Question

You roll a die. If it comes up a 1 or 2​, you win ​$200. If​ not, you get to roll again. If you get a 1 or 2 the second​ time, you win ​$100. If​ not, you lose. ​a) Create a probability model for the amount you win. ​b) Find the expected amount​ you'll win.

Answer 1

To create a probability model for your scenario, assign probabilities to each potential winning outcome and use these to calculate the expected value. After performing the calculation, the expected amount you'll win each time you play the game is approximately $88.89.

Step-by-step explanation:

Probability Model and Expected Value Calculation

To answer your question, first, we need to create a probability model for the amount you win, and then we'll calculate the expected amount you'll win when rolling a die with the conditions specified. Let’s consider the outcomes:

• 
  
• Winning $200 if you roll a 1 or 2 on the first roll
• 
  
• Winning $100 if you don't roll a 1 or 2 on the first roll but do on the second roll
• 
  
• Losing (or winning $0) if you don't roll a 1 or 2 on either roll

For part a), the probability model is as follows:

• 
  
• P(Win $200) = P(Rolling a 1 or 2 on the first roll) = 2/6 or 1/3
• 
  
• P(Win $100) = P(Not a 1 or 2 on the first roll) * P(Rolling a 1 or 2 on the second roll) = 4/6 * 1/3 = 4/18 or 2/9
• 
  
• P(Win $0) = P(Not a 1 or 2 on both rolls) = 4/6 * 4/6 = 16/36 or 4/9

For part b), the expected value E(X) is calculated by multiplying each outcome by its probability and adding the results:

• 
  
• E(X) = ($200 * 1/3) + ($100 * 2/9) + ($0 * 4/9)
• 
  
• E(X) = $66.67 + $22.22 + $0
• 
  
• E(X) = $88.89 approximately

So, the expected amount you'll win each time you play the game is approximately $88.89.

Answer 2

On the first roll, you have a 1/3 probability of rolling a 1 or 2 and thus winning $200.

There's a 2/3 probability of not rolling a 1 or 2 on the first roll. On the second roll, there is again a 1/3 probability of rolling a 1 or 2, and 2/3 probability otherwise, so that there is a 1/3*2/3 = 2/9 probability of getting a 1 or 2 and thus winning $100, and 2/3*2/3 = 4/9 probability of losing.

(a) Let
`W` be a random variable representing the winnings from playing the game. It has the probability mass function

`P(W=w)=\begin{cases}\frac13&\text{for }w=\$200\\\frac29&\text{for }w=\$100\\\frac49&\text{otherwise}\end{cases}`

(b) Compute the expected value of
`W`:

`E[W]=\displaystyle\sum_w w\,P(W=w)=\$200\cdot\frac13+\$100\cdot\frac29+\$0\cdot\frac49\approx\$88.89`