How to calculate how often I win a middle

[guthrybill]

Thank you for your advice. Really useful

[Carni]

I'm not sure if more information is required regarding the probability distribution, but if I had to come up with a number I get 5.4% as well.

Total ------ Under - Over
215.5 ------ 47.8% - 52.2%
210.5 ------ 42.4% - 57.6%
Difference -- 5.4% -- 5.4%

Thanks for chiming in Blackswan. I've done it a few different ways and that's the result I get. It makes sense when I draw it out diagrammatically.

[blackswan]

I'm not sure if more information is required regarding the probability distribution, but if I had to come up with a number I get 5.4% as well.

Total ------ Under - Over
215.5 ------ 47.8% - 52.2%
210.5 ------ 42.4% - 57.6%
Difference -- 5.4% -- 5.4%

[Carni]

I've got a model that gives me totals for basketball, and with it I use it to calculate the probabilities of winning overs and unders. Let's ignore odds and say I've got a profitable middle where I'm betting:
Over 210.5 (% chance of being over 210.5 = 57.6%)
Under 215.5 (% chance of being under 215.5 = 47.8%)

(As I said, let's ignore odds, but if the over was paying 1.99, and the under 1.99, this would obviously be a +EV middle with some risk (i.e. no pure arbitrage)).

My question is: given the probabilities I have above, how do I calculate the % chance I win both bets? I feel like intuitively the answer is: (1-(1-P(over)-(1-P(under)) = 5.4% [equivalently (1-P(lose over)-P(lose under))], but I'm not at all sure.

Anyone have better ideas?