When multiclassing or starting as a monster in pathfinder and AD&D3.5 is the BAB combined between all the classes to follow the standard progression or is it totalled up separately?


Does a Rogue 4 (BAB 3) Fighter 4 (BAB 4) have a BAB of 7 or +7/+2 ?

And would an Monster (of BAB 4) Fighter 3 (BAB 4) have a BAB of 8 or +8/+3 ?

  • 6
    \$\begingroup\$ D&D 3.5e PHB; Pg 59 "Multiclass Characters", "Class Level and Features"; Paragraph 3, "Base Attack Bonus", Line 9: "A [total] base attack bonus of +6 allows a second attack with a bonus of +1 . . . even though neither the +4 from the rogue levels nor the +2 from the wizard levels normally allows an extra attack." -1 for lack of research \$\endgroup\$
    – CatLord
    Commented Jun 25, 2012 at 17:03
  • 1
    \$\begingroup\$ (And assuming you don't have the book, both d20SRD and d20PFSRD have the info in the first search result for "Multiclass") \$\endgroup\$
    – CatLord
    Commented Jun 25, 2012 at 17:10

1 Answer 1


The two expressions are equal.

Every time the attack bonus reaches +6, you gain another attack at your full BAB -5 (that is: +1); but you must use the full attack action. You gain an additional attack at +11 (the additional attack is at your full BAB -10) and +16 (full BAB -15), independently by how you sum up our BAB (source).

Stating that a creature has a BAB +13, inherently implies that it has 3 attacks with the full attack action:

  • the 1st at full attack bonus (+13),
  • the 2nd at full attack bonus -5 (+8), and
  • the 3rd at full attack bonus -10 (+3)

So, BAB +13 = BAB +13/+8/+3.

  • 1
    \$\begingroup\$ Great, thanks; I was about 95% sure that was the case but just couldn't find the source for it. \$\endgroup\$
    – Rob
    Commented Jun 25, 2012 at 8:23
  • 2
    \$\begingroup\$ Note, however, that monsters using natural weapons don't gain the additional attacks (aka "Iterative Attacks"). You only gain iterative attacks with manufactured weapons. \$\endgroup\$
    – Lokathor
    Commented Jul 1, 2012 at 14:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .