If it were not possible for there to be ties, the solution required
would be to just iterate over all possible pairs of die values:

for (int dv1 : die1)
   for (int dv2 : die2)
      if (dv1 > dv2)
         win1++
      else
         win2++
return win1 / (win1 + win2)

and then print the ratio of win1 to the sum of win1 and win2.  With
ties, though, we need to do a bit of algebra.

Let's split the 36 pairs of die values into three categories:

   Player 1 wins  win1
   Player 2 wins  win2
   They tie       ties

Let x be the final probability that x wins.

   x = (win1 / 36) + ??

The ?? above is the probability of a win that starts with a tie.
If they tie, then we simply repeat the procedure.  So the ?? in
the above function is just

   (ties / 36) * x

So

   x = (win1 / 36) + (ties / 36) * x

   36 * x = win1 + ties * x

   (36 - ties) * x = win1

   x = win1 / (36 - ties)

   x = win1 / (win1 + win2)

so the exact same expression works.  The revised code, then, to
exclude ties, is simply

for (int dv1 : die1)
   for (int dv2 : die2)
      if (dv1 > dv2)
         win1++
      else if (dv2 > dv1)
         win2++
return win1 / (win1 + win2)