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Suppose that you roll a single die. If an odd number (1,3,0 5) comes up, you win the amount of your roll (\$1, \$3, or \$5 respectively). If an even number \((2,4,\) or 6\()\) comes up, you have to pay the house the amount of your roll \((\$ 2, \$ 4,\) or \(\$ 6\) respectively). (a) Find the expected payoff for this game. (b) Is this a fair game? Explain.

Short Answer

Expert verified
The expected payoff of the game is \$0, hence it could be considered a fair game as there is no anticipated gain or loss over time.

Step by step solution

01

Determine payout possibilities

The results of rolling a die are the numbers from 1 to 6. This means there are six possible outcomes, each with a probability of \(\frac{1}{6}\). If an odd number (1,3,5) is rolled, the payout is the rolled amount. If an even number (2,4,6) is rolled, the payout is the rolled amount, but as a loss.
02

Compute Expected Payoff

The expected payoff is computed using the sum of the product of each outcome and its probability. Hence multiply each payoff by its probability and sum them up. The expected payoff \(\E\) is given by: \[\E = \frac{1}{6}(\$1) + \frac{1}{6}(\$3) + \frac{1}{6}(\$5) - \frac{1}{6}(\$2) - \frac{1}{6}(\$4) - \frac{1}{6}(\$6)\]
03

Simplify expected payoff

Now simplify the equation to find the expected payoff: \[\E = \$1/6 + \$1/2 + \$5/6 - \$1/3 - \$2/3 - \$1 = 0\]
04

Determine if the game is fair

Since a game is fair if the expected value is 0, and here it is found to be 0, thus this game is fair. There's neither anticipated gain or loss over time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value
In probability theory, the expected value is a fundamental concept that gives a measure of the "center" of a probability distribution. It's essentially the average outcome you'd expect from a random experiment if you could repeat it infinitely many times. To calculate the expected value for our die-rolling game, we multiply each possible payout by its probability and sum these products. This results in:\[ \E = \frac{1}{6}(1) + \frac{1}{6}(3) + \frac{1}{6}(5) - \frac{1}{6}(2) - \frac{1}{6}(4) - \frac{1}{6}(6) \]This equation simplifies to:- for odd outcomes (1, 3, and 5): - \( \frac{1}{6}(1) = \frac{1}{6} \) - \( \frac{1}{6}(3) = \frac{1}{2} \) - \( \frac{1}{6}(5) = \frac{5}{6} \)- for even losses (2, 4, and 6): - \( \frac{1}{6}(2) = \frac{1}{3} \) - \( \frac{1}{6}(4) = \frac{2}{3} \) - \( \frac{1}{6}(6) = 1 \)After simplifying, each component contributes to an expected value of 0, indicating that over time, you'd neither win nor lose money on average.
Fair Game
A game is considered "fair" if the expected value of the payouts from that game is zero. In a fair game, players do not have an advantage or disadvantage. Over time, the average net gain for a participant is zero. In our exercise, we determine fairness by calculating the expected value of the game, which we found to be zero. This implies:
  • There is no net gain or loss on average over many plays.
  • The probabilities and payouts are balanced perfectly, so neither the player nor the house stands to systematically win.
Thus, given this evaluation, we conclude that the described die-rolling game qualifies as a fair game.
Probability Distribution
Understanding probability distribution is key to analyzing games of chance. A probability distribution lists all possible outcomes of an experiment and assigns a probability to each one.In our dice game, there are six possible outcomes when rolling a die:
  • 1, where you win \\(1
  • 2, where you lose \\)2
  • 3, where you win \\(3
  • 4, where you lose \\)4
  • 5, where you win \\(5
  • 6, where you lose \\)6
Each outcome has a probability of \( \frac{1}{6} \).By examining this probability distribution, compounded by the odds, you can effectively calculate the expected value and determine the fairness of the game. This systematic approach helps to decide if engaging in such a game is strategically advantageous or neutral in the long run.

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