Okay, so I need to find the value of n, which is the number of fair 6-sided dice rolled simultaneously. The probability that exactly two of them show a number other than 1 is given as 25/216. Hmm, let's think about how to approach this.

First, I remember that when rolling multiple dice, each die is independent. The probability of getting a specific outcome on one die is 1/6 for each number. Since we're dealing with "other than 1," that means each die has a 5/6 chance of showing a number other than 1, and a 1/6 chance of showing 1.

The problem states that exactly two dice show a number other than 1. So, this is a binomial probability problem. The formula for the probability of exactly k successes in n trials is C(n, k) * p^k * (1-p)^(n-k). Here, a "success" would be rolling a number other than 1, so p = 5/6, and the probability of failure (rolling a 1) is 1 - p = 1/6.

In this case, k is 2 because exactly two dice show a number other than 1. So plugging into the formula, the probability should be C(n, 2) * (5/6)^2 * (1/6)^(n-2). And this is equal to 25/216.

So the equation is:

C(n, 2) * (5/6)^2 * (1/6)^(n-2) = 25/216

I need to solve for n. Let me write that equation again:

C(n, 2) * (25/36) * (1/6)^(n-2) = 25/216

First, notice that 25/36 is a common factor on both sides. Let me divide both sides by 25/36 to simplify:

C(n, 2) * (1/6)^(n-2) = (25/216) / (25/36)

Dividing the right side: (25/216) ÷ (25/36) = (25/216) * (36/25) = (36/216) = 1/6

So now we have:

C(n, 2) * (1/6)^(n-2) = 1/6

Hmm, so C(n, 2) * (1/6)^(n-2) = 1/6

Let me write C(n, 2) as n(n-1)/2. So substituting that in:

(n(n-1)/2) * (1/6)^(n-2) = 1/6

Multiply both sides by 2 to eliminate the denominator:

n(n-1) * (1/6)^(n-2) = 2/6 = 1/3

So now the equation is:

n(n-1) * (1/6)^(n-2) = 1/3

Hmm, this seems a bit tricky. Let's see if we can manipulate this further. Let's take both sides and write them in terms of exponents with base 1/6. Let me see.

First, note that (1/6)^(n-2) can be written as (1/6)^n * 6^2 = 36*(1/6)^n. Wait, let's check:

(1/6)^(n-2) = (1/6)^n * (1/6)^(-2) = (1/6)^n * 6^2 = 36*(1/6)^n. Yes, that's correct.

So substituting back into the equation:

n(n-1) * 36 * (1/6)^n = 1/3

Then, divide both sides by 36:

n(n-1) * (1/6)^n = (1/3)/36 = 1/108

So now:

n(n-1) * (1/6)^n = 1/108

Hmm, this equation is still a bit complicated. Maybe we can try plugging in integer values for n to see which one satisfies the equation.

Since n must be an integer greater than or equal to 2 (because we have exactly two dice showing numbers other than 1, so n must be at least 2). Let's start testing n=2,3,4,...

For n=2:

Left side: 2*1*(1/6)^2 = 2*(1/36) = 2/36 = 1/18 ≈ 0.0555

Right side: 1/108 ≈ 0.009259

Not equal.

n=3:

Left side: 3*2*(1/6)^3 = 6*(1/216) = 6/216 = 1/36 ≈ 0.02778

Right side: 1/108 ≈ 0.009259

Not equal.

n=4:

Left side: 4*3*(1/6)^4 = 12*(1/1296) = 12/1296 = 1/108 ≈ 0.009259

Right side: 1/108 ≈ 0.009259

Oh! That's equal. So n=4.

Wait, let me verify again:

For n=4, the probability is C(4,2)*(5/6)^2*(1/6)^2 = 6*(25/36)*(1/36) = 6*(25/1296) = 150/1296 = 25/216, which matches the given probability. So yes, n=4.

But let me check n=5 to be thorough.

n=5:

Left side: 5*4*(1/6)^5 = 20*(1/7776) = 20/7776 ≈ 0.002572

Which is much less than 1/108 ≈ 0.009259. So n=5 is too big.

Therefore, n=4 is the solution.

But let me check again how the equation simplified. Starting from:

C(n, 2)*(5/6)^2*(1/6)^(n-2) = 25/216

Then we divided both sides by (25/36):

C(n, 2)*(1/6)^(n-2) = 1/6

Then substituted C(n,2) = n(n-1)/2:

n(n-1)/2 * (1/6)^(n-2) = 1/6

Multiply both sides by 2:

n(n-1)*(1/6)^(n-2) = 1/3

Then converted (1/6)^(n-2) to 36*(1/6)^n:

n(n-1)*36*(1/6)^n = 1/3

Divide both sides by 36:

n(n-1)*(1/6)^n = 1/108

Then testing n=4:

4*3*(1/6)^4 = 12*(1/1296) = 12/1296 = 1/108. Correct.

So yes, n=4 is the answer.

I think that's thorough. I considered the steps, checked the binomial formula, simplified the equation, and tested possible integer values for n. Since n must be an integer and the equation only holds for n=4, that's the answer.
Therefore, the value of $n$ is $\boxed{4}$.
</think>