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I was asked a question today which was What are the odds of the world ending on any date in the calendar in the future ? I have assumed it will end on one day I gave my answer as 364.25/1 on any day and 1460/1 on february 29th I was told that the odds are lower for december 4th at any time in the future than december 2nd in the future I cant see how this right Can anyone help please as i think i am right

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That would be because December 4th is the day of the feast for Saint Ada, and Ada is a programming language designed specifically with a view to support the construction of reliable software. It would be too tragic an irony for the universe to sustain if the eventual crash of the world were to happen on here feast day. – Henning Makholm Dec 3 '12 at 17:46
@HenningMakholm The world of reliable software is, alas, one of tragic irony. – WimC Dec 3 '12 at 19:51
Whoever stated this is probably of the impression that the earth is like a porcelain vase whose lifetime has an exponential distribution. – WimC Dec 3 '12 at 19:53

Suppose that at midnight, Australian Eastern Daylight Time, the probability that the world will end during the next 24 hours is $p$ where $0\lt p\lt1$. So, the probability the world ends today is $p$. The probability the world ends tomorrow is $p(1-p)$; the world can only end tomorrow if it didn't already end today. In general, the probability the world ends on day $n$ is $p(1-p)^n$ (where today is day zero).

For $0\le n\le364$, the probability the world ends on date $n$ is (ignoring leap years) $$Q_n=p(1-p)^n+p(1-p)^{n+365}+p(1-p)^{n+730}+\cdots=p\ {(1-p)^n\over 1-(1-p)^{365}}$$ Now it's clear that $$Q_0\gt Q_1\gt Q_2\gt\cdots\gt Q_{364}$$

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