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A game involves rolling a number cube. You get one point if you roll an even number and negative one point if you roll an odd number.




Roll		123456

Probability		$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{6}$

Point(s)		-11-11-11






Which statements are true regarding this scenario? Check all that apply.


  • The sample space is $\{1,2,3,4,5,6\}$.
  • The expected value for the game is 0.
  • The expected value for the game is $\frac{1}{2}$.
  • The game is fair because the expected value is equal to 0.
  • The game is unfair because the expected value does not equal 0.

Verified Answer :

  • The sample space is \(\{1, 2, 3, 4, 5, 6\}\).
  • The expected value for the game is \(\boxed{0}\).
  • The game is fair because the expected value is equal to \(0\).
  • True statements: "The sample space is {1, 2, 3, 4, 5, 6}.", "The expected value for the game is 0.", and "The game is fair because the expected value is equal to 0."

Explanation

  1. Analyze the sample space

The scenario involves rolling a standard six-sided number cube. The set of all possible outcomes for one roll is \(\{1, 2, 3, 4, 5, 6\}\), which confirms that the sample space is indeed \(\{1, 2, 3, 4, 5, 6\}\). Each of these 6 outcomes occurs with an equal probability of \(\frac{1}{6}\).

  1. Calculate the expected value

To find the expected value (EV) of the game, we multiply each possible point outcome by its probability and sum these products:
\[E[X] = \sum x_i P(x_i)\]
Given the points for odd numbers (1, 3, 5) are \(-1\) and for even numbers (2, 4, 6) are \(+1\), we calculate:
\[E[X] = \left(-1 \times \frac{1}{6}\right) + \left(1 \times \frac{1}{6}\right) + \left(-1 \times \frac{1}{6}\right) + \left(1 \times \frac{1}{6}\right) + \left(-1 \times \frac{1}{6}\right) + \left(1 \times \frac{1}{6}\right)\]
\[E[X] = -\frac{1}{6} + \frac{1}{6} - \frac{1}{6} + \frac{1}{6} - \frac{1}{6} + \frac{1}{6} = 0\]
Thus, the expected value for the game is \(0\).

  1. Evaluate fairness and conclude true statements

A game is defined as "fair" if its expected value is equal to \(0\). Since we have calculated the expected value to be \(0\), we conclude that the game is fair. Any statement claiming the expected value is \(\frac{1}{2}\) or that the game is unfair is false.

Examples

This concept is fundamental in game theory and probability. For instance, if you are designing a carnival game, calculating the expected value helps you determine if the game will be profitable for the operator or if it's "fair" for the players. If a player pays \$1 to play and the expected payout is \$1, the game is fair, meaning that over the long run, neither the player nor the operator gains money.

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