Math exercise - What is my probability of winning?

Hmmm... I still don't know if you're agreeing with my answer or disputing it.  I think you're disputing it but it's not clear what you believe the answer is.

.39 x .45 does equal .1755, but this equation is irrelevant to the stated problem.  It's obvious I had a much better chance than 17.5% of winning.  And that number subtracted from 1 (.8245) was also not my chance of winning. 

.39 x .45 is irrelevant to the problem.

Yes.  Exactly.

Thank you.

There's some bad math in this thread:

He needs one of two, not two of two, so assuming the probabilities are:

Rams win 39% and the total hits 45%, the probability of hitting at least one of two is:

1- the probability of two losers or 1-(.61*.55) = .66

Not so hard

Yes - typo in my post - should say 66 not 64.

not sure why you're not getting this. this method agrees with your results. OR-probability, the chance of A or B occurring, is calculated by:

Ah.  Your post initially confused me.  My mistake.  Thanks.