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I'm sure most people are familiar with word grid games like Boggle and the newer digital versions Scramble with Friends and Ruzzle.

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For anyone not familiar, the idea is to find words by using adjacent tiles. You start from any cell and try to spell a word by dragging up, down, left, right, or diagonal. The board doesn't wrap, and you can't reuse letters you've already selected.

I'm trying to figure out the likelihood of a word appearing on a board given the likelihood of knowing how often individual letters appear. For example, if I know that the letter A appears 9.8% of the time, what is the probability of seeing the word AA?

I know this is fairly simple, but it's been too long since college stats class. (I do have two hokey models, but I'd like to hear from the experts.) I could run a simulation of a million boards and come up with an empirical answer--in fact, someone has--but I'd rather understand why that is. In order to make things simpler, I'd like to ignore two rules that add to the complexity:

  • We can ignore the constraints of how many times a letter can appear on a board. e.g., don't worry about whether there will be enough E's to make the word ELECTEE.
  • We can ignore the fact that letters have to be adjacent. This means we don't have to figure out the likelihood that a letter is a neighbor of another letter that we need for the word

So with that being said, and with the following probabilities

  • A: 0.098
  • E: 0.146
  • R: 0.079
  • S: 0.102
  • T: 0.098
  • C: 0.021

What is the likelihood of a random board containing the word SEE? What about SET? TEAR? What about the rarer CREATES?

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  • $\begingroup$ Is the board 4 X 4? Don't we have to know the probabilities of the other letters as well? $\endgroup$ – Geoffrey Critzer Jun 6 '14 at 0:09
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I have the probability that a 16 letter board contains at least one S and at least two E's given the frequency probabilities above is: .565135

nn = 16; nn! Coefficient[ Series[(Exp[.102 x] - 1) (Exp[.146 x] - 1 - .146 x) Exp[.752 x], {x, 0, nn}], x^nn]

I ran this code in Mathematica which I believe simulates 1000000 boards and returns the number of "good" boards.

Length[Select[ RandomChoice[{.102, .146, .752} -> {s, e, o}, {1000000, 16}], MemberQ[#, s] && Count[#, e] > 1 &]]

The agreement is very good.

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