Question

Consider a random poker hand of 5 cards. Let X be the number of Hearts in this hand, and let y be the number of Diamonds in this hand. Find Cov(X, Y). (Hint: rewriting in terms of indicator functions may be helpful.)

Answer 1

There are 13 hears, 13 diamonds and 26 cards that are neither hearts or diamonds in a deck of 52 cards.

Therefore, for a 5 card poker hand, we first compute the expected values for X and Y ehre as:

E(X) = (Number of heart cards / Total cards) * 5 = (13 /52)*5 = 5/4 = 1.25
E(Y) = (Number of Diamond cards / Total cards) * 5 = (13 /52)*5 = 5/4 = 1.25

Now the expected value of EY here is computed as by first getting the probability distribution function for XY here as:

P(XY = 1) = P(X = 1)P(Y = 1) = Number of ways to draw 1 hearts * Number of ways to draw 1 diamond card * Number of ways to draw 3 non heart non diamond card / total ways to draw 5 cards from 52 cards

P(XY = 1) = 13*13*(26c3) / (52c5)

Similarly,
P(XY = 2) = P(X = 2, Y = 1) + P(X = 1, Y = 2)
P(XY = 3) = P(X = 3, Y = 1) + P(X = 1, Y = 3)
P(XY = 4) = P(X = 4, Y = 1) + P(X = 1, Y = 4) + P(X = Y = 2)
P(XY = 6) = P(X = 3, Y = 2) + P(X = 2, Y = 3)
P(XY = 0) = 1 - P(XY = 1) - P(XY = 2) - ... and so on..

This is computed as:

X	Y	p(X, Y)	XY
1	1	0.16906763	1
1	2	0.12680072	2
2	1	0.12680072	2
3	1	0.03719488	3
1	3	0.03719488	3
4	1	0.00357643	4
1	4	0.00357643	4
2	2	0.06086435	4
3	2	0.00858343	6
2	3	0.00858343	6
		0.5822429

Using the above table, the PDF for XY here is obtained as:

XY	p(XY)	xy*P(xy)
1	0.169067627	0.16906763
2	0.253601441	0.50720288
3	0.074389756	0.22316927
4	0.068017207	0.27206883
6	0.017166867	0.1030012
0	0.417757103	0
	1	1.2745098

Therefore now the covariance here is computed as:
Cov(X, Y) = E(XY) - E(X)E(Y) = 1.2745 - 1.25² = -0.2880

Therefore -0.2880 is the required covariance here.