How much would you pay to play this fair coin flipping game?

[docnet]

Homework Statement

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Relevant Equations

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I feel like there is something missing in my solutions, because the answers are coming out weird.

(a)

The reward for getting a head is $2^n$ dollars, the number of flips $n$ it takes to get the first heads.

the probability to get a head in the $n$th flip is $\frac{1}{2^n}$

so the expected value of the game is

$$E(X)=\frac{1}{2}\cdot 2+\frac{1}{4}\cdot 2^3+\frac{1}{8}\cdot 2^3+\frac{1}{16}\cdot 2^4+\cdots$$

$$E(X)=\sum_{i=1}^\infty 1 \Rightarrow \text{ the sum diverges}$$

Who would want to pay an indefinite amount of money to play this game...?

(b)

$$E(X)=\sum_{i=1}^{40} 1 = 40$$

By this logic, I would pay no more than 40 dollars to play this game

The probability of going more than 40 rounds is

$$P(X>40)=\sum_{i=41}^\infty\frac{1}{2^i}\approx 0$$

Yet, the expectation value drastically differs from the value in (a)

(c)

$$E(X)=\sum_{i=1}^{40} 1 + \sum_{i=1}^\infty \frac{1}{2^n} =40+1 =41$$

I'd pay 41 dollars to play this game

(d)

no one has that much monay. that's too many zeros.

It turns out this is famous problem called St. Petersburg Paradox. - it seems to be a rare example of when probability theory isn't rational for predicting the outcome, since the player is probably only going to win a modest amount of money in this game.

 

[Office_Shredder]

I think all of your answers are correct here.