Okay, so I need to solve this probability problem involving an unfair six-sided die. Let me first try to understand the problem step by step.

The die has faces numbered 1 through 6. For this die, the probability of rolling face F is greater than 1/6. The probability of rolling the face opposite to F is less than 1/6. All other faces have a probability of exactly 1/6. Also, the sum of opposite faces is 7. So, opposite faces add up to 7. That means 1 is opposite 6, 2 is opposite 5, and 3 is opposite 4.

Now, when two such dice are rolled, the probability of getting a sum of 7 is 47/288. I need to find the probability of obtaining face F, which is m/n, and then compute m + n where m and n are coprime.

Let me start by recalling that when two dice are rolled, the probability of their sum being 7 is the sum over all pairs (i, j) where i + j = 7 of the product of their probabilities. Since each die is identical in terms of their unfairness (except for face F and its opposite), the probability distribution for each die is the same.

First, let's define the probabilities for each face on a single die. Let's denote:

- Let p be the probability of rolling face F.
- Then, the probability of rolling the face opposite to F is q, where q < 1/6.
- The other four faces each have probability 1/6.

But wait, the total probability must sum to 1. So, p + q + 4*(1/6) = 1. Let me write that equation:

p + q + 4/6 = 1 ⇒ p + q = 1 - 4/6 = 2/6 = 1/3. So, p + q = 1/3.

That's one equation.

Now, when two dice are rolled, the probability that their sum is 7. Let's think about all possible pairs (i, j) where i + j = 7. The pairs are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). But since the dice are identical, some of these pairs might have the same probability.

But in this case, the die is unfair, so the probability of each face depends on whether it's F, its opposite, or another face.

Let me denote face F as, say, 1. Then its opposite face would be 6. But the problem doesn't specify which face is F, just that F and its opposite have different probabilities. Wait, but the problem mentions "face F", so maybe F is a specific face. Let me check the problem statement again.

Wait, the problem says "the probability of obtaining face F is greater than 1/6, the probability of obtaining the face opposite face F is less than 1/6, the probability of obtaining each of the other faces is 1/6". So F is a particular face, not any face. So, for example, if F is 1, then 6 is opposite. If F is 2, opposite is 5, etc.

But since all dice are the same, the specific face F is fixed for a die. So when two dice are rolled, the possible combinations that sum to 7 will involve pairs where one die shows a face and the other shows its opposite, or other combinations.

Wait, no. Let's think: for two dice, the sum of 7 can be achieved in several ways:

- Die 1 shows 1, Die 2 shows 6.
- Die 1 shows 2, Die 2 shows 5.
- Die 1 shows 3, Die 2 shows 4.
- Die 1 shows 4, Die 2 shows 3.
- Die 1 shows 5, Die 2 shows 2.
- Die 1 shows 6, Die 2 shows 1.

But since the dice are identical, some of these pairs are symmetric. However, the probability of each pair depends on the probabilities of the individual faces.

But since the die is unfair, the probability of F and its opposite are different. Let's suppose that face F is, say, face k. Then its opposite is face (7 - k). Let's denote:

- Let p = probability of face k.
- Then, probability of face (7 - k) is q = probability of opposite face.
- The rest of the faces (which are not k or 7 - k) each have probability 1/6.

So, for two dice, the probability that their sum is 7 is the sum over all pairs (i, j) where i + j = 7 of P(die1 = i) * P(die2 = j).

So, let's compute this probability.

First, the possible pairs (i, j) where i + j = 7 are:

(1,6), (2,5), (3,4), (4,3), (5,2), (6,1).

Now, depending on whether each face is F, opposite, or neither, the probabilities will vary.

But since F is a specific face, say k, then the opposite face is (7 - k). Let's note that:

- If k is one of the numbers from 1 to 6, then (7 - k) is also from 1 to 6, and they are distinct unless k = 3.5, which is impossible, so they are distinct.

Therefore, when considering the pairs (i, j) summing to 7, some of these pairs may involve F or its opposite, and others may not.

So, let's suppose that F is some face, say, face a. Then its opposite is face (7 - a). Therefore, the pairs (a, 7 - a) and (7 - a, a) are the pairs where one die shows F and the other shows its opposite. The other pairs (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) but excluding (a, 7 - a) and (7 - a, a) may involve other faces, which have probabilities 1/6.

Wait, but if F is, say, 1, then 6 is its opposite. Then the pairs (1,6) and (6,1) would involve F and its opposite. Similarly, if F is 2, then the pairs (2,5) and (5,2) involve F and its opposite. Wait, but in that case, the pair (2,5) would be F and opposite? Wait, no. Wait, if F is 2, then the opposite face is 5, so yes, (2,5) and (5,2) would be pairs where one is F and the other is its opposite.

