Die-Morphosis

I feel certain that somewhere out there is a beautiful way to solve this that is currently eluding me. Until I think of it, here is the "brute force" solution:

The probability that you lose in exactly 1 turn is $\frac{1}{6}$ .

The probability that you lose in exactly 2 turns is $\sum_{j=2}^{6} \mbox{Prob( first roll is } j \mbox{ ) } \times \mbox{Prob( second roll is 1)} = \frac{1}{6} \sum_{j=2}^{6} \frac{1}{j}$

The probability that you lose in exactly 3 turns is $\sum_{j=2}^{6} \mbox{Prob( first roll is } j \mbox{ ) } \times \sum_{k=2}^{j} \big( \mbox{Prob( second roll is } k \mbox{ )} \times \mbox{Prob( third roll is 1)} \big) = \frac{1}{6} \sum_{j=2}^{6} \sum_{k=2}^{j} \frac{1}{jk}$

Putting this all together and using a script, the probability that you lose before the fourth turn is the sum of the above probabilities, which turns out to be $\boxed{0.62449}$