Question

During the 1992 football season, the Los Angeles Rams (now the St. Louis Rams) had a bizarre streak of coin-toss losses. In fact, they lost the call 11 weeks in a row. \({ }^{8}\) a. The Rams' computer system manager said that the odds against losing 11 straight tosses are 2047 to 1 . Is he correct? b. After these results were published, the Rams lost the call for the next two games, for a total of 13 straight losses. What is the probability of this happening if, in fact, the coin was fair?

Short answer

Answer: The odds against losing 11 straight tosses are 2047 to 1, and the probability of losing 13 straight tosses if the coin is fair is approximately 0.0122% or 1/8192.

Step-by-step solution

Calculate the probability of losing one toss

Assuming the coin is fair, there are two possible outcomes: heads (win) and tails (lose). Since each outcome is equally likely, the probability of losing one toss is 1/2.

Calculate the probability of losing 11 straight tosses

Each toss is an independent event, so to find the probability of losing 11 straight tosses, simply multiply the probability of losing one toss 11 times. Mathematically, this is represented as: \((1/2)^{11}\)

Convert probability to odds

Odds are the ratio of the probability of an event occurring to the probability of the event not occurring. In this case, the odds against losing 11 straight tosses are the ratio of the probability of not losing all 11 tosses to the probability of losing all 11 tosses. That is: \(\frac{1 - (1/2)^{11}}{(1/2)^{11}}\) Now, simplify the expression: \(\frac{2^{11} - 1}{2^{11}}\) The odds against losing 11 straight tosses are indeed 2047 to 1. #b. Calculate the probability of losing 13 straight tosses if the coin is fair#

Calculate the probability of losing 13 straight tosses

Similar to the previous part of the problem, we will multiply the probability of losing one toss 13 times to find the probability of losing 13 straight tosses if the coin is fair: \((1/2)^{13}\) The probability of losing 13 straight tosses if the coin is fair is: \(1/8192 \approx 0.000122\) or \(0.0122\%\)

Key concepts

Independent events

In probability, independent events refer to scenarios where the outcome of one event does not affect the outcome of another. In other words, each event occurs with no influence from the outcomes of previous events.
For example, when flipping a fair coin, the probability distribution remains constant across flips. Each time you flip the coin, the chances of landing on heads or tails is always 50% (or 1/2), independent of previous results.
In the context of the Rams' streak of coin toss losses, each toss is independent. This independence means that the outcome of the 11th toss does not rely on the outcome of the preceding 10 tosses. Every time the coin is tossed, it's a fresh event, resetting the probability chances.
Understanding independent events is crucial in calculating probabilities over several trials, like in the Rams' case, where each coin toss had the same probability of losing despite prior losses.

Odds calculation

Odds serve as a way to express the likelihood of a particular outcome occurring compared to not occurring. It is different from probability, which gives the rate of an event happening against all possible events.
To calculate the odds of the Rams losing 11 straight tosses, we first need to determine the probability of this losing streak. With a fair coin, the probability of losing one toss is 1/2. To find the odds of the streak:

• First, calculate the probability of losing 11 times consecutively, which is \((1/2)^{11}\).
• Then, calculate the probability of not losing all 11 times, which is \(1 - (1/2)^{11}\).
• Finally, establish the odds by making a ratio of not losing to losing, giving \(\frac{1 - (1/2)^{11}}{(1/2)^{11}}\).

After simplification, we get the odds as 2047 to 1. This expression means there is only one chance out of 2048 that the event does occur, showcasing how rare such a streak is.

Probability of consecutive events

The probability of consecutive events, such as losing multiple coin tosses in a row, relies on the concept of independent events and compound probability. When dealing with fair coin tosses, the probability of both outcomes (heads or tails) is 1/2.
To find the probability of losing a specific number of consecutive tosses, multiply the probability of one event by itself for the number of desired consecutive outcomes. For example:

• The probability of losing 11 consecutive tosses is calculated as \((1/2)^{11}\), representing a highly unlikely event.
• For 13 consecutive losses, extend this calculation to \((1/2)^{13}\), resulting in \(1/8192\) or just a 0.0122% chance.

This calculation captures the sheer improbability of sustained streaks, highlighting events that, while possible, are incredibly rare in practice. Each coin toss resets the conditions for the next, aligning with the nature of probability and randomness.

Fair coin toss

A fair coin toss is a classic example used to illustrate basic probability principles. A 'fair' coin refers to one that has equal chances of landing on heads or tails when tossed. This fundamental property makes it an ideal model for exploring independent events and probability calculations.
In probability theory, a fair coin is assumed to have \(P( ext{heads}) = 1/2\) and \(P( ext{tails}) = 1/2\). These probabilities ensure that the coin has no bias, thus portraying an equal distribution of possible outcomes.
Using a fair coin to analyze events, like the Rams' streak, underscores essential probability concepts, as each toss is expected to be unbiased and independent. Understanding how a fair coin behaves is critical to grasping why certain outcomes, like 13 consecutive losses, are statistically intriguing due to their low probability under fair conditions.