91Ó°ÊÓ

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

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 $(H,T)=\frac{1}{2}×\frac{1}{2}={\left(\frac{1}{2}\right)}^2$.

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

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:

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

$=\underset{n=1}{\overset{\mathrm{∞}}{∑}} \left\{2^n\left(\frac{1}{2^n}\right)\right\}$

$=\underset{n=1}{\overset{\mathrm{∞}}{∑}} 1$

$=\mathrm{∞}$

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.

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

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

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,

$2^nâ‰\yen 1000000$

(Apply log on both sides)

${\log }_{10}^{2^n}â‰\yen {\log }_{10}^{\left(1000000\right)}$

$n{\log }_{10}^2â‰\yen 6$

$nâ‰\yen \frac{6}{{\log }_{10}^2}$

$nâ‰\yen 19.93$

$n≈20$

The probability of winning a million is,

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

$=0.0000009537$

$≈0$

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.