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You play a game in a casino: you roll two dice and if the sum of the spots equals seven, you win $5$€.
In every other case, you lose $1$€. You decide to play this game $120$ times.
What is approximately the distribution of your total earning?

So the probability that the sum of two dice $X$ and $Y$ is $7$ can be calculate with: $P(X+Y=7)=\frac{6}{36}=\frac{1}{6}$ considering all the cases: ((1,6), (2,5), (3,4), (4,3), (5,2), (6,1)).
Then the probability that the sum of the two dice is not $7$ is $1-\frac{1}{6}=\frac{5}{6}$.
So I think that the distribution should be $Bin(120,\frac{1}{6})$ but from the teacher answer it is $N(0,600)$, where I'm wrong? Why is not Binomial but Normal?

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    $\begingroup$ I'm guessing that teacher approximated binomial distribution with the central limit theorem. $\endgroup$ – Jakobian Jan 26 at 19:44
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    $\begingroup$ $\mathrm{Bin(120,6)}$ is the distribution of successful gambles, not the distribution of your total earning. $\endgroup$ – zahbaz Jan 26 at 20:00
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$X \sim \text{Bin}(120, \frac{1}{6})$ is the number of times you win. But the winnings themselves are $5X - (120 - X) = 6X - 120$.

This has a distribution, and you could compute its mass function if you like, but it's not one of the standard distributions and so it'll be tedious. What your teacher is getting at, however, is that you can use normal approximation. By the central limit theorem, the sum of a lot of iid random variables is approximately normally distributed.

We can compute $\mathbb{E}[6X-120] = 0$, and $\text{Var}[6X-120] = 36\text{Var}[X] = 36 \times 120 \times \frac{1}{6} \times \frac{5}{6} = 600$, so the normal approximation will be $N(0, 600)$.

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  • $\begingroup$ So to calculate the expectation and variance you use the formula from the binomial distribution? $\mu=np$ $\sigma^2=np(1-p)$? $\endgroup$ – Mark Jacon Jan 26 at 20:43
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    $\begingroup$ I calculate the true mean and variance (as I said, it's not exactly binomial). $X$ is binomial, but I need to adjust since I have $6X-120$ instead of just $X$. Then I take a normal that has the same mean and variance. $\endgroup$ – Todor Markov Jan 26 at 21:00

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