One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another

Q: Wait, why can't I just use the standard probability formula $1/6$ for each face?

That is a great question! The standard formula assumes a fair, six-sided die where every face is equally likely. Here, the dice are non-standard; some numbers appear more frequently than others. We must account for the specific frequency of each face to get the correct probability.

Q: Do I really need to list every single pair? It seems tedious.

It might feel tedious, but in probability problems involving discrete outcomes, systematic listing is the most robust way to avoid errors. By breaking it down into cases (Sum = 4 and Sum = 5), you ensure you don't miss any combinations.

Q: Is there a faster way using generating functions?

Yes, you could use polynomial multiplication (generating functions) where the coefficient of $x^n$ represents the number of ways to get sum $n$. However, for a simple problem like this, manual counting is often faster and less prone to algebraic mistakes during an exam.

Question

One die has two faces marked 1, two faces marked 2, one face marked 3 and one face marked 4. Another die has one face marked 1, two faces marked 2, two faces marked 3 and one face marked 4. The probability of getting the sum of numbers to be 4 or 5, when both the dice are thrown together, is

Options:

(A) $\frac{2}{3}$

(B) $\frac{1}{2}$

(C) $\frac{4}{9}$

(D) $\frac{3}{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are holding two peculiar dice. One die has faces marked $\{1, 1, 2, 2, 3, 4\}$, and the other has faces $\{1, 2, 2, 3, 3, 4\}$.

When you roll them, the standard rules of probability—where every outcome is equally likely—do not apply. We must look at the underlying structure of these dice to move from intuition to systematic counting.

### The Grid of Possibilities

Before we calculate the probability of specific sums, we must define our sample space. Since each die has six faces, the total number of possible outcomes when both are thrown is:

$$6 	imes 6 = 36$$

Think of this as a $6 	imes 6$ grid where each cell represents a unique combination of the two dice. Even though some numbers repeat, we treat each face as a distinct entity to ensure our counting remains accurate.

### Case 1: The Quest for Sum 4

We are looking for the probability that the sum is $4$ or $5$. Let's start with the sum of $4$. The pairs $(x, y)$ from our dice that produce this sum are $(1, 3)$, $(2, 2)$, and $(3, 1)$.

We calculate the number of ways for each:

*   For $(1, 3)$: Die 1 has two $1$s and Die 2 has two $3$s, so there are $2 	imes 2 = 4$ ways.
*   For $(2, 2)$: Die 1 has two $2$s and Die 2 has two $2$s, so there are $2 	imes 2 = 4$ ways.
*   For $(3, 1)$: Die 1 has one $3$ and Die 2 has one $1$, so there is $1 	imes 1 = 1$ way.

Adding these up, we find $4 + 4 + 1 = 9$ favorable outcomes for a sum of $4$.

### Case 2: The Quest for Sum 5

Now, let's turn our attention to the sum of $5$. The pairs that satisfy this are $(1, 4)$, $(2, 3)$, $(3, 2)$, and $(4, 1)$.

Again, we calculate the frequency for each:

*   For $(1, 4)$: Die 1 has two $1$s and Die 2 has one $4$, giving $2 	imes 1 = 2$ ways.
*   For $(2, 3)$: Die 1 has two $2$s and Die 2 has two $3$s, giving $2 	imes 2 = 4$ ways.
*   For $(3, 2)$: Die 1 has one $3$ and Die 2 has two $2$s, giving $1 	imes 2 = 2$ ways.
*   For $(4, 1)$: Die 1 has one $4$ and Die 2 has one $1$, giving $1 	imes 1 = 1$ way.

Summing these, we get $2 + 4 + 2 + 1 = 9$ favorable outcomes for a sum of $5$.

### The Final Synthesis

We have successfully navigated the two cases. The total number of favorable outcomes is the sum of the ways to get $4$ and the ways to get $5$:

$$9 + 9 = 18$$

The probability is the ratio of favorable outcomes to total outcomes:

$$P = \frac{18}{36}$$

Simplifying this fraction, we arrive at the elegant result of **$\frac{1}{2}$**. This problem teaches us that probability is not just about formulas; it is about the patience to map out the reality of the situation.

Select Answer:

Visualized Solution

The Non-Standard Dice

Total Possible Outcomes

Case 1: Sum equals 4

Analyzing Pair $(1, 3)$

Analyzing Pair $(2, 2)$

Analyzing Pair $(3, 1)$

Total Ways for Sum 4

Case 2: Sum equals 5

Analyzing Pairs $(1, 4)$ and $(2, 3)$

Analyzing Pairs $(3, 2)$ and $(4, 1)$

Total Ways for Sum 5

Total Favorable Outcomes

Final Probability Calculation

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Grid of Possibilities

Case 1

Case 2

The Final Synthesis

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