No. That is not correct.
Consider the case where you believe the "true" total to be 8 runs:
First (based on some combination of either the historical record or a run distribution model) estimate how likely you believe the game is to end with a total of each of 8, 9, and 10 runs. Let's say you come up with values of 7.3%, 10.2%, and 6.6%, respectively.
If you believe the "true" total to be 8 runs, then this would imply that that the actual total will be <8 with probability (1-7.3%)/2 = 46.35%, will be exactly 8 with probability 7.3%, will be exactly 9 with probability 10.2%, and will be exactly 10 with probability 6.6%.
This means you'll win the under 10 with probability 46.35%+7.3%+10.2% = 63.85%, will push with probability 6.6%, and will lose with probability 1-63.85%-6.6% = 29.55%.
Hence, your win probability conditioned on not pushing is 63.85%/(1-6.6%) = 68.36%.
So:
- at a money line of -110 your edge would be (1+100/110) * 68.36% - 1 = 30.51%
- at a money line of -120 your edge would be (1+100/105) * 68.36% - 1 = 25.33%
- at a money line of -216.06 your edge would be (1+100/216.06) * 68.36% - 1 = 0.00%
