# 82% of an event in a persons life, what is the daily chance?

If there is an 82% chance that within your average lifetime, lets assume 70 years (25567 days), that "E" event will happen: what is the percent chance that it will happen on any given day?

I'm assuming that it'll be something like .02347% or something small. My problem is I've only taken up through Advanced Algebra and that was six years ago. I would have searched for this on google to solve it myself, but I'm not even sure what to call this beyond "math I don't know".

So if you could help me identify the solution, how to solve it, and what this is (i.e. I imagine terms like 'graphing polonomials' 'deductive statistics' or some such name that would give me an idea of what exactly it is that I should be learning to handle similar questions on my own). Thanks for the help in advance.

If the probability that it happens in a specific day is $p$, then the probability that it does not happen is $(1-p)$ (complimentary events). Thus the probability that it will not happen in $25567$ days is $(1-p)^{25567}$ (product rule, independent events). Therefore the probability that it will happen in a lifetime is $1 - (1 - p)^{25567}$ (again, complimentary events), and that is your $82\%$.
So, we have that \begin{align} 0.82 &= 1 - (1-p)^{25567} \\ 0.18 &= (1-p)^{25567}\\ \sqrt{0.18} &= 1 - p\\ 0.999933 &= 1-p\\ 0.000067 &= p \end{align} so the probability of it happening in a given day (assuming it doesn't change with day of the week, time of the year, age, and so on) is $0.0067\%$.
This is a bit backwards from the standard textbook example (it would give you the daily probability and ask you to calculate the lifespan probability), but it's more or less elementary probability, as they should teach in any high school. The two main concepts used in setting up the first equation was complimentary probability and the product rule. Once the equation was set up, however, we basically forget all about the probability, and focus on the algebra of isolating and calculating $p$. When we have that, we go back to the probability interpretation, and say what it means.
• The formatting and the $symbol rather than say using "" quotations made this a bit hard to follow. Perhaps you could reformat it? Use the bb 'code' blocks or something? Otherwise, thanks for the help. Basically, I found a statistic that said if you live to be 70, your chance of being the victim of violent crime is 82%. So that just made me wonder what the chance is on any given day. However, if we take into account that I've lived say 8765 days of my life and never experienced a viloent crime, does the chance per day grow exponentially? – Arammil Oct 10 '13 at 18:13 • @Arammil It doesn't mean that. The chance per day is (in an ideal mathematical world) the same no matter what happened to you yesterday, or the 70 years before. This is an intuition that many people struggle to be rid of. They think that because something hasn't happened yet, it's "due" in some way. Think of a roulette wheel, if you always go for the number$10$. If you play a hundred times, and it never lands on your number, you'd think it was due some time soon, but that's not how it works. Each time there's a one in 37 (or 38) chance that it lands right, no matter if it missed a thousand. – Arthur Oct 11 '13 at 7:27 • @Arammil So you'd think that if you play a thousand times, you're supposed to win about 27 times. So if you play a thousand times and do not win at all, do you have a greater chance of winning if you continue? The answer is no, and in an ideally simplified world, the same goes for the daily odds of being the victim of a crime. That being said, it wouldn't surprise me if older people were slightly more likely to be victims of violent crime, but in that case it has more to do with how violent crimers choose their targets than you being due as a target. – Arthur Oct 11 '13 at 7:32 • Okay, That makes sense. I just assumed, erroneously, that reality worked differently. The lotto example works better than my distorted reality. Also, I guess all the '$' signs I was seeing was just my computers way of saying that the page was not being displayed correctly... the answer makes much more sense. Thanks. – Arammil Oct 28 '13 at 13:29