Suppose you are going to roll a die 60 times and record the proportion of times that a 1 or 2 is showing, s. The sampling distribution of s should be centered about....?

The die has six sides, equally likely.

Two of the six are yes values.
So the probability of a yes is 2/6 or 1/3
So if you roll 60 times, the mean result should be 20 out of 60 if the die is fair.

Now we could do a binomial or more likely a normal distribution for n = 60 to figure the probabilities of less than 18 or whatever :)

so my final answer would be 1/3 right? it seems too simple to be true...

YES

the mean of the sample should be the same as the mean of the population.

To determine the center of the sampling distribution of the proportion, we need to consider the theoretical probability of rolling a 1 or 2 on a fair six-sided die.

Since there are 6 equally likely outcomes when rolling a die, and there are 2 favorable outcomes (rolling a 1 or 2), the probability of rolling a 1 or 2 is 2/6 or 1/3.

In a sequence of 60 rolls, we can expect the observed proportion of rolling a 1 or 2 to be centered around the theoretical probability of 1/3.

Therefore, the sampling distribution of s should be centered around 1/3.