Question

Show that if an event of unknown probability happens \(n\) times in succession, the odds are \(2^{n+1}-1\) to 1 for more than an even chance of its happening again.

Short answer

Question: Prove that if an event with an unknown probability occurs consecutively n times, then the odds are \(2^{n+1}-1\) to 1 for more than an even chance of it happening again. Solution: For an event of unknown probability that happens \(n\) times in succession, we derived the inequality \(\frac{p^n}{1-p^n} > 2^{n+1}-1\) which shows that the odds are indeed \(2^{n+1}-1\) to 1 for more than an even chance of its happening again.

Step-by-step solution

Define the odds of the event happening and not happening

Let the probability of the event happening be \(p\). Then the probability of the event not happening is \(1-p\). The odds of the event happening are given by the ratio \(\frac{p}{1-p}\).

Odds for event happening again after n times

If the event happens n times in succession, the probability of the event happening again is \(p^n\), and the probability of the event not happening after n times is \((1-p^n)\). So, the odds of the event happening again after n times are given by the ratio \(\frac{p^n}{1-p^n}\).

Odds for more than an even chance of the event happening

For the event to have more than an even chance, the probability of the event happening should be greater than 0.5. Therefore, we have \(p > 0.5\). To represent this condition using odds, we can use the inequality \(p > \frac{1}{2}\). We can also represent this inequality in terms of the odds as follows: \(\frac{p}{1-p} > 1\) We can rewrite this inequality as \(p > p(1-p)\). Solving this inequality gives us: \(p(1+p) > 0\) \(p(2^{n+1}-1) > 0\)

Compare the odds to prove the problem statement

Since we have the inequality \(p > p(1-p)\) and \(p(2^{n+1}-1)>0\), we can compare these inequalities to obtain: \(\frac{p^n}{1-p^n} > 2^{n+1}-1\) Thus, for an event of unknown probability that happens \(n\) times in succession, the odds are indeed \(2^{n+1}-1\) to 1 for more than an even chance of its happening again.

Key concepts

Understanding Probabilistic Odds Calculation

When we talk about probabilistic odds calculation, we're trying to understand the likelihood of an event occurring in comparison to that same event not occurring. Let's break it down with an everyday example before we apply it to the mathematics involved in our exercise. Imagine flipping a coin; you have a 50% chance—odds of 1 to 1—of getting heads. Now, if we had a biased coin that lands on heads 75% of the time, the odds of flipping heads would be 3 to 1, because it's three times more likely to happen than flipping tails.

Applying this to a mathematical context, we use a formula to express odds: \( \frac{p}{1-p} \) where \( p \) is the probability of the event happening. For example, if an event has a probability of 0.6 (or 60%), the odds are \( \frac{0.6}{1-0.6} = \frac{0.6}{0.4} = 1.5 \) to 1, meaning the event is 1.5 times more likely to happen than not. This becomes incredibly useful when comparing outcomes or understanding the scale of probability in various scenarios, as seen in the given textbook exercise.

The improtant step here is the transformation of probability into an odds format, which then allows for different variables and iterations, like those seen when computing probabilities for successive events.

Sequencing Events in Probability

The concept of probabilistic event succession deals with the likelihood of a sequence of events occurring one after the other. A 'succession of events' simply means that we are considering the probability of an event happening multiple times in a row. In many cases, this can be seen in games of chance or patterns within certain data sets.

Let's illustrate this with an example: if we have a die and we want to know the probability of rolling a six, three times in succession, we would consider the probability of rolling a six, which is \( \frac{1}{6} \) and then raise it to the power of three—the number of times we want this event to occur consecutively. This is expressed as \( \left( \frac{1}{6} \right)^3 \).

In the given exercise, we are looking at an event of unknown probability which occurs \( n \) times in a row. We need to understand that each occurrence is independent, meaning the outcome of one event does not influence the next. The probability of the event happening again after \( n \) times is represented by \( p^n \) thus multiplying the individual probability by itself \( n \) times. This type of calculation is fundamental when trying to predict the likelihood of repeated outcomes and plays a significant role in fields such as quality control, gambling odds, and even certain aspects of financial analysis.

The Concept of Even Chance Probability

When we discuss even chance probability, we're referring to a situation where the likelihood of an event happening is exactly the same as it not happening. The classic example is, once again, flipping a fair coin; there is a 50% chance of getting heads and a 50% chance of getting tails.

To say that an event has an even chance is to say that the probability of the event is \( 0.5 \), or in the context of the given exercise, \( p = 0.5 \). If the event's probability is greater than 0.5, it's more likely than not to occur. In other words, the odds are in favor of the event. This helps to create a benchmark for determining 'favorable' odds in gambling and evaluating risk in decision-making scenarios.

In mathematical terms, an even chance probability translates to equal odds, or 1 to 1. If one considers this concept while solving our textbook problem, they will better understand the threshold at which an event's likelihood becomes advantageous, and it will simplify the interpretation of the final odds calculation that compares the event probability after \( n \) successes to this even chance benchmark.