Letting the deck cycle through helps. Think of every jack as having a 4-card neighborhood. If the jacks are sufficiently far enough apart so that none of the neighborhoods overlap, then the maximum number of cards that can belong to a jack neighborhood is 16. That means the minimum number of cards that do not belong to a jack neighborhood is 32. Let's call these cards the "outskirts."
The probability that none of the 6s are in a jack neighborhood is equal to the probability that all of them are in the outskirts. When there are 32 cards in the outskirts, this is
(32 choose 4)/(48 choose 4) = 1798/9729 ≈ 18.48%
And so the probability that at least one 6 is in a jack neighborhood is about 81.52%. The reason why the denominator is not (52 choose 4) is because the four jacks are like anchors, neither in a neighborhood or the outskirts. A 6 cannot occupy the same position as a jack.
Now, there are cases when the jacks are too close together and some neighborhoods overlap, which makes the neighborhood size less than 16, which makes the probability of success lower. In the most extreme case, all four jacks are next to each other and the total neighborhood is only four cards. It's a lot of work to analyze all these sub cases, so think of 81.52% as an upper bound. I can write a script later to simulate a couple million random shuffles and give you a better estimate of the true probability, but I hope this has been of help.
Thanks for the food for thought. May the free booze keep flowing.