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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 \(f(p)\).

Short Answer

Expert verified
The probability that the first player wins given a head probability of \(p\) is: \[ f(p) = \frac{p}{2p - p^2} \]

Step by step solution

01

(a) Analyzing Monotonicity of \(f(p)\)

Let's consider what happens as \(p\) varies. - If the probability of getting heads is low (\(p\) is close to 0), then it will be challenging for any of the players to get heads on their first flip. The first player would still have a higher probability of winning since they play first. - If the probability of getting heads is high (\(p\) is close to 1), then it is more likely that both players will get heads on their first flip. However, the first player still has an advantage since they play first. Since the chances of the first player winning increase with an increase in the probability of getting heads, we can conclude that \(f(p)\) is a monotonically increasing function.
02

(b) Limit of \(f(p)\) as \(p \rightarrow 1\)

As \(p \rightarrow 1\), the probability of obtaining a head on any given toss approaches certainty. Since the first player flips the coin first, they will almost surely win if \(p=1\). Thus, we can deduce that: \[ \lim_{p \rightarrow 1} f(p) = 1 \]
03

(c) Limit of \(f(p)\) as \(p \rightarrow 0\)

As \(p \rightarrow 0\), the probability of obtaining a head on any given toss becomes very low. However, the first player still has the advantage of flipping the coin first. Since both players will have a tough time obtaining heads, the game will likely go on for longer turns. But the first player will always have a higher probability of winning compared to the second player. Therefore, as the probability of getting heads approaches 0, the probability of the first player winning will still be greater than that of the second player: \[ \lim_{p \rightarrow 0} f(p) > 0 \] Now let's find the exact limit. Let \(P_n\) be the probability that the first player wins in the \(n^{th}\) round. Then, the probability that they have not won by the end of the \((n - 1)^{th}\) round would be: \[ P_n = p(1 - p)^{2n - 2} \] The probability of the first player winning the game is the sum of the probabilities of winning in all possible rounds: \[ f(p) = \sum_{n = 1}^{\infty} P_n \] Plugging in \(p = 0\), we find: \[ \lim_{p \rightarrow 0} f(p) = 0 \]
04

(d) Deriving \(f(p)\)

We will now calculate the function \(f(p)\) by summing the probabilities \(P_n\). \[ f(p) = \sum_{n = 1}^{\infty} P_n = \sum_{n = 1}^{\infty} p(1 - p)^{2n - 2} \] This is an infinite geometric series with the first term \(a = p\) and the common ratio \(r = (1-p)^{2}\). The sum of this series converges to: \[ f(p) = \frac{a}{1 - r} = \frac{p}{1 - (1 - p)^2} = \frac{p}{2p - p^2} \] So, the probability that the first player wins given a head probability of \(p\) is: \[ f(p) = \frac{p}{2p - p^2} \]

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Key Concepts

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

Monotone Function
In mathematics, a monotone function is one that either never increases or never decreases as its argument is varied. In the given problem, we consider the probability function denoted by \(f(p)\), representing the likelihood that the first player wins a game of coin flips given the probability \(p\) of flipping heads. When approaching such a scenario, it's essential to grasp the behavior of \(f(p)\) in relation to \(p\). Intuitively, if \(p\) is low, the chance for either player to win on the first flip is small, yet the first player still has an upper hand by flipping first. Conversely, if \(p\) is high, the likelihood of winning on the first flip increases notably for the first player.

From this observation, we infer that as \(p\) grows, so does \(f(p)\). This relationship categorizes \(f(p)\) as a monotonically increasing function of \(p\). This means the function \(f(p)\) satisfies the criteria for being monotone because the probability of the first player winning either stays the same or increases as \(p\) increases; it never decreases. Recognizing the monotonicity of functions is key in mathematical analysis as it helps us understand how one variable influences another and can simplify solving problems considerably.
Probabilistic Limits
Within the realms of probability and calculus, we often talk about the concept of limits, which reveal the behavior of functions as they approach a particular point. In probabilistic terms, when we address questions like the value of \(\lim _{p \rightarrow 1} f(p)\) or \(\lim _{p \rightarrow 0} f(p)\), we are essentially considering how the probability function behaves as the likelihood of an event (flipping heads) becomes either very likely or very unlikely.

As per the given solution, the limit of \(f(p)\) as \(p\) approaches 1 is 1, which concurs with the logical prediction that if a head is certain to come up on a flip, the first player will win for sure by flipping heads on their very first turn. On the other hand, as \(p\) tends towards 0, the outcome becomes intriguing. Although the probability of flipping heads is close to negligible, the mere fact that the first player goes first grants them a consistent edge over their opponent, hence \(\lim_{p \rightarrow 0} f(p) > 0\). This kind of analysis involving probabilistic limits is critical for understanding long-term behaviors and trends in stochastic processes, allowing predictions about outcomes based on changing probabilities.
Geometric Series
The calculation that leads to the probability function \(f(p)\) seamlessly ties into the concept of a geometric series. An infinite geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In our problem, the probability of the first player winning after any even number of flips can be seen as a geometric series with the probability \(p\) and the common ratio \((1-p)^2\).

To find the total probability of the first player winning (\(f(p)\)), we sum the infinite series of probabilities of them winning in each round. The sum of an infinite geometric series when the absolute value of the common ratio is less than one is given by \(\frac{a}{1 - r}\), where \(a\) is the first term and \(r\) is the common ratio. Applying this formula, we derive that \(f(p) = \frac{p}{1 - (1 - p)^2}\) which simplifies to \(f(p) = \frac{p}{2p - p^2}\), revealing the elegant relationship between the initial flipping probability and the overall chance of the first player winning. Understanding how the sum of a geometric series works is incredibly useful in various fields of study, including finance, computer science, and, as illustrated here, probability theory.

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