True probabilities in self-referential statements

On December 8, Éric Angelini asked in an online MathFun forum for a solution to a self-referential sentence in which randomly picking four letters from the statement yielding the four letters F, O, U, and R, was a probability 𝜶/𝜷, where 𝜶 and 𝜷 were the English number names for the numeric quantities that allowed the sentence to be true. After determining that picking one letter depleted the letter availability of the next pick by one, I settled on my template being: "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are 𝜶 out of 𝜷."

Mathematica has a function for converting integers into English words so all I had to do was run through a bunch of them and test the resulting sentences for truthfulness. Here are nine solutions:

1. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are two out of two hundred nineteen thousand six hundred eighty-seven."

2. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are three out of two hundred ninety-two thousand nine hundred sixteen."

3. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are three out of one hundred ninety thousand six hundred fifty."

4. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are six out of seven hundred twenty-one thousand seven hundred ninety-one."

5. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are twelve out of one million seven hundred sixty-six thousand six hundred twenty-two."

6. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are eighteen out of one million six hundred thousand two hundred."

7. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are twenty-one out of two million ninety-seven thousand twenty-four."

8. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are thirty-five out of two million six hundred sixty-seven thousand four hundred twenty."

9. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are seventy-one out of nine million nine hundred twenty thousand two hundred sixty-two."

I believe that this exhausts the possibilities for 𝜶 up to 100 and 𝜷 up to 10000000. But only for this particular template. One may easily alter the template by adding/subtracting/changing words without altering the essential thrust of what the sentence is saying. Each of these new statements would have its own set of solutions.

Of the above nine solutions, the first two are special: The ratio 𝜶/𝜷 is in its lowest terms. In the other seven the two numbers share a common factor, so they can be reduced. Of course we mustn't do that because that would rob the statements of their truthfulness.

If you are interested in verifying the nine statements, this will help:

 1.  {2,219687}  {5,11,6,8}  <132>   5*11*6*8/(132*131*130*129) = 2/219687
 2.  {3,292916}  {5,11,6,9}  <132>   5*11*6*9/(132*131*130*129) = 3/292916
 3.  {3,190650}  {7,10,6,9}  <126>   7*10*6*9/(126*125*124*123) = 1/63550
 4.  {6,721791}  {5,11,6,8}  <135>   5*11*6*8/(135*134*133*132) = 2/240597
 5. {12,1766622} {5,12,6,8}  <145>   5*12*6*8/(145*144*143*142) = 2/294437
 6. {18,1600200} {5,12,6,8}  <128>   5*12*6*8/(128*127*126*125) = 1/88900
 7. {21,2097024} {6,13,5,7}  <130>   6*13*5*7/(130*129*128*127) = 7/699008
 8. {35,2667420} {7,12,7,10} <147>  7*12*7*10/(147*146*145*144) = 1/76212
 9. {71,9920262} {5,13,6,8}  <146>   5*13*6*8/(146*145*144*143) = 1/139722

After {𝜶, 𝜷} are the letter counts of {F,O,U,R}, followed by the statement's <total-letter-count>, followed by the probability calculation and 𝜶/𝜷 in its lowest terms.