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If there was a 10% chance of an event occurring over a year, how would I work out how likely this event is to happen per month, per day etc?

Now because if, using per month as an example, if it occurs in January it cannot happen again for that particular instance, which takes away simple division.

desertnaut
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sxthomson
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2 Answers2

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If the event has probability $m$ to happen in a month, during each month it does not happen with probability $1-m$ hence (here an assumption is hidden, will you recognize it? Events must be independent) it does not happen during a whole year with probability $(1-m)^{12}$ and the probability $y$ that it happens during a year is $y=1-(1-m)^{12}$.

Note that when $m\ll1$, $y\sim12m$, and that $y\lt12m$ for every $m$ in $(0,1]$.

futtetennista
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Did
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  • This is an awesome answer. What if the period goes the other direction, for example, the probability is determined for a year, but you want to see the probability of it happening over 50 years? – Neil Ticktin Sep 11 '13 at 16:44
  • $1-(1-y)^{50}$ (thought this was obvious from the post). – Did Sep 11 '13 at 20:03
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Well, I think I have the answer to the ORIGINAL question. y=1−(1−x)^(1/365) For odds on any given day

y=1−(1−x)^(1/12) For odds on any given months, (assuming months are all the same length...)

Edit. Incidentally, if it happens in January it doesn't mean it WOULDN'T have happened in February. This doesn't find the most likely day it happens. December doesn't become practically never, just because it likely happened already in January, (or February, or March).

SeanR
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