<Question>
$n$ fair 6-sided dice are simultaneously rolled. The probability that exactly two of them show a number other than 1 is $\frac{25}{216}$. Find $n$.
</Question>

<think>
<Plan_1>
Okay, let me try to figure out this problem. So, the question is: If we roll n fair 6-sided dice all at once, the probability that exactly two of them show a number other than 1 is 25/216. We need to find n. Hmm, alright. Let's break this down step by step.

First, I remember that when dealing with probabilities involving multiple dice, each die is independent, so we can use combinations and probabilities multiplied together. The general formula for such problems often involves binomial coefficients because we're looking for exactly k successes in n trials.

<prompt_1.1> Set up the problem using the binomial probability formula and identify the components needed. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
In this case, a "success" might be defined as rolling a number other than 1. Wait, but the problem says "exactly two of them show a number other than 1". So, that means out of n dice, exactly 2 show 2-6, and the remaining n-2 show 1. Right?

So, the probability of a single die showing a number other than 1 is 5/6, since there are 5 outcomes (2 through 6) out of 6 possible. Conversely, the probability of rolling a 1 is 1/6.

Therefore, the probability of exactly two dice showing numbers other than 1 would be the number of ways to choose those two dice multiplied by the probability of each of them showing 2-6, and the remaining n-2 dice showing 1. So, the formula should be:

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

And this is given as 25/216. So, we can set up the equation:

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

Our job is to solve for n here.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me write that out more formally:

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

First, let's simplify the left-hand side (LHS). Let's compute each part step by step.

<prompt_2.1> Simplify the left-hand side of the equation by computing each part step by step. </prompt_2.1>
-----

Wait, but let me check if n=5 makes sense here. Let's verify the calculation again.

<prompt_2.2> Re-verify the calculation for n=5 to ensure the probability is correct. </prompt_2.2>
-----

But wait, is there another possible n? Let's see. Suppose n=3.

<prompt_2.3> Explore other possible values of n, such as n=3 or n=4, to see if they could also satisfy the equation. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
First, C(n, 2) is the combination formula n choose 2, which is n(n-1)/2. Then, (5/6)^2 is 25/36. Then, (1/6)^{n-2} is just 1 divided by 6^{n-2}. So putting it all together:

[n(n - 1)/2] * (25/36) * (1/6)^{n - 2} = 25/216

Hmm, let's see if we can simplify this equation. First, notice that 25 appears on both sides, so maybe we can divide both sides by 25 to cancel that out. Let's try that:

[n(n - 1)/2] * (1/36) * (1/6)^{n - 2} = 1/216

Wait, because 25/25 is 1, so that cancels. Then, 36 is 6^2, and 6^{n-2} multiplied by 1/36 would be 6^{n - 2} / 6^2 = 6^{n - 4}. Wait, maybe another approach. Let's write all the denominators as powers of 6.

Let me rewrite the equation:

[n(n - 1)/2] * (25/6^2) * (1/6)^{n - 2} = 25/6^3

Because 216 is 6^3. So, 25/216 is 25/6^3. So, substituting that in:

Left-hand side (LHS): [n(n - 1)/2] * (25/6^2) * (1/6)^{n - 2} = [n(n - 1)/2] * 25 / 6^{n}

Right-hand side (RHS): 25 / 6^3

So, setting them equal:

[n(n - 1)/2] * 25 / 6^{n} = 25 / 6^3

We can divide both sides by 25 to cancel that term:

[n(n - 1)/2] / 6^{n} = 1 / 6^3

Multiply both sides by 6^{n}:

[n(n - 1)/2] = 6^{n} / 6^3 = 6^{n - 3}

So, simplifying:

n(n - 1)/2 = 6^{n - 3}

Therefore, the equation reduces to:

n(n - 1) = 2 * 6^{n - 3}
</execution_2.1>

<execution_2.2>
C(5,2) is 10. (5/6)^2 is 25/36. (1/6)^{3} is 1/216. So, multiplying them together: 10 * 25/36 * 1/216. Wait, 25/36 * 1/216 is 25/(36*216). 36*216 is 7776. So, 25/7776. Then 10 * 25/7776 is 250/7776. Simplify that: divide numerator and denominator by 2: 125/3888. Wait, 125/3888 is approximately 0.03215, and 25/216 is approximately 0.1157. Hmm, that's not matching. Wait, that can't be right.
</execution_2.2>

<execution_2.3>
Then C(3,2)=3. The probability would be 3*(25/6^2)*(1/6)^1=3*(25/36)*(1/6)=3*(25/216)=75/216, which is 25/72, which is more than 25/216. So nope.

n=4: C(4,2)=6. Probability:6*(25/36)*(1/6)^2=6*(25/36)*(1/36)=6*(25/1296)=150/1296=25/216. So that works. Wait, so n=4 also gives 25/216? Wait, but earlier when I thought n=5, but here n=4 also works. Wait, hold on, maybe I made a mistake here.

Wait, let's re-examine. If n=4, then exactly two dice show numbers other than 1. So, the probability is C(4,2)*(5/6)^2*(1/6)^2. Wait, hold on, (1/6)^{n-2}. So if n=4, that's (1/6)^2. So, yes, (5/6)^2*(1/6)^2. So that's (5^2 * 1^2)/(6^4). Which is 25/1296. Then multiplied by C(4,2)=6. So 6*25/1296=150/1296=25/216. So, yes, n=4 gives 25/216.

But earlier, when I considered n=5, I thought that if (1/6)^{n-2} is (1/6)^3, which would be 1/216. But if n=4, then (1/6)^2=1/36. So, in that case, the term (1/6)^{n-2} would be (1/6)^2 when n=4. So, in the equation, if we have (1/6)^{n-2} and the other term is (5/6)^2, then the product is (25/36)*(1/6)^{n-2}. So, if n=4, then (1/6)^2, so 25/36*1/36=25/1296, times 6 gives 25/216.

