Some rather obsessive nerdiness follows:

Assuming you see the river, the chance of hitting quads is actually higher than 1/4165. If you don't have a pocket-pair, then you can hit quads with either one of your two different ranks by seeing a five-card board with the three remaining cards of the given rank and two more from the 47 remaining cards in the deck, so the probability is

2*C(47,2)/C(50,5) ~ 0.00102, or one in 980.

If you have a pocket-pair, then you have to see the two remaining cards of that rank and three more from the 48 remaining cards in the deck, so the probability is

C(48,3)/C(50,5) ~ 0.00816, or one in 122.5.

One in seventeen hands is a pocket-pair, so the overall probability of hitting quads on a given hand (call it 'q') is

q = 16/17 x 1/980 + 1/17 x 1/122.5 ~ .001441, or about one in 694.17.

In a session of 3,000 hands, the probability of never hitting quads at all is

(1-q)^3000 ~ (1 - 1/694.17)^3000 ~ 0.0132, or about one in 75.6, which is quite low. (Although note we're still assuming you see the river every hand.)

In any two consecutive hands, the probability that you hit quads on both hands is

q^2 ~ (1/694.17)^2 ~ 0.00000208, or about one in 482,000.

To calculate the probability that in a session of 3,000 hands you never hit quads on two hands in a row, first work out the probability that you don't hit quads in both the 1st and 2nd, or both 3rd and 4th, or ... or both 2999th and 3000th hands (call this "the first condition"). This doesn't mean you couldn't hit quads twice in a row on the 2nd and 3rd, or 4th and 5th, or ... or 2998th and 2999th hands, so now work out the probability that you also don't hit quads in any of these ways, given the first condition.

The probability of the first condition is

(1-q^2)^1500. So now we know there are no "pairs of quads" in the 1st and 2nd, or 3rd and 4th, or ... 2999th and 3000th hands.

What's the probability you don't hit quads in both the 2nd and 3rd hands? We already know that the 1st and 2nd hands are not both quads, so the conditional probability that the 2nd hand is quads (call it q') is, by Bayes' theorem,

q' = (1-q) x q / (1-q^2) ~ 0.001438, or about one in 695.17. (This is slightly lower than q, which makes sense.)

We also know that the 3rd and 4th hands are not both quads, so by the same calcuation the conditional probability that the 3rd hand is quads is also q'. Therefore the probability you don't hit quads in both the 2nd and 3rd hands is 1-q'^2, which means the probability you don't hit quads twice in a row on the 2nd and 3rd, or 4th and 5th, or ... or 2998th and 2999th hands is (1-q'^2)^1499.

Putting this all together, the probability of never hitting quads two hands in a row, in a session of 3,000 hands, is

(1-q^2)^1500 x (1-q'^2)^1499 ~ .9938; or in other words, the probability of

hitting quads on two consecutive hands at some point is roughly 0.00620, or about one in 161.

Extending these calculations for the probability of hitting quads on three consecutive hands at some point in a session of 3,000 hands, it turns out to be

1 - (1-q^3)^1000 x (1-(1-q)(1-q^2)q^2/(1-q^3)^2)^1000 x (1-(1-q^3)q^3/(1+q+q^2-2q^3-q^4))^1000 ~ 0.002074, or about one in 482.

This formula doesn't obviously extend to runs of four consecutive quads, and there's almost certainly an existing solution for these kinds of problems that I don't know about.

It's possible I've made an error here, or used a hidden mistaken assumption, but the formula indicates that in one million hands that go to the river, a given player should have about a 49.98% chance of being dealt quads in three consecutive hands at least once.

I think this illustrates the observed fact that even unlikely events are bound to happen sooner or later.

(Sheesh! That took a lot of effort that could have been used to play more poker.)