# Duolingo Leagues

Duolingo, the language learning app, places users in groups of 50 and assigns a league to each user to encourage competition. The leagues are

• Bronze, Silver, Gold, Sapphire, Ruby, Emerald, Amethyst, Pearl, Obsidian, and Diamond (in that increasing order)

# What proportion of Duolingo users are in each league?

The rules are

• everyone starts in the Bronze League
• the top 15 percent of each group gets promoted to the next league up (measured weekly)
• the bottom 10 percent of each group is related downward

In this post, I will try out some stochastic processes calculations to answer that question.

leagues <- c("Bronze", "Silver", "Gold", "Sapphire", "Ruby",
"Emerald", "Amethyst", "Pearl", "Obsidian", "Diamond")

transition_matrix <- matrix( rep(0, 100), 10)

# trying shortcuts
diag(transition_matrix[-10,-1]) <- 15 #top 15 percent of each group gets promoted
diag(transition_matrix[-1,-10]) <- 10 #bottom 10 percent of each group is relegated
diag(transition_matrix)         <- 75 #the rest stay where they are

# fix the corners
transition_matrix[1,1]    <- 85
transition_matrix[10, 10] <- 85

# make row-stochastic (i.e. so each row adds up to one)
transition_matrix <- transition_matrix/100

# R allows user to label rows and columns
rownames(transition_matrix) <- leagues
colnames(transition_matrix) <- leagues

print(transition_matrix)
##          Bronze Silver Gold Sapphire Ruby Emerald Amethyst Pearl Obsidian
## Bronze     0.85   0.15 0.00     0.00 0.00    0.00     0.00  0.00     0.00
## Silver     0.10   0.75 0.15     0.00 0.00    0.00     0.00  0.00     0.00
## Gold       0.00   0.10 0.75     0.15 0.00    0.00     0.00  0.00     0.00
## Sapphire   0.00   0.00 0.10     0.75 0.15    0.00     0.00  0.00     0.00
## Ruby       0.00   0.00 0.00     0.10 0.75    0.15     0.00  0.00     0.00
## Emerald    0.00   0.00 0.00     0.00 0.10    0.75     0.15  0.00     0.00
## Amethyst   0.00   0.00 0.00     0.00 0.00    0.10     0.75  0.15     0.00
## Pearl      0.00   0.00 0.00     0.00 0.00    0.00     0.10  0.75     0.15
## Obsidian   0.00   0.00 0.00     0.00 0.00    0.00     0.00  0.10     0.75
## Diamond    0.00   0.00 0.00     0.00 0.00    0.00     0.00  0.00     0.10
##          Diamond
## Bronze      0.00
## Silver      0.00
## Gold        0.00
## Sapphire    0.00
## Ruby        0.00
## Emerald     0.00
## Amethyst    0.00
## Pearl       0.00
## Obsidian    0.15
## Diamond     0.85

Knowing some about stochastic processes, we can either implement an initial distribution and employ matrix-vector multiplication over many iterations, or we can find an eigenvector.

findStatDist <- function(P){
# This function will compute the stationary distribution for a transition matrix
# Input: row-stochastic matrix P
# Output: row vector pi_vec

# obtain left-eigenvector for lambda = 1
x <- eigen(t(P))\$vectors[,1]

# normalize the eigenvector in the one-norm
pi_vec <- x / sum(x)
pi_vec #return this vector
}

# R allows user to label rows and columns

print(answer)
##          percentage
## Bronze            3
## Silver            4
## Gold              5
## Sapphire          7
## Ruby              9
## Emerald          11
## Amethyst         13
## Pearl            15
## Obsidian         16
## Diamond          17