**– Develop a probability model for the amount you win.**

**– Find the expected amount you win.**

This problem aims to find the **probability** of getting a **particular number,** say $6$, by **rolling** **a dice** and creating a **probability model** for our outcomes. The problem requires the knowledge of **probability model creation** and the **expected value formula**.

## Expert Answer

The **predicted amount** of the problem is equal to the **sum of the products** of each trial and its **probability.** As in the problem, the **loss** is not specified if you don’t score $6$ at any **roll, **but this is required for the **computation.** For this problem, we are going to assume that a **loss** has an impact of $0$, and a **win** has an impact of $100$.

The **probability** that there will be a $6$ on a certain **roll** is **equal to the probability** that there is $6$ on the **first roll** plus the probability that there will be a $6$ on the $2^{nd}$ roll. Every **rolling die** has $6$ **sides,** so there is a $1$ side out of $6$ that will **probably win,** so the probability of hitting a $6$ on the first trial is $\dfrac{1}{6}$

So the probability of getting a $6$ is $\dfrac{1}{6}$.

Probability of not $6$ is $1 – \dfrac{1}{6} = \dfrac{5}{6} $.

#### Part One

For **winning** $100$, it is compulsory to **score** a $6$ in the **first trial**, and the **probability** of $6$ is $\dfrac{1}{6}$.

For **succeeding** $50$, it is required **not** to **score** $6$ in the **first roll** and $6$ in the **second roll,** and the probability of not getting a $6$ is $\dfrac{5}{6}$ and the probability of $6$ is $\dfrac{1}{6}$, so the probability, in this scenario, would be $\dfrac{1}{6} \times \dfrac{5}{6}$, which is equals to $\dfrac{5}{36}$.

For $0$, it’s required not to score $6$ in both the rolls, so the probability, in this circumstance, becomes $\dfrac{5}{6} \times \dfrac{5}{6}$, that is equal to $\dfrac{25}{36}$.

#### Probability Model

Part b:

**Formula for expected value** is given as:

\[E(x) = \sum Value.P(x) \]

\[ = (100)(\dfrac{1}{6}) + (50)(\dfrac{5}{36}) + (0)(\dfrac{25}{36}) \]

## Numerical Result

The **expected amount** is:

\[E(x) = \$23.61 \]

## Example

You **roll** a **die.** If it comes up a $6$, you **win** $100$. If not, you get to roll again. If you get $6$ the $2^{nd}$ time, you win $50$. If not, you get to roll again. If you get a $6$ the $3^{rd}$ time, you win $25$. If not, you lose. Find the **Expected amount** you win.

For **winning** $100$, **P(x)** is $\dfrac{1}{6}$

For **winning** $50$, **P(x)** is $\dfrac{1}{6} \times \dfrac{5}{6} = \dfrac{5}{36}$

For **winning** $25$, **P(x)** is $\dfrac{1}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} = \dfrac{25}{216}$

For **winning** $0$, **P(x)** is $\dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} = \dfrac{125}{216}$

In the end, the **expected amount** is the sum of the multiplication of the results and their probabilities:

\[E(x) = \sum Value.P(x)\]

\[= (100)(\dfrac{1}{6}) + (50)(\dfrac{5}{36}) + (25)(\dfrac{25}{216}) + (0)(\dfrac{125}{216})\]

This is the **expected amount** after the given number of trials:

\[ E(x) = \$25.50 \]

*Images/mathematical drawings are created with GeoGebra.*