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
$${\cal I}_i = \left\{\begin{array}{ccc} 1 & {\rm if} & i \; {\rm wins} \\ 0 & & {\rm otherwise}\end{array}\right.$$
Show that ${\cal I}_1$ and ${\cal I}_2$ are positively correlated.
Answer: I understand that I have to show that cov(${\cal I}_1, {\cal I}_2) = E[{\cal I_1}{\cal I}_2] - E[{\cal I_1}]E[{\cal I_2}] > 0$. I understand how to find each expected value, however I am unaware of how to find the joint probability distribution for the situation. For example, how do I go about calculating $P({\cal I}_i)$, $P({\cal I}_1 = 1, {\cal I}_2 = 1)$, etc?
Right up front, we can predict that these will be positively correlated because if you tell me that player $1$ has won, that says something about the bank number tending to be lower than its usual distribution, and that in turn increases the chances that player 2 will have won as well. But that might be hand-waving; let's look at the joint distribution and the actual comparisons.
Note that the bank's number is among $\{12, 11, 10, \cdots , 2\}$ with probabilities $$\{\frac1{36}, \frac2{36}, \frac3{36}, \frac4{36}, \frac5{36}, \frac6{36}, \frac5{36}, \frac4{36}, \frac3{36}, \frac2{36}, \frac1{36}\}$$ and that the probability of each of the players rolling better than the bank will be $$ \left\{\frac{0}{36}, \frac{0+1}{36}, \frac{0+1+2}{36}, \cdots, \frac{0+1+2+3+4+5+6+5+4+3+2}{36} \right\} $$
So the probability, for a given bank number ranging from $12$ down to $2$, of one specified player winning will be $$\frac{1}{36}\left\{0 , 1, 3, 6, 10, 15, 21, 26, 30, 33, 35 \right\} $$ and multiplying by the probability of each of those bank numbers we get: $${\cal I}_1 = \frac{1}{36} \frac{1\cdot 0 + 2 \cdot 1 + 3 \cdot 3 + \cdots +6 \cdot 15 + 5\cdot 21 \cdots + 2 \cdot 33 + 1 \cdot 35}{36} = \frac{575}{1296} $$
To find the joint probability distribution, it will suffice to find $P_{11}$, the probability of both players winning, and $P_{00}$, the probability of both players losing, since pretty obviously $P_{10} = P_{01} = \frac{1-P_{11}-P_{00}}2$ and that covers all four possible joint outcomes.
To find $P_{11}$ we need the probability, for a given bank number ranging from $12$ down to $2$, of both players winning, which will be will be $$\frac{1}{36^2}\left\{0^2 , 1^2, 3^2, 6^2, 1^2, 10^2, 15^2, 21^2, 26^2, 30^2, 33^2, 35^2 \right\} $$
Then multiplying by the probability for each bank number we get $$P_{11} = \frac{1}{36}\frac{1\cdot 0^2 + 2 \cdot 1^2 + 3 \cdot 3^2 + \cdots +6 \cdot 15^2 + 5\cdot 21^2 \cdots + 2 \cdot 33^2 + 1 \cdot 35^2}{36^2} = \frac{13035}{46656} $$
Similarly, to find $P_{00}$ we need the probability, for a given bank number ranging from $12$ down to $2$, of neither player winning, which will be will be $$P_{00} = \frac{1}{36}\frac{1\cdot 36^2 + 2 \cdot 35^2 + 3 \cdot 33^2 + \cdots +6 \cdot 21^2 + 5\cdot 15^2 \cdots + 2 \cdot 3^2 + 1 \cdot 1^2}{36^2} = \frac{18291}{46656} $$ and from these we can calculate $$P_{10}=P_{01}= \frac{7665}{46656}$$
Finally, to answer your question, $E({\cal I}_1) = E({\cal I}_2) = \frac{575}{36^2}$ while $E({\cal I}_1 {\cal I}_2) = P_{11} = \frac{13035}{36^3}$.
$$ E({\cal I}_1 {\cal I}_2) = \frac{13035}{36^3} = \frac{469260}{36^4} > \frac{330625}{36^4} = \left( \frac{575}{36^2} \right)^2 = E({\cal I}_1) E({\cal I}_2) $$ so ${\cal I}_1$ and ${\cal I}_2$ are positively correlated.