Question

Suppose that Joanne tosses a fair coin with probability 1/2 of heads, and Jaden tosses a biased coin with probability p(≠1/2) of heads, at the same time. Joanne will get one point if both coins show the same outcome, and lose one point otherwise, in each trial where Joanne and Jaden toss their own coins simultaneously. That is, Joanne's point is +1 if both coins show the same outcome, and -1 otherwise, in each trial.
Let Y be the number of tosses until both coins simultaneously show the same outcome. Find the expected value of Y^2, E(Y^2).​

Answer 1

Answer:

$E[Y^2]=\frac{2-q}{q^2}$  where $q = \frac{p}{2}+\frac{1-p}{2}$.

Step-by-step explanation:

Since both Joanne and Jaden toss the coins at the same time, we will assume that this tosses are independent. Consider the event A that they get the same outcome. This means that they either get tails or heads. Since we consider each toss independent, the probability of each outcome is obtained by multiplying the probabilities. Then the probability of having both heads is p/2 and the probability of having both tails is (1-p)/2. Then

$P(A) = \frac{p}{2}+\frac{1-p}{2}$

In this case, we can model the random variable Y as a geometric random variable with success probability $q = \frac{p}{2}+\frac{1-p}{2}$, since we want to count the number of trials until both get the same outcome. We will use the following

$E[Y^2] = Var(Y)+(E[Y])^2$

In this case, we will take as a fact that

$Var(Y) = \frac{1-q}{q^2}, E[Y] = \frac{1}{q}$.

Then,

$E[Y^2] = \frac{1-q}{q^2}+ \frac{1}{q^2} = \frac{2-q}{q^2}$

where $q = \frac{p}{2}+\frac{1-p}{2}$.