Grading quizzes... except words this time!

Question

So, you may remember that I’m a professor of Awesomeness at the prestigious Ad Hoc University! This time, I assigned my students a quiz where they have to determine how I scored a word. Once again, I overestimated their abilities and they all failed the quiz! I’m sure it can’t be impossible, right (especially since I’m giving you all the individual parts of the score)? You tell me! Here’s the quiz I gave them:

magic = 0.1 + 1.0 + 0.77 = 1.87

dinosaur = 0.08 + 0.31 + 3.73 = 4.12

puzzling = 0.08 + 1.23 + 4.42 = 5.73

albatross = 0.18 + 0.08 + 4.08 = 4.34

seagull = 0.14 + 1.46 + 2.23 = 3.83

hunger = 0.06 + 0.62 + 2.5 = 3.18

famine = 0.06 + 0.46 + 1.62 = 2.14

Can you do it?

Note: I think this one is easier than my grid one (linked above).

Hint 1:

The values are all rounded to the nearest hundredth... Think of some of the letters' positions in the alphabet...

Hint 2:

The score consists of a "bonus," the first letter score, and the rest of the word score. (These are shown in order above.)

Hint 3:

The letters' positions in the alphabet are divided by 26 to get their scores...

Answer 1

I think I've gotten about halfway there but I can't finish it out, so maybe this can help someone else.

As you said, you've given the individual parts of the scores, there are three parts.

#1

Seems to based on the length of the word, $\ell$. If $\ell \equiv 0 \ (\text{mod} \ 2)$, the first score is $0.01\ell$. Otherwise, the first score is $0.02\ell$. This can be written concisely using the modulo operator $\%$ as $$0.01\ell(\ell\%2 + 1)$$

#2

Seems to be based on the numerical value of the initial letter. Define $\#(\alpha)$ as the numerical value of a letter $\alpha$,so $\#(a) = 1, \ \#(b) = 2, \dots, \#(z) = 26$. Notably, the second score is a strictly increasing function of $\#$ of the first letter in each word. Even more notably, for each $\alpha$ such that $\#(\alpha) \equiv 1 \ (\text{mod} \ 3)$ we have the second score is $\frac{1}{300}(1+23\#(\alpha))$. We only have one $\#(\alpha) \equiv 0 \ (\text{mod} \ 3)$, but we do have that for that one, $\frac{1}{300}(23\#(\alpha))$, as a linear function, this goes through the origin if we graph it, so it seems reasonable. However, the pattern doesn't continue. If we make a third parallel line for $\#(\alpha) \equiv 2 \ (\text{mod} \ 3)$, we get $\frac{1}{300}(44+23\#(\alpha))$, which is nice and round, but I was hoping it would be a $2$ instead of a $44$. If that were the case, we would have been able to write the second score as $$\frac{1}{300}(\#(\alpha)\%3+23\#(\alpha))$$ but this isn't the case. I could be way off on this one, but the fact that 5 points are colinear and have the same remainder seems too much a coincidence.

#3

I don't really have a clue about this one. It's generally increasing with respect to the length of the words, but not exactly. It's almost strictly increasing with respect to the sum of the $\#$ values for all the letters in the word, but "hunger" and "seagull" are flipped. It could be some variation of this, where maybe vowels and consonants are worth different values, but I couldn't find anything.

And lastly,

If what I have so far is correct, then "voldemortswrath" should be 0.3 + 1.69 + 7.35 = 9.34, further suggesting that the third score is somehow related to the word length/numerical values.

Answer 2

First part

$0.01ℓ(ℓ\%2+1)$
where $ℓ$ is the length of the word, as figured out by AlexanderJ93

Second part

$x/13$
where $x$ is the numerical value of the first character within the alphabet

Third part

Total sum of applying the following to each character except the first one
$x/26$
where $x$ is the numerical value of the character within the alphabet