Yet, another math question?

If someone has 2 handicappers on the same same game and both cappers have a long history of winning 55% and each capper picks opposite teams, is it best not to place a bet?

Thanks for any input too!!

What criteria would you use to choose? There
isn't any.

If that's your only information, it's no information at all; it's 50/50. Your expected loss is half the commission, if any.

Bet on the side whose line has moved to your favor.

When there are separate (and possibly multiple or no) probabilities given for each of a set of multiple outcomes, the probability of any particular outcome happening is in proportion to the product of (the product of the accuracy probabilities of those who predicted the correct event) and (the product of the inaccuracy probabilities - i.e. 1 minus each accuracy probability - of those who predicted the incorrect event. For those outcomes that nobody predicted, assume "the product of the accuracy probabilities" is 1.

In this case, if handicapper A picks team X and handicapper B picks team Y:
P'(X winning) = P(A is correct) x P(B is incorrect) = 0.55 x 0.45 = 99/400
P'(Y winning) = P(B is correct) x P(A is incorrect) = 0.55 x 0.45 = 99/400
Since they are the same, each team has a 0.5 probability of winning based on those two handicappers.

Now, if there is a third handicapper, C, with an accuracy of 60%, who picks team X:
P'(X winning) = P(A is correct) x P(C is correct) x P(B is incorrect) = 0.55 x 0.6 x 0.45 = 297/2000
P'(Y winning) = P(B is correct) x P(A is incorrect) x P(C is incorrect) = 0.55 x 0.45 x 0.4 = 198/2000
The ratio of P'(X) to P'(Y) is 3/2, so the probability of X winning is 3/(3+2) = 3/5, and the probability of Y winning is 2/(3+2) = 2/5.

Gee I hope you are not paying for their picks !