#TI-BASIC, 13 12 bytes
TI-BASIC, 13 12 bytes
I found some 12 byte solutions
Ans-1:sum(Ansnot(fPart(Ans/2
The previous is the same byte count as below due to the line separator costing a byte
sum((Ans-1)2fPart(Ans/2
The above solutions use NfPart(Ans/N as a modulus operation to judge even/odd, giving us a list of dice that can have petals. Multiplying that by the original list -1 restores the petal counts to the list which can then be summed
2sum(Ans=3)+4sum(Ans=5
The original 13 byte solution simply compared the list once against 3 and separately against 5 before summing those results independently to get the petal count
All the above solutions take input as a list in Ans