A level 5 rogue will quite probably have a thievery dc of 13, if they invest in it and max dex. The average lock has a dc of 25 and requires 4 successes. It takes a roll of 12 or better to have a single success, and will average about 9 rolls to rack up those 4 successes. With 9 rolls wherein you crit fail on a 2 or lower, the likelihood of breaking a pick is ~61%.
Should a level 5 rogue take a minute to open the average lock, and more likely than not break a pick in the process?
And let’s look at a good Lock: DC 30, requiring 5 successes. The level 5 rogue will only succeed on a 17, meaning it will take on average 20 attempts to get those 5 successes. On one attempt in a thousand our Lvl 5 rogue will open this lock before breaking a pick, and will typically break 3 in the process.
Am I missing something?
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Thieve tools are 3Gp but replacement picks are 3sp. Also, the tools are light bulks, and the replacement pick don’t even list a bulk.
Well that’s a game changer. Thanks for the info!
Merwyn already pointed out that replacement picks are cheap and weightless. Regarding the number of dice, every PF2 campaign I’ve been a part of (which is two, so a pretty small sample size) has just had the rogue roll for each tumbler right up front. Need 5 successes, roll 5 dice. If any of them are a critical success, it offsets a critical failure (in addition to counting for two tumblers as normal); otherwise, each critical failure = -1 replacement pick. It’s meant on average 2-3 sets of rolls per lock, rather than 7-8, and keeps things moving at a fine clip.