A dice is weighted such that the probability of rolling the face numbered n is proportional to `n^(2)` (n = 1, 2, 3, 4, 5, 6). The dice is rolled twice, yielding the number a and b. The probability that `a gt b` is p then the value of `[2//p]` (where [ . ] represents greatest integer function) is _______.

Explanation

To solve the problem step-by-step, we need to find the probability \( p \) that \( a > b \) when a weighted die is rolled twice. The probability of rolling a face numbered \( n \) is proportional to \( n^2 \).

### Step 1: Determine the probabilities for each face of the die
The probabilities for each face \( n \) (where \( n = 1, 2, 3, 4, 5, 6 \)) are given as:
\[ P(n) \propto n^2 \]
Thus, we can express the probabilities as:
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