Similarly, if F is 3, then 4 is its opposite. So the pairs (3,4) and (4,3) would involve F and its opposite.

Wait, but the problem says that the probability of F is greater than 1/6, the probability of its opposite is less than 1/6, and the rest are 1/6. Therefore, for example, if F is 1, then face 1 has probability p > 1/6, face 6 has probability q < 1/6, and faces 2,3,4,5 have probability 1/6 each.

Therefore, when two dice are rolled, the probability of getting a sum of 7 is:

P(sum=7) = P(die1=1 and die2=6) + P(die1=6 and die2=1) + P(die1=2 and die2=5) + P(die1=5 and die2=2) + P(die1=3 and die2=4) + P(die1=4 and die2=3).

But depending on whether 1 and 6 are F and its opposite or not, their probabilities will vary.

Wait, but since in the problem, the die is such that F is a specific face and its opposite is another. So, for example, if the die has face 1 with probability p, face 6 with probability q, and the rest with 1/6. Then, when considering the sum 7, the pairs (1,6) and (6,1) have probabilities p*q and q*p respectively. The pairs (2,5) and (5,2) have probabilities (1/6)*(1/6) each, since neither 2 nor 5 is F or its opposite. Similarly, pairs (3,4) and (4,3) also have probabilities (1/6)*(1/6) each.

But wait, hold on. If F is, say, 1, then faces 2,3,4,5,6 are the other faces. But face 6 is the opposite of 1, so it has probability q. Therefore, when considering pairs (2,5) and (5,2), since neither 2 nor 5 is F (which is 1) or its opposite (which is 6), so their probabilities are 1/6 each. Similarly for (3,4) and (4,3). Therefore, the total probability for sum 7 would be:

P(sum=7) = 2*p*q + 4*(1/6)*(1/6)

Wait, let's check:

- For pair (1,6): probability is p * q
- For pair (6,1): probability is q * p
- For pair (2,5): probability is (1/6)*(1/6)
- For pair (5,2): same as above
- For pair (3,4): same, (1/6)*(1/6)
- For pair (4,3): same

Therefore, total probability is 2*p*q + 4*(1/36) = 2pq + 4/36 = 2pq + 1/9.

But the problem states that this probability is 47/288. So:

2pq + 1/9 = 47/288

Let me compute 1/9 in 288 denominator:

1/9 = 32/288

So:

2pq + 32/288 = 47/288 ⇒ 2pq = 15/288 ⇒ pq = 15/576 = 5/192.

Also, from earlier, we have p + q = 1/3.

So we have two equations:

1. p + q = 1/3
2. pq = 5/192

These are the sum and product of p and q. Therefore, p and q are the roots of the quadratic equation x² - (1/3)x + 5/192 = 0.

Let me solve this quadratic equation.

Multiply through by 192 to eliminate denominators:

192x² - 64x + 5 = 0

Using quadratic formula:

x = [64 ± sqrt(64² - 4*192*5)] / (2*192)

Compute discriminant:

64² = 4096

4*192*5 = 3840

So sqrt(4096 - 3840) = sqrt(256) = 16

Therefore,

x = [64 ± 16]/384

So two solutions:

x = (64 + 16)/384 = 80/384 = 5/24

x = (64 - 16)/384 = 48/384 = 1/8

So the roots are 5/24 and 1/8.

But we know that p > 1/6 (since probability of F is greater than 1/6) and q < 1/6 (since probability of opposite face is less than 1/6). Let's check the values:

1/8 = 0.125, 1/6 ≈ 0.1667, 5/24 ≈ 0.2083.

So 5/24 is greater than 1/6 (since 1/6 ≈ 0.1667 and 5/24 ≈ 0.2083), and 1/8 = 0.125 < 1/6. Therefore, p = 5/24 and q = 1/8.

Therefore, the probability of face F is 5/24. Then m = 5, n = 24. They are coprime since 5 and 24 share no common divisors other than 1. Therefore, m + n = 5 + 24 = 29.

Wait, but let me double-check. Let's see:

Given p = 5/24 ≈ 0.2083 > 1/6 ≈ 0.1667, which is correct.

q = 1/8 = 0.125 < 1/6 ≈ 0.1667, which is also correct.

Then, p + q = 5/24 + 1/8 = 5/24 + 3/24 = 8/24 = 1/3, which matches the first equation.

And pq = (5/24)(1/8) = 5/192, which matches the second equation.

Therefore, the calculations are correct.

Therefore, the probability of face F is 5/24, so m + n = 5 + 24 = 29.

**Final Answer**
\boxed{29}
Therefore, the probability of obtaining face $F$ is $\boxed{29}$.
</think>