If the 5th algo in the chain is found to be broken, you'd still have to come up with an input of the 4th algo that generates the output you need to feed into the 5th algo; so you'd have to come up with an input for the 3rd algo that generates the correct output to feed into the 4th algo to generate the output you need for the 5th algo; and then you'd have to come up with an input for the 2nd algo that generates the correct output to feed into the 3rd algo.... and so on. So basically you'd need rainbow tables for all of the preceding algorithms in the chain, in order to do anything meaningful.
This is what I was originally thinking. Rainbow tables do exist... But this does make it far-fetched... To put a number on it is more than 1, less than 11.
The argument now becomes circular.
Yes, more than one potential attack vector is worse than just one. But, if that 1 vector is exposed, you are bork just as bad...
Bork or not bork is the resulting boolean interest and I'm still not convinced that 11 hashes = 11 vectors via baddw's example.
Some hashes have had their bork streamlined to break in seconds on a Core i5...