If we roll a dice 10000 times and we gain 1€ if the result is odd, 2€ if it's 2 or 4 and 5€ if it's 6, what is the prob of gain 22000€ or more?

Question

In the end I modeled a random variable as

   { 1 = 3/6;
     2 = 2/6; 
     5 = 1/6 }

And the situation is that we roll a dice 10,000 times and we want to know what is the probability of gain 22000€ or more assuming that roll a dice is gratis/free.

I was wondering if could I use the Central Limit Theorem and approximate the probability distribution as a Gaussian one. But I don't find the right parameters since I'm a beginner.

What kind of strategy should I use to answer this question?

Answer 1

The expectation for the money gained by rolling a single dice is $$\mu=\frac{3}{6}\times1+\frac{2}{6}\times2+\frac{1}{6}\times5=2$$ and the variance $$\sigma^2=6-2^2=2$$

As 10000 is quite a large number, by the central limit theorem, the sum of the money gained by rolling a dice 10000 times shall follow the normal distribution $N(10000\mu,10000\sigma^2)$

Hence, the answer to your problem shall be $1-\Phi(\frac{22000-10000\mu}{100\sigma})=1.044\times10^{-45}$, which is a very unlikely case.

Answer 2

You have the probabilities and the payouts, so the next step would be finding the mean and variance. Mean is $$ \mu=1\cdot3/6+2\cdot2/6+5\cdot1/6=2 $$ and variance is $$ \sigma^2=(1-2)^2\cdot3/6+(2-2)^2\cdot2/6+(5-2)^2\cdot1/6=2 $$ if I have not made any mistakes.

Now you can apply the Central Limit Theorem.

Answer 3

Let $X_i$ denote the result of the $i$-th dice roll. Although you don't explicitly state this, I assume that the $X_{i}$ are independent. In order to use the Central Limit Theorem, you need to compute some quantities related to $X_{i}$.

$E[X_{i}] = 1 \cdot \frac{3}{6} + 2 \cdot \frac{2}{6} + 5 \cdot \frac{1}{6} = 2$

$E[X_{i}^{2}] = 1^2 \cdot \frac{3}{6} + 2^2 \cdot \frac{2}{6} + 5^2 \cdot \frac{1}{6} = 6$

$Var[X_{i}] = E[X_{i}^{2}]-E[X_{i}]^{2} = 6 - 2^{2} = 2$

You wish to compute:

\begin{align*} P(\sum_{i=1}^{10000}{X_{i}} > 22000) &= P(\sum_{i=1}^{10000}{(X_{i}-E[X_{i}])} > 22000-10000 \cdot 2) \\ &= P\left(\frac{\sum_{i=1}^{10000}{(X_{i}-E[X_{i}])}}{\sqrt{10000 \cdot Var[X_{i}]}} > \frac{2000}{100 \cdot \sqrt{2}} \right) \end{align*}

Finally, the CLT states that:

\begin{align*} \frac{\sum_{i=1}^{n}{(X_{i}-E[X_{i}])}}{\sqrt{n Var[X_{i}]}} &\approx N(0,1) \end{align*}

Hence,

\begin{align*} P(\sum_{i=1}^{10000}{X_{i}} > 22000) &\approx P(Z > 20/\sqrt{2}), \end{align*}

where $Z$ follows a standard gaussian distribution.