In each of the sets, the total given is evidently about how many evens and how many odds they are in the interactivity. In Set B, the odds outnumber the evens, and an odd add an even is an odd. However, with an odd and an odd, it is good for winning because it produces even. With a higher denomination of odd numbers, you are more likely to encounter two odds, therefore making it more likely to win.
Set C has lots of even numbers, so is more likely than Set A to score, but Set D has the most odd numbers while maintaining the lowest amount of even numbers
I worked out the many possibilities for Set A:
Please note: W=win L=lose and each combination is in inverse
2+3=5=L
2+4=6=W
2+5=7=L
2+6=8=W
We notice a pattern as we go systematically through them, as we increase the last number we are adding by 1, it goes through a multitude of odd, even, odd, even etc.
3+2=5=L
3+4=7=L
3+5=8=W
3+6=9=L
Here, because 3 is an odd number there are three loses and a win instead of an equal number of each set.
4+2=6=W
4+3=7=L
4+4=8=W
4+5=9=L
An equal number on the sets. It seems like on an even is equal. This makes sense as an odd add an even is an odd and an odd add an even is an odd too. This means it is higher probability of you getting an odd result.
5+2=7=L
5+3=8=W
5+4=9=L
5+6=11=L
Already, we can see the probability is higher to get an odd result than an even one by a long shot.
6+2=8=W
6+3=9=L
6+4=10=W
6+5=11=L
Again, an equal number.
The mean of winning when I played 100 times:
0.380, 0.426, 0.390, 0.400, 0.440
Overall mean: 0.4072
Set B
The mean of winning when I played 100 times:
0.570, 0.500, 0.640, 0.600, 0.560
Overall mean: 0.574
Set C
The mean of winning when I played 100 times:
0.460, 0.470, 0.420, 0.480, 0.440
Overall mean: 0.454
Set D:
The mean of winning when I played 100 times:
0.690, 0.670, 0.590, 0.630, 0.670
Overall mean: 0.650
I would probably play with D to make sure I had the most chance of winning.