Re: No fancy name (KK with an Ace on board)
 

I understand that you never took a stats class, so it's ok. But counting the times that they both have you beat is important so that you don't ignore some events. It's the same as the reason that when you are figuring out how often you hit a flush, you have to be more careful than saying "I hit on the turn x% of the time and I hit on the river y% of the time, so I must hit my flush x+y% of the time".

Re: No fancy name (KK with an Ace on board)
 

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So if MP alone has you beat 1/4 of the time and CO alone has you beat 1/2 of the time and they each have you beat maybe 1/8th of the time, then you are beat approximately:
6/24 + 12/24 + 3/24 = 21/24th of the time. I think I may be overestimating how often CO alone has you beat, but not by much. So you are winning 3/24th or 1/8th of the time. Getting 8:1 pot odds, you have to be winning 1/9th of the time to make the call. So, you should call but there's not a lot of value here.

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Whether or not both Villains have us beat is irrelevant unless there's a consolation prize for second place.

I never took a statistics class or anything, but if Villain 1 has us beat 1/2 of the time and Villain 2 has us beat 1/4 of the time, then absent any other information we can go with 1/2 as the chances we're beat.

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Both wrong, assuming the the 1/2 and 1/4 numbers are correct the way to look at it is: Vil 1 beats us say 4/8, vil 2 beats us 2/8. But half the time vil 2 beats us, we were already beat by vil 1, so it's only 1/8 incremental times, so 4/8 + 1/8= 5/8 of the time we are beat. That's the intuitive approach, the mathematical formula is (P = probability):

P1 + P2 - (P1*P2) = P total

4/8 + 2/8- (4/8*2/8) = 6/8 - 1/8 = 5/8