# Scrabble Racks

Scrabble is a very popular word game, and it has been estimated that over 150 million physical games have been sold worldwide (never mind the proliferation of online variants enjoyed around the globe).
The standard English language version ships with 100 tiles in the following distribution:
A×9, B×2, C×2, D×4, E×12, F×2, G×3, H×2, I×9, J×1, K×1, L×4, M×2, N×6, O×8, P×2, Q×1, R×6, S×4, T×6, U×4, V×2, W×2, X×1, Y×2, Z×1, [BLANK]×2
From this set, each player draws seven tiles, unseen, for their initial rack. The first player uses their letters to make a word which must straddle the center square on the board. This first move is unique because it does not build on any tiles already on the board (all subsequent moves must use at least one tile already on the board).
This begs the question, is it possible to get a really bad starting rack: A rack so poor that it is impossible to form any starting word? Yes, sadly it is, for example the following hellish rack would not butter any parsnips:
The next question to answer is how often this nightmare might occur?

## Let me count the ways …

First of all, we need to calculate all the possible starting racks from the bag of letters. How many different ways can we select seven scrabble tiles from the bag of a hundred letters? Because there are multiple copies for many of the letters, and these letters are indistinguishable from each other, there are a lot of duplicates. An easy way to think about removing duplicates is to think about each rack rearranged into alphabetic order and simply counting the distincts. I actually wrote a few lines of recursive code to generate all the solutions and determined there were 3,199,724 possible starting racks (not necessarily equally likely).
Mathematicians will calculate this answer using generating functions.
There are five letters of frequency one, ten with frequency two, just one with frequency three, four with frequency four, none with frequency five, three with frequency six, and four with frequency seven or more; Even though there are more than seven of most of the vowels, you cannot take more than seven in any rack.
Thankfully the Wolfram Alpha calculation of this confirms my recursive piece of code (Trust but Verify).
Using these 3,199,724 possible starting racks it’s possible to check to see if it is possible to make any words with this collection of letters. To do this, we need a dictionary file. To be legit, I used the SOWPODS dictionary file. We need to be careful to remember that the blank tiles are wildcards and can be used for any letter.
A quick calculation shows that there are 261,487 bogus starting racks.
(Just as I was writing up this article, I checked the rules and, according to the official Scrabble rules, the starting word needs to be at least two letters. My calculation allowed for the possibility of single letter words {I,A,O}, which meant that any rack containing at least one blank tile was never bogus, and also allowed for more possibilities to make many of the official two letter words. However, if there are two blank tiles, it will always be possible to make a two letter word, and of all the two letter words possible, every letter but the letter 'V' is represented, so if there is just one blank tile, it's never possible to have the rest of the rack entirely 'V', so again it will always be possible to make at least a two letter word. As long as you start with a blank tile, you will always be able to move).
This means that 8.17% of all possible starting racks in Scrabble are bogus and cannot make any word.