Confusion about the definition of "expected number" in probability When you roll a dice $6$ times, the "expected number" of $6$'s that should come up is $1$. I used to think that this is because by the time you have rolled the dice $6$ times, the chance of having rolled a $6$ is over $50\%$. However, further inquiry has led me to believe that this is not in fact what the "expected number" means.
The chance of rolling a $6$ after $n$ rolls = $1 - (5/6)^n$
After $4$ rolls, the chance of rolling a $6$ is $1 - (5/6)^4 = 51.8\%$ (1 d.p.)
So by then, the probability has already risen to over $50\%$. So what does the "expected number" of $6$'s actually mean?
 A: The expected value is just a weighted average of the values of the outcomes. For instance, the expected value when rolling a fair dice is $3.5$, as:
$$
\frac{1}{6}\cdot 1 +\frac{1}{6}\cdot 2 +\frac{1}{6}\cdot 3 +\frac{1}{6}\cdot 4 +\frac{1}{6}\cdot 5 +\frac{1}{6}\cdot 6 = 3.5
$$
Each outcome is a realization of a random variable (eg, ''six'' when rolling a dice), and the weights are the corresponding probabilities (note that the sum of the probabilities for all possible outcomes should be exactly one, that's why the coefficients of the outcomes are actually weights. 
The expected number of sixes when you roll six fair dice at once (or six times a fair dice) is exactly 1. To see why, consider an equivalent experiment with smaller numbers: rolling four 4-side dice (tetrahedral shape). The expected number of fours is exactly 1 as well. Why? Well, when you roll four dice, there are 256 possible outcomes: $(1,1,1,1), (1,1,1,2), \dots (4,4,4,3), (4,4,4,4)$. Here, the number of fours is the outcome (not four itself, but the number of them), and the probabilities are as follows:


*

*Probability that the number of fours is zero is 81/256 $\approx$ 0.316

*Probability that the number of fours is one is 108/256 $\approx$ 0.422

*Probability that the number of fours is two is 54/256 $\approx$ 0.211

*Probability that the number of fours is three is 12/256 $\approx$ 0.047

*Probability that the number of fours is four is 1/256 $\approx$ 0.004


Therefore, the expected value of the number of fours is the weighted average:
$$
0.316 \cdot 0 + 0.422\cdot 1+ 0.211 \cdot 2 + 0.047 \cdot 3 + 0.004 \cdot 4 = 1.000
$$
Note that, because the random variable is the number of fours, the result wouldn't change if you compute the expected value of the number of twos or the number of threes. You will always get exactly 1.
A: Any time you have a random number, its expected value is a type of average over the outcomes. If you have a few random numbers, say $X_1, ..., X_6$, their sum $X_1 + \cdots + X_6$ is also a random number. Since averaging is an additive procedure, the expected value of the sum is equal to the sum of the expectations, or in symbols,
$$
\mathbb E(X_1+\cdots +X_6)=\mathbb EX_1+\cdots + \mathbb EX_6,
$$
where $\mathbb E$ denotes expectation. This formula is called "linearity of expectation".
Let's apply this to your case: let $X_i$ equal $1$ if a $6$ came up on the $i^{th}$ roll, and $0$ if not. Then $\mathbb EX_i=\frac{1}{6}$, since the procedure for finding the expected value takes into account probabilities, meaning that $1$ is given a weight of $\frac{1}{6}$ and $0$ is given a weight of $\frac{5}{6}$ (since there are $5$ ways to not roll a $6$ on any given turn). Therefore, by linearity of expectation,
$$
\mathbb E(X_1+\cdots +X_6)=\frac{1}{6}+\cdots+\frac{1}{6}=1.
$$
But let's think about what $X_1+\cdots X_6$ is: the sum of $N$ ones, where $N$ is the number of times a $6$ was rolled. Or in other words, $X_1+\cdots +X_6=N$, so that $\mathbb EN=1.$ In plain english, this says that
$$
\text{the expected number of times a $6$ is rolled in $6$ rolls equals $1$}.
$$
A: The expected number of something is pretty subtle. If you have a function $X$ which outputs numbers $\{x_1, \dots, x_n\}$, then its mathematical expectation is defined to be: 
$$E(X) = x_1 Pr(X = x_1) + x_2 Pr(X = x_2) + \dots + x_n Pr(X = x_n)$$
In your case, $X$ would be the "number of sixes in a roll of six dice". You'd get: 
$$E(X) = 1 Pr(X = 1) + 2 Pr(X = 2) + \dots + 6 Pr(X = 6)$$
The probability of rolling ``$k$ sixes'' would be: $Pr(X = k) = \binom{6}{k} (5/6)^{6-k} (1/6)^{k}$. If you calculated out that sum, you get one. (Here it is on WolframAlpha.)
It might help to imagine what happens if you roll seven dice. In that case, you get $E(X) = 1.1666.$ You can never quite get a fractional number as an output, but you would "expect" that many outputs of six.
Does that help?
