Yeah, it's unlucky, but not colossally so -- this is exactly why this algorithm has that huge warning -- single bit changes in the input, especially late in the string, have very little effect on the final output (the avalanche effect is very small in the last two rounds).
In your example, you have pretty much the worst-case-scenario for collissions: two strings of the same length, which are identical up until the last two 32-bit blocks ("et_1" and "536" vs "et_2" and "560"). Even worse, the difference in the second-to-last block is only in the lowest two bits, so the hashes at that point only difffer by 4 bits out of 32.
The unlucky part is that one of the differences in the final block (the "3" vs "6") exactly cancels out two bits of that, and then the last difference ("6" vs "0") takes care of the rest, so the hashes become identical again. -- I'm not sure what the probability of that is, but I'm sure it's far greater than you'd expect from hashing a random pair of uncorrelated strings.
You might improve your odds against this even by reversing the string before hashing it: making the strings differ as early as possible might help, but that's just a guess. You might end up with the same situation, just with the hashes becoming identical earlier in the algorithm, and then staying in lockstep from there out.