PhD physicist. It is not identical to the Monty Hall problem, and it is indeed 50/50 in this case. Because the boxes are removed at random, and not by supreme directive, your odds scale proportionally with the number of boxes remaining from 1/N to 1/2.

Back when I was first learning R, I wrote a script to simulate many, many games of Deal or No Deal in which people ended up with the $1M box somewhere in the mix at the end (this very situation) to see what fraction of times they would end up with the box. You can run it and see the histogram of results centered around 0.5.

Fair warning, this code is absolute shit and is R written with a strong C accent. I should update it and write it as R is intended to be:

result <- c()
a <- 0
b <- 0
for (j in 1:100) {
  for (i in 1:100) {
    q <- 0
    while (q == 0) {
      vec <- c(0.01,1,5,10,25,50,75,100,200,300,400,500,750,1000,5000,10000,25000,50000,75000,100000,200000,300000,400000,500000,750000,1000000)
  x <- sample(vec,1)
  vec <- setdiff(vec,x)

  for (j in 1:24) {
    y <- sample(vec,1)
    vec <- setdiff(vec,y)
  }
  if (1000000 %in% vec || 1000000 %in% x) {
    q <- 1
  }
}
if (vec == 1000000) {
  a <- a + 1
} else
  b <- b + 1
  }
  ratio <- (a / (a + b))
  #print(ratio)
  result <- c(result,ratio)
}
hist(result,breaks = 10)