdots \\ a_{y} &{} \cdots &{} a_{y-1} \end{pmatrix} \end{aligned}$$

We denote $K\left( c_{00},c_{01},\ldots ,c_{n-1m-1}\right)$ for the vector $c_{00},c_{01},\ldots ,c_{n-1m-1}$ the matrix corresponding to $\mathcal {C} _{00}$ times $c_{00}$ and $\mathcal {C}_{01}$ times $c_{01}$, etc... $K\left( c_{00},c_{01},\ldots ,c_{n-1m-1}\right)$ is a $d\times d$ matrix.

The correlated equilibrium distribution that matches the outcome of the unsatisfactory mediation is:

$$\begin{aligned} Q_{\left( 1.0\right) }= & {} \left[ \begin{array}{ccc} {\frac{1}{4}} &{} 0 &{} {\frac{1}{4}} \\ 0 &{} 0 &{} 0 \\ {\frac{1}{4}} &{} 0 &{} {\frac{1}{4}} \end{array} \right] = \begin{pmatrix} c_{00} &{} c_{01} &{} c_{02} \\ c_{10} &{} c_{11} &{} c_{12} \\ c_{20} &{} c_{21} &{} c_{22} \end{pmatrix}\\ Q_{\left( 1.1\right) }= & {} \left[ \begin{array}{ccc} {\frac{1}{4}} &{} 0 &{} {\frac{1}{4}} \\ 0 &{} 0 &{} 0 \\ {\frac{1}{4}} &{} 0 &{} {\frac{1}{4}} \end{array} \right] = \begin{pmatrix} c_{00} &{} c_{01} &{} c_{02} \\ c_{10} &{} c_{11} &{} c_{12} \\ c_{20} &{} c_{21} &{} c_{22} \end{pmatrix} \end{aligned}$$

In this case, $n\times m=3\times 3$ (cf. Q), and $d\times d=4\times 4$. If we set $\mathcal {C}_{00}=\mathcal {C}_{02}=\mathcal {C} _{20}=\mathcal {C}_{22}=1$, then:

$$\begin{aligned} K\left( o_{1},o_{2},o_{3},o_{4},o_{5},o_{6},o_{7},o_{8},o_{9}\right) = \begin{pmatrix} o_{1} &{} o_{3} &{} o_{7} &{} o_{9} \\ o_{3} &{} o_{7} &{} o_{9} &{} o_{1} \\ o_{7} &{} o_{9} &{} o_{1} &{} o_{3} \\ o_{9} &{} o_{1} &{} o_{3} &{} o_{7} \end{pmatrix} \end{aligned}$$

for $K\left( c_{00},c_{01},c_{02},c_{10},c_{11},c_{12},c_{20},c_{21},c_{22}\right) :=K\left( o_{1},o_{2},o_{3},o_{4},o_{5},o_{6},o_{7},o_{8},o_{9}\right)$. We denote x modulo n and w modulo m by $x\left( n\right)$ and $w\left( m\right)$ respectively. Therefore, in the cell $\left( x,w\right)$ of the grand matrix appears the matrix

$$\begin{aligned} K\left( c_{\mathtt {i}+x\left( n\right) ,\mathtt {j}+w\left( m\right) }\right) _{0\le \text {\texttt {i}}\le n-1,0\le \text {\texttt {j}}\le m-1} \end{aligned}$$

with $0\le x\le n-1,0\le w\le m-1$. Players’ signals are in this case $S_{i}^{0}=\left\{ 0,\ldots ,n-1\right\} \times \left\{ 1,\ldots ,d_{i}\right\}$ and $S_{j}^{0}=\left\{ 0,\ldots ,m-1\right\} \times \left\{ 1,\ldots ,d_{j}\right\}$ with $\left( x,d_{i}\right) \in S_{i}^{0}$ and $\left( w,d_{j}\right) \in S_{j}^{0}$. The announcement becomes $k\left( \left( x,d_{i}\right) ,\left( w,d_{j}\right) \right)$ and describes the cell $\left( d_{i},d_{j}\right)$ of the matrix corresponding to the cell $\left( x,w\right)$ of the grand matrix. For the cell $\left( 0,0\right) ,$ we obtain $K\left( c_{\mathtt {i},\mathtt {j} }\right) _{\text {\texttt {i}},\text {\texttt {j}}}:=K\left( c_{0,0}\right) _{0,0}$ of the grand matrix:

$$\begin{aligned} K\left( c_{00},c_{01},c_{02},c_{10},c_{11},c_{12},c_{20},c_{21},c_{22}\right) = \begin{pmatrix} c_{00} &{} c_{02} &{} c_{20} &{} c_{22} \\ c_{02} &{} c_{20} &{} c_{22} &{} c_{00} \\ c_{20} &{} c_{22} &{} c_{00} &{} c_{02} \\ c_{22} &{} c_{00} &{} c_{02} &{} c_{20} \end{pmatrix} \end{aligned}$$

