Problem 164 - An Introduction to Mathematical Statistics and Its Applications

Question

A hand of six cards is dealt from a standard poker deck. Let \(X\) denote the number of aces, \(Y\) the number of kings, and \(Z\) the number of queens. (a) Write a formula for \(p_{X, Y, Z}(x, y, z)\). (b) Find \(p_{X, Y}(x, y)\) and \(p_{X, Z}(x, z)\).

Step-by-Step Solution

Answer

The joint probability mass function is \(p_{X, Y, Z}(x, y, z) = \frac{{C(4,x) \cdot C(4,y) \cdot C(4,z) \cdot C(40,6-x-y-z)}}{C(52,6)}\), and the marginal probability mass functions are \(p_{X, Y}(x, y) = \sum_{z=0}^{6-x-y} p_{X,Y,Z}(x,y,z)\) and \(p_{X, Z}(x, z) = \sum_{y=0}^{6-x-z} p_{X,Y,Z}(x,y,z)\).

Step 1: Write formula for \(p_{X, Y, Z}(x, y, z)\)

This represents the probability of drawing \(x\) aces, \(y\) kings, and \(z\) queens. We find this probability by calculating the number of possible combinations. This can be expressed as the product of the probabilities of picking aces, kings, and queens, and the probability of all other cards, all divided by the number of ways to pick any 6 cards from 52: \(p_{X, Y, Z}(x, y, z) = \frac{{C(4,x) \cdot C(4,y) \cdot C(4,z) \cdot C(40,6-x-y-z)}}{C(52,6)}\), where \(C(a, b)\) denotes 'a choose b', the number of combinations of a items taken b at a time, and 0 ≤ x, y, z ≤ 4 and \(x + y + z ≤ 6\).

Step 2: Compute \(p_{X, Y}(x, y)\)

This is the marginal probability mass function of \(X\) and \(Y\), and it can be found by summing the joint probability mass function over all possible values of \(Z\). We can do this by allowing \(Z\) to extend over all its possible values, but we must ensure that \(x + y + z ≤ 6\): \(p_{X, Y}(x, y) = \sum_{z=0}^{6-x-y} p_{X,Y,Z}(x,y,z)\)

Step 3: Compute \(p_{X,Z}(x, z)\)

This can be calculated similarly to the previous step, but now we sum over all possible values of \(Y\), ensuring \(x + y + z ≤ 6\): \(p_{X, Z}(x, z) = \sum_{y=0}^{6-x-z} p_{X,Y,Z}(x,y,z)\)

Key Concepts

Combinatorics

Combinatorics is the branch of mathematics that studies combinations, permutations, and the counting of objects. In our card problem, we use combinatorics to understand how many different ways we can draw a specific number of aces, kings, and queens from a deck. For example, if you want to know how many ways you can draw exactly 2 aces out of the 4 available in the deck, combinatorics provides a formula for this. This is commonly known as "combinations" and is represented by the formula \(C(n, k)\), which reads as 'n choose k'.

• \(C(4, 2)\) would mean choosing 2 aces out of the 4.
• The general formula for combinations is \(\frac{n!}{k!(n-k)!}\), where \(!\) denotes a factorial, meaning the product of all positive integers up to that number.

This approach lets you calculate various probabilities based on drawing cards under specific conditions, by calculating the number of favorable outcomes versus total possible outcomes.

Joint Probability

Joint probability refers to the likelihood of two or more events happening at the same time. In the context of our example, the joint probability \(p_{X, Y, Z}(x, y, z)\) is the probability of drawing exactly \(x\) aces, \(y\) kings, and \(z\) queens in a single hand of 6 cards.
The formula given:\[p_{X, Y, Z}(x, y, z) = \frac{C(4,x) \cdot C(4,y) \cdot C(4,z) \cdot C(40,6-x-y-z)}{C(52,6)}\]helps to calculate this by considering all possible combinations of these events:

• \(C(4, x)\) calculates the ways to choose \(x\) aces from 4.
• \(C(4, y)\) calculates the ways to choose \(y\) kings from 4.
• \(C(4, z)\) calculates the ways to choose \(z\) queens from 4.
• \(C(40, 6-x-y-z)\) handles the remaining cards being selected from the other 40 cards in the deck.

The denominator, \(C(52, 6)\), represents all possible ways to draw any 6 cards from the whole 52-card deck.

Marginal Probability

Marginal probability finds the probability of a subset of events while summing over the extra, uninteresting variables. In our card problem, this involves deriving probabilities for subsets like \(p_{X, Y}(x, y)\) and \(p_{X, Z}(x, z)\).
For \(p_{X, Y}(x, y)\), you sum over all possible outcomes of \(Z\) to see the cumulative probability of drawing \(x\) aces and \(y\) kings, regardless of the number of queens:\[p_{X, Y}(x, y) = \sum_{z=0}^{6-x-y} p_{X,Y,Z}(x,y,z)\]

• This sums the joint probabilities for all possible \(z\) values, ensuring \(x + y + z \leq 6\).

Similarly, for \(p_{X, Z}(x, z)\), the sum is over all possible \(Y\) outcomes:\[p_{X, Z}(x, z) = \sum_{y=0}^{6-x-z} p_{X,Y,Z}(x,y,z)\]

• Again, this ensures the condition \(x + y + z \leq 6\) holds, capturing the broader picture of how often \(x\) aces and \(z\) queens appear together in a hand of cards.

Marginal probabilities provide a way to summarize outcomes for a selected number of variables, giving an overview of certain results in complex probability scenarios.