### Every vicious cycle is also a virtuous cycle

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 $C_n$ 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

$P(C_{n+1} = 0 | C_n = 0) > P(C_{n+1} = 0)$.

And we’ll call the scenario a “virtuous cycle” if making a good choice makes the next choice more likely to be good:

$P(C_{n+1} = 1 | C_n = 1) > P(C_{n+1} = 1)$.

It turns out that these two conditions are equivalent! In general, for two events A and B we have $P(A | B) > P(A)$ (B is evidence for A) if and only if $P(A | B^c) < P(A)$ (not-B is evidence against A) if and only if $P(A^c | B^c) > P(A^c)$ (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”

1. 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.

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1. 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.

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