Optimizing Spin-and-Win: Expected Value and Impact on Owner with 1000 Customers

Click For Summary
SUMMARY

The discussion focuses on calculating the expected value of the "Spin-and-Win" game under new management, which charges $1 per spin. Two trials were presented: Trial 1 incorrectly calculated the expected value for the player as -$0.50, while Trial 2 correctly determined it as -$0.25. Consequently, the owner expects to make a profit of $250 per week from 1000 customers based on Trial 2's calculations. The correct expected value indicates that the game is profitable for the owner despite the negative expected value for players.

PREREQUISITES
  • Understanding of expected value calculations in probability.
  • Familiarity with basic game theory concepts.
  • Knowledge of profit and loss analysis.
  • Ability to interpret statistical outcomes in a business context.
NEXT STEPS
  • Research advanced probability concepts in game design.
  • Learn about customer behavior analysis in gaming environments.
  • Explore profit maximization strategies for game operators.
  • Study the impact of game mechanics on player engagement and retention.
USEFUL FOR

Game designers, business analysts, and operators of gaming establishments looking to optimize game profitability and understand player dynamics.

Glen Maverick
Messages
14
Reaction score
0
Expected value: urgent help needed

Homework Statement



a1cb48cf8a3bcbef8c6b2657cc60cc3f.jpg

"Spin-and-Win" is under new management. The new management has created the spinner to the right and now charges $1 per spin. (The person spinning wins the amount pointed to by the spinner.)

a) What is the expected value for this game?

b) What does your answer in part 'a' mean for the owner/manager, if there are 1000
customers who play the game in a week?

Homework Equations





The Attempt at a Solution



I tried and produced two different solutions, and I don't know which one is right!

Trial 1:
a) expected value for the player
= 0.25*1/2 + 0.50*1/4 + 1*1/4 - 1
= (-) $0.50

b) for the owner, the expected value is (+) $0.50
so , assuming each customer plays once,
he will gain 1000*0.50
= $500/wk

Trial 2:

Customers expected winnings = (1/2)*$0.25 + (1/4)*$0.5 + (1/4)*$2 - $1 = - $0.25

This means the owner makes a profit of 25c per customer on average, so $250 in an average week of 1000 customers.
---------------------------------------------
Could you please check for me? I am so confused!
 
Physics news on Phys.org


Glen Maverick said:
Trial 1:
a) expected value for the player
= 0.25*1/2 + 0.50*1/4 + 1*1/4 - 1
= (-) $0.50

Your (a) is wrong - you wrote 1*1/4 but it should be 2*1/4 .
 


Thank you. I can see that trial 2 is correct.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 15 ·
Replies
15
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K