Math Problem Statement

Assume that it costs $3 to play a state's daily number. The player chooses a three-digit number between 000 and 999, inclusive, and if the number is selected that day, then the player wins $850 (this means the player's profit is $850 - $3 = $847). If the player buys 5 chances on 5 different numbers, what is the expected value for the player?

Solution

The image shows a problem where the player buys 5 chances to play a state’s daily number, with each number costing $3. If the player wins, they receive $850, but they lose the $3 spent to enter the game. The goal is to find the expected value for the player when purchasing 5 chances.

To solve this, let’s break down the steps:

  1. The probability of winning:
    There are 1000 possible three-digit numbers (from 000 to 999 inclusive), so the probability of winning on a single play is: P(win)=11000P(\text{win}) = \frac{1}{1000}

  2. The profit for one play:
    If the player wins, they receive $850, but they paid $3 to play. So, the net profit for a win is: Profit for one win=8503=847\text{Profit for one win} = 850 - 3 = 847 If they lose, the profit is negative $3, as they paid $3 to play and did not win.

  3. Expected value for one play:
    The expected value EE for one play is calculated using the formula: E(one play)=P(win)×Profit for one win+P(lose)×Loss for one playE(\text{one play}) = P(\text{win}) \times \text{Profit for one win} + P(\text{lose}) \times \text{Loss for one play} Where: P(lose)=1P(win)=9991000P(\text{lose}) = 1 - P(\text{win}) = \frac{999}{1000} Substituting the values: E(one play)=11000×847+9991000×(3)E(\text{one play}) = \frac{1}{1000} \times 847 + \frac{999}{1000} \times (-3) E(one play)=0.8472.997=2.15E(\text{one play}) = 0.847 - 2.997 = -2.15 So the expected value per play is approximately -$2.15.

  4. Expected value for 5 plays:
    Since the player is purchasing 5 chances, the expected value for 5 plays is: E(5 plays)=5×E(one play)=5×(2.15)=10.75E(\text{5 plays}) = 5 \times E(\text{one play}) = 5 \times (-2.15) = -10.75 Thus, the expected value for the player is approximately -$10.75.

Let me know if you'd like further details or have any questions!

Here are some related questions to explore:

  1. How would the expected value change if the player purchased more than 5 chances?
  2. What if the prize amount was doubled to $1700—how would that affect the expected value?
  3. How can the expected value be used to inform the player's strategy in this situation?
  4. What is the difference between the probability of winning and the expected value of the game?
  5. How does the number of chances purchased affect the variance of the player's winnings?

Tip: The expected value is a powerful tool to understand long-term outcomes in probability-based games. It helps quantify whether a game is favorable or not.

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Math Problem Analysis

Mathematical Concepts

Probability
Expected Value
Basic Arithmetic

Formulas

Expected Value = Σ (Probability of Outcome × Value of Outcome)
P(win) = 1/1000
Profit = Winnings - Cost

Theorems

Law of Total Expectation

Suitable Grade Level

Grades 10-12