Start with $1 and toss a coin. If it comes up heads your winnings double, if tails you lose all your winnings. What are the expected winnings? You start with 1 dollar and toss a coin. If it comes up heads your winnings double, if it comes up tails you lose all your winnings. What are the expected winnings?
This seems a variation of St. Petersburg's Paradox to me. Intuitively you are bound to lose everything and therefore end up with 1 dollar less than you started.
Mathematically, I thought about it in two different ways:

*

*Let X be the accumulated winnings, then on the next toss you stand to have a net gain of X with probability 1/2 or a net loss of -X with 1/2 i.e 1/2(x)-1/2(x)=0. Therefore your expected gains are zero, you are indifferent and would play the game


*Let X be the accumulated winnings, then on the next toss you stand to increase your winnings to $2\times$ with probability 1/2, or lose everything (-X) with probability 1/2 i.e $1/2(2x)-1/2(x)=1/2X$. Hence the expected value of the game tends to infinity and you would play the game
I am not confident with any of the above. Any hints or comments would be helpful.
 A: The expected winnings after $n$ turns of the game are
$$\sum x p(x)$$
where $p(x)$ is the probability of winning $x$ dollars. In this case you have $\frac12^n$ chance of winning $2^n$ dollars, and $(1-\frac12^n)$ chance of losing everything, so the expected amount you win is:
$$
\frac12^n \times $2^n + (1-\frac12^n)\times $0 = $1
$$
This result is counterintuitive, since you cannot ever win exactly one dollar playing this game.
A: Suzu's answer shows that if you agree to stop after $n$ turns (where $n$ is determined in advance), then your expected final fortune is $1$.
Second version.  I will never stop.  Then (according to the Borel-Cantelli lemma) I eventually get tails, and so I eventually lose everything.
A: The sample space for this situation is $\Omega=\{0,1\}^{\mathbb N}$, (0 for loosing, 1 for winning) with Bernoulli's measure $P$, while the random variable $X$ is the constant function zero, except that
$$
X(1,1,1,\ldots)=\infty.
$$
Therefore
$$
E(X)=\int_\Omega X(\omega)\,dP(\omega)=0.
$$
In case you decide to stop after $n$ turns, then the relevant random variable is
$$
X_n = 2^n 1_{C_n},
$$
where $1_{C_n}$ is the characteristic function of the cylinder
$$C^n=\{1\}\times\{1\}\times\cdots\times\{1\}\times\{0,1\}^\infty.
$$
Even though $X_n\to X$ almost everywhere, the integrals do not converge, that is,
$$
\lim_n\int_\Omega X_n(\omega)\,dP(\omega)
\neq\int_\Omega X(\omega)\,dP(\omega),
$$
mainly because Lebesgue's dominated convergence Theorem doesn't apply.
