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Gerolamo Cardano in his book, The Gambling Scholar, written in the early 1500 s, considers the following carnival game. There are six dice. Each of the dice has five blank sides. The sixth side has a number between 1 and 6 -a different number on each die. The six dice are rolled and the player wins a prize depending on the total of the numbers which turn up. (a) Find, as Cardano did, the expected total without finding its distribution. (b) Large prizes were given for large totals with a modest fee to play the game. Explain why this could be done.

Short Answer

Expert verified
(a) The expected total is 3.5. (b) Large prizes are justified because average outcomes are low, making large wins rare.

Step by step solution

01

Understanding the Dice

The game involves six dice, each with one numbered side and five blank sides. Each numbered side shows a different number from 1 to 6.
02

Expectation of One Die

For each die, the expected value can be calculated by noting that it has a 1/6 probability of landing on its numbered side. Therefore, the expected value for each die is its numbered side times the probability, i.e., \(\frac{1}{6} \times 3.5 = \frac{3.5}{6} = \frac{7}{12}\), assuming the die throws its numbered side.
03

Calculate Total Expected Value

For six dice, since each die's expected value is independent of the others, simply multiply the expected value of one die by the number of dice. This gives the total expected value as \(6 \times \frac{7}{12} = \frac{42}{12} = 3.5\).
04

Rationale Behind Large Prizes

In the game, the design allows for large variability in outcomes because a player can either hit a blank (for no points) or land on a number, leading to an overall low average score (3.5). Offering large prizes for rare outcomes (high totals) attracts players despite a small chance of winning such prizes. The game can still yield a profit because the average total is not high enough to frequently award the larger prizes.

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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, expected value is a fundamental concept that serves as a measure of the center of a probability distribution. It provides the long-term average result of a random event if it were to be repeated numerous times. In essence, it helps us anticipate what to expect.
For a single die in Cardano's game, each die has a unique number and five blank sides. The probability of landing on the number is \(\frac{1}{6}\), which is derived from the single numbered side among the six possible outcomes. Hence, the expected value of a die is calculated by multiplying the value of the numbered side by this probability.
Given that each die averages a return of \(\frac{7}{12}\) when it is included in consideration, and there are six such dice, the total expected value is the sum of the expected values of all dice involved: \(6 \times \frac{7}{12} = 3.5\). This demonstrates how expected value can forecast the average payoff of complex games over time.
Probability Distribution
Probability distribution describes how the probabilities are allocated among various possible outcomes in a random experiment. It represents the likelihood of each possible outcome and provides a complete picture of how randomness plays out.
In the context of Cardano's dice game, each die's sides have defined probabilities: five sides with zero numbers (probabilities of zero) and one with a number (probability \(\frac{1}{6}\)).
  • Five blank sides: Probability of occurrence \(\frac{5}{6}\).
  • One numbered side: Probability of occurrence \(\frac{1}{6}\).
The distribution for each die reflects this, and the combination of six such dice results in a broader probability distribution reflecting the cumulative outcomes possible from all the dice.
The probability distribution is pivotal as it helps determine not only the expected value but also provides an understanding of the range and likelihood of all outcomes in the carnival game. Analyzing this distribution can guide decisions, like setting game rules or prizes.
Independent Events
In probability, events are considered independent if the occurrence of one does not affect the probability of the occurrence of another. This means their outcomes do not influence each other.
  • In Cardano's game, each die acts independently.
  • The outcome of rolling one die does not influence the next.
  • Independence allows us to simply add probabilities or expected values across dice.
Because the dice rolls are independent, we easily compute the total expected value by summing up the expected values of individual dice. This independence is critical in many probability calculations since it simplifies the mathematical treatment of combined events.
Understanding independent events sets the foundation for more complex probability concepts and helps in defining the behavior of systems with multiple random variables, such as this multi-dice game.
Variance
Variance measures how much the outcomes deviate from the expected value, reflecting the spread of a probability distribution. It provides insight into the degree of variability or volatility associated with random events.
In Cardano’s dice game, variance helps us understand why players are occasionally rewarded with large prizes for high totals despite the low expected average score.
  • The variance of one die is calculated based on its probabilities and possible deviations from the expected value \(3.5\).
  • The total variance across the dice can help anticipate such high payouts.
The presence of high variance explains the allure of the game. Players are tempted by the possibility of rare, high-paying outcomes without realizing that high variability often results in lower average earnings over time.
Variance not only influences the perception of risk and reward but is also a key factor in designing games and determining the sustainability and attractiveness of payoffs.

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Most popular questions from this chapter

A coin is tossed until the first time a head turns up. If this occurs on the \(n\) th toss and \(n\) is odd you win \(2^{n} / n,\) but if \(n\) is even then you lose \(2^{n} / n\). Then if your expected winnings exist they are given by the convergent series $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$ called the alternating harmonic series. It is tempting to say that this should be the expected value of the experiment. Show that if we were to do this, the expected value of an experiment would depend upon the order in which the outcomes are listed.

Find the expected value and the variance for the number of boys and the number of girls in a royal family that has children until there is a boy or until there are three children, whichever comes first.

Show that, if \(X\) and \(Y\) are random variables taking on only two values each, and if \(E(X Y)=E(X) E(Y),\) then \(X\) and \(Y\) are independent.

A number is chosen at random from the set \(S=\\{-1,0,1\\}\). Let \(X\) be the number chosen. Find the expected value, variance, and standard deviation of \(X .\)

Let \(X\) and \(Y\) be random variables. The covariance \(\operatorname{Cov}(\mathrm{X}, \mathrm{Y})\) is defined by (see Exercise 6.2 .23\()\) $$ \operatorname{cov}(\mathrm{X}, \mathrm{Y})=\mathrm{E}((\mathrm{X}-\mu(\mathrm{X}))(\mathrm{Y}-\mu(\mathrm{Y}))) $$ (a) Show that \(\operatorname{cov}(\mathrm{X}, \mathrm{Y})=\mathrm{E}(\mathrm{XY})-\mathrm{E}(\mathrm{X}) \mathrm{E}(\mathrm{Y})\) (b) Using (a), show that \(\operatorname{cov}(X, Y)=0,\) if \(X\) and \(Y\) are independent. (Caution: the converse is not always true.) (c) Show that \(V(X+Y)=V(X)+V(Y)+2 \operatorname{cov}(X, Y)\).

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