how many Scrabble racks are there?

I've been playing a lot of Scrabble on Facebook lately. While Googling around for some stuff on Scrabble strategy (I'm not the type to memorize words, mostly because that crosses some sort of invisible line between "fun" and "work", but knowing about things like how to try to keep a good mix of vowels and consonants on my rack makes the game more interesting just because I'm less likely to get stuck with a bad rack, which is frustrating), I found a couple things of interest.

First, this handout from the MIT Scrabble club. (I actually knew the guy who started it. He lived down the hall from me my senior year, when he was a freshman.) On that handout they quote Joel Sherman, who says:

Accept the idea that Scrabble is a math game just as much as it is a word game. All the strategic theory of the game is based on statistical analysis, probabilities, spatial relationships on the board, maximizing the value of small-numbered tiles by playing bingos (using all 7 tiles on your rack to earn the 50-point bonus), and large-numbered# tiles by causing them to interact with the colored premium squares on the board, or with other words on the board. (I.e.: you can score just as much for placing a 4-letter word in a place where it lies parallel to 4 other letters or whole words without hitting any premium squares as you might for the same word hitting a double or even triple word square but forming only one or two words on the play.)

Indeed, the conventional wisdom is that people who are good at math do well at Scrabble, and that having a large vocabulary is not particularly useful. (I suspect a lot of this is because large vocabularies tend to consist of long words, and words much longer than seven letters are quite rare in Scrabble.) The same thing is said to be true of crossword puzzles, although there it's not as obvious why there should be a correlation, and my belief is that it's not as pronounced.)

I also found at Scrabble on the Brain a question: "How many games of Scrabble would you have to play before you see every possible rack?"

How to answer this question is not obvious, because the racks one gets in a game aren't independent. I'll answer the easier question of how many games of Scrabble you'd have to play before you see every possible rack on the opening play. This is an instance of the "coupon collector's problem", which says: if we pick a sequence of elements from {1, 2, ..., n} at random, with replacement, how long will it take until we've picked each of the n integers at least once? The standard version of the problem has one picking uniformly at random, so this models, for example, the number of rolls of a standard six-sided die that would be necessary to have each number come up at least once. The answer in this case is n(1 + 1/2 + 1/3 + ... + 1/n), which is asymptotically equal to n(log n + γ), where γ = 0.5772156649... is the Euler-Mascheroni constant. I suspect that in the case where the distribution is non-uniform, as it is for Scrabble racks, it takes longer to observe all possiblilities;

and so the important thing is to compute the number of possible Scrabble racks. (If this has been done before, I can't find it!)

If Scrabble tiles could be differentiated (so instead of having nine tiles labeled A, we had tiles labeled A₁ through A₉), then the answer would just be the number of ways of picking seven tiles from 100, which is C(100,7) = 16,007,560,800. But they're not. I can't see a nice way to compute the actual number (although that doesn't mean it's not there, and I'd appreciate knowing it!) But the following sampling procedure should yield an estimate. Let X be a random variable defined as follows: pick a rack R of seven Scrabble tiles uniformly at random from all 7-sets of (distinguished) Scrabble tiles, and let X(R) be the number of different 7-sets of distinguished Scrabble tiles that collapse to the same set of non-distinguished tiles. So, for example, if we picked the tiles EXAMPLE, there are 12 E's, 1 X, 9 A's, 2 M's, 2 P's, and 4 L's in the tile set, so the number of possible sets of distinguished tiles reading EXAMPLE is C(12,2) C(1,1) C(9,1) C(2,1) C(2,1) C(4,1) = 9504. Then the number of possible racks of undistinguished tiles is the sum of 1/X(R) over all possible racks of distinguished tiles, or C(100,7) times the expectation of 1/X(R).

(Compare if we were trying to find the number of possible one-letter racks, that is, letters; we'd have, say, 9 terms which are each 1/9 corresponding to the labeled tiles A₁ through A₉, 2 terms each 1/2 corresponding to B₁ and B₂, and so on. The terms corresponding to each tile type sum to 1, so the sum is just the number of tile types, or 27 including the blank.)

And the expectation of 1/X(R) can be found by sampling. I generated a million racks at random and evaluated 1/X(R) (with some hacked-together Maple code); the expectation of 1/X(R) calculated from this sample is 0.0002014090957, and I approximate the number of possible Scrabble racks to be this quantity times C(10,7), or 3224068. (I'm not interested enough to try to calculate the standard error on this -- and I hope I can get an exact answer.) The number of games of Scrabble you'd have to play before you see every possible rack on the opening play is probably at least 3224068 log 3224068, or around fifty million.