Discrete Probability Distribution

[elove]

Okay, my online class has posed another word problem and I cannot seem to understand this week's material or how to formulate a solution.

Here it is:
Imagine you are in a game show, a money give-away! There are 4 prizes hidden on a game board with 16 spaces. One prize is worth \$4000, another is worth \$1500, and two are worth \$1000.

But, wait! You are also told that, in the rest of the spaces, there will be a bill of \$50 that you have to pay to the host as a penalty for not making the "wise" choice.

Choice #1: You are offered a sure prize of $400 cash, and you just take the money and walk away. Period. No question asked...
Choice #2: Take your chance and play the game...

What would be your choice? and why?

I cannot understand how to make this problem into a legitimate formula. It seems like opinion.

what would a sample look like? any guidance would be great. I am in communication with my student group now but no one has a starting point.

 

[Greg]

I know the title of this thread is "Discrete Probability Distribution" but wouldn't expected value apply here?

By the way, you can correct the formatting of your post by preceding all \$ signs by a \.

 

[elove]

I think that is why I am so confused. This week's material covers Continuous Distribution, Normal Distribution, Z-Score, The Standard Normal Distribution, Application of Normal Distributions, Normal as Approximation to Binomial Distribution, and The Central Limit Theorem. None of these seem to apply to the word problem, which was posted with the title "Discrete Probability Distribution ?" - maybe the question marks literally mean it's not discrete probability distribution problem.

 

[Siron]

I will suppose you play one time. Let $X$ be your gain after one play. Since there are $16$ different spaces you have the following distribution for $X$
$$X = \left \{ \begin{array}{lllll} 4000, \quad p = \frac{1}{16} \\ 1500, \quad p = \frac{1}{16} \\ 1000, \quad p = \frac{2}{16} \\ -50, \quad p = \frac{3}{4} \end{array} \right.$$
Let us now compute $\mathbb{E}[X]$, that is,
$$\mathbb{E}[X] = 4000\left(\frac{1}{16}\right)+1500 \left(\frac{1}{16}\right)+ 1000 \left(\frac{2}{16}\right)-50 \left(\frac{3}{4}\right) = 431.25$$
What would you choose? When you walk away from the game you have a certain profit of $400 \$$. However, when you play the game, you expect to win $431.25 \$$.

 

[elove]

Thank you! This expected value equation makes much more sense now! Our group was halfway there with determining the probability but didn't follow through with multiplying the value to the probability.