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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, and the two immediately preceding it lands tails.) Let \(X\) denote the number of flips made, and find \(E[X]\).

Short Answer

Expert verified
The expected value of the number of coin flips made until the TTH pattern appears is \(E[X] = \frac{3(1-p)^2 p}{1-(1-p)^3}\).

Step by step solution

01

Note the probabilities

First, let's clearly define the probabilities for the event. \(P(H) = p\) \(P(T) = 1 - p\) Now, let's define the random variable \(X\) as the number of flips made until the pattern TTH appears. #Step 2: Calculate the probability for 3 flips#
02

Calculate the probability for the first 3 flips

After one flip, the outcome doesn't matter, so we will focus on calculating the probability for the first 3 flips. The probability of getting the TTH pattern in the first 3 flips is: \(P(X=3) = (1-p)^2 p\) #Step 3: Calculate the probability for more than 3 flips#
03

Calculate the probability for more than 3 flips

Now, let's consider the case where the pattern TTH occurs after more than 3 flips. We can divide this case into 2 sub-cases: Sub-case 1: The last flip is tail (TTT) Probability: \((1-p)^3\) Sub-case 2: The last flip is head (HTH or TTH) Probability: \((1-p)^2 p\) #Step 4: Write down the total probability in terms of conditional probabilities#
04

Express the conditional probabilities

We can write the total probability for the pattern TTH happening after \(n+1\) flips conditionally based on when TTH happened on the \(n^{th}\) flip: \(P(X = n+1 | X = n) =\) \(P(\text{Sub-case 1}) \times P(\text{Sub-case 2})=\) \((1-p)^3 \times (1-p)^2 p = (1-p)^5 p\) #Step 5: Calculate the expected value of X#
05

Apply the total expectation theorem

Now we can apply the total expectation theorem to find the expected value of the number of flips made: \(E[X] = \sum_{n=3}^{\infty} n \times P(X = n)\) Using the conditional probabilities derived in step 4, the expected value becomes: \(E[X] = 3(1-p)^2 p + \sum_{n=3}^{\infty} (n+1)(1-p)^5 p\) By evaluating the infinite sum, we get: \(E[X] = \frac{3(1-p)^2 p}{1-(1-p)^3}\) This is the expected value of the number of coin flips made until TTH appears.

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Key Concepts

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

Conditional Probability
Understanding conditional probability is essential when analyzing situations where one event's occurrence depends on another. In the context of our exercise, conditional probability allows us to calculate the likelihood of the pattern 'TTH' occurring, given that a certain number of flips have already been made.

To put it simply, if we flip a coin several times, getting a tails (T) followed by another tails and then a heads (H) on the third flip doesn't just happen out of the blue—it depends on what came before. So, if we want to know the probability of TTH happening on the 4th, 5th, 6th flip, and so on, we analyze it knowing what has already occurred; this is where conditional probability steps in. It's like asking 'What are the odds now, given what we've seen so far?'

The formula used in probability theory for conditional probability is typically written as
\( P(A|B) = \frac{P(A \cap B)}{P(B)} \)
where \( P(A|B) \) is the probability of event A given that B has occurred, and \( P(A \cap B) \) is the probability of both A and B happening.
Random Variable
A random variable is a backbone concept in probability theory. It's a rule that assigns a numerical value to each potential outcome in a space of possible outcomes. Think of it as a way to keep score in the game of chance—how many flips till we see 'TTH'? With each toss of the coin, the random variable, denoted by \( X \) in our exercise, keeps track of the flip count until we hit our pattern.

There are two types of random variables: discrete and continuous. Our exercise deals with a discrete random variable because the number of coin flips counts in whole numbers—you can't flip a coin halfway. To calculate the expected value (or mean) of such a variable, we multiply each possible value by its probability, then sum all these products. It's like figuring out the average number of times you'll need to flip that coin to see 'TTH' pop up.
Probability Theory
Probability theory gives us a systematic way to study uncertainty and randomness. It's the mathematical framework that underpins games of chance, statistics, and even machine learning. In this exercise, we use probability theory to tackle questions like: 'What's the chance of flipping a tail?', 'What's the likelihood of flipping a head after two tails?', and 'How many flips on average to get the TTH sequence?'.

Fundamental to this theory are the concepts of events, outcomes, and probabilities. An 'event' is just a set of outcomes—for example, getting 'TTH' is an event that can occur in various ways over multiple coin flips. Each possible sequence of flips leading to 'TTH' is an outcome, and probability theory allows us to assign each outcome a numerical probability, so we can make predictions about how likely different sequences are.
Total Expectation Theorem
The Total Expectation Theorem, sometimes called the Law of Total Expectation, is a powerful tool that simplifies the calculation of expected values by breaking down complex random processes into simpler ones. It's particularly useful when a problem involves several stages or layers of random events.

In practical terms, you first identify all the possible scenarios or conditions your random variable can experience. For each, you calculate the expected value given that scenario. Then, you weigh each expected value by the probability of its corresponding scenario and add them up to get the overall expected value.

For example, in the coin flip exercise, we don't just consider the chance of getting 'TTH' in a vacuum; we look at the different ways in which flips can progress and conditionally calculate the expected number of flips until 'TTH' happens. This theorem incredibly simplifies the process by allowing us to treat each potential path to 'TTH' separately before bringing them together for the big picture.

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Most popular questions from this chapter

Two players alternate flipping a coin that comes up heads with probability \(p\). The first one to obtain a head is declared the winner. We are interested in the probability that the first player to flip is the winner. Before determining this probability, which we will call \(f(p)\), answer the following questions. (a) Do you think that \(f(p)\) is a monotone function of \(p ?\) If so, is it increasing or decreasing? (b) What do you think is the value of \(\lim _{p \rightarrow 1} f(p) ?\) (c) What do you think is the value of \(\lim _{p \rightarrow 0} f(p) ?\) (d) Find \(f(p)\).

The number of coins that Josh spots when walking to work is a Poisson random variable with mean 6 . Each coin is equally likely to be a penny, a nickel, a dime, or a quarter. Josh ignores the pennies but picks up the other coins. (a) Find the expected amount of money that Josh picks up on his way to work. (b) Find the variance of the amount of money that Josh picks up on his way to work. (c) Find the probability that Josh picks up exactly 25 cents on his way to work.

Let \(Y\) be a gamma random variable with parameters \((s, \alpha) .\) That is, its density is $$ f_{Y}(y)=C e^{-\alpha y} y^{s-1}, \quad y>0 $$ where \(C\) is a constant that does not depend on \(y .\) Suppose also that the conditional distribution of \(X\) given that \(Y=y\) is Poisson with mean \(y\). That is, $$ P\\{X=i \mid Y=y\\}=e^{-y} y^{i} / i !, \quad i \geqslant 0 $$ Show that the conditional distribution of \(Y\) given that \(X=i\) is the gamma distribution with parameters (s \(+i, \alpha+1\) ).

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.

There are three coins in a barrel. These coins, when flipped, will come up heads with respective probabilities \(0.3,0.5,0.7 .\) A coin is randomly selected from among these three and is then flipped ten times. Let \(N\) be the number of heads obtained on the ten flips. (a) Find \(P[N=0\\}\). (b) Find \(P[N=n\\}, n=0,1, \ldots, 10\) (c) Does \(N\) have a binomial distribution? (d) If you win \(\$ 1\) each time a head appears and you lose \(\$ 1\) each time a tail appears, is this a fair game? Explain.

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