In my probability class, we did the following problem regarding expected values/variance:
Consider an experiment where you roll a fair, 6-sided dice until you see a 6. Let $T$ be a random variable denoting the total sum, including the final 6. Also, consider an experiment where you roll a fair, 6-sided dice until you see a 1. Let $S$ be a random variable denoting the total sum, including the final 1.
a) Which is larger, $E[T]$ or $E[S]$?
We showed that $E[T] = E[S] = 21$, by conditioning $T$ and $S$ on a random variable $N$ denoting the total number of rolls.
b) Which is larger, $Var[T]$ or $Var[S]$?
Here, we showed that $Var[T] > Var[S]$, by again conditioning on $N$.
However, I took a different approach to (b). I noted that when we consider the variance of a random variable, we are really concerned with the deviation between a value the random variable can take and it's expectation, not that actual numbers themselves. Also, note each roll is independent of the others. Therefore, let $T = T_1 + T_2 + ... + T_N$, where $T_i$ is the value rolled on the i-th roll, and define $S_i$ similarly. I believe $Var[T_i] = Var[S_i]$ on any given roll before the final one, as in both experiments, the deviations can take the values $-2, 1, 0, 1, 2$ with probability $\frac{1}{5}$. Furthermore, on the last roll of each experiment, the deviations between the values rolled and the expectation are identical, taking the value $3$. If all of the deviations are identical then, how are the variances different?
Note: I do understand how to compute the variances by conditioning on $N$, and accept that my answer was wrong. However, I'm specifically looking for where the logic breaks down in my argument, as it seems that the deviations and probabilities are the same for each experiment, which to me implies that the variances should be equal as well.