2

There is a strategy called (highest) resource strategy in Catan. That is, build your initial settlements on hexes where the sum of the favorable die rolls is a maximum. That assumes 1) that all resources are about equal in value, and 2) an abundant resource is as valuable (or nearly so) as a scarce resource. But both of these assumptions seem counterintuitive.

In this video, a you tuber opined that "one ore > two sheep." That is, he would build his second settlement on a site with less ore than sheep, because the former is more valuable. This person has been blogging and playing for some time, so I assume that he is at least "somewhat" expert. His view also corresponds with my own rough calculations that the relative value of resources is: ore, 5, wheat 4, wood and brick, 3, and sheep 2. Taking the base of 3 for brick wood, I come up with value coefficients of 1.6, 1.3, and 0.7 for ore, wheat, and sheep respectively. I then multiply this vector by the values of ore, wheat and sheep to come up with a "dot product" that gives me an adjusted resource value. In this example, an ore hex with two die rolls is worth more than a sheep hex with four die rolls, because 2 x 1.6 (=3.2) >4x0.7 (=2.8).

Do any systems or experts "re-weight" resources using an algorithm similar to mine and then choose intersections for early settlements accordingly?

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.