You may already be familiar with algorithms to compute the majority using multiple queries. The way these problems work is that you are given a bin of $n$ marbles of binary colors, whose colors you cannot see, and an oracle will tell you some information about a group of different marbles of $k$ size if you choose them.
In this case, $k=4$. $n$ is irrelevant, because I'm curious only about a subset of the bin of size $k$. In this case, our oracle will only tell you the difference between the numbers of the two colors $a$ and $b$, i.e. $|\sum b - \sum a|$, which $=2$ if there are $3$ of one color and $1$ of the other.
Thus, the following permutations of marbles colored 1 and 0 will return $2$:
My question is, is it possible to determine which color is in the majority using only one additional query? I know that it is possible to do so in 1.5 additional queries on average, by using a "control marble," which you know is in the majority (by knowing whether the other $n-4$ marbles are the same color).
If you replace the first marble in the group with the control marble, and then ask the oracle again, then one of the 8 possible permutations will result in a $4$, 3 of them will result in a $0$, and the remaining 4 will result in a $2$ (again). You can then replace the second marble in the group instead for the queries that returned $2$ again, and get a definitive answer as to whether the control marble is in the majority within that group or not.
So, is it possible to reduce this to one query to the oracle that will tell you what marble is in the majority within the group? If not, why?