If the probability of an event happening in 1 second is 10%, what is the probability that it happens in 0.1s? How do I convert probabilities between time scales?
My current understanding is that I can use the following:
$p_{\text{new}} = 1 - (1 - p_{\text{orig}})^{\text{int}_{\text{new}} / \text{int}_{\text{orig}}}$
so I would get:
$p_{\text{new}} = 1 - (1 - 0.1)^{0.1/1}$
Is this accurate, and if so, what is the name of this formula/theory?
I've also looked at the Poisson distribution, which looks like it may also apply, but I'm not sure how that differs from the above method.
EDIT: I should probably mention that I'm talking about stochastic mechanisms here that are completely independent.
 A: Under appropriate assumptions, this is correct.
Specifically, for this calculation to work out, we must assume that the following events are independent and equally likely (where we have a stopwatch reading out in seconds):

*

*An event occurs when the stopwatch reads between $0$ and $0.1$.


*An event occurs when the stopwatch reads between $0.1$ and $0.2$.
...

*

*An event occurs when the stopwatch reads between $0.9$ and $1.0$.

If we let $p_{\text{new}}$ be the probability of any individual event occurring, then, since we assumed they are independent, the probability of none of them occurring is
$$(1-p_{\text{new}})^{10}$$
so the probability of at least one of them occurring is
$$p_{\text{original}}=1-(1-p_{\text{new}})^{10}.$$
Solving for $p_{\text{new}}$ would give your formula
$$p_{\text{new}}=1-(1-p_{\text{original}})^{1/10}.$$
Note that you usually need to be careful about the assumption of independence - some events are appropriate to measure that way (e.g. "a raindrop hit a piece of paper") and some are definitely not (e.g. "a single randomly chosen time in the next 10 seconds was in this period" or, as suggested in a comment, "the probability that the seconds hand of a clock moves in the next 0.1 seconds").

As you suggest, this formula can be generalized to say that, under similar assumptions, to say that
$$p_{\text{new}}=1-(1-p_{\text{original}})^{d_{\text{new}}/d_{\text{original}}}.$$
where $d_{\text{new}}$ is the new duration and $d_{\text{original}}$ is the original one. Proving this is basically just repeating the above argument when the quotient in the exponent is rational and then doing a little analysis to handle the other cases.

The connection to the Poisson distribution is that the Poisson distribution considers a similar setup. It measures how many times some event occurs, assuming that the events "it occurs at least once between time $x$ and $y$" are independent of each other (so long as the time intervals do not overlap) and that their probability only depends on the duration between $x$ and $y$.
The Poisson distribution is, given the expected number of occurrences in some duration and these assumptions of independence, the distribution of the count of events in a duration. So, for instance, if you expected $1.5$ events to happen each second (on average), it would tell you things like "the probability of $5$ events happening in that second is $\frac{(1.5)^5e^{-1.5}}{5!}$."
You could solve the question using these distributions by noting that, under these assumptions, if we expected $x$ events to happen per second, then the probability of no events happening is $e^{-x}$, per the Poisson distribution's definition. We'd expect $\frac{1}{10}x$ events to happen per $0.1$ seconds under the given assumptions - which would yield the probability of none of them happening to be $e^{-\frac{1}{10}x}$ - which aligns with your calculations.
A: You're on the right track.
Here's a hint that might help clarify your thinking: Suppose the probability that the event happens in some interval of .1 seconds is $p$. Then the probability that it doesn't happen in some interval of .1 seconds is $1-p$. The probability that it doesn't happen in ten consecutive intervals of .1 seconds is $(1-p)^{10}$, and this has to equal what?
