[asdfasdf32]

I googled this question and got this response:

"Lets look at a simple 2-team parlay using a real example, the Monday Night Football game from this past weekend. St Louis –10.5 over the Rams and a game total of 43.5. Parlaying a side and the total gives us four combinations:

a) St Louis –10.5 and Over 43.5
b) St Louis –10.5 and Under 43.5
c) Chicago +10.5 and Over 43.5
d) Chicago +10.5 and Under 43.5

The odds of any one of these plays being a winner are 1-in-4 so the actual odds would be 3/1. In actual fact, most books pay 2.6/1 (you see it commonly written as 13/5) assuming all bets are at stand payoffs (-110). "

Okay, seems fair enough, but...from what I've seen on this site 60% is the highest attainable win rate against the spread & o/u over the long haul. This number is probably too high, but let's go with it anyways. If someone is 60% to win against the spread and 60% to win against the over/under, they'll win both on a single game 36% of the time. By using the same math, that makes them 12.96% to win against the spread & the o/u on both games. This makes the odds ~6.7 to 1 instead of the 3 to 1 quoted in the article. Is my logic goofed up?

Straighten me up fellas.