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Not quite. I think this is conventional wisdom: If you take a risk and co-found your own startup with 0 years of experience, then even if you tank 2 years lat
by leelin 16y ago
Not quite. I think this is conventional wisdom:
If you take a risk and co-found your own startup with 0 years of experience, then even if you tank 2 years later, you are likely to wind up close to the 5 years of experience point of the hockey stick when applying for a job.
Basically, the startup gives you a small chance of huge upside, and on the downside, you become a normal W-2 salaried employee but begin at a point farther along than someone who played it safe the entire time.
Whether this happens in practice is probably highly dependent on whether the risk taker demonstrated some goodness during the failed startup.
- azanar 16y agoIf you take a risk and co-found your own startup with 0 years of experience, then even if you tank 2 years later, you are likely to wind up close to the 5 years of experience point of the hockey stick when applying for a job. That's probably closer to the conventional wisdom around HN, but it is a far cry from conventional wisdom amongst the masses. The wisdom amongst the masses is that if you co-found or join a startup, you are playing roulette with your financial security. But the article's point -- which I agree with -- is that even if you accelerate this process of getting to that 5 year threshold, you are doing yourself a disservice fiscally by settling in at the plateau unless you have a good reason forcing you into settling. You're likely still learning more, and becoming more valuable, but it becomes much more difficult to extract that value through salary. So, it becomes beneficial to join companies where that only makes up a part of the total compensation package.
- Dylanlacey 16y agoI certainly don't think that. If you're incompetent you can get lucky, have a moderately successful failure, and still know 4/5ths of fuckall. And you may then become that most dangerous of individuals, the ignorant 'expert'.