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Chapter 3: Problem 54

Suppose that we want to generate the outcome of the flip of a fair coin but that all we have at our disposal is a biased coin which lands on heads with some unknown probability \(p\) that need not be equal to \(\frac{1}{2}\). Consider the following procedure for accomplishing our task. 1\. Flip the coin. 2\. Flip the coin again. 3\. If both fli?s land heads or both land tails, retum to step 1 . 4\. Let the result of the last flip be the result of the experiment. (a) Show that the result is equally likely to be either heads or tails. (b) Could we use a simpler procedure that continues to flip the coin until the last two flips are different and then lets the result be the outcome of the final flip?

Short Answer

Expert verified
In summary, the given procedure of flipping a biased coin twice and considering the second flip as the experiment result, if the outcomes of both flips are different, results in a fair coin flip with equal probabilities of getting heads or tails (\(P(\text{Heads}) = P(\text{Tails})\)). Additionally, a simpler procedure that continues to flip the coin until the last two flips are different and then considers the outcome of the final flip as the result also yields a fair coin flip with equal probabilities of getting heads or tails.

Step by step solution

01

Probability of each outcome for the given procedure

Let's consider all possible outcomes for the first two flips. Each coin flip result is either Heads (H) or Tails (T). So there are 4 possibilities: HH, HT, TH and TT. We can use a probability tree to represent this. \[ \require{cancel} \begin{array}{cccccc} & & - & - - & H(1) - & - -| \\ & / / / & & & edge1/ \\ & p & H(1) & edge2/(p) & edge3/ \Rightarrow & edge4/(1-p) \\ 1 / & ht & & / / & T(1)- & / / \\ \\ edge0/ \\ & \cancel{\cancel{\cancel{\cancel{( }}} \\ 0 \ & - & - -& - - & H(2)- & / \\ & p & T(2) & edge5/(1-p) & T(2)edge6/\Rightarrow & edge7/ (p) \\ & \cancel{\cancel{\cancel{t}}} & & / & T(2)- & \\ \end{array} \] Based on the procedure, we can eliminate the cases where both flips are the same (HH and TT) and go back to step 1. Therefore, we are left with the desirable outcomes of HT (outcome in the first flip H, outcome in the second flip T) and TH (outcome in the first flip T, outcome in the second flip H). The probability of each outcome is: - HT: \(p(1-p)\) - TH: \((1-p)p\) Now, we need to show that the result of the application is equally likely to be heads or tails.
02

Probabilities of heads and tails results

The probability of getting heads as the result (from the HT outcome) is: \(P(\text{Heads}) = p(1-p)\) The probability of getting tails as the result (from the TH outcome) is: \(P(\text{Tails}) = (1-p)p\) Since both probabilities are equal, we can conclude that the given procedure will result in a fair coin flip with equal chances of getting heads or tails: \(P(\text{Heads}) = P(\text{Tails})\) So, the result of the given procedure is equally likely to be either heads or tails. (b) Now, we have to analyze if it's possible to use a simpler procedure that keeps flipping the coin until the last two flips are different, and then consider the outcome of the final flip as the result. Let's consider the probability of this new procedure having different flips: \(P(\text{different}) = p(1-p) + (1-p)p = 2p(1-p)\). Now let's analyze the last flip: - If last flip is H: \(P(\text{H}|\text{different}) = \frac{p(1-p)}{2p(1-p)} = \frac{1}{2}\) - If last flip is T: \(P(\text{T}|\text{different}) = \frac{(1-p)p}{2p(1-p)} = \frac{1}{2}\) Thus, this alternative procedure would also yield a fair coin flip since the probabilities of getting heads or tails are equal. Therefore, we could use this simpler procedure to flip the biased coin fairly.

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Key Concepts

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

Biased Coin
A biased coin is one where the probability of landing heads is not equal to the probability of landing tails. This can happen due to imperfections in the coin or other influencing factors. Unlike a fair coin, where each outcome (heads or tails) has an equal probability of occurring, a biased coin might favor one side more than the other. For example, if the chance of landing heads is 0.7 and tails is 0.3, the coin is biased. Here, the key challenge is to generate equal likelihood outcomes from such a biased setup, where initially the probability is represented by an unknown variable \(p\) for heads and \(1-p\) for tails.
Fair Coin Toss
A fair coin toss is one where the probabilities of getting heads or tails are equal, i.e., each has a probability of \(\frac{1}{2}\). This is the ideal scenario people often rely on for random outcomes, like deciding which team starts a game. When using a biased coin to simulate a fair coin toss, the goal is to ensure that despite the coin's bias, the end result mimics a fair situation. The procedure discussed in the exercise accomplishes this by using multiple flips and conditions to finally derive an outcome. Thus, through specific sequential flips and retrials, we achieve a result that is equally likely to land as heads or tails.
Probability Tree
A probability tree is a crucial tool in visualizing and solving problems involving sequential events, like coin tosses. It breaks down events into branches, making calculation easier and helping to understand the flow of possible outcomes.
In the exercise solution, a probability tree is suggested to map out the outcomes of the first two flips: HH, HT, TH, and TT. The path that leads to equally desirable outcomes (HT and TH) is isolated by eliminating paths where both flips result in heads or both in tails. Each branch of the tree holds a probability, which helps in calculating the likelihood of reaching the desired result. It's an excellent way to simplify complex probability computations into manageable, visual steps.
Equal Likelihood
Equal likelihood means that two or more outcomes have the same probability of occurring. In the context of the coin-flipping exercise, the aim is to create a situation where the outcomes of heads or tails are equally probable.
This is achieved by constructing a method where the probability of getting heads (HT sequence) and getting tails (TH sequence) from our coin flips is exactly the same: \(p(1-p)\) for heads and \((1-p)p\) for tails. By ensuring that both sequences have the same probability, we achieve equal likelihood for heads or tails, thus simulating a fair coin flip using a biased coin. This process employs the mathematics of conditional probability and expected value to validate the fairness of the method.

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Most popular questions from this chapter

Let \(Q_{n}\) denote the probability that in \(n\) tosses of a fair coin no run of 3 consecutive heads appears. Show that $$ \begin{aligned} &Q_{n}=\frac{1}{2} Q_{n-1}+\frac{1}{4} Q_{n-2}+\frac{1}{8} Q_{n-3} \\ &Q_{0}=Q_{1}=Q_{2}=1 \end{aligned} $$ Find \(Q_{8}\). HINT: Condition on the first tail.

Stores \(A, B\), and \(C\) have 50,75 , and 100 employees and, respectively, 50 , 60 , and 70 percent of these are women. Resignations are equally likely among all employees, regardless of sex. One employee resigns, and this is a woman. What is the probability that she works in store \(C\) ?

Suppose that an ordinary deck of 52 cards is shuffled and the cards are then turned over one at a time until the first ace appears. Given that the first ace is the 20 th card to appear, what is the conditional probability that the card following it is the (a) ace of spades; (b) two of clubs?

Consider a collection of \(n\) individuals. Assume that each person's birthday is equally likely to be any of the 365 days of the year and also that the birthdays are independent. Let \(A_{i, j}\) denote the event that persons \(i\) and \(j\) have the same birthday, \(i \neq j\). Show that these events are pairwise independent. That is, \(A_{i j}\) and \(A_{r, s}\) are independent but the \(\left(\begin{array}{l}n \\ 2\end{array}\right)\) events \(A_{i j}, i \neq j\) are not independent.

What is the probability that at least one of a pair of fair dice lands on 6, given that the sum of the dice is \(i, i=2,3, \ldots, 12 ?\)
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