What I have to share today is a simple idea that I’ve found really motivating. When I feel like I’m stuck in a loop of choices I regret, each leading to the next in a vicious cycle, I remind myself that every vicious cycle can also be a virtuous cycle.
For example, if I don’t make time to regularly go on long walks, I find that the joints in my toes can sometimes get painfully stiff. That makes it a little more painful to walk on them, so I’m less likely to go for a long walk at all, and so on: once this cycle gets going it’s hard to stop. But viewed another way, when I do go on long walks regularly, my feet feel better, and I’m more likely to want to go for a walk — a virtuous cycle. Remembering the possibility of this virtuous cycle makes it easier to choose to go for a walk despite the pain in my feet.
In fact, probability theory tells us that every vicious cycle hides a virtuous cycle in this way, if we define those terms as follows. Let’s say we have a sequence of choices to make between a good option and a bad option, and we’ll encode these choices by a sequence of 1’s (for the good options) and 0’s (for the bad options). We’ll call this scenario “a vicious cycle” if making a bad choice makes the next choice more likely to be a bad choice too, i.e. if
And we’ll call the scenario a “virtuous cycle” if making a good choice makes the next choice more likely to be good:
It turns out that these two conditions are equivalent! In general, for two events A and B we have (B is evidence for A) if and only if (not-B is evidence against A) if and only if (not-B is evidence for not-A). So if a bad choice today makes a bad choice tomorrow more likely, it must also be the case that a good choice today makes a good choice tomorrow more likely too.
2 thoughts on “Every vicious cycle is also a virtuous cycle”
That’s really nice, thank you! I find it kind of surprising that one gets to carry the strict inequality across; I was somehow expecting it to become non-strict; I guess because this had the feeling of negating somehow. But in fact (as you point out) they are just exactly equivalent.
Yes, it’s quite nice! The only thing you have to assume is that the probability of B is neither 0 nor 1, in order to make sense of the conditional probabilities.