The question is ill-posed because there are many different ways of choosing two spots “at random”. BUT if we assume that the cuts are made independently with each chosen according to a uniform probability per unit length then the answer is the fraction of the big square that is not shaded in the diagram below . (ie p = 1-(17/20)^2)
Since this is tagged with “Puzzles and Trick Questions” it may be that I am missing something. But my answer would be [math]\Sigma_{m=0}^{n-1}p^{2n-1-m}(1-p)^m[/math] .
This follows the pattern of the best 2 out of 3 case where Alice has to win either two or three games – which happens in cases lww,wlw,wwl or www with probability [math]3p^2(1–p)+p^3=3p^2–2p^3[/math] (where the fact that the game may be stopped when she wins twice just corresponds to the fact that [math]pp(1-p)+ppp=p^2[/math] , and the same answer is obtained by taking the complement of the cases where Bob wins either 2 or 3 games).
This question has been around for a while and has some decent answers. But I want to suggest a simpler and more intuitive version. (And it will be easier to follow if I replace the variable [math]x[/math] by [math]t[/math] so as not to confuse it with the real part of the complex function value.)
First, to define [math]f(z)=e^z[/math] without using trigonometry or ever mentioning trig functions, we can use either the power series or the complex differential equation [math]f’=f[/math] with [math]f(0)=1[/math]. And either way we get [math]\frac{d}{dt}e^{it}=ie^{it}[/math].
Now multiplication by [math]i[/math] just rotates the complex plane by a right angle, so the curve in the plane given parametrically by [math](x(t),y(t))[/math] with [math]x(t)+iy(t)=e^{it}[/math] has a tangential velocity vector which is always perpendicular to its position vector and equal in magnitude.
Since it starts at [math]t=0[/math] at [math](x,y)=(1,0)[/math] the curve is just the unit circle centred at the origin.
And since its velocity vector is always of length 1, if we think of the parameter [math]t[/math] as representing time, then the point moves with speed 1 and so the time taken to complete a circuit, ie the period of [math]e^{it}[/math], is just the same as the circumference of the unit circle (commonly denoted by [math]2\pi[/math] ).