2 other players (I'm assuming the three players left includes you, if only to avoid having to extend the following working even further), they hold 4 'mystery' cards.
There are four 'gutshot' cards to choose from remaining deck of 46 (52, less 4 board cards, less your two cards)
Let's call the cards in villain's hands A,B,C, and D, and assume we're working on a completely 'random hand' basis.
P(A is a 6) = 4/46 = 0.087
P(A is not a 6) = 1-0.087 = 0.913
| = "given that", for those who aren't acquainted with stat notation.
P(B is a 6 and A is not a 6) = 0.913*(4/45) = 0.913*0.089 = 0.081
P(B is a 6 | A is not a 6) = P(B is a 6 and A is not a 6) / P(A is not a 6)
P(B is a 6 | A is not a 6) = 0.081 / 0.913
P(B is a 6 | A is not a 6) = 0.089
P(A and B are not 6s) = 0.913*(1-0.089) = 0.831
P(C is a 6 and A and B are not 6s) = 0.831*(4/44) = 0.831*0.091 = 0.076
P(C is a 6 | A and B are not 6s) = P(C is a 6 and A and B are not 6s) / P (A and B are not 6s)
P(C is a 6 | A and B are not 6s) = 0.076 / 0.831
P(C is a 6 | A and B are not 6s) = 0.092
P(A, B, and C are not 6s) = 0.831*(1-0.092) = 0.755
P(D is a 6 and A, B, and C are not 6s) = 0.755*(4/43) = 0.755*0.093 = 0.070
P(D is a 6 | A, B, and C are not 6s) = P(D is a 6 and A, B, and C are not 6s) / P(A, B, and C are not 6s)
P(D is a 6 | A, B, and C are not 6s) = 0.070 / 0.755
P(D is a 6 | A, B, and C are not 6s) = 0.093
P(A, B, C, or D are a 6) = P(A is a 6) + P(B is a 6 | A is not a 6) + P(C is a 6 | A and B are not 6s) + P(D is a 6 | A, B, and C are not 6s)
P(A, B, C, or D are a 6) = 0.087 + 0.089 + 0.092 + 0.093
P(A, B, C, or D are a 6) = 0.361 = 36%
This doesn't account for situations where more than one of A,B,C,or D are a 6, but the probability of that is low and won't affect the outcome much. I've just realised after working all this I could have accounted for it by working Bombjack's way, but I'm too lazy to go through it again. I wouldn't think it would change the result more than a couple of percent though.
I might have gone wrong somewhere (it has been a few years since I did my maths course, after all!), but the result is around about what I expected, so perhaps not.