Why K8S HA chooses 3 instead of 5..6..7 as the size of masters?

Theory: In order to take any action, there has to be n/2 + 1 members reach the consensus.

Algorithm: RAFT

Cluster size	Majority	Fault Tolerance
1	1	0
2	2	0
3	2	1
4	3	1
5	3	2
6	4	2
7	4	3
8	5	3
9	5	4

cluster_size / 2 + 1 = majority

fault_tolerance = cluster_size - majority - In order to let the system work properly on failures , the number of majority has to be reached. So the number of fault_tolerance is the cluster size minus the required number of majority.

Why choosing 3:

If all other factors are the same, the larger cluster size, the more expensive. Choosing 4 does not increase the number of fault tolerance. However, if choosing 9, one needs to consider the cost of the cluster.

Sometime people choose 5 as well in production:

You have 2 fault tolerances. That means you could have one fails, one is down for maintainance.