If I code as 1 the times the correct turn is to the right and as 0 the times the correct turn is to the left, this is my maze:

Code:

` turn_right <- c(1,0,0,0,1,0,1,0,0,0,0,1,1,0,0,0,1,1,1,0,1,1,0,1,1,1,1,0,1,0,0,1)`

These are my signs, where a 1 means that the sign points to the correct direction, and a 0 means that the sign points to the wrong direction.

Code:

` sign <- c(1,0,1,0,0,1,1,1,0,1,0,1,0,1,0,1,0,0,1,1,1,1,0,0,0,0,0,1,1,0,0,1)`

Code:

`trust_sign <- c(0,0,0,0,0,0,0,0,1,1,1,0,0,1,1,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0)`

Since we have binary choices, I thought I could model the participant's choices (trust_sign) with a beta distribution:

Code:

```
maze <- data.frame(turn_right, sign, trust_sign)
sum32 <- sum(maze$trust_sign[1:32])
curve(dbeta(x, sum32, 32- sum32),add=TRUE,lty="solid",ylim=c(0,6),ylab="Probability Density",las=1)
```

I can also calculate the likelihood of a sign being correct given the actual maze:

Code:

```
k = 16 # number of times a sign is correct
n = 32 # total number of intersections
numSteps = 200 ## x-axis for plotting
x = seq(0, 1, 1 / numSteps)
L = x^k * (1 - x)^(n - k) ## Likelihood function
L = L / sum(L) * numSteps ## Just normalize likelihood
plot(x, L, type = 'l', lwd = 3, ylim = c(0,6),
main = "Bernoulli Likelihood",
xlab = expression(theta), ylab = "pdf")
```

Given that likelihood and the behavior seen before, what is the belief of my participant prior to entering the maze? Is this the right framework for this question or am I missing something?