# You roll a die. If it comes up a 6 you win 100. If not, you get to roll again. If you get a 6 the second time, you win 50. If not, you lose.

– 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.

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 $\dfrac{25}{36}$.

Figure 1

#### 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.