Method Man said:
What kind of odds you giving on that?
We have a 1/4, 1/3 and 2/3 chance to win our remaining games per the (poor) ESPN game predictor. That's gives us a crude version of expected value at 1.25 wins.
Let's say, for conversation sake, we give us the LSU game - the probability of stealing one of the first two games with those odds is… about 42%
Probability of winning the first game and losing the second: Probability of winning first game (1/4) multiplied by the probability of losing the second game (2/3) = (1/4) * (2/3) = 2/12.
Probability of losing the first game and winning the second: Probability of losing the first game (3/4) multiplied by the probability of winning the second game (1/3) = (3/4) * (1/3) = 3/12.
Combined probability of winning one game out of two:
Add the probabilities from both scenarios: (2/12) + (3/12) = 5/12