Calculating Probability of Second Die Higher than First

This problem involves calculating the probability of a specific outcome when rolling a pair of fair dice. A fair die has six sides, numbered 1 through 6. When rolling two fair dice, each die's outcome is independent of the other.

Total Possible Outcomes

When rolling a pair of fair dice, the outcome of the first die can be any value from 1 to 6, and the outcome of the second die can also be any value from 1 to 6. The total number of possible outcomes is the product of the number of outcomes for each die.

Total outcomes = (Outcomes for die 1) × (Outcomes for die 2)

Total outcomes = \(6 \times 6 = 36\)

We can visualize these outcomes as pairs (die 1 value, die 2 value). Here's a table showing all 36 possible outcomes:

Favorable Outcomes: Second Die Higher than First

We are interested in the outcomes where the value on the second die is higher than the value on the first die. Let (d1, d2) represent the outcome where the first die shows d1 and the second die shows d2. We need to find all pairs where d2 > d1.

Let's list the favorable outcomes systematically:

• If the first die is 1, the second die can be 2, 3, 4, 5, 6 (5 outcomes: (1,2), (1,3), (1,4), (1,5), (1,6))
• If the first die is 2, the second die can be 3, 4, 5, 6 (4 outcomes: (2,3), (2,4), (2,5), (2,6))
• If the first die is 3, the second die can be 4, 5, 6 (3 outcomes: (3,4), (3,5), (3,6))
• If the first die is 4, the second die can be 5, 6 (2 outcomes: (4,5), (4,6))
• If the first die is 5, the second die can be 6 (1 outcome: (5,6))
• If the first die is 6, there are no values on the second die that are higher than 6 (0 outcomes)

The total number of favorable outcomes is the sum of the outcomes for each case:

Total favorable outcomes = \(5 + 4 + 3 + 2 + 1 + 0 = 15\)

Calculating the Probability

The probability of an event is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.

Probability (Second die higher than first) = \(\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}\)

Probability = \(\frac{15}{36}\)

This fraction can be simplified by dividing both the numerator and the denominator by their greatest common divisor, which is 3.

Probability = \(\frac{15 \div 3}{36 \div 3} = \frac{5}{12}\)

So, the probability that the second die lands on a higher value than the first is \(\frac{5}{12}\).

Revision Table: Dice Probability

Additional Information: Symmetry in Dice Rolls

When rolling two fair dice, consider the three types of outcomes:

1. The second die is higher than the first (d2 > d1).
2. The first die is higher than the second (d1 > d2).
3. The two dice are equal (d1 = d2).

Due to the symmetry of fair dice, the probability that the second die is higher than the first is equal to the probability that the first die is higher than the second. Let P(d2 > d1) and P(d1 > d2) denote these probabilities.

The outcomes where the two dice are equal are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). There are 6 such outcomes.

The probability that the two dice are equal is P(d1 = d2) = \(\frac{6}{36} = \frac{1}{6}\).

The sum of the probabilities of these three disjoint events must equal 1 (since they cover all possible outcomes):

P(d2 > d1) + P(d1 > d2) + P(d1 = d2) = 1

Since P(d2 > d1) = P(d1 > d2), we can write:

\(2 \times \text{P(d2 > d1)} + \text{P(d1 = d2)} = 1\)

\(2 \times \text{P(d2 > d1)} + \frac{1}{6} = 1\)

\(2 \times \text{P(d2 > d1)} = 1 - \frac{1}{6} = \frac{5}{6}\)

\(\text{P(d2 > d1)} = \frac{5/6}{2} = \frac{5}{12}\)

This confirms the result obtained by counting the favorable outcomes. This alternative method uses the symmetry of the problem and the probability of equal outcomes.