How does a decision tree pick the exact number to split a column on?

Take the six rows $x=1,2,3,4,5,6$ with labels $0,0,1,0,1,0$. The parent group has two positives and four negatives, so its Gini is $2\cdot2\cdot4/6^2=16/36=4/9$. At cut 2.5 the left group is $0,0$, with $p=0,n=2$, so $G_L=0$. The right group is $1,0,1,0$, with $p=2,n=2$, so $G_R=2\cdot2\cdot2/4^2=8/16=1/2$. Weighted: $(2/6)\cdot0+(4/6)\cdot(1/2)=1/3$. At cut 1.5 the weighted score is $(5/6)\cdot(12/25)=2/5$, and at cut 3.5 it is $4/9$. The smallest is $1/3$ at 2.5, so the split sends $x=1$ and $x=2$ left and $x=3,4,5,6$ right. The reduction is $4/9-1/3=1/9$.

The table shows why the best cut can sit in an unremarkable place. Cut 4.5 also looks plausible because it separates the last two rows, but its weighted score is $5/12=0.4167$, worse than $1/3$. Cut 5.5 isolates one zero on the right and scores $2/5$. The winner is not the cut that makes one side pure. Cuts 1.5 and 5.5 each make one side pure and still lose, because the other side is large and mixed. The criterion is weighted, so a pure small group does not compensate for a mixed large one.

The natural guess is to take the mean, the median, or a gap that looks wide. That fails because the score is not monotone in the cutoff. Moving the cut from 2.5 to 3.5 puts $x=3$ on the left, and $x=3$ is a positive, making the left group $0,0,1$ and the right group $0,1,0$. Both groups then have Gini $4/9$ and the weighted score rises from $1/3$ to $4/9$. A cutoff placed exactly on an observed value is not wrong, but it needs an explicit rule for equality. The midpoint convention avoids relying on that rule without changing any partition.

There are only $k-1$ candidates when the column has $k$ distinct values. The score can change only when a row crosses from one side to the other, and that happens at each observed value. Between two adjacent observed values every cutoff gives the same two groups. For values 2 and 3, the cuts 2.1, 2.5 and 2.9 all send every row with value 2 left and every row with value 3 right. The implementation reports 2.5 because it is the midpoint, not because 2.5 has any special meaning in the data. If the column has duplicates, use adjacent distinct values, not adjacent rows.

If two cuts tie, any deterministic rule works. Implementations usually keep the first in column order or the lowest cut. The tie means the data does not distinguish them under the chosen mixedness measure, not that the cutoff is arbitrary. At prediction time the case's value is compared with the stored cut using the same inequality. Because the cut sits in a gap between training values, a new value in that gap can land on either side depending on which midpoint was stored. Two trees grown on the same data with different midpoint conventions can therefore differ on a new case that falls between the same two training values. Using one convention for growing and prediction removes most of that sensitivity.