Dont count cases please!

We know that if we roll two dice then no.of ways of getting 2 as sum is 1way and 3 as sum is 2ways so on...And finally we get no.of ways of getting sum as 2,3,4,...,12 as 1,2,3,4,5,6,5,4,3,2,1. So use the same technique here to solve this kind of problems I.e. as follows, (a+c) to be 2 no.of ways is 1 and to be 3 no.of ways is 2.And for 4 it is 3 and so on... We get 1,2,3,4,5,6,7,8,9,10,9,8,7,6,5,4,3,2,1 as no.of ways of getting sum from 2 to 20 So now no.of ways of getting same sum by picking 2 no'.s of 2 groups from [1,10] is 1²+2²+3²+...+9²+10²+9²+8²+...+1² which is equal to 670 . So , therefore probability is 670/100² =0.067

We have the probabilities $\mathbb{P}[a+b=j] \; = \; \frac{10 - |j-11|}{100} \hspace{2cm} 2 \le j \le 20$ and so $\begin{aligned} \mathbb{P}[a+b=c+d] & = \sum_{j=2}^{20} \mathbb{P}[a+b=j]^2 \; = \; \frac{1}{10^4}\sum_{j=2}^{20}\big(100 - 20|j-11| + (j-11)^2\big) \\ & = \frac{1}{10^4}\Big[ 1900 - 40\sum_{j=1}^9 j + 2\sum_{j=1}^9 j^2\Big] \; = \; \frac{1}{10^4}\big[1900 - 40\times45 +2\times285\big] \\ & = \frac{67}{1000} \end{aligned}$ making the answer $\boxed{0.067}$

An alternative proof is to note that the required probability is the coefficient of $x^0$ in the expression $\begin{aligned} \left(\frac{x + x^2 + \cdots x^{10}}{10}\right)^2 \times \left(\frac{x^{-1} + x^{-2} + \cdots x^{-10}}{10}\right)^2 & = \frac{x^{-18}(1 - x^{10})^4}{10^4 (1-x)^4} \\ & = \frac{1}{10^4x^{18}}\big(1 - 4x^{10} + \cdots\big) \times \sum_{n=0}^\infty \binom{n+3}{n} x^n \end{aligned}$ This is because the first term in the product is the probability generating function of the sum $a+b$ , and hence the whole product is the generating function of $a+b-c-d$ . Thus the desired probability is equal to $\frac{1}{10^4}\left\{ \binom{21}{18} - 4\binom{11}{8}\right\} \; = \; \frac{670}{10^4} \; = \; 0.067$

I am assuming that a,b,c,d need not be distinct. First we find total number of favourable choices of a,b,c,d.Since, a,b,c,d belong to {1,2,....,10}, the sum of any two of a,b,c,d will be {2,3,.....,20}.So if we have the equation x+y=r (r=2,3,...10) where x,y belong to {1,2,.....10} ; then the total number of ordered pairs (x,y) will be (r -1).But for r=11,12,...20 if we have the equation x+y=r where x,y belong to {1,2,....10} then we cannot have (r -1) ordered pairs of (x,y) as this includes those ordered pairs also where any one out of x and y is greater than 10.So,we use the substitution x=(10 - x') and similarly for y, that is y= (10 -y') ; hence x' and y' belong to {0,1,...,9} and we have x'+y'=(20-r).Since, (20-r)=0,1,...9 , now we will not have any unfavourable ordered pair.So,for r=11,12,...20 total number of ordered pairs of (x,y) will be (21-r).Hence,if r=2,3,....10 ; then total number of ways of selecting ordered pairs (a,b) and (c,d) is (r -1)² as r goes from 2 to 10. So total number of ways of selecting ordered pairs (a,b) and (c,d) is 1²+2²+...9²=285.And, if r=11,12,....20 then total number of ways of selecting ordered pairs (a,b) and (c,d) is (21-r)² as r goes from 11 to 20.So, total number of ways of selecting ordered pairs (a,b) and (c,d) is 1²+2²+3²+...+10²=385. Hence, as r goes 2 to 20 total number of ways of selecting ordered pairs (a,b) and (c,d) is 285+385=670. Hence, total number of ways of selecting a,b,c,d from {1,2,...,10} such that a+b=c+d is 670.Also, total number of ways of choosing a,b,c,d from {1,2,...10} is 10⁴=10000, as each of a,b,c,d have 10 choices. Hence, the required probability is favourable outcomes divided by total number of outcomes=670/10000=0.067.