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This is my homework question, so I don't want straightforward answer of-course! I just don't understand how to write "probability of making overall profit?"

This is from st petersburg paradox, I'm trying to calculate after how many games this profit should be more than .5

Exact que: Suppose you play instead the following game: At the beginning of each game you pay an entry fee of $F. A coin is tossed until a head appears, counting n = the number of tosses it took to see the first head. Your reward is 2^n. After about how many games (estimate) the probability of making a profit overall is bigger than 50%

If someone can help me how to start, like is it calculating expected value of "something" or Defining this probability, that would be great!

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Expected profit $E[\pi]$ is the expected payout of the game $E[X]$ minus the entry cost $F$. $$E[\pi] = E[X] - F $$

It doesn't matter how many times you play the game, the expectation will not change. And you will notice that the expected payout of the game diverges, going to infinity.

$$E[X] = \sum_{n=1}^\infty p_n * 2^n = \sum_{n=1}^\infty (1/2)^n * 2^n = \sum_{n=1}^\infty 1 = \infty $$

where $p_n$ is the probability the the nth flip being heads while the others are tails. So, you should on average expect an infinite profit for any finite entry cost F.


To find out the probability of being profitable (without regard to the magnitude of that profit), we could compute the following.

Choose $ N > \frac{\log F}{\log 2}$. This will give $ F < 2^N $. This is the profitability condition. Then, to find the overall probability that any one game would fall in this state, we compute

$$P_f = \sum_{n=N}^\infty p_n = \sum_{n=N}^\infty (1/2)^n = 1 - \sum_{n=1}^{N-1} (1/2)^n$$

, where $P_f$ is the probability of a game resulting in a profit.

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But we are asked the number of plays, not the expected profit from one play. If $F$ is $10$, say, we will lose money most of the time, but if we play long enough we will hit a big enough win to come out ahead. How long is the question. – Ross Millikan Feb 27 '13 at 17:47
but each time I play game I have to pay entry fee F right? so my expected value will be 2^n - F*n where n is nth game – code muncher Feb 27 '13 at 17:49
@RossMillikan yeah agree ... – code muncher Feb 27 '13 at 17:49
how I think is find n such that 2^n - n * F > 1.5 * n * F. But I don't know how to use probability thingy here! – code muncher Feb 27 '13 at 17:51
@codemuncher: no your expected value, whether from one game or many is infinite, as jmbejara shows. – Ross Millikan Feb 27 '13 at 17:54

You have $\frac 12$ chance of winning $1-F, \frac 14$ chance of winning $2-F, \frac 18$ chance of $4-F, \ldots \frac 1{2^n}$ chance of $2^{N-1}-F$

For small $F$ it is easy. For $F \lt 1$ we win every time. For $1 \lt F \lt 1.5$ we win in one game if the first flip is tails, $\frac 12$ chance and in two games unless both end in one flip, so with chance $\frac 34$. For $F$ a little over $2$ we need two tails in the first game, so only win with probability $\frac 14$ For two games we need two tails in either game, so win with probability $\frac 7{16}$ For three games we either need a set of three tails or a set of two and a single one. We only lose if the numbers of tails are $(2,0,0),(1,1,1),(1,1,0),(1,0,0),(0,0,0)$ in any order. These chances are $\frac 3{32},\frac 1{64}, \frac 3{32}, \frac 3{16},\frac 18$, for a total chance to lose of $\frac {33}{64}$. I haven't checked, but it looks like the chance of winning goes over $50\%$ at four games.

As $F$ grows it gets harder. Let's say $F=3$ and we are going to play $6$ games. That costs $18$ to play, so we win if we get one game at $16$, two games at $8$, one at $8$ and two at $4$ and so on. This will be a mess to compute. I think I will make a computer model.

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