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I have never been able to wrap my head around probability, and I often find that my intuition is wrong. In this case, I don't even have intuition as to where to begin.

If I have two trials, each with a 1/16 chance of success, what are the chances that either or both of them result in success? How, mathematically, do you arrive at the correct probability? How, intuitively, can I understand this number?

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    @Hans: Thanks very much -- that's quite informative. So I'll post answers as answers in the future.2011-04-07

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The probability that neither trial is successful is $(15/16)^2$ (assuming that the trials are independent), and the chances that at least one trial is successful is one minus that: $1-(15/16)^2 = 31/256$.

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It also might be helpful to think of this in terms of geometric distribution.

Geometric distribution represents the number of failures before you get a success and is defined as:

$(1-p)^{n-1} \times p$

In your case, what we have is two geometric distributions added together:

$\left(\frac{15}{16}^{0} \times \frac{1}{16}\right)+\left(\frac{15}{16}^{1} \times \frac{1}{16}\right) = \frac{31}{256}$

The reason we add it together is the first term assumes that you had no failure and your first trial was a success, your second term assumes that you had one failure and your second trial was a success.