Probability of getting the exact same letters in Scrabble 2 turns in a row? I would guess that the odds are astronomical that when you play 7 letters (as I did with LAUHTER added to the G on the board to play the bingo: LAUGHTER) and then get 7 new letters that are exactly the same as you just played. Well, it just happened to me.

 A: The first thing you need to know is that when you ask a statistician a question, the probability that the first two words they say will be
#"It depends"
is $P>0.95$. This question is of particular interest to me, because in my family, our idea of a good time over the holidays is to dust off the Scrabble board and argue about what words are acceptable for several hours at a stretch. Thanksgiving is next week, so arguing about the underlying mathematics is sure to spice up this time-honored tradition!
First, let's consider the probability of drawing a hand that has a specific composition.
A typical way to approach this is using a classical probability model. Classical probability typically considers random (that is, uniform probability of selection for all units) selection of balls from urns, or like conceptual frameworks, although in my entire life I've never actually selected a ball from an urn. In this case, we have tiles in a bag, but that's immaterial to the mathematics.
First we have to define the sample space, which in this example is the contents of the bag. Scrabble tiles have a known composition. I assume we're using a standard English-language Scrabble set.
Each hand is generated as a random selection of the remaining tiles left in the bag. For the first hand drawn, this is a simple counting exercise using the fundamental theorem of counting: define the string that you're interested in, and count how many times it can occur as a fraction of the number of possible permutations.
Computations for subsequent hands must account for the fact that some tiles have been removed and are therefore no longer available to be drawn. This brings us to the first complexity:
How do we consider the contents of the opponent's hand?
Option 1: Perfect Information. If we know (say, because we are playing an exposition match, or we spied their tile holder while refilling our brandy snifter) the contents of opposing hands, then we have perfect information. With some bookkeeping, we just need to apply the fundamental theorem of counting again.
Option 2: Probabalistic Inference. This option is more complicated, but tractable. In the case where we don't know the content's of the opponent's hand exactly, but (under some assumptions) we can assign a probability to each possible hand. As an example: suppose you draw your hand, then your opponent draws his hand. If you drew Z, the probability that your opponent has Z is $0$, as there is only one Z tile. But if you didn't draw Z, there is some small probability your opponent will draw it.
Rack management is an important component of higher-level Scrabble playing, though, so against a skilled component, s/he will tend to try and keep a balanced load of consonants and vowels. In any case, their playing of tiles won't be strictly random, since, for example, ZQFMGB is not a word. All this is to point out that there may be additional sources of non-random behavior in the allocation of tiles to the board versus one's hand, so depending on how you want to approach the problem, there are some decisions to be made.

What event are we interested in?
Consider these two alternative statement of this problem.
Question 1:

I play a cromulent word using any number of tiles. What's the probability that I draw those same tiles when I replenish my hand?

Contrast this question to
Question 2:

Given this particular game state (tiles already played and tiles in my opponent's hand), what's the probability that when I play LAUGHTER, I will draw LAUGHTER when I replenish my hand?"

It should be apparent that these questions have radically different answers. The second question is answerable, if tedious, but the first question may be under-specified. For example, if the bag is empty (or has fewer than the required number of tiles), the probability must be 0.
A canonical way to think about the difference is the "Texas sharp-shooter." A Texan claims to be a crack shot with a rifle, and as proof, shows you the side of his barn, where there's a bulls-eye and a bullet-hole squarely in the center of the target. Under the assumption that the bulls-eye was there before he fired, this provides some evidence that he's handy with a rifle. But what would you think of his skills if I told you he shot the barn, then painted the bulls-eye afterwards?
The Texas sharp-shooter example is directly analogous to what has happened in your particular game: you're probably rather accomplished in Scrabble, or else you wouldn't have seen the bingo play, so you've played a large number of games. In this game, something happened that made you think "Wow! That's unusual!" And yes, it is unusual, but to quantify how improbable it is, we need to carefully define what we consider "unusual."
So the answer to this question hinges critically on what question, precisely, you're interested in answering.
(Nice job on LAUGHTER, by the way!)
A: Sycorax is correct that the question is underspecified. One way to interpret it is: given that all the letters showing on the board have been removed from the bag, what are the chances of drawing LAUHTER (in any order)? The remaining tiles are:
" " 2  "E" 3  "A" 1  "I" 6  "O" 4  "N" 2  "R" 1  "L" 2  "S" 2 
"U" 3  "D" 2  "G" 1  "B" 1  "C" 1  "M" 1  "V" 2  "W" 2  "Y" 1
"K" 1  "J" 1  "X" 1  "Q" 1  "Z" 1

That's 42 tiles. Not accounting for identical tiles, there are 42 choose 7 = 26,978,328 different racks of 7 you could draw. To get the number of these combinations that equal LAUHTER, use the multiplication principle and simply multiply all the counts of the corresponding tiles: $2 \cdot 1 \cdot 3 \cdot 0 \cdot 0 \cdot 3 \cdot 1$. You will now notice that since there are no Hs or Ts left, it actually isn't possible to draw LAUHTER (so this turn of events had to be a bug a priori). To keep things interesting, let's ask about drawing both of the blank tiles in place of H and T. The product is then $2 \cdot 1 \cdot 3 \cdot 1 \cdot 3 \cdot 1 = 18$. Dividing by 26,978,328 yields 1 over 1,498,796.
