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Chapter 3: Problem 194

Odd Man Out. Three people play a game called "Odd Man Out." In this game, each player flips a fair coin until the outcome (heads or tails) for one of the players is not the same as that for the other two players. This player is then "the odd man out" and loses the game. Find the probability that the game ends (i.e., either exactly one of the coins will fall heads or exactly one of the coins will fall tails) after only one toss by each player. Suppose one of the players, hoping to reduce his chances of being the odd man out, uses a two- headed coin. Will this ploy be successful? Solve by listing the sample points in the sample space.

Short Answer

Expert verified
Using a two-headed coin reduces the chance of being the odd man out to 1/2.

Step by step solution

01

Define the Sample Space

Each player flips a fair coin. Thus, the possible outcomes for one flip of each coin are: HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT. These represent all the possible outcomes of the first toss for all three players. H stands for heads and T for tails.
02

Identify 'Odd Man Out' Scenarios

For a player to be the 'odd man out', their outcome must differ from the other two. This occurs if exactly one player gets either heads while the other two get tails or tails while the other two get heads. The events representing this scenario are: HHT, HTH, THH, TTH, THT, and HTT.
03

Count Valid Outcomes

From the sample space, we identified 6 events where there is an odd man out: HHT, HTH, THH, TTH, THT, and HTT. These are the outcomes where the game ends after one toss by each player.
04

Calculate the Probability

Each outcome is equally likely since the coins are fair, except for one player who might use a two-headed coin. Thus, without the unfair coin, the probability of each outcome is \( \frac{1}{8} \). There are 6 successful outcomes where there is an odd man out, so the probability is \( \frac{6}{8} = \frac{3}{4} \).
05

Evaluate the Two-Headed Coin Scenario

If one player uses a two-headed coin, he will always show heads. The remaining possible outcomes are HHT and HTH, instead of the 6 fair outcomes. Therefore, the probability of the game ending becomes \( \frac{2}{4} = \frac{1}{2} \).
06

Compare Probabilities

Compare the probabilities: with fair coins the game ends after one toss with a probability of \( \frac{3}{4} \), while with a two-headed coin, it drops to \( \frac{1}{2} \). This analysis shows that using a two-headed coin indeed lowers the chance of being the odd man out to 1/2 compared to using three fair coins.

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Key Concepts

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

Sample Space
In probability, the sample space represents all possible outcomes of a random experiment. When three players each flip a fair coin, the sample space comprises all the combinations of heads (H) and tails (T) they can achieve in one toss. To visualize this, imagine each player flipping their coin simultaneously. The possible results are:
  • HHH: All three show heads.
  • HHT: First two show heads, third shows tail.
  • HTH: First and third show heads, second shows tail.
  • HTT: First shows head, last two show tails.
  • THH: First shows tail, others show heads.
  • THT: First and third show tails, second shows head.
  • TTH: First two show tails, third shows head.
  • TTT: All show tails.
Each of these represents a point in the sample space, illustrating all possible outcomes of the coin flips.
Fair Coin
A fair coin is one that has an equal probability of landing on heads or tails when tossed. In mathematical terms, this means that for each toss, the probability of obtaining heads (H) is \( \frac{1}{2} \) and the same applies for tails (T), making it an unbiased coin. In the 'Odd Man Out' game, each player starts with a fair coin, implying every outcome within the sample space is initially equally likely.Using fair coins, no player has a statistical advantage over the others. Hence, each of the eight potential outcomes listed in the sample space has an equal probability of occurrence, specifically, \( \frac{1}{8} \) for each individual outcome.
Two-Headed Coin
A two-headed coin is a coin designed to show heads on both sides, ensuring the outcome is always heads (H) when flipped. In games involving probability, like 'Odd Man Out', a two-headed coin can significantly alter the outcomes. It introduces a bias where one result is guaranteed, skewing the fair chance characteristic necessary for an unbiased game. When a player opts for a two-headed coin in the game, their flips will always be heads. This reduces the sample space and alters the probability calculations because certain outcomes are no longer possible with its use. Specifically, only scenarios where other players' fair coins produce a tail alongside two heads (i.e., HHT and HTH) remain feasible for ending the game.
Probability Calculation
Calculating probability involves determining how likely an outcome is, given the sample space. For 'Odd Man Out', we want to find the probability of the game ending after the first coin flips.With all fair coins, we see 6 out of 8 outcomes result in an odd man out (HHT, HTH, THH, TTH, THT, HTT). Thus, the probability is \( \frac{6}{8} = \frac{3}{4} \). This means there's a 75% chance the game ends immediately.However, introduce a two-headed coin, and the odds change. The player guaranteed to show heads alters the sample space outcomes, leaving only HHT and HTH viable. Thus, the new probability calculation becomes \( \frac{2}{4} = \frac{1}{2} \), reducing the likelihood of concluding the game after a single toss to 50%.

