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
Ii={1 if i wins, 0 otherwise}


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

Answer 1

Answer:

they are positively correlated.

Step-by-step explanation:

We can calculate the individal expectations first. FIrst player will win if that player's roll is greater than the bank's roll. There are (6 possible rolls of player 1 * 6 possible rolls of bank =) 36 total possible rolls, out of which player 1 will win in 15 cases.

$\therefore E(I_i) = 1\cdot \frac{15}{36} + 0 \cdot \frac{21}{36} = \frac{5}{12} \approx 0.4167$

For the joint expectation, there are (6 possible rolls of player 1 * 6 possible rolls of player 2 * 6 possible rolls of bank =) 216 total possible rolls.

Cases where both players win: Expectation = $2.

If bank rolls 1, both players will win in 5*5 = 25 cases. P1 is one of {2,3,4,5,6}, P2 is one of {2,3,4,5,6}

If bank rolls 2, both players will win in 4*4 = 16 cases.

If bank rolls 3, both players will win in 3*3 = 9 cases.

If bank rolls 4, both players will win in 2*2 = 4 cases.

If bank rolls 5, both players will win in 1*1 = 1 cases.

If bank rolls 6, both players will win in 0*0 = 0 cases.

Total cases = 25+16+9+4+1+0 = 55 cases.

Cases where player 1 wins $1 and player 2 loses: Expectation = $1.

If bank rolls 1, player 1 will win and player 2 will lose in 5*1 = 5 cases. P1 is one of {2,3,4,5,6}, P2 is {1}

If bank rolls 2, player 1 will win and player 2 will lose in 4*2 = 8 cases.

If bank rolls 3, player 1 will win and player 2 will lose in 3*3 = 9 cases.

If bank rolls 4, player 1 will win and player 2 will lose in 2*4 = 8 cases.

If bank rolls 5, player 1 will win and player 2 will lose in 1*5 = 5 cases.

If bank rolls 6, player 1 will win and player 2 will lose in 0*6 = 0 cases.

Total cases = 5+8+9+8+5+0 = 35

Cases where player 2 wins $1 and player 1 loses: Expectation = $1.

This is the same as above with player 1 and 2 exchanged.

Total cases = 35

Cases where both players lose: Expectation = $0.

If bank rolls 1, both players will lose in 1*1 = 1 cases. P1 is {1}, P2 is {1}

If bank rolls 2, both players will lose in 2*2 = 4 cases.

If bank rolls 3, both players will lose in 3*3 = 9 cases.

If bank rolls 4, both players will lose in 4*4 = 16 cases.

If bank rolls 5, both players will lose in 5*5 = 25 cases.

If bank rolls 6, both players will lose in 6*6 = 36 cases.

Total cases = 1+4+9+16+25+36 = 91 cases.

Total of all cases (we expect this to be 216 as mentioned above) = 55+35+35+91=216

So, joint expectation is:

$E(I_1I_2) = \frac{2\cdot 55 +1\cdot 35+1\cdot 35+0\cdot 91}{216} = \frac{180}{216}= \frac{5}{6} \approx 0.8333$

So, the covariance is given by:

$\texttt{Cov}(I_1I_2) =E(I_1I_2) -E(I_1)\cdot E(I_2)= \frac{5}{6}-\frac{5}{12}\cdot\frac{5}{12}=\frac{95}{144} \approx 0.6597$

As this is greater than 0 and closer to 1, they are positively correlated.

The reason why this result is expected is because the same bank roll is being used for both players. So, it is very likely that both players will win if the bank roll is 1 or even 2. Also, it is very likely that both players will lose if the bank roll is 6, 5, or even 4. This shows positive correlation between the events.