It's the least I could do ....

Introduce a fifth random variable $v$ chosen in the same manner as the other four. For fixed $w,x,y,z$ the probability that $v \lt \min(w,x,y,z)$ is just $\min(w,x,y,z).$ So when we average over all choices of $w,x,y,z$ , the expected value of $\min(w,x,y,z)$ is the probability that, once $v,w,x,y,z$ are chosen, we end up with $v \lt \min(w,x,y,z).$ But as the five variables $v,w,x,y,z$ are chosen in the same manner, the probability that any one of them is the least is just $\frac{1}{5}$ , and so $a + b = 1 + 5 = \boxed{6}.$

Let us deal with the particular case when $x<y<z<w$ .The probability and expectation for all other cases will be the same by symmetry.The probability density function is $f(x,y,z,w) = 1$ for $0<x<1\,\, , \,0<y<1\,\, , \,0<z<1,\,\,\,0<w<1$ and $0$ elsewhere. So in this case $\min(x,y,z,w)=x$ .

The expectation will be given by:-

$\displaystyle \int_{0}^{1}\int_{x}^{1}\int_{y}^{1}\int_{z}^{1}x \,dw\,dz\,dy\,dx=\frac{1}{120}$ .

There are $4!=24$ such ordered arrangements of $x,y,z,w$ like $\left\{(x<y<z<w),(x<z<y<w),.....\right\}$ . All of them will have equal probabilities and expectations by Symmetry. Hence by Total Probability we have the expectation as :- $\displaystyle \frac{24}{120} = \frac{1}{5}$ .

Note that this method can be generalized for n-variables. Then the expected value will be just $\displaystyle \frac{n!}{(n+1)!}=\frac{1}{n+1}$

Python:

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from random import random as r
trials = 10000000
total = 0.0
for trial in xrange(trials):
    total += min((r(),r(),r(),r()))
print "Answer:", round(total/trials,2)