First of all, observe that every possible outcome for the first spinner claims (if possible), exactly one outcome for the second spinner, in order to obtain 52 as the sum:
[tex]\begin{array}{c|c}\text{1st spinner}&\text{2nd spinner}\\1&\text{impossible}\\2&50\\3&49\\4&48\\\vdots&\vdots\\49&3\\50&2\end{array}[/tex]
As you can see, if you spin a 1, you can't get a sum of 52, because the second spin will give at most 50, and you'll have a sum of 52.
All the other first spins "require" the second spin to output a precise value (namely, 52-x, where x is the first spin result).
This table has 50-2+1=49 rows, which are the 49 possible combinations that give 52 as sum.
The total number of possible outcomes is 50^2=2500, so the probability of getting 52 is 49/2500.
