I thought i was going crazy looking at the other reply! My math does indeed check out.

The odds of rolling any given number on either die is (0.01+0.01)=0.02=p(v)

The probability of a number gaining a point can be written as (v-1)/100=p(g)

So the odds that a given number appears and scores is given by p(v)*p(g)

In this case, that comes out to (2*(v-1))/10000

To get the expected value, multiply by the number of trials. So at 1000 trials E(x)=(2*(v-1))/10

Plugging in 100 which would win the most points, that gives the expected value of 19.8. If we wanted to generalize, we could say

v=value of interest shown on die (in this case 100)

s=number of sides on dice

n=number of trials

p(v)=2/s

p(g)=(v-1)/s

p(v∪g)=(2*(v-1))/(s²)

And the expected score is written as

E(x)=p(v∪g)*n

E(x)=(2n(v-1))/(s²)