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A coin having probability \(p\) of coming up heads is successively flipped until two of the most recent three flips are heads. Let \(N\) denote the number of flips. (Note that if the first two flips are heads, then \(N=2\).) Find \(E[N]\).

Short Answer

Expert verified
The expected number of flips, E[N], to have two of the most recent three flips be heads for a coin with probability p of coming up heads is given by the expression: \(E[N] = \frac{2p^2 + (1-p)p + (1-p)p^2}{1 - (1-p)^2 - (1-p)p}\)

Step by step solution

01

Set up a recursive equation

Consider the general case where the probability of having two heads out of the most recent three flips is given by E[N]. Since the flips are independent events, we can think of E[N] as the expectation when we flip the coin, considering the outcomes for the first two flips. There are three cases: 1. The first two flips are heads (HH), which happens with probability \(p^2\) and stops the flips with N = 2. 2. The first flip is tails (T) and the second flip is heads (H). This happens with the probability \((1-p)p\) and takes us to the case with one head in the first two flips. 3. The first two flips are tails (TT), which happens with probability \((1-p)^2\). In this case, we're back to the original problem since we still have to find two of the most recent three flips be heads. Therefore, we can set up the following recursive equation: \(E[N] = p^2 \cdot 2 + (1-p)p\cdot (1 + E[N']) + (1-p)^2\cdot (1 + E[N])\) Here, E[N'] is the expectation in the case where there is one head in the first two flips.
02

Find the expectation for the case with one head in the first two flips

For the case with one head in the first two flips (E[N']), and knowing that the third flip will be either a head with probability p or a tail with probability 1-p, we have three possible outcomes: 1. The third flip is a head (H), which happens with probability p. In such a scenario, we have two heads (2 out of 3), and N' = 1. 2. The third flip is a tail (T), which will happen with a probability (1-p). This case takes us back to the original problem with E[N], where we need to find two of the most recent three flips be heads. So, we have the following equation for E[N']: \(E[N'] = p \cdot 1 + (1-p) \cdot (1 + E[N])\)
03

Combine the equations and solve for E[N]

Now, we can substitute the E[N'] in the E[N] equation, and get: \(E[N] = p^2 \cdot 2 + (1-p)p \cdot (1 + (p \cdot 1 + (1-p) (1+E[N]))) + (1-p)^2 \cdot (1 + E[N])\) Next, we can simplify and rearrange to solve for E[N]: \(E[N] - (1-p)^2 E[N] - (1-p)p E[N] = 2p^2 + (1-p)p + (1-p)p^2\) Now, factor out E[N] and find the final expression for E[N]: \(E[N] = \frac{2p^2 + (1-p)p + (1-p)p^2}{1 - (1-p)^2 - (1-p)p}\) The above expression gives the expected number of flips to have two of the most recent three flips be heads for a coin with probability p of coming up heads.

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Key Concepts

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

Expected Value
The concept of Expected Value plays a crucial role when dealing with probabilities, like the chance of a coin landing heads. In simple terms, Expected Value (denoted as \( E[X] \)) is the average or mean value you would anticipate over a large number of trials. It's a way to predict what to expect in the long run.
For example, when you flip a fair coin, which has equal probability of heads or tails, the expected number of heads you get is around half of the number of flips.
In the context of this exercise, the expected value \( E[N] \) represents the average number of coin flips required until two out of the last three are heads. This gives us insight into how many attempts are usually needed, on average, emphasizing the significance of expected value in predicting outcomes in random processes.
Markov Chains
Markov Chains serve as a mathematical system that undergoes transitions between states, following certain probabilistic rules. Markov Chains are "memoryless," which means that the next state depends solely on the current state and not on the sequence of events prior to it. This is a property known as the "Markov property."
In the coin flipping exercise, the states can be the sequences of heads and tails in the recent flips. Each flip transitions from one state to another, and these transitions can be described with probabilities. For example, from a state of 'one head in the last two flips', there are probabilities to flip another head or tails, transitioning to states with 'two heads' or 'returns to zero heads'.
This concept provides a structured way to model and analyze random processes where the future is independent of the past, once you know the present. By leveraging Markov Chains, we can understand the transitions of getting a certain sequence of heads and, ultimately, the expectation of steps to reach the condition required.
Stochastic Processes
Stochastic Processes encompass collections of random variables that evolve over time, often used to model uncertainties and systems evolving randomly. Each state or event is part of a set of possible outcomes, and stochastic processes help in predicting future probabilities based on randomness and inherent unpredictability.
In terms of coin flipping, this exercise is a type of stochastic process where each flip of the coin is a random event influenced by probability \( p \), the likelihood of landing on heads. As you continue to flip the coin, you observe a sequence of random outcomes that form the stochastic process, which could be visualized through a flow of random events over time.
This concept is valuable in analyzing how the coin transitions among different states of heads and tails configurations over various trials. It provides a broad framework under which Markov Chains are specialized to capture these random transitions. Understanding stochastic processes can significantly enhance insights into systems and phenomena dictated by random occurrences.

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Most popular questions from this chapter

A coin, having probability \(p\) of landing heads, is continually flipped until at least one head and one tail have been flipped. (a) Find the expected number of flips needed. (b) Find the expected number of flips that land on heads. (c) Find the expected number of flips that land on tails. (d) Repeat part (a) in the case where flipping is continued until a total of at least two heads and one tail have been flipped.

Let \(A\) and \(B\) be mutually exclusive events of an experiment. If independent replications of the experiment are continually performed, what is the probability that \(A\) occurs before \(B ?\)

A prisoner is trapped in a cell containing three doors. The first door leads to a tunnel that returns him to his cell after two days of travel. The second leads to a tunnel that returns him to his cell after three days of travel. The third door leads immediately to freedom. (a) Assuming that the prisoner will always select doors 1,2, and 3 with probabilities \(0.5,0.3,0.2\), what is the expected number of days until he reaches freedom? (b) Assuming that the prisoner is always equally likely to choose among those doors that he has not used, what is the expected number of days until he reaches freedom? (In this version, for instance, if the prisoner initially tries door 1 , then when he returns to the cell, he will now select only from doors 2 and 3.) (c) For parts (a) and (b) find the variance of the number of days until the prisoner reaches freedom.

A coin that comes up heads with probability \(p\) is continually flipped until the pattern \(\mathrm{T}, \mathrm{T}\), \(\mathrm{H}\) appears. (That is, you stop flipping when the most recent flip lands heads, , ind the two immediately preceding it lands tails.) Let \(X\) denote the number of flips made, and find \(E[X]\).

A manuscript is sent to a typing firm consisting of typists \(A, B\), and \(C\). If it is typed by \(A\), then the number of errors made is a Poisson random variable with meap \(2.6\); if typed by \(B\), then the number of errors is a Poisson random variable with mean 3 ; and if typed by \(C\), then it is a Poisson random variable with mean 3.4. Let \(X\) denote the number of errors in the typed manuscript. Assume that each typist is equally likely to do the work. (a) Find \(E[X]\). (b) Find \(\operatorname{Var}(X)\).

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