Let \(\epsilon=\dfrac{1}{N}\).\[\]
Choose a number at random between 0 and 1.

Choose a second number between \(\epsilon\) and \(1+\epsilon\).

Choose a third number between \(2\epsilon\) and \(1+2\epsilon\).

Continue in this way until you choose the \(N^{\text{th}}\) number between \(1-\epsilon\) and \(2-\epsilon\).\[\] If the probability that the first number you chose is the smallest is \(\bigg (\dfrac{a}{b\times N}\bigg )^c\), find \(\lfloor 100c\times(a+b) \rfloor\).\[\]**Details and Assumptions**:
###### This is part of Ordered Disorder.

Choose a second number between \(\epsilon\) and \(1+\epsilon\).

Choose a third number between \(2\epsilon\) and \(1+2\epsilon\).

Continue in this way until you choose the \(N^{\text{th}}\) number between \(1-\epsilon\) and \(2-\epsilon\).\[\] If the probability that the first number you chose is the smallest is \(\bigg (\dfrac{a}{b\times N}\bigg )^c\), find \(\lfloor 100c\times(a+b) \rfloor\).\[\]

Assume \(N\) is very large.

You might wish to make small approximations.

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