Poker Outcomes and Variance: Calculating Likelihood of an Observed Outcome

1. The probability of observing this difference is, as the calculator says, 0.1029 or 10.29%. I will explain below. 

2. The Winrate follows a normal distribution (we call it $X$). You only need to know what parameters to use. The expected value of the $X$ is exactly the Winrate expected value, here 20 BB/100. The standard deviation (SD) of $X$ needs some calculation. We know that the SD is 500 BB/100, i.e. 500 BB in 100 hands. The SD is proportional to the inverse of the square root of the number of hands. So it follows that the SD of 25000 hand, which is 250 times 100 hands, is $$ \frac{500}{\sqrt{250}}=\frac{500}{15.81}=31.62\; BB/100$$ 
If you use this SD in the normal distribution, you will see that the probability of going below -20 is 10.29%. You can use the following calculator to compute it for example 

https://onlinestatbook.com/lms/calculators/normal_dist.html

In general, if the expected value is $\mu$ BB/100 and SD is $\sigma$ BB/100 and the number of hands is $N$, then the Winrate follows a normal distribution with mean $\mu$ and SD equal to $$\frac{\sigma}{\sqrt{N/100}} $$ There is also a mathematical formula for computing the probability of observing Winrate less than or equal to say $m$, which is $$\int _{-\infty} ^m \frac{\sqrt{N/100}}{\sigma \sqrt{2\pi}} \exp \frac {-(x-\mu)^2N/100}{2\sigma^2}\,dx$$ Although to actually compute this you need a calculator anyway.