Question

Consider a tetrahedron Die that has four integers 1, 2, 3 and 4 written on its faces. Roll the Die 2000 times and for each i, $1 \leq i \leq 4$ let $f(i)$ represent the number of times $i$ is written on the bottom face.

Let $A$ denote total sum of the numbers on the bottom face for these 2000 rolls, if $A^4 = 6144f(1)f(2)f(3)f(4)$, then

The value of $\left| \frac{f(1)-f(2)}{f(4)} \right|$ is equal to

Answer 1

2

Answer 2

The problem involves a tetrahedral die with faces labeled 1, 2, 3, 4. The die is rolled 2000 times. Let $f(i)$ be the number of times face $i$ lands on the bottom. The total number of rolls is $N = \sum_{i=1}^4 f(i) = 2000$.

The total sum of the numbers on the bottom face for these 2000 rolls is $A = 1 \cdot f(1) + 2 \cdot f(2) + 3 \cdot f(3) + 4 \cdot f(4)$.

We are given the condition $A^4 = 6144f(1)f(2)f(3)f(4)$.

Let $p_i$ be the experimental probability of face $i$ landing on the bottom, so $p_i = f(i)/N$. Then $f(i) = N p_i$. Substituting these into the given equation: $A = \sum_{i=1}^4 i \cdot N p_i = N \sum_{i=1}^4 i p_i$. Let $S = \sum_{i=1}^4 i p_i = p_1 + 2p_2 + 3p_3 + 4p_4$. So, $A = N S$.

The given equation becomes: $(NS)^4 = 6144 (N p_1)(N p_2)(N p_3)(N p_4)$ $N^4 S^4 = 6144 N^4 p_1 p_2 p_3 p_4$ $S^4 = 6144 p_1 p_2 p_3 p_4$.

Consider the terms $x_1 = p_1$, $x_2 = 2p_2$, $x_3 = 3p_3$, $x_4 = 4p_4$. The sum of these terms is $S = x_1+x_2+x_3+x_4$. The product of these terms is $P = x_1 x_2 x_3 x_4 = p_1 (2p_2) (3p_3) (4p_4) = 24 p_1 p_2 p_3 p_4$.

By the AM-GM inequality, for non-negative numbers $x_1, x_2, x_3, x_4$: $\frac{x_1+x_2+x_3+x_4}{4} \ge \sqrt[4]{x_1 x_2 x_3 x_4}$ $\frac{S}{4} \ge (24 p_1 p_2 p_3 p_4)^{1/4}$ Raising both sides to the power of 4: $(\frac{S}{4})^4 \ge 24 p_1 p_2 p_3 p_4$ $\frac{S^4}{256} \ge 24 p_1 p_2 p_3 p_4$ $S^4 \ge 256 \times 24 p_1 p_2 p_3 p_4$ $S^4 \ge 6144 p_1 p_2 p_3 p_4$.

The given condition is $S^4 = 6144 p_1 p_2 p_3 p_4$. This means that the equality condition for the AM-GM inequality must hold. The equality in AM-GM holds if and only if all the terms are equal: $x_1 = x_2 = x_3 = x_4$ So, $p_1 = 2p_2 = 3p_3 = 4p_4$.

Let this common value be $k$. $p_1 = k$ $p_2 = k/2$ $p_3 = k/3$ $p_4 = k/4$

We know that the sum of probabilities must be 1: $p_1 + p_2 + p_3 + p_4 = 1$ $k + k/2 + k/3 + k/4 = 1$ $k \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right) = 1$ $k \left(\frac{12+6+4+3}{12}\right) = 1$ $k \left(\frac{25}{12}\right) = 1$ $k = \frac{12}{25}$.

Now we can find the individual probabilities: $p_1 = \frac{12}{25}$ $p_2 = \frac{12/25}{2} = \frac{6}{25}$ $p_3 = \frac{12/25}{3} = \frac{4}{25}$ $p_4 = \frac{12/25}{4} = \frac{3}{25}$

The frequencies $f(i)$ are $N p_i$, where $N=2000$: $f(1) = 2000 \times \frac{12}{25} = 80 \times 12 = 960$ $f(2) = 2000 \times \frac{6}{25} = 80 \times 6 = 480$ $f(3) = 2000 \times \frac{4}{25} = 80 \times 4 = 320$ $f(4) = 2000 \times \frac{3}{25} = 80 \times 3 = 240$

We need to find the value of $\left| \frac{f(1)-f(2)}{f(4)} \right|$. $f(1)-f(2) = 960 - 480 = 480$. $f(4) = 240$. $\left| \frac{f(1)-f(2)}{f(4)} \right| = \left| \frac{480}{240} \right| = |2| = 2$.

The final answer is $\boxed{2}$.