Let me check this out a sec:

Let n = the number of times you slip up in 30 days.

n = 0, means you have succeeded.

Assume n-1 slip ups in 30 days yields success. This would imply that you have gotten back on track n times. If you then slip up an nth time, getting back on track an n+1th time will still yield success.

Therefore, success is guaranteed for all finite values of n.

Yep. Seems sound to me.

Seriously, though. Good luck with everything. I hope your job interviews go well.

helllooooo math heads!

Best wishes for today and this moment. I find if I can be present and realistic in the here and now, things are not so impossible. Right now I am breathing. Right now I feel the earth below me. Right now I can do this.

Let us know how the interviews went!

n!

:-)

I'll leave y'all to decide whether that's n factorial or just me being exclamatory ... I just like to see factorials written, because they seem so enthusiastic.

I think the idea of trying to determine the minimum value of n that you can personally manage is a good one. So, find the gradient of f(n) and head downwards!

It makes perfect sense that alcohol would reduce the racing thoughts, since it reduces all thoughts. It's little wonder that people use it to self-medicate. But clearly, we need to keep some. Another optimization problem in multiple variables ...

We're pulling for you, Ralph.

Well if n represents the number of failures, the original blog entry was just an attempt to define n in objective terms. I have no useful model to explain what variables might be controlled in order to minimize n. That is the current exploration, however I am unsure it would be worthwhile to gather a sufficiently large data set as I suspect that error terms would be more significant than the independent variables.