Dear List, I'm trying to understand the following example from LYAH import Data.List (all) flipThree :: Prob Bool flipThree = do a <- coin b <- coin c <- loadedCoin return (all (==Tails) [a,b,c]) Where import Data.Ratio newtype Prob a = Prob { getProb :: [(a,Rational)] } deriving Show and data Coin = Heads | Tails deriving (Show, Eq) coin :: Prob Coin coin = Prob [(Heads,1%2),(Tails,1%2)] loadedCoin :: Prob Coin loadedCoin = Prob [(Heads,1%10),(Tails,9%10)] The result: ghci> getProb flipThree [(False,1 % 40),(False,9 % 40),(False,1 % 40),(False,9 % 40), (False,1 % 40),(False,9 % 40),(False,1 % 40),(True,9 % 40)] See http://learnyouahaskell.com/for-a-few-monads-more#making-monads. My understanding of what's going on here is sketchy at best. One of several explanations that I am considering is that all combination of a, b and c are evaluated in (==Tails) [a,b,c] but I cannot explain how the all function creates 'fuses' the list [f a, f b, f c]. I know that all f xs = and . map f xs (the definition on hackage is a lot more complicated) but, again, I cannot explain how the and function 'fuses' the list [f a, f b, f c]. If I'm on the right track I realize that I'm going to have to study the list the between list comprehensions and the do-notation in order how all the return function create one Prob. Regards, - Olumide