When a certain biased die is rolled, a particular face occurs with probability 1/6 - x and its oppos

Q: Wait, why do we multiply by 2 in the probability equation?

That is a fantastic observation! When we roll a die twice, the outcome $(1, 6)$ is distinct from $(6, 1)$. Since the problem asks for the sum to be 7, we must count both permutations. Multiplying by 2 accounts for this symmetry, ensuring we capture every possible way to hit that target sum.

Q: Why are the probabilities of 2, 3, 4, and 5 just $1/6$?

The problem specifies that only one particular face and its opposite are biased. By elimination, the remaining four faces must share the remaining probability mass equally to ensure the sum of all probabilities equals 1. Since the bias $x$ is only applied to the pair $(1, 6)$, the others remain fair.

Q: Why can't we just use energy conservation or some other shortcut?

Probability problems are rarely about 'shortcuts' and almost always about 'accounting.' The only shortcut here is recognizing the symmetry of the pairs. Stick to the Law of Total Probability—it is the most robust tool in your arsenal for these types of discrete problems.

Question

When a certain biased die is rolled, a particular face occurs with probability $\frac{1}{6} - x$ and its opposite face occurs with probability $\frac{1}{6} + x$. All other faces occur with probability $\frac{1}{6}$. Note that opposite faces sum to 7 in any die. If $0 < x < \frac{1}{6}$, and the probability of obtaining total sum = 7, when such a die is rolled twice, is $\frac{13}{96}$, then the value of $x$ is:

Options:

(A) $\frac{1}{16}$

(B) $\frac{1}{8}$

(C) $\frac{1}{9}$

(D) $\frac{1}{12}$

Sigma Percentile · Accepted Answer

### The Mystery of the Biased Die

Welcome, future engineer. Today, we are going to peel back the curtain on a classic probability puzzle. It is not just about numbers; it is about understanding the subtle shift in a system when we introduce a bias.

Imagine you are holding a die. It looks normal, but it has a secret. Two of its faces—let us say $1$ and $6$—are not behaving like the others. This is the heart of our problem.

### Phase 1: Decoding the Bias

We are told that the probability of rolling a $1$ is $P(1) = \frac{1}{6} - x$, and the probability of rolling its opposite, $6$, is $P(6) = \frac{1}{6} + x$.

Notice the elegance here? The bias $x$ is subtracted from one and added to the other. This is a beautiful constraint because it ensures the total probability remains conserved.

The other faces—$2, 3, 4,$ and $5$—are perfectly fair, each with a probability of $P(i) = \frac{1}{6}$. We are operating under the constraint $0 < x < \frac{1}{6}$, which keeps our probabilities positive and valid.

### Phase 2: The Geometry of the Sum

When we roll this die twice, we want the total sum to be $7$. Let us list the pairs $(a, b)$ that satisfy $a + b = 7$.

We have:
1. $(1, 6)$ and $(6, 1)$
2. $(2, 5)$ and $(5, 2)$
3. $(3, 4)$ and $(4, 3)$

These are our winning scenarios. Notice that the first pair involves our biased faces, while the others involve the fair faces. This distinction is the key to the entire calculation.

### Phase 3: The Algebraic Dance

To find the total probability $P(S=7)$, we sum the probabilities of these independent events. Since order matters, we multiply by $2$ for each pair.

Our master equation becomes:

$$P(S=7) = 2 	imes \left[ P(1)P(6) + P(2)P(5) + P(3)P(4) ight]$$

Now, let us substitute our values. For the biased pair, we have $(\frac{1}{6} - x)(\frac{1}{6} + x)$. For the fair pairs, we have $(\frac{1}{6})(\frac{1}{6})$.

Substituting these in:

$$P(S=7) = 2 	imes \left[ \left( \frac{1}{6} - x ight) \left( \frac{1}{6} + x ight) + \left( \frac{1}{6} ight) \left( \frac{1}{6} ight) + \left( \frac{1}{6} ight) \left( \frac{1}{6} ight) ight]$$

Look at that first term. It is a classic difference of squares: $(a-b)(a+b) = a^2 - b^2$. This simplifies beautifully to $\frac{1}{36} - x^2$.

Our equation now looks like this:

$$P(S=7) = 2 	imes \left[ \frac{1}{36} - x^2 + \frac{1}{36} + \frac{1}{36} ight]$$

Combining the constants, we get $\frac{3}{36}$, which is $\frac{1}{12}$. So, our expression simplifies to:

$$P(S=7) = 2 	imes \left[ \frac{1}{12} - x^2 ight] = \frac{1}{6} - 2x^2$$

### The Final Resolution

The problem tells us that $P(S=7) = \frac{13}{96}$. Now, we simply equate and solve:

$$\frac{1}{6} - 2x^2 = \frac{13}{96}$$

Rearranging for $2x^2$:

$$2x^2 = \frac{1}{6} - \frac{13}{96} = \frac{16}{96} - \frac{13}{96} = \frac{3}{96} = \frac{1}{32}$$

Dividing by $2$, we find $x^2 = \frac{1}{64}$. Taking the square root, we get **$x = \frac{1}{8}$**.

Since $\frac{1}{8}$ is indeed less than $\frac{1}{6}$, our solution is valid. You have successfully navigated the bias and found the truth hidden in the numbers.

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Visualized Solution

$U n d er s t an d in g t h e B ia se d D i e$

$P r o babi l i t i eso f O t h er F a ces$

$I d e n t i f y in g O u t co m es f or S u m = 7$

$G r o u p in g b y P r o babi l i t y$

$S e tt in g u pt h e P r o babi l i t y E q u a t i o n$

$S u b s t i t u t in g t h e P r o babi l i t i es$

$E x p an d in g t h e B ia se d P r o d u c t$

$S im pl i f y in g t h e E x p r ess i o n$

$F ina l S im pl i f i e d P r o babi l i t y$

$E q u a t in g t o G i v e nV a l u e$

$S o l v in g f or 2 x^{2}$

$F in d in g x^{2}$

$T h e F ina l V a l u eo f x$

The Sigma Insight: Classical Definition of Probability

The Mystery of the Biased Die

Phase 1

Phase 2

Phase 3

The Final Resolution

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } UnderstandingtheBiasedDie

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ProbabilitiesofOtherFaces

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Findingx2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheFinalValueofx