Consider the following dice game, as played at a certain gambling casino: Players$1$and $2$roll a pair of dice in turn. The bank then rolls the dice to determine the outcome according to the following rule: Player$i,i=1,2,$wins if his roll is strictly greater than the banks. For$i=1,2,$let

$I_i=\left\{\begin{array}{l}1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{if}i\text{wins}\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{otherwise}\end{array}\right.$

and show that $I1$and $I2$are positively correlated. Explain why this result was to be expected.

Given that a dice game, as played at a certain gambling casino: Players$1$and localid="1647263046589" $2$roll a pair of dice in turn. The bank then rolls the dice to determine the outcome according to the following rule: Player$i,i=1,2,$wins if his roll is strictly greater than the banks.

Characterize arbitrary factors $X_1,X_2$and$Y$that mark the results of player one, player two, and the bank. We need to show that

$\mathrm{Cov}\left(I_1,I_2\right)=E\left(I_1{I}_2\right)-E\left(I_1\right)E\left(I_2\right)>0$

see that,

$E\left(I_1{I}_2\right)=P\left(I_1=1,I_2=1\right)=P\left(X_1>Y,X_2>Y\right)$

$=P\left(X_1\ge X_2>Y\right)+P\left(X_2\ge X_1>Y\right)$

Utilizing the law of the all-out probability, we have that

$P\left(X_1\ge X_2>Y\right)=\sum _{y=1}^6 P\left(X_1\ge X_2>Y\mid Y=y\right)P(Y=y)$

$=\sum _{y=1}^6 P\left(X_1\ge X_2>y\right)P(Y=y)$

$=\frac{1}{6}\sum _{y=1}^6 \frac{1}{(6-y{)}^2}\frac{(6-y)(6-y+1)}{2}$

$=\frac{1}{6}\sum _{y=1}^5 \frac{(y+1)}{2y}=\frac{437}{720}$

So, because of the symmetry, we have that

$E\left(I_1{I}_2\right)=\frac{437}{360}$

Presently, we have that

$E\left(I_1\right)E\left(I_2\right)={\left(\frac{15}{36}\right)}^2=\frac{25}{144}$

so we see that

$\mathrm{Cov}\left(I_1,I_2\right)>0$

This outcome is instinctive - realizing that player one has dominated his match infers that the bank could have an exceptionally low outcome on its die, so it passes on more noteworthy space for player two to win. In this way, these factors are decidedly connected.

Show that $E\left(I_1{I}_2\right)>E\left(I_1\right)E\left(I_2\right)$utilizing the law of the complete probability.