## Tuesday, December 31, 2019

## Thursday, December 19, 2019

### True probabilities in self-referential statements

On December 8, Γric 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 πΆ/π·, where πΆ and π· 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 πΆ out of π·."

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

2. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are

3. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are

4. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are

5. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are

6. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are

7. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are

8. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are

9. "The odds of randomly picking four letters from this statement and having them be F, O, U, and R, are

I believe that this exhausts the possibilities for πΆ up to 100 and π· 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 πΆ/π· 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 {πΆ, π·} are the letter counts of {F,O,U,R}, followed by the statement's <total-letter-count>, followed by the probability calculation and πΆ/π· in its lowest terms.

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 πΆ up to 100 and π· 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 πΆ/π· 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 {πΆ, π·} are the letter counts of {F,O,U,R}, followed by the statement's <total-letter-count>, followed by the probability calculation and πΆ/π· in its lowest terms.

## Friday, December 13, 2019

### Sapphire

Today is the sixth Friday, December 13th anniversary of our marriage. The photograph of the somber newlyweds was taken by my friend, Alfy Marcuzzi.

## Thursday, December 12, 2019

### A330365

The idea for OEIS sequence A330365 came from Γric Angelini who asked me if I was willing to check and extend it. I was happy to. On the OEIS page for the sequence one can click "graph" and get (at the bottom) a logarithmic scatterplot. I have a much prettier version:

Click on the picture for a larger display of it. The blue points are the values at odd indices; the red points, values at even indices. Green lines connect adjacent values.

Click on the picture for a larger display of it. The blue points are the values at odd indices; the red points, values at even indices. Green lines connect adjacent values.

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