Question

Game A has you roll a fair die once and receive the number of dollars that is equal to the value on the top face. Game \(B\) has you roll a fair die twice and receive the number of dollars that is the maximum of the two values that show on the top face. It costs \(\$ 3\) to play game \(A\) and \(\$ 4\) to play game \(B\). Which game would you choose?

Short answer

Choose Game A, as it has a higher expected profit of $0.50 compared to Game B's $0.47.

Step-by-step solution

Calculate Expected Value for Game A

For Game A, you roll a fair six-sided die and receive the number of dollars equal to the top face. Thus, the outcomes are 1, 2, 3, 4, 5, and 6. Since each outcome is equally likely, the probability for each is \( \frac{1}{6} \). The expected value (EV) of Game A is calculated as: \( \text{EV}_A = \frac{1}{6}(1+2+3+4+5+6) = \frac{1}{6} \times 21 = 3.5 \).

Calculate Profit for Game A

To find the expected profit for Game A, subtract the cost of playing from the expected payout: \( \text{Profit}_A = 3.5 - 3 = 0.5 \). Game A yields an expected profit of $0.50.

Calculate Expected Value for Game B

For Game B, you roll the die twice and choose the maximum of the two values. The possible outcomes are pairs (1,1), (1,2), ..., (6,6). We calculate the probability of each maximum value and then find the expected value. The probability that the maximum is \(n\) takes into account all combinations where at least one die shows \(n\) and none shows a higher value. Compute each separately and find \( \text{EV}_B = \sum_{i=1}^{6} i \cdot P(\max = i) \, = \frac{6}{36} \times 1 + \frac{11}{36} \times 2 + \frac{15}{36} \times 3 + \frac{19}{36} \times 4 + \frac{23}{36} \times 5 + \frac{36}{36} \times 6 = 4.47 \).

Calculate Profit for Game B

Subtract the cost of playing Game B from its expected value to find the expected profit: \( \text{Profit}_B = 4.47 - 4 = 0.47 \). Game B yields an expected profit of $0.47.

Compare the Profits of Both Games

Now compare the profit from Game A and Game B. Game A has an expected profit of $0.50, and Game B has an expected profit of $0.47. Since $0.50 is greater than $0.47, Game A offers a higher expected profit.

Key concepts

Probability

Probability is a fundamental concept in understanding games of chance, such as rolling dice games. It expresses the likelihood of an event happening in numerical terms, typically between 0 and 1. In the context of this problem, we are dealing with fair six-sided dice. Hence, each face has the same chance of landing face up.
For Game A, the probability of rolling any specific number (1 through 6) is \( \frac{1}{6} \). This makes it relatively straightforward to calculate outcomes.Game B, however, is a little more complex as you roll two dice and take the maximum. Here, finding the probability involves calculating the chances of different pairs, since more than one combination might result in the same maximum value.
For example, both dice showing "3" or one showing "3" while the other shows a lower number both result in "3" as the maximum. Each possible maximum involves a slightly different calculation of probability.

Decision Making

In decision making, especially involving games or bets, expected value plays a crucial role. It helps in forecasting long-term outcomes and making informed choices.
In our exercise, deciding between Game A and Game B involves comparing their expected profits. Each game's cost and potential return need evaluation. Decision making here means choosing the option with a higher expected profit since it promises better returns in a statistical context. Steps to making such decisions often involve:

• Identifying possible outcomes.
• Calculating the probabilities of these outcomes.
• Weighing these probabilities to find expected values and deducting any costs.

Such an analysis helps avoid reliance on luck or intuition alone, promoting a more strategic approach.

Expected Profit

Expected profit is what really counts when evaluating which game might be preferable in the long run. It considers both the expected value of the game and the cost of playing. For Game A, the expected value calculation is quite straightforward: the average of all possible outcomes is $3.50, but since it costs $3 to play, the expected profit becomes $$3.50 - $3 = $0.50.
For Game B, it's slightly more intricate due to factoring in two dice. The expected value here turns out to be $4.47, and after considering the $4 cost, the expected profit becomes $$4.47 - $4 = $0.47.
So, when deciding solely based on expected profit, Game A offers $0.50 compared to Game B's $0.47, thus making Game A the slightly better choice, revealing the importance of thorough calculation in decision contexts.