Building a Distribution from a Sample Space

Count Outcomes to Build a Distribution

In Probability Distribution of a Discrete Random Variable, you learned that a distribution pairs each possible value of a random variable with its probability. The earlier tutorial on Finding a Missing Probability in a Distribution used the fact that the probabilities must add to 1. In this tutorial, we will build a distribution directly from the outcomes of a chance process: rolling a fair six-sided die twice and recording the larger value.

Let \(M\) be the larger of the two numbers rolled. For instance, if the rolls are 2 and 5, then \(M=5\). If both rolls are 4, then \(M=4\). The possible values of \(M\) are 1, 2, 3, 4, 5, and 6.

To count carefully, represent the outcome as an ordered pair \((a,b)\), where \(a\) is the first roll and \(b\) is the second. The order matters for describing outcomes: \((2,5)\) and \((5,2)\) are different outcomes, even though both give \(M=5\). With two independent rolls of a fair six-sided die, there are \(6\cdot6=36\) equally likely ordered pairs. Each has probability \(1/36\).

Count the Outcomes for Each Maximum

One direct way to count the outcomes with maximum \(m\) is to list pairs that include \(m\) but do not include any number larger than \(m\). For example, when \(m=4\), at least one roll must be 4, and the other roll can be 1, 2, 3, or 4. That gives seven ordered pairs: \((4,1)\), \((4,2)\), \((4,3)\), \((4,4)\), \((1,4)\), \((2,4)\), and \((3,4)\). The pair \((4,4)\) is counted only once.

There is also a useful counting method based on cumulative totals. The number of ordered pairs with both rolls at most \(m\) is \(m^2\): there are \(m\) choices for the first roll and \(m\) choices for the second. Of these, \((m-1)^2\) pairs have both rolls below \(m\). Subtracting leaves the pairs whose maximum is exactly \(m\).

For \(m=1\), the count is \(1^2-0^2=1\), the outcome \((1,1)\). For \(m=6\), it is \(6^2-5^2=11\). Dividing each count by 36 gives the probability of that maximum.

Worked Example: Construct the Full Distribution

Why Larger Maximums Have More Outcomes

The count increases as the maximum increases. A maximum of 1 requires both rolls to be 1, so there is only one matching pair. A maximum of 6 can occur whenever at least one roll is 6; the other roll can be any value from 1 through 6, giving 11 pairs altogether. Thus, the possible maximums are not equally likely, even though each individual die face is equally likely on a single roll.

The counts \(1,3,5,7,9,11\) also have a cumulative interpretation. For example, there are \(4^2=16\) outcomes with maximum at most 4, because each roll must be one of 1, 2, 3, or 4. The exact-maximum count for 4 is the difference between the number with maximum at most 4 and the number with maximum at most 3: \(16-9=7\). This “cumulative count, then subtract the previous cumulative count” approach is useful whenever outcomes can be organized by whether a value is at most a given cutoff.

Use the Distribution to Find Event Probabilities

Once the distribution is built, probabilities for events involving \(M\) can be found by adding the probabilities for the values that meet the event. For example, the event “the larger roll is at least 5” includes \(M=5\) and \(M=6\). These are separate possible values of the random variable, so their probabilities can be added.

A cumulative count can also check the result. “At least 5” is the complement of “at most 4.” There are \(4^2=16\) pairs with both rolls at most 4, so \(36-16=20\) pairs have a maximum of at least 5. The distribution and the direct count should agree.

Common Mistakes and AP Exam Tips

• Assuming the six maximums are equally likely. The die faces are equally likely on one roll, but the maximum comes from a pair of rolls. Count the ordered pairs for each maximum; the counts are not all the same.
• Forgetting that order creates different outcomes. The outcomes \((2,5)\) and \((5,2)\) are distinct ordered pairs. Count both when appropriate, but count \((5,5)\) only once.
• Counting pairs with a roll equal to \(m\) without restricting the other roll. If \(m=4\), pairs such as \((4,6)\) do not have maximum 4. The other roll must be at most \(m\).
• Dividing by the wrong total. For two rolls, the sample space contains \(6\cdot6=36\) equally likely ordered pairs, not 6 outcomes.
• Giving only a table without checking it. A complete answer should show that the listed values cover all possible maximums and that the probabilities total 1. Here, the counts total 36, matching the size of the sample space.

For full-credit communication, define the random variable, identify the equally likely outcomes, show how the count for the target value is obtained, and divide by the total number of outcomes. Finish by interpreting the result in context. For a complete distribution, include every possible value and verify the probabilities sum to 1.

Key Takeaway

To build a distribution from a sample space, assign the random variable’s value to each outcome and count how often each possible value occurs. For the larger value from two fair six-sided die rolls, the count for maximum \(m\) is \(m^2-(m-1)^2=2m-1\), out of 36 equally likely ordered pairs.