For the cell $\left( 0,1\right)$, we obtain¹⁴ $K\left( c_{\mathtt {i},\mathtt {j} +1\left( 3\right) }\right) _{\texttt {i},\texttt {j}}:=K\left( c_{0,1}\right) _{0,0}$ of the grand matrix:

$$\begin{aligned} K\left( c_{01},c_{02},c_{00},c_{11},c_{12},c_{10},c_{21},c_{22},c_{20}\right) = \begin{pmatrix} c_{01} &{} c_{00} &{} c_{21} &{} c_{20} \\ c_{00} &{} c_{21} &{} c_{20} &{} c_{01} \\ c_{21} &{} c_{20} &{} c_{01} &{} c_{00} \\ c_{20} &{} c_{01} &{} c_{00} &{} c_{21} \end{pmatrix} \end{aligned}$$

For the cell $\left( 0,2\right)$, we have¹⁵ $K\left( c_{\mathtt {i},\mathtt {j} +2\left( 3\right) }\right) _{\text {\texttt {i}},\text {\texttt {j}}}:=K\left( c_{0,2}\right) _{0,0}$ of the grand matrix:

$$\begin{aligned} K\left( c_{02},c_{00},c_{01},c_{12},c_{10},c_{11},c_{22},c_{20},c_{21}\right) = \begin{pmatrix} c_{02} &{} c_{01} &{} c_{22} &{} c_{21} \\ c_{01} &{} c_{22} &{} c_{21} &{} c_{02} \\ c_{22} &{} c_{21} &{} c_{02} &{} c_{01} \\ c_{21} &{} c_{02} &{} c_{01} &{} c_{22} \end{pmatrix} \end{aligned}$$

For instance, in the position $\left( \text {\texttt {i}},\text {\texttt {j}} \right) =\left( 0,1\right)$, for the cell $\left( 0,0\right)$ we obtain $K\left( c_{\mathtt {i},\mathtt {j}}\right) _{\text {\texttt {i}},\text {\texttt {j} }}:=K\left( c_{0,1}\right) _{0,1}$; and for the cell $\left( 0,1\right)$, we get¹⁶ $K\left( c_{\mathtt {i},\mathtt {j}+1\left( 3\right) }\right) _{\text {\texttt {i} },\text {\texttt {j}}}:=K\left( c_{0,2}\right) _{0,1}$. In other words, the last case tell us that we obtain $c_{0,1}$ instead of $c_{0,0}$, $c_{0,2}$ instead of $c_{0,1}$ and $c_{0,0}$ instead of $c_{0,2}$, and so forth... through permutations. We proceed in the same way for the cells $\left( 1,0\right) ,\left( 1,1\right) ,\left( 1,2\right)$ and $\left( 2,0\right) ,\left( 2,1\right) ,\left( 2,2\right)$ of the grand matrix $\forall \theta$ and arrive at the following $en\mathcal {C}oding matrix$:

