Question

I am doing a kids fishing game. right now I am doing it with 15 fish and number them 1-5. prize one the lowest value and prize 5 the highest value. What do I need to do for this?

You can work with a partner or by yourself. I want you to invent a game with at least 12 different possible monetary outcomes. These outcomes need to include prizes other than money. You are going to charge people money (you decide how much) to play the game and then you decide how much each outcome is worth (win or lose). Make sure that you are able to find the probability of each outcome (Hint: use dice, deck of cards, etc.) Your goal is to come up with an appealing game that people will want to play, but that you will be the one making money in the end. NO RAFFLES.

Title your game, explain the rules, and list the prizes.

Construct a probability distribution in chart form for your game, your random variable x should represent the player’s monetary outcome (there should be at least 12 different x values). Be sure to subtract what you charged to play from their winnings.

Find the player’s expected value of your game. The expected value needs to be between $-2.00 and $0.00.

Explain how this game is going to make you money, but still be appealing enough for people to try and play.

Answer 1

The price value 1,2,3,4,5 are randomly assigned to 15 fish.

so the probability of getting 1 is 1/5, similarly probability of getting 2,3,4,5 is also 1/5.

Let thecharge for playing game is $5.

the player wins if he catches 1,3,&5 numbered fish.

the player loses if he catches 2 &4.

if the player wins 1 it contains stationary item worth $1.

if the player wins 3 it contains ball worth $2

if the player wins 5 it contains toy $3.

the game name is CATCH ME UP.

The game follows binomial distribution,

the probability of wining the game is 3/5 ( probability of getting 1,3,5)

the probability of losing the game is 2/5 (probability of getting 2 &4)
let X be the random variable ( players monetary value)

if the player win the game the x value will be $3,$2,$1

if the player lose the game the x value will be $5.

expected value= = 1(1/5) + 0(1/5)+ 3(1/5)+0(1/5)+5(1/5)

=1/5+3/5+5/5 = 9/5

=1.8

even if the player wins we can get $2 or $3 or $4 . if the player loses the game we can get $5.

this is my suggestion only please ignore if this is wrong.