# What is approximately the distribution of your total earning?

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?

• I'm guessing that teacher approximated binomial distribution with the central limit theorem. – Jakobian Jan 26 at 19:44
• $\mathrm{Bin(120,6)}$ is the distribution of successful gambles, not the distribution of your total earning. – zahbaz Jan 26 at 20:00

$$X \sim \text{Bin}(120, \frac{1}{6})$$ is the number of times you win. But the winnings themselves are $$5X - (120 - X) = 6X - 120$$.
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 $$N(0, 600)$$.
• So to calculate the expectation and variance you use the formula from the binomial distribution? $\mu=np$ $\sigma^2=np(1-p)$? – Mark Jacon Jan 26 at 20:43
• 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. – Todor Markov Jan 26 at 21:00