A recent commenter asked whether the following is true: "ETAERIO is the most likely seven-letter word to get in scrabble."
As all scrabble players know, if you use all 7 of your letters, you get a bonus. However, except for the first turn, you'd need an 8-letter word to achieve this--ETAERIO would not do. Thus, I am going to try to answer the question: "What are the chances of getting the letters ETAERIO on your initial turn in scrabble (you also have to hope you are first)?" I have no hope of finding out whether this is the most likely seven letter word, because I can't automatically check all letter combinations, but I will try to give some guidance there as well.
To find the chances of gettting ETAERIO, we need the number of combinations that produce these letters divided by the total number of combinations. In other words, we have to go back to 12-th grade math, where we all learned (or sort-of learned) permutations and combinations.
There are 100 tiles in scrabble and we are choosing 7. Thus, there are 100 ways to to choose the first tile, 99 ways to choose the second, and so forth down to 94. If we chose them in order, we'd have 100*99*98*97*96*95*94 permutations. However, we don't care about the order, so we have to take the above product and divide by the number of ways we can permute the 7 tiles, which is 7*6*5*4*3*2*1. The shorthand way to express this number of combinations is "100 choose 7" or
By dividing the two ratios above ((100*99*98*97*96*95*94) / (7*6*5*4*3*2*1)), we come up with 16,007,560,800. Since most letters appear multiple times, the number of possible letter combinations is far less, and to know the chances of getting ETAERIO, I need to know how many times each letter appears.
Thus, I found the letter distributions on Wikipedia (counting our own scrabble pieces would probably not do with a three-year old distributing them around the house). The most common ones as follows:
E - 12 tiles
A, I - 9 tiles
O - 8 tiles
N, R, T - 6 tiles
D, L, S, and U - 4 tiles
other letters - 3 or less, but not relevant here
To figure out the chances of getting ETAERIO, we need to know the number of combinations that produce it. We need 2 E's, 1 T, 1 A, 1 R, 1 I, and 1 O. It turns out that the number of ways is the product of each of these implied combinations. Thus, it is "12 choose 2" (E's) times "6 choose 1" (T) times "9 choose 1" (A) and so forth. This comes out to 1,539,648 ways to get the letters in ETAERIO. If we divide this by the total number of combinations (16,007,560,800), we find that there is about a 1 in 10,000 chance of getting ETAERIO as your first 7 letters. Of course, from there, you have to know it is a word and figure out that you can make that word from those letters, since they are not likely to appear in that order.
I could not find another word with a higher probability, but I did find TREASON and TRAINED (both about 1 in 20,000). However, It's clear from the distribution of letter tiles that in order to find a word that beats ETAERIO, you can only use letters appearing in 6 or more tiles.
Now that we all remember the mechanics of combinations (or at least, we are on the subject of them), let's investigate another oft-asked question around here: what's the chance of being dealt a 7 card suit in bridge? This would be 4 (number of suits) times "13 choose 7" (ways to choose 7 from a suit) times "39 choose 6" (ways to choose the other 6 from the other 3 suits) divided by "52 choose 13" (ways to choose 13 cards from 52). This comes out to about 3.5%, or 3 or 4 times in every 100 hands.
For an 8-card suit, it is 1 in about 200. For a 9-card suit, it is about 1 in 2,700. Of course, my kids are always asking about the chances of being dealt a 10 card suit or even a 13-card suit:
10-card suit: 1 in 60,738
11-card suit: 1 in 2,746,693 (less than 1 in million)
12-card suit: 1 in 313,123,057 (less than 1 in 300 million)
13-card suit: 1 in 158,753,389,900 (less than 1 in 150 billion)
The chances aren't too great, but with some really poor shuffling, they've managed the 13-card suit once or twice.