Animated Solution for Mathematics - Probability: A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. Then the probability that 5 comes before 7 is .........

When you roll two fair, six-sided dice, the total number of outcomes is $6\times 6=36$. Visualize this as a $6\times 6$ grid where each cell represents an ordered pair $(d_1,d_2)$.

Every single one of these 36 outcomes is equally likely, with a probability of $\frac{1}{36}$.

We define Event $A$ as rolling a sum of 5. The favorable outcomes are $(1,4),(2,3),(3,2),$ and $(4,1)$.

There are exactly 4 favorable outcomes. Thus, the probability of rolling a 5 on any single throw is:

Event $B$ is defined as rolling a sum of 7. The favorable outcomes are $(1,6),(2,5),(3,4),(4,3),(5,2),$ and $(6,1)$.

There are 6 favorable outcomes. Thus, the probability of rolling a 7 on any single throw is:

The problem states we roll until we get a 5 or a 7. Any roll that results in a sum other than 5 or 7 is essentially a 'reset' button.

These 'neutral' rolls do not advance the game and do not favor either 5 or 7. Consequently, they effectively vanish from our calculation, leaving us with a reduced sample space where only the outcomes of 5 and 7 matter.

Since we only care about the relative likelihood of $A$ versus $B$, we can use the ratio of their probabilities. The probability that $A$ occurs before $B$ is given by:

This formula is a powerful shortcut. It bypasses the need to sum an infinite geometric series, which would otherwise be expressed as:

Now, we substitute our calculated values into the ratio:

Since the denominator of 36 appears in every term, it cancels out perfectly. This leaves us with:

Reducing this fraction, we arrive at our final answer: