One question regarding Basic Probability theory

Question

If I throw a fair dice $12$ times, the expected number of $6$ is $2$ i.e $6$ is expected to appear $2$ times when the dice is thrown $12$ times. But the probability of getting $6$ exactly $2$ times is ${12}\choose{2}$$(1/6)^{2} (5/6)^{10} $ which is less than $1$.

Now my question is How can you expect the face value six to appear for two times , when the possibility of that appearing for two times is very low?

I am tying to give an analogy..If you are participating a game where you can win , lose or remain undecided. How can you expect to win When you know the possibility of winning the game is very low?

Can anyone please make me understand where I am getting wrong? I am really trying hard to understand.

Answer 1

The expected value is an average. On any given throw of a fair die you can get any of the values from $1$ to $6$ with equal probability. If we throw the die many times and record the results we expect each number to be thrown one sixth of the time.

But each individual throw is independent. That means if we throw $10$ sixes in a row which is improbable but not impossible, this does not mean that the odds of throwing an eleventh six has a different probability than if we hadn't rolled any sixes.

Answer 2

Actually, you should not expect to get exactly 2 sixes over every sequence of 12 throws. What you can expect is that, if you throw your dices hundreds or thousands of times, on average you will get approximately 1 six out of every 6 throws. On average. And approximately. In fact, if you go on and throw your dices millions of times, you are "almost sure" to have about $\frac{1}{6}$ of these yields a six.

Now this is a very loose explanation. The underlying theory, that makes very accurate predictions when it is applied to large series of events (if they are independent, and maybe identically distributed), is called the Law of large numbers and quantifies exactly (well, in a proababilistic sense) what is the chance that on average over a long run you get the expected number of events.

Answer 3

Let $N$ be the number of times you roll a 6 on your 12 attempts. Then, compute the following probabilities:

\begin{align} \text{Prob}\left(N=n\right)= \binom{12}{n} \frac{1}{6}^n \frac{5}{6}^{12-n} \ , \end{align}

for $n=0,\ldots,12$.

Instead of repeating this experience (i.e. throwing the dice 12 times) over and over again, you can think of (independent) parallel universes where you are running this experience. Each universe is governed by the same set of probabilities (described above), but the outcomes of the experience are independent. Then the outcome of $N$ in parallel universe $i$ is $N_i$.

In this scenario, having $\mathbf E \left[N\right]=2$ means that if you compute the average value of $N$ over those parallel universes you would get the number $2$, i.e.

\begin{align} \lim\limits_{U\to\infty} \frac{1}{U} \sum\limits_{i=1}^U N_i = 2 \ . \end{align}

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But it is just a coincidence that $\mathbf E\left[N\right]$ is a valid outcome for $N$.

Imagine you were flipping a fair coin. Then let $X$ be a random variable for which $X=1$ when you get heads, and $X=-1$ when you get tails. The expected value of $X$ is $0$, despite the fact that $X\in\{-1,1\}$.

PS: I'm clearly explaining what @JohnDouma and @lmsteffan in other words. But this seems to be one of those questions that asks for different points of view about the same stuff.