Just for the record ..... I dont get a chance to use my math degree too much anymore ...but in this instance the better play is to parlay the two instead of betting both individually (at those prices).
Assuming both favorites are -220, then they each have a 68.75% chance of winning, 31.25% chance of losing.
Using those probabilities, your expected return in a 2-team parlay is $0.22 on every $100 wagered.
Your expected return if you bet both individually is $0.04 (ignoring minimal round-off error).
For the parlay: you have a .4727
(.6875 x.6875) chance of winning $112. You have a .4296 chance of losing $100 (2x.6875x.3125) and you have a .0976 chance of losing $100 (.3125x.3125). Therefore,
your expected return is:
$112(.4727)-$100(.4296)-$100(.0976) = $0.22
For the individual bets, your return is computed as:
$200(.4727)-$120(.4296)-$440(.0976) = $0.04
So, mathematically, the smarter bet would be the 2-teamer. Positive expected value return either way though. Dont see that too often in gambling.
Assuming both favorites are -220, then they each have a 68.75% chance of winning, 31.25% chance of losing.
Using those probabilities, your expected return in a 2-team parlay is $0.22 on every $100 wagered.
Your expected return if you bet both individually is $0.04 (ignoring minimal round-off error).
For the parlay: you have a .4727
(.6875 x.6875) chance of winning $112. You have a .4296 chance of losing $100 (2x.6875x.3125) and you have a .0976 chance of losing $100 (.3125x.3125). Therefore,
your expected return is:
$112(.4727)-$100(.4296)-$100(.0976) = $0.22
For the individual bets, your return is computed as:
$200(.4727)-$120(.4296)-$440(.0976) = $0.04
So, mathematically, the smarter bet would be the 2-teamer. Positive expected value return either way though. Dont see that too often in gambling.
