Meh.
∂γ∂c=b0xcln(x)∑r=0∞(c+y−1)(c+α)r(c+β)r(c+1)r(c+γ)rxr+
+b0xc∑r=0∞(c+γ−1)(c+α)r(c+β)r(c+1)r(c+γ)r(1c+γ−1+
+∑k=0r−1(1c+α+κ+1c+β+κ+1c+1+κ−1c+γ+κ))xr
=b0xc∑r=0∞(c+γ−1)(c+α)r(c+β)r(c+1)r(c+γ)r(ln x+1c+γ−1+
+∑k=0r−1(1c+α+κ+1c+β+κ−1c+1+κ−1c+γ+κ))xr
.
It's been a while since I looked at how I built my project. And my math may be entirely incorrect below, or correct. I may have it backwards. Honestly, I forget. BUT! Using only binomial combination, without taking into account blank tiles which throws the entire thing out of whack. The simple combination solution without wild.
I asked these questions myself, and built my own scrabble words probability dictionary because of it. You don't need a dictionary of possible words pulled out, only the math behind it and available letters based on letters in tile bag. The array of English rules is below. I spent weeks developing the math just to answer this question for all English words that can be used in a game, including words that can not be used in a game. It may all be incorrect.
The probability of drawing a given word from a bag of letters in Scrabble, requires how many letters are available in the bag, for each letter ( A-Z ) and, whether we're using the wild card as an addition to the math. The blank tiles are included in this math - assuming 100 tiles, 2 of which are blank. Also, how many tiles are available differs based on language of the game, and game rules from around the world. English scrabble differs from Arabic scrabble, obviously. Just alter the available letters, and the math should do the work.
If anyone finds errors, I will be sure to update and resolve them.
Boot: The probability of Boot in a game of scrabble is 0.000386% which is a chance of 67 out of 173,758 hands as shown on the word page for boot.
English Tiles
all is the array of letters in the bag. count is the array of available tiles for that letter, and point is the point value of the letter.
// All arranged by letter, number of letters in scrabble game, and point for the letter.
$all = array("a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z");
$count = array("9", "2", "2", "4", "12", "2", "3", "2", "9", "1", "1", "4", "2", "6", "8", "2", "1", "6", "4", "6", "4", "2", "2", "1", "2", "1");
$point = array("1", "3", "3", "2", "1", "4", "2", "4", "1", "8", "5", "1", "3", "1", "1", "3", "10", "1", "1", "1", "1", "4", "4", "8", "4", "10");
There are 100 tiles in an English scrabble game (i.e., the sum of $count). It does not matter how the tiles are pulled, so it's not a permutation.
The Math I Used
Determine how many letters are in the word and what letters are in the word, how many of those letters are available in the tile bag ( count for each letter, unique and allchars ). Binomial coefficient of each, divided by binomial coefficient of length word.
Determine the binomial combinations available
let C(n,r) be binomial coefficient: n!/[n!(n-r)!], or 0 if r > n
Foreach letter, what is the binomial coefficient.
There is 1 "B". There are 2 available, a 2% chance of pulling the b.
There is 2 "O". There are 8 available, a 8% chance of pulling the o.
There is 1 "T". There are 6 available, a 6% chance of pulling the t.
BOOT is a 4 letter word, being taken from a 100 tile set with blanks, 98 without.
n = 98. The number of tiles without blank in the English set
B=(21)=2!2!(2−1)!
O=(82)=8!8!(8−2)!
T=(61)=6!6!(6−1)!
B×O×T
divided by the binomial coefficient of tilecount
98!98!(98−length)!