Probaballistic fun

Let $\epsilon\equiv\frac{1}{N}$ Choose a number at random between 0 and 1. Choose a second number between $\epsilon$ and $\epsilon+1$ . Choose a third number between $2\epsilon$ and $1+2\epsilon$ . Continue this process, until you choose an Nth number $1-\epsilon$ between and $2-\epsilon$ . What is the probability that the first number you choose is the smallest of all the numbers? Assume that N is very large, and make suitable approximations. if the answer is in the form $P\thickapprox\sqrt{\frac{a}{b^{2}N^{\frac{\sqrt{2}}{b}}}}$ then find the value of a upto two digits