Question 1178011: In a lottery game, a player picks six numbers from 1 to 30. If the player matches all six numbers, they win $40,000 dollars. Otherwise, they lose $1.
The draw is WITHOUT REPLACEMENT.
So the ticket machine only allows the player to choose a particular number once.
Keep in mind that the ORDER of the numbers DOES NOT matter.
You will need to dust off your knowledge of binomial coefficients to complete this problem. (This is 'n choose k'.)
What is the expected value of this game
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to calculate the expected value of the lottery game:
**1. Calculate the Total Number of Combinations**
* We need to find the number of ways to choose 6 numbers from 30, where order doesn't matter. This is a combination problem, denoted as "30 choose 6" or C(30, 6).
* C(30, 6) = 30! / (6! * 24!) = 593,775
**2. Calculate the Probability of Winning**
* There's only one winning combination.
* The probability of winning is 1 / 593,775.
**3. Calculate the Probability of Losing**
* The probability of losing is 1 - (probability of winning).
* Probability of losing = 1 - (1 / 593,775) = 593,774 / 593,775.
**4. Calculate the Expected Value**
* Expected value = (probability of winning * winnings) + (probability of losing * loss)
* Expected value = (1 / 593,775 * $40,000) + (593,774 / 593,775 * -$1)
* Expected value = (40,000 / 593,775) - (593,774 / 593,775)
* Expected value = (40,000 - 593,774) / 593,775
* Expected value = -553,774 / 593,775
* Expected value ≈ -$0.9326
**Therefore, the expected value of this game is approximately -$0.93. This means that on average, a player can expect to lose about $0.93 for each ticket they buy.**
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