The linkage map of X-chromosome of fruitfly has 66 units, with yellow body gene (y) at one end and bobbed hair (b) gene at the other end. The recombination frequency between these two genes (y and b) should be : -

Map distances are obtained by adding the recombination frequencies between closely placed genes. So the total length of a map (here 66 units) can be more than 50.
However, the recombination frequency actually observed between two genes can never exceed 50 percent. This is because multiple crossovers between distant genes cancel each other's effect (a double crossover restores the parental combination), and crossing over involves only two of the four chromatids.
A recombination frequency of 50 percent means the genes behave as if they assort independently.
So between y and b, at opposite ends of the chromosome, the recombination frequency should be $\leq$ 50%.