Question

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

\(f(p)\) is an increasing function, and the limits are \(\lim_{p \rightarrow 1} f(p)=1\) and \(\lim_{p \rightarrow 0} f(p)=\frac{1}{2}\). The function \(f(p)\) is given by \(f(p)=\frac{p}{1-(1-p)^2}\).

Step-by-step solution

Determine the probability of first player winning

If the first player flips a head on their first turn, they win. This occurs with probability \(p\). If both players flip tails on their first turn, the probability of this happening is \((1-p)^2\). In this case, the game resets, and the probability that the first player wins is again \(f(p)\). So we can write \(f(p)\) as: \[f(p)=p + (1-p)^2f(p)\]

Solve for \(f(p)\)

Next, we will solve the equation for \(f(p)\). First, we need to isolate \(f(p)\) on one side: \[f(p)-(1-p)^2f(p)=p\] \[f(p)(1-(1-p)^2)=p\] Now, divide both sides by \((1-(1-p)^2)\): \[f(p)=\frac{p}{1-(1-p)^2}\]

Analyze the monotonic property of \(f(p)\)

Now, since \(p\) is a probability, it ranges from 0 to 1. If we analyze the derivative of \(f(p)\), we can determine whether it is increasing or decreasing on the given interval. The derivative of \(f(p)\) with respect to \(p\) is: \[f'(p)=\frac{1}{(1-(1-p)^2)^2}2(1-p)\] From this derivative, we can see that \(f'(p)>0\), for all \(p\in (0,1)\). Therefore, \(f(p)\) is an increasing function on the interval \((0,1)\).

Calculate the limit of \(f(p)\) as \(p\) approaches 1

We need to calculate the limit of the function as \(p\) approaches 1: \[\lim_{p \rightarrow 1} f(p)=\lim_{p \rightarrow 1} \frac{p}{1-(1-p)^2}\] As \(p\) approaches 1, the denominator approaches 1-0 = 1, and the limit becomes: \[\lim_{p \rightarrow 1} f(p)=\frac{1}{1}=1\]

Calculate the limit of \(f(p)\) as \(p\) approaches 0

Now, calculate the limit of the function as \(p\) approaches 0: \[\lim_{p \rightarrow 0} f(p)=\lim_{p \rightarrow 0} \frac{p}{1-(1-p)^2}\] As \(p\) approaches 0, both the numerator and the denominator approach 0. Thus, we can apply L'Hopital's Rule to evaluate the limit: \[\lim_{p \rightarrow 0} f(p)=\lim_{p \rightarrow 0} \frac{1}{2(1-p)}\] As \(p\) approaches 0, the limit becomes: \[\lim_{p \rightarrow 0} f(p)=\frac{1}{2}\] Now, we have answered all the questions: (a) \(f(p)\) is an increasing function. (b) \(\lim_{p \rightarrow 1} f(p)=1\). (c) \(\lim_{p \rightarrow 0} f(p)=\frac{1}{2}\). (d) \(f(p)=\frac{p}{1-(1-p)^2}\).