Mutually exclusive dices? Suppose I have 3 dice. Each has some mechanism that can prevent other dice being in 2 if itself is 2 when they are rolled together. Now I roll the 3 dice at the same time. Then what is the probability of having a 2 appearing?
Is the answer 1/6 + 1/6 + 1/6 = 1/2 since the event are mutually exclusive and each has a probability of 1/6?
But without the mechanism the probability of at least one 2 appearing is 1 - (5/6)^3 = 91/216 < 1/2?
Why should the probability decrease?
 A: If the mechanism only applies to 2, then they are just like ordinary dice until a 2 appears.  Hence the probability of no 2s in three rolls is still $\left(\frac56\right)^3$.  So the probability of at least one 2 in three rolls is $\frac{91}{216}$ just like ordinary dice.
A: It is reasonable to suppose by default that the mysterious mechanism does not apply when a number other than $2$ is tossed. 
The probability of not getting a $2$ is then as usual $\left(\frac{5}{6}\right)^3$, so the probability of getting (at least) one $2$ is the same as usual.
Note that the probability of multiple $2$'s is now $0$, so for example the probability of at least one $6$ is greater than usual. 
Remarks: $1.$ The problem now is to produce a physically reasonable model, to put flesh on the bones of the mysterious mechanism. I have not succeeded in doing so. 
One possibility that is contrary to the wording is that the dice are rolled sequentially. If the first toss is a $2$, the machinery kicks in, and we end up with one $2$, probability $\frac{1}{6}$. If we don't get a $2$, the probability of a $2$ on the next toss is $\frac{1}{6}$, and then the machinery kicks in and we again get a success. And of course we can have success if we get a non-$2$ twice, then a $2$. That gives probability $\frac{1}{6}+\frac{5}{6}\cdot\frac{1}{6}+\frac{5^2}{6^2}\cdot\frac{1}{6}$. That gives probability $\frac{91}{216}$. 
$2.$ Below, verbatim, is my earlier (wrongly analyzed) suggestion.  The error was pointed out by Did. 
We supply an explicit non-magical mechanism that satisfies the OP's condition. Toss three dice. If we get more than one $2$, toss again until we get one or fewer $2$'s. In any round of this game, the probability of getting no $2$'s is unchanged from the usual independent fair dice assumption. 
A: Due to this mechanism, there's only two possibilities for how many twos appear. Either no twos appear or one two appears. Let $A$ be the event that no twos appear. Let $B$ be the event that one two appears. 
So $P(B) = 1-P(A)$. 
Since $P(A) = {125 \over 216}$, 
$P(B) = 1 - {125 \over 216} = {91 \over 216}$.
The trick is to disregard events where more than one two appears since that is impossible in this scenario.