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Most popular questions from this chapter

Colors of M\&Ms candies. Originally, M\&Ms Plain Chocolate Candies came in only a brown color. Today, M\&Ms in standard bags come in six colors: brown, yellow, red, blue, orange, and green. According to Mars Corporation, today \(24 \%\) of all M\&Ms produced are blue, \(20 \%\) are orange, \(16 \%\) are green, \(14 \%\) are yellow, \(13 \%\) are brown, and \(13 \%\) are red. Suppose you purchase a randomly selected bag of M\&Ms Plain Chocolate Candies and randomly select one of the M\&Ms from the bag. The color of the selected M\&M is of interest. a. Identify the outcomes (sample points) of this experiment. b. Assign reasonable probabilities to the outcomes, part a. c. What is the probability that the selected M\&M is brown (the original color)? d. In \(1960,\) the colors red, green, and yellow were added to brown M\&Ms. What is the probability that the selected M\&M is either red, green, or yellow? e. In \(1995,\) based on voting by American consumers, the color blue was added to the M\&M mix. What is the probability that the selected M\&M is not blue?

State the additive rule of probability for any two events.

Forensic analysis of JFK assassination bullets. Following the assassination of President John F. Kennedy (JFK) in \(1963,\) the House Select Committee on Assassinations (HSCA) conducted an official government investigation. The HSCA concluded that although there was a probable conspiracy, involving at least one additional shooter other than Lee Harvey Oswald, the additional shooter missed all limousine occupants. A recent analysis of assassination bullet fragments, reported in the Annals of Applied Statistics (Vol. 1,2007 ), contradicted these findings, concluding that evidence used to rule out a second assassin by the HSCA is fundamentally flawed. It is well documented that at least two different bullets were the source of bullet fragments used in the assassination. Let \(E=\\{\) bullet evidence used by the HSCA\\}, \(T=\\{\) two bullets used in the assassination \(\\},\) and \(T^{C}=\) \(\\{\) more than two bullets used in the assassination \(\\}\). Given the evidence \((E),\) which is more likely to have occurred \(-\) two bullets used \((T)\) or more than two bullets used \(\left(T^{C}\right) ?\) a. The researchers demonstrated that the ratio, \(P(T \mid E) / P\left(T^{C} \mid E\right),\) is less than \(1 .\) Explain why this result supports the theory of more than two bullets used in the assassination of JFK. b. To obtain the result in part a, the researchers first showed that \(P(T \mid E) / P\left(T^{C} \mid E\right)=[P(E \mid T) \cdot P(T)] /\left[P\left(E \mid T^{C}\right) \cdot P\left(T^{C}\right)\right]\) Demonstrate this equality using Bayes's theorem.

Chance of rain. Answer the following question posed in the Atlanta Journal- Constitution: When a meteorologist says, "The probability of rain this afternoon is \(4, "\) does it mean that it will be raining \(40 \%\) of the time during the afternoon?

Florida license plates. In the mid-1980s, the state of Florida ran out of combinations of letters and numbers for its license plates. Then, each license plate contained three letters of the alphabet, followed by three digits selected from the 10 digits \(0,1,2, \ldots, 9\) a. How many different license plates did this system allow? b. New Florida tags were obtained by reversing the procedure, starting with three digits followed by three letters. How many new tags did the new system provide? c. Since the new tag numbers were added to the old numbers, what is the total number of licenses available for registration in Florida?
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