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Finding the best dog treat with statistics

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Posted on June 19, 2026 by Adam Wespiser

Bebop, my 83lb, 33 inch tall, Greyhound, loves three things: running fast, following me around the house, and treats. Whether it’s a chew treat, pizza out of a child’s hand who strayed too far from a party, or a small tray of cat food, he has a nose for what he likes and the athleticism to give him a fair shot at getting it. I’ve watched him eat for years, so it was upsetting to realize I don’t know what his favorite snack is, and can’t easily ask him.

Fortunately for Bebop’s palate, the Bradley-Terry model gives us a way to figure out a “strength” of treat from pairwise comparisons. The model assigns each competitor (or treat) (i) a positive strength score p i .

Given two competitors i and j, the probability that i beats j is:

Pr(i > j) = p i /(p i + p j )

Equivalently, if we write each strength as an exponential score,

p i = eβ i

then the same probability can be written as:

Pr (i > j) = eβ i /(eβ i + eβ j )

So the model is saying: the difference between two competitors’ latent strengths determines the log-odds that one beats the other.

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