Probability statistics I pick sports for fun with my friends. we pick 5 sports picks everyday. what is the probability of going at least 3/5 for 50 days in a row? How would you set this up

Well you haven't told us much in your question formulation so I am going to assume that the chance of  picking correctly is the same as the chance of picking incorrectly and so all events have equal probability of $0.5$.
Now, to solve this problem, let's think about what is the probability of picking at least 3/5 sports correctly in one day. Define $X=$ the number of sports correctly picked in a day. Then, the probability of picking at least 3 out of the 5 sports correctly is the following: 
\begin{align*}
Pr(X\geq3) &= Pr(X=3) + Pr(X=4) + Pr(X=5)\\
&={5\choose 3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^2+{5\choose 4}\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^1+{5\choose 5}\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^0\\
&=10\left(\frac{1}{2}\right)^5+5\left(\frac{1}{2}\right)^5+1\left(\frac{1}{2}\right)^5\\
&=\left(\frac{1}{2}\right)^5(10+5+1)\\
&=\left(\frac{1}{2}\right)^5(16)\\
&=0.5
\end{align*} 
And so, the probability of selecting 3 out of the 5 sports correctly in one day is 0.5. This shouldn't come as a surprise since we could have picked X={0,1,2,3,4,5}  sports correctly and only half (i.e., 0.5) of that list is at least 3 or greater.  
Now, ultimately, you want to know what is the probability of picking 3 out of the 5 sports correctly for 50 days in a row.  And so, this probability is the following: 
\begin{align*}
Pr(X\geq 3 \text{ for }50 \text{ days in a row})&=Pr(X\geq 3 \text{ on day 1 and day 2 ... and day 50})\\
&=Pr(X\geq 3 \text{ on day 1})\times\cdots\times Pr(X\geq 3 \text{ on day 50})\\
&=\left(\frac{1}{2}\right)^5(16)\times\cdots\times \left(\frac{1}{2}\right)^5(16)\\
&=\left(\left(\frac{1}{2}\right)^5(16)\right)^{50}\\
&=(0.50)^{50}\\
&=8.881784e-16
\end{align*}
So practically a 0% chance of it occurring.
