It depends on how many checks will be needed...
The more jumps, the more probable is failure.
For example, if the rogue has +7 jump, and the dc is 10, he falls if he rolls 1 or 2 (which is 10% probability of falling) and 90% probability of success.
Like that, if he has to jump 6 pillars with a 10 dc, then the probability of jumping each of them without falling, is:
(9/10)^6=0.53
Which means that 47% he will fall (if I did the maths right).
Now with 14 dc he fails with 1,2,3,4,5 or 6 on the dice, which is a 30% probability to fall. If, in case he falls you allow him a climb check to get a hold and not drop, and we assume the rogue has a +7 climb, with a Climb dc of 17, then 50% of that time he will still survive. This brings the probability of failure down to 15%. And again, for 6 jumps we have
(85/100)^6=0.377149516 (37% probability of success) [ok now I get why they use the P(x) symbolism

, for not writing all the time]
Then again, if you allow a jump, a climb to hold, and a second climb to stop somewhere lower, the probability of success goes up...
DC 15 jump (35% failure)
DC 15 climb (12% failure)
DC 15 climb (4% failure)
and the actual climb dc to climb up for the next jump
DC 15 climb (5.5% failure) [from the 8% who will fail the first climb and manage the second,15% will simply fall, by failing their climb by 5 or more

)
Now for 6 jumps,
(94.5/100)^6=0.71 71% probability to success, 29% to fall.
Because for each additional "save" you allow and for each additional jump you ask you multiply, there is a great difference between making 4, 6 or 10 jumps...(80%, 71%, 56%)
And the point of my post is that the more you roll, the more you fail
Because your pc's will use rope
PS: If I totally failed the math, well, shame on me! But they will still use rope
PS2:
If the jump DC is 15, and if they succeed the DC to not fall is 15, then 57% they will fall...and even with your "save" only 80% will pass. (For just 1 jump)