Question

g Suppose I will flip a fair coin until it lands Heads for the first time, and you will receive $2 if the game lasts for 1 round, $4 if the game lasts for 2 rounds, $8 if the game lasts for 3 rounds, and in general, $2n if the game lasts for n rounds. How much would you be willing to pay to play this game once

Answer 1

Answer:

Step-by-step explanation:

The main idea is that I would like to pay less than what I'm expecting to win, so in that way, I get a profit out of playing this game. Let X be the number of tosses until I get a Heads. By definition, this is a geometric random variable with parameter p = 1/2.

Let Y the amount I received for playing. So, we want to calculate the expected value of Y.

We can calculate it as follows

$E[Y] = 2 P(X=1)+ 4 P(X=2)+ 8 P(X =3) + \dots = \sum_{n=1}^infty 2^n P(X=n)$

Since X is a geometric random variable, we have that $P(X=n) = (\frac{1}{2})^{n-1}\frac{1}{2}$

Then,

$E[Y] = \sum_{n=1}^\infty 2^{n} (\frac{1}{2})^{n-1} \frac{1}{2} = \sum_{n=1}^\infty 2^{n-1} \cdot \frac{1}{2^{n-1}} = \sum_{n=1}^\infty 1 = \infty$

So, we expect to have an infinite amount. Given this, we can pay as much as we want to play the game.