Question

Question: Consider the following game. A person flips a coin repeatedly until a head comes up. This person receives a payment of \({2^n}\) dollars if the first head comes up at the \(nth\) flip.

a) Let \(X\) be a random variable equal to the amount of money the person wins. Show that the expected value of \(X\) does not exist (that is, it is infinite). Show that a rational gambler, that is, someone willing to pay to play the game as long as the price to play is not more than the expected payoff, should be willing to wager any amount of money to play this game. (This is known as the St. Petersburg paradox. Why do you suppose it is called a paradox?)

b) Suppose that the person receives \({2^n}\) dollars if the first head comes up on the \(nth\) flip where \(n < 8\) and \({2^8} = 256\) dollars if the first head comes up on or after the eighth flip. What is the expected value of the amount of money the person wins? How much money should a person be willing to pay to play this game?

Short answer

Answer

a) The expected value of random variable \({\rm{X}}\) does not exist.

b) The expected value of the amount of money that a person win is \(512\).

Step-by-step solution

Given data

Person receives \({2^n}\) dollars if the first head comes up on the \({n^{{\rm{th }}}}\) flip where \(n < 8\) and \({2^8} = 256\) dollars if the first head comes up on or after the eighth flip.

Concept of Probability

Probability is simply how likely something is to happen. Whenever we’re unsure about the outcome of an event, we can talk about the probabilities of certain outcomes—how likely they are. The analysis of events governed by probability is called statistics.

For example:

The probability of getting head/tail \( = \frac{1}{2}\).

Calculation for the expected value of random variable

The condition of winning the game is the player who & get a head in \(n\) trials precisely after \((n - 1) + \) Nils.

Therefore, the probability that one head comes after \((n - 1)\) tail is \(\frac{1}{2}{\left( {\frac{1}{2}} \right)^{n - 1}} = \frac{1}{{{2^n}}}\).

The expectation of the game is \({2^n} \times \frac{1}{{{2^n}}} = 1\).

The expected value of winning is \(1\) irrespective of money paid.

It goes to infinity when \(n\) goes from \(1\) to infinity.

Therefore, the amount of money that a person wins can go to infinity.

The expected value of random variable \(x\) does not exist.

Calculation for the expected value of the amount of money

Person flips a coin repeatedly until the head comes up and receives a payment of \({2^n}\) dollars if the first head comes up at the \({n^{{\rm{th }}}}\) flip.

If a head comes in \({9^{{\rm{th }}}}\) flip, then the expected value of the amount of money that a person wins is given as:

\(\begin{aligned}{}{2^n} = {2^9}\\{2^n} = 512\end{aligned}\)

Therefore, the expected value of the amount that a person win is \(512\).