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A six faced fair die is tossed a large number of times (tending to infinity). What is the mean of outcomes of the die?

My try: let the die be tossed $n$ times then the mean of outcomes

$$=\frac{(\frac 16+\frac 16+\frac 16+\ldots+upto \ n \ times )}{n}=\frac{\frac n6}{n}=\frac 16$$ But I am not if my answer is correct. please explain If I am wrong & give right answer.

Thanks

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    Are you searching for the expected value over $n $ tosses? Or for the observed mean? – 2017-02-04
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    for the observed mean – 2017-02-04
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    Ok. So why are you summing $n $ times the quantity $1/6$ in the numerator, instead of summing the truly observed $n $ values? – 2017-02-04

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The possible outcomes are $1,2,3,4,5,$ and $6$. Each have a probability of $1/6$, hence the theoretical mean is $$\frac{1}{6}\times 1 +\frac{1}{6}\times 2+\frac{1}{6}\times 3+\frac{1}{6}\times 4+\frac{1}{6}\times 5+\frac{1}{6}\times 6 = 3.5 $$

The observed mean is

$$\frac{\text{Roll 1}+\text{Roll 2}+\text{Roll 3 }+...+\text{ Roll }n}{n} $$ where $n$ is the number of rolls (and "Roll $i$" means the number of eyes on roll number $i$).

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    According to what stated in the comments, he is searching for the observed mean, not the expected value. – 2017-02-04
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    Fair enough, I added it to my answer :). – 2017-02-04