Talk:Proficiency Purchase (3.5e Variant Rule)

This is kewl. Too bad the campaign I'm running has already applied my system... Will you be horribly offended if I use this in my next D&D campaign? Teh Storm 23:58, May 15, 2010 (UTC)

Not at all. While I have been using this in a few of my games I never published it, after viewing your stuff I figured, eh, why not. I should add it on there. Use away. -- Eiji Hyrule 00:01, May 16, 2010 (UTC)

Looking back over it, when combining the gold and experience cost it is really expensive. That can be a good thing, but what about a cost in gold based on the weapon. Maybe weapon price x10?Teh Storm 17:59, May 17, 2010 (UTC)

Ratings

	Aarnott likes this article and rated it 3 of 4.
	Because spending feats on proficiencies has always been stupid. If I used it I might have a weighting system based on the ratio of your BAB to level (as a multiplier to those costs). That way a fighter can pick them up twice as easily than a wizard and a rogue is in the middle. Fractional BAB would be required for that. But, either way, I really like this idea.

That's... not a bad idea. If you could give me some numbers, I'd totally add it as an alternative on the bottom. -- Eiji-kun 05:24, 30 July 2011 (UTC)

Take the costs exactly as you have them, but then multiply them by character level divided by Fractional BAB. So a character with all wizard levels will multiply the costs by 2 (it's a ratio, so it will end up being 1 / 0.5 = 2). Similarly, a rogue will multiply all the costs by 1.5. A rogue 6/wizard 3/fighter 3 will have a fractional BAB of 6 * 0.75 + 3 * 0.5 + 3 = 9 and a character level of 12, which gives a ratio of 12/9 = 1.33.

That's the more "accurate" way of doing it (it counts every level you have). A good approximation is just to count whichever BAB progression you have the most levels of and base the cost on that. So if you had 4 rogue levels and 2 fighter levels, it would cost 1.5x as much. If you have the same amount in 2 different progressions, you'd choose the better one. I'd probably go with this method since it is easier to figure out and effectively amounts to the same thing. --Aarnott 16:22, 30 July 2011 (UTC)