But the problem states that exactly two dice show numbers other than 1. So, the number of such dice is 2, regardless of what they show, as long as they are not 1. So, each of those two dice can be 2-6, which is 5 possibilities each. But in the problem statement, are we told that the two numbers are specific, like exactly two show 2 and the rest show 1? Or is it just that they show numbers other than 1, regardless of which ones?


-----

But wait, let me verify once again to ensure there's no mistake. Let's re-express the original equation:

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

Let me write 25/216 as (25/36) * (1/6)^{n - 2 + 2 - 2} Wait, maybe that's not helpful. Alternatively, let's express both sides with the same base.
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Hmm, this seems tricky. Let's see. Maybe we can try plugging in small integer values for n to see which one satisfies the equation.

<prompt_3.1> Try small integer values for n to see which one satisfies the equation. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Since n must be at least 2 (because we need to choose 2 dice out of n), let's start testing n=2, 3, 4, etc.

But before that, maybe simplify the equation more. Let's write 6^{n - 3} as 6^{n}/6^3, so:

n(n - 1)/2 = 6^{n - 3}

Alternatively, multiply both sides by 8 (since 6^3 is 216, and 216/8=27, but maybe not helpful). Wait, 6^3 is 216, right? So 6^{n - 3} is 6^n / 6^3 = 6^n / 216. So, the equation is:

n(n - 1)/2 = 6^n / 216

Multiply both sides by 216:

108n(n - 1) = 6^n

So, 6^n = 108n(n - 1)

Now, we need to find integer n such that 6^n equals 108n(n - 1). Let's compute 6^n and 108n(n - 1) for various n.

Starting with n=2:

Left: 6^2 = 36

Right: 108*2*1 = 216

36 ≠ 216. Not equal.

n=3:

Left: 6^3 = 216

Right: 108*3*2 = 648

216 ≠ 648

n=4:

Left: 6^4 = 1296

Right: 108*4*3 = 1296

1296 = 1296. That works! So n=4 is a solution.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
Let's check n=5 to see if there are other solutions.

n=5:

Left: 6^5 = 7776

Right: 108*5*4 = 2160

7776 ≠ 2160

n=6:

Left: 6^6 = 46656

Right: 108*6*5 = 3240

Not equal.

n=1:

<prompt_4.1> Check other values of n, starting from n=5 and moving upwards, to see if they satisfy the equation. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
Left: 6^1=6

Right: 108*1*0=0

Nope.

n=0: Not possible, since we have dice.

So, only n=4 satisfies the equation.

Wait, but earlier when I tried n=5, I thought it gave 25/216, but actually, when I recalculated, it's n=4. Let me check again.

Wait, if n=4, then the probability is C(4,2)*(5/6)^2*(1/6)^2 = 6*(25/36)*(1/36) = 6*(25)/(36*36) = 150/1296 = 25/216. Yes, that's correct. So n=4 is the solution.

But wait, when I first tried n=5, I thought:

C(5,2)*(5/6)^2*(1/6)^3 = 10*(25/36)*(1/216) = 10*(25)/(36*216) = 250/7776 ≈ 0.032, which is 125/3888. But 25/216 is approximately 0.1157, so they are different. Therefore, n=5 does not work. So n=4 is the correct answer.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
But let me confirm once more. Let's plug n=4 into the original probability formula.

<prompt_5.1> Confirm the solution by plugging n=4 back into the original probability formula. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
The probability should be C(4,2)*(5/6)^2*(1/6)^2. Compute each part:

C(4,2) = 6

(5/6)^2 = 25/36

(1/6)^2 = 1/36

Multiply them all together: 6*(25/36)*(1/36) = (6*25)/(36*36) = 150/1296. Simplify this fraction: divide numerator and denominator by 6: 150 ÷ 6 = 25; 1296 ÷ 6 = 216. So, 25/216. Exactly matches the given probability. Therefore, n=4 is the correct answer.
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
But wait, let me check if there's another possible n. For example, n=3:

<prompt_6.1> Explore other possible values of n, such as n=3 or n=5, to see if they could also satisfy the equation. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
C(3,2)=3. Probability: 3*(5/6)^2*(1/6)^1 = 3*(25/36)*(1/6) = 75/216 = 25/72 ≈ 0.347, which is higher than 25/216 ≈ 0.1157. So nope.

n=5: As before, 10*(25/36)*(1/216) = 250/1296 ≈ 0.192, which is still higher than 25/216 ≈ 0.1157. Wait, 25/216 is approximately 0.1157, and 250/1296 is approximately 0.192, which is higher. So n=5 is too high.

n=6: C(6,2)=15. Probability:15*(25/36)*(1/6)^4=15*(25/36)*(1/1296)=15*25/(36*1296)=375/46656≈0.008, which is way lower. So, the probability decreases after a certain point. So, n=4 is the only solution.

Alternatively, maybe there's a mathematical way to solve 6^n = 108n(n - 1). Let's see. For n=4: 6^4=1296, 108*4*3=1296. For n=5: 6^5=7776 vs 108*5*4=2160. So, 7776 > 2160. For n=3: 6^3=216 vs 108*3*2=648. 216 < 648. So, the function 6^n grows exponentially, while 108n(n - 1) grows quadratically. Therefore, after n=4, 6^n will outpace 108n(n - 1) rapidly. So, n=4 is the only solution.
</execution_6.1>
</Execution_6>

<Final_answer>
Therefore, the value of $n$ is $\boxed{4}$.
</Final_answer>
</think>