Ken has a six sided die. He rolls  the die, and if the result is not even, he rolls the die one more time. Find the probability that he ends up with an even number.

[i]Proposed by Gabriel Wu[/i]
1. **Case 1: He gets an even number on the first roll.**
   - A six-sided die has three even numbers (2, 4, 6) and three odd numbers (1, 3, 5).
   - The probability of rolling an even number on the first roll is:
     \[
     P(\text{even on first roll}) = \frac{3}{6} = \frac{1}{2}
     \]

2. **Case 2: He gets an even number on the second roll.**
   - This case occurs if the first roll is odd, which has a probability of:
     \[
     P(\text{odd on first roll}) = \frac{3}{6} = \frac{1}{2}
     \]
   - If the first roll is odd, he rolls the die again. The probability of getting an even number on the second roll is:
     \[
     P(\text{even on second roll}) = \frac{3}{6} = \frac{1}{2}
     \]
   - Therefore, the combined probability of getting an odd number on the first roll and an even number on the second roll is:
     \[
     P(\text{odd on first roll and even on second roll}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}
     \]

3. **Total Probability:**
   - The total probability that Ken ends up with an even number is the sum of the probabilities of the two cases:
     \[
     P(\text{even number}) = P(\text{even on first roll}) + P(\text{odd on first roll and even on second roll}) = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}
     \]

The final answer is \(\boxed{\frac{3}{4}}\).