Q.7.46 - A First Course in Probability

Question

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.

Step-by-Step Solution

Answer

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

Step 1:Given Information

Given that a 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. 

Step 2:Explanation

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,

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

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

Step 3:Characterise Arbitrary Factors

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

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

Step 4:Calculated Probability

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.

Step 5:Final Answer

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.