First consider what goes into rolling 2d6 10 times and adding the result.

twoDsix <- function(n) {
  rolls <- rep(NA, n)
  for(i in 1:n) {
    rolls[i] <- sum(sample(1:6, 2, replace = T))
    }
  sum(rolls)
  }

sim_results1 <- rep(NA, 50000)
for (j in 1:50000){
  sim_results1[j] <- twoDsix(10)
}

Now consider what goes into rolling 2d6 once, and multiplying that result by 10.

sim_results2 <- rep(NA, 50000)
for (i in 1:50000){
  sim_results2[i] <- twoDsix(1) * 10
}

We can compare the two distributions:

par(mfrow = c(2, 1))
#sim_results1 <- c(sim_results1, 0, 140)
#sim_results2 <- c(sim_results2, 0, 140)
#barplot(table(sim_results1), main = "Outcomes when rolling 2d6 10 times", col = "steelblue")
hist(sim_results1, main = "Outcomes when rolling 2d6 10 times", col = "steelblue", xlim = c(20, 120))
barplot(table(sim_results2), main = "Outcomes when rolling 2d6 once and multiplying by 10", col = "steelblue")

#hist(sim_results2, main = "Outcomes when rolling 2d6 once and multiplying by 10", col = "steelblue", xlim = c(0, 140))
var(sim_results1)

## [1] 58.32

var(sim_results2)

## [1] 577.9

Thus consolidating the randomness into a single roll instead of 10 increases the variance by a factor of one hundred.