Question

A person tosses a fair coin until a tail appears for the first time. If the tail appears on the n th flip, the person wins 2n dollars. Let X denote the player's winnings. Show that $E\left[X\right]=+\infty$. This problem is known as the St. Petersburg paradox.

(a) Would you be willing to pay \( 1 million to play this game once?

(b) Would you be willing to pay \) 1 million for each game if you could play for as long as you liked and only had to settle up when you stopped playing?

Short answer

(a) The individual is not ready to pay $ 1 million to play this game once.

(b) The individuals will never be willing to pay a $ 1 million to play this game.

Step-by-step solution

Given information (Part a)

A person tosses a fair coin until a tail appears for the first time. If the tail appears on the n th flip, the person wins 2n dollars

Solution (Part a)

From the given data, a person throws a fair coin until a tail occurs for the first time. If a tail appears on the first toss, the person wins 2 dollars.

If a tail appears on the second toss, the person wins $2^2$dollars.

If a tail appears on $n^{\text{th}}$toss, the person wins $2^n$dollars.

Then,

From the comprehended information, the probability that obtaining tail is $\frac{1}{2}$.

The probability that tail on first toss $\left(T\right)=\frac{1}{2}$.

The probability that tail on second toss $(H,T)=\frac{1}{2}\times \frac{1}{2}={\left(\frac{1}{2}\right)}^2$.

The probability that tail on n th toss, $(H,H,\dots ,T)=\frac{1}{2}\times \frac{1}{2}\times \dots \times \frac{1}{2}={\left(\frac{1}{2}\right)}^n$.

Let X be the event that player's winnings.

Here, the range of X is, X=1,2,....

So the expected winning of the player will be:

$E\left(X\right)=\left(2\times \frac{1}{2}\right)+\left(2^2\times \frac{1}{2^2}\right)+\left(2^3\times \frac{1}{2^3}\right)+\dots .$

$=\sum _{n=1}^{\mathrm{\infty }} \left\{2^n\left(\frac{1}{2^n}\right)\right\}$

$=\sum _{n=1}^{\mathrm{\infty }} 1$

$=\mathrm{\infty }$

Solution (Part a)

The above-mentioned information tells that, the probability of winning 2 dollars is 0.5,4 dollars is 0.25,8 dollars is 0.125, and so on.

So, the probability of winning $ 1 million is very minor. Existing a risk element, an individual would pay $ 1 million because there is a possibility for loss.

Final answer (Part a)

Thus, the individual is not ready to pay $ 1 million to play this game once.

Given information (Part b)

A person tosses a fair coin until a tail appears for the first time. If the tail appears on the n th flip, the person wins 2n dollars

Solution (Part b)

If an individual is willing to pay $ 1 million for per game, then the probability of winning a million is ${\left(\frac{1}{2}\right)}^n$.

So, the amount that the person will win for each game will be,

$2^n\ge 1000000$

(Apply log on both sides)

${\log }_{10}^{2^n}\ge {\log }_{10}^{\left(1000000\right)}$

$n{\log }_{10}^2\ge 6$

$n\ge \frac{6}{{\log }_{10}^2}$

$n\ge 19.93$

$n\approx 20$

The probability of winning a million is,

${\left(\frac{1}{2}\right)}^n={\left(\frac{1}{2}\right)}^{20}$

$=0.0000009537$

$\approx 0$

Final answer (Part b)

So, the probability of succeeding a million is almost equal to zero. Thus, the possibility that people willing to pay $ 1 million to play this game is almost 0. Thus, the individuals will never be willing to pay a $ 1 million to play this game.