$$\begin{aligned}&1.0 \begin{pmatrix} \begin{array}{cccc} o_{1} &{} o_{3} &{} o_{7} &{} o_{9} \\ o_{3} &{} o_{7} &{} o_{9} &{} o_{1} \\ o_{7} &{} o_{9} &{} o_{1} &{} o_{3} \\ o_{9} &{} o_{1} &{} o_{3} &{} o_{7} \end{array} &{} \begin{array}{cccc} o_{2} &{} o_{1} &{} o_{8} &{} o_{7} \\ o_{1} &{} o_{8} &{} o_{7} &{} o_{2} \\ o_{8} &{} o_{7} &{} o_{2} &{} o_{1} \\ o_{7} &{} o_{2} &{} o_{1} &{} o_{8} \end{array} &{} \begin{array}{cccc} o_{3} &{} o_{2} &{} o_{9} &{} o_{8} \\ o_{2} &{} o_{9} &{} o_{8} &{} o_{3} \\ o_{9} &{} o_{8} &{} o_{3} &{} o_{2} \\ o_{8} &{} o_{3} &{} o_{2} &{} o_{9} \end{array} \\ &{} &{} \\ \begin{array}{cccc} o_{4} &{} o_{6} &{} o_{1} &{} o_{3} \\ o_{6} &{} o_{1} &{} o_{3} &{} o_{4} \\ o_{1} &{} o_{3} &{} o_{4} &{} o_{6} \\ o_{3} &{} o_{4} &{} o_{6} &{} o_{1} \end{array} &{} \begin{array}{cccc} o_{5} &{} o_{4} &{} o_{2} &{} o_{1} \\ o_{4} &{} o_{2} &{} o_{1} &{} o_{5} \\ o_{2} &{} o_{1} &{} o_{5} &{} o_{4} \\ o_{1} &{} o_{5} &{} o_{4} &{} o_{2} \end{array} &{} \begin{array}{cccc} o_{6} &{} o_{5} &{} o_{3} &{} o_{2} \\ o_{5} &{} o_{3} &{} o_{2} &{} o_{6} \\ o_{3} &{} o_{2} &{} o_{6} &{} o_{5} \\ o_{2} &{} o_{6} &{} o_{5} &{} o_{3} \end{array} \\ &{} &{} \\ \begin{array}{cccc} o_{7} &{} o_{9} &{} o_{4} &{} o_{6} \\ o_{9} &{} o_{4} &{} o_{6} &{} o_{7} \\ o_{4} &{} o_{6} &{} o_{7} &{} o_{9} \\ o_{6} &{} o_{7} &{} o_{9} &{} o_{4} \end{array} &{} \begin{array}{cccc} o_{8} &{} o_{7} &{} o_{5} &{} o_{4} \\ o_{7} &{} o_{5} &{} o_{4} &{} o_{8} \\ o_{5} &{} o_{4} &{} o_{8} &{} o_{7} \\ o_{4} &{} o_{8} &{} o_{7} &{} o_{5} \end{array} &{} \begin{array}{cccc} o_{9} &{} o_{8} &{} o_{6} &{} o_{5} \\ o_{8} &{} o_{6} &{} o_{5} &{} o_{9} \\ o_{6} &{} o_{5} &{} o_{9} &{} o_{8} \\ o_{5} &{} o_{9} &{} o_{8} &{} o_{6} \end{array} \end{pmatrix}\\&1.1 \begin{pmatrix} \begin{array}{cccc} o_{1} &{} o_{3} &{} o_{7} &{} o_{9} \\ o_{3} &{} o_{7} &{} o_{9} &{} o_{1} \\ o_{7} &{} o_{9} &{} o_{1} &{} o_{3} \\ o_{9} &{} o_{1} &{} o_{3} &{} o_{7} \end{array} &{} \begin{array}{cccc} o_{2} &{} o_{1} &{} o_{8} &{} o_{7} \\ o_{1} &{} o_{8} &{} o_{7} &{} o_{2} \\ o_{8} &{} o_{7} &{} o_{2} &{} o_{1} \\ o_{7} &{} o_{2} &{} o_{1} &{} o_{8} \end{array} &{} \begin{array}{cccc} o_{3} &{} o_{2} &{} o_{9} &{} o_{8} \\ o_{2} &{} o_{9} &{} o_{8} &{} o_{3} \\ o_{9} &{} o_{8} &{} o_{3} &{} o_{2} \\ o_{8} &{} o_{3} &{} o_{2} &{} o_{9} \end{array} \\ &{} &{} \\ \begin{array}{cccc} o_{4} &{} o_{6} &{} o_{1} &{} o_{3} \\ o_{6} &{} o_{1} &{} o_{3} &{} o_{4} \\ o_{1} &{} o_{3} &{} o_{4} &{} o_{6} \\ o_{3} &{} o_{4} &{} o_{6} &{} o_{1} \end{array} &{} \begin{array}{cccc} o_{5} &{} o_{4} &{} o_{2} &{} o_{1} \\ o_{4} &{} o_{2} &{} o_{1} &{} o_{5} \\ o_{2} &{} o_{1} &{} o_{5} &{} o_{4} \\ o_{1} &{} o_{5} &{} o_{4} &{} o_{2} \end{array} &{} \begin{array}{cccc} o_{6} &{} o_{5} &{} o_{3} &{} o_{2} \\ o_{5} &{} o_{3} &{} o_{2} &{} o_{6} \\ o_{3} &{} o_{2} &{} o_{6} &{} o_{5} \\ o_{2} &{} o_{6} &{} o_{5} &{} o_{3} \end{array} \\ &{} &{} \\ \begin{array}{cccc} o_{7} &{} o_{9} &{} o_{4} &{} o_{6} \\ o_{9} &{} o_{4} &{} o_{6} &{} o_{7} \\ o_{4} &{} o_{6} &{} o_{7} &{} o_{9} \\ o_{6} &{} o_{7} &{} o_{9} &{} o_{4} \end{array} &{} \begin{array}{cccc} o_{8} &{} o_{7} &{} o_{5} &{} o_{4} \\ o_{7} &{} o_{5} &{} o_{4} &{} o_{8} \\ o_{5} &{} o_{4} &{} o_{8} &{} o_{7} \\ o_{4} &{} o_{8} &{} o_{7} &{} o_{5} \end{array} &{} \begin{array}{cccc} o_{9} &{} o_{8} &{} o_{6} &{} o_{5} \\ o_{8} &{} o_{6} &{} o_{5} &{} o_{9} \\ o_{6} &{} o_{5} &{} o_{9} &{} o_{8} \\ o_{5} &{} o_{9} &{} o_{8} &{} o_{6} \end{array} \end{pmatrix} \end{aligned}$$

each of the symbols appearing exactly twice in each corresponding row and column. There is the same conditional probability between players: $\frac{1 }{2}$. If we refer to relation (2.1), we can verify that $L\left( b_{1}\mid b_{2}\right) =Q\left( a_{1}\mid a_{2}\right) =\frac{{\frac{1}{4}}}{{\frac{1}{4}}+0+{\frac{1}{4}}}={ \frac{1}{2}}$.