# Resolving a paradox concerning an expected value

We have a coin that has a probability $p>1/2$ of coming up heads (and probability $1-p$ of coming up tails). We now play the following game:

2. We toss the coin. If it comes up heads, we double our current fortune, and we repeat this step (step 2). However, if it comes up tails, we lose all of our fortune, i.e. go broke, and stop playing.

Note that after $n$ plays, our fortune is $2^n$ with probability $p^n$, and zero with probability $1-p^n$.

This means that the expected value of our fortune after $n$ plays is $2^np^n = (2p)^n$. Since $p>1/2$, we have $2p>1$, which tells us that our expected value after $n$ plays approaches infinity as $n$ approaches infinity.

On the other hand, our probability of going broke after $n$ plays, $1-p^n$, approaches one as $n$ approaches infinity. In other words, we know for sure that we will eventually go broke.

My question is this: How can we, in the infinity, both possess an infinite fortune and yet be completely broke? How can this paradox be resolved?

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Check out St. Petersburg paradox – Marc van Leeuwen Dec 12 '11 at 16:01
This isn't exactly the St. Petersburg paradox. (Not that you were necessarily implying that; just to clarify.) – joriki Dec 12 '11 at 16:11
This is much like a sequence of functions on $[0,1]$, all of whose integrals are $1$, but whose pointwise limit is the $0$ function. It is more a problem with how we think of limits. – Thomas Andrews Dec 12 '11 at 16:39
We don't "know for sure that we will eventually go broke". This an example of where the phrase "almost surely" can be used. – Henry Dec 12 '11 at 17:24

How can we, in the infinity, both possess an infinite fortune and yet be completely broke?

First, as already explained by @joriki, this statement should be corrected as one possesses a mean infinite fortune and one is broke almost surely. Second, rephrasing things like I just did hints as an occurrence of the interplay between different convergence modes.

Indeed, let $X_n$ denote the random fortune after $n$ plays and $X$ the random fortune after infinitely many plays. Both exist, in particular $X_n\to X$ almost surely and $X=0$ almost surely. Hence, in a sense, the mean fortune after infinitely many plays is $\mathrm E(X)=0$. On the other hand, the mean fortune after $n$ plays is $\mathrm E(X_n)=(2p)^n$, which does not go to zero if $p\geqslant\frac12$.

All this means that we are facing a case where the expectation of the limit is not the limit of the expectations, that is, $\mathrm E(X_n)\not\to\mathrm E(X)$ although $X_n\to X$ (almost surely). Well, these things just happen...

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Thanks! You seem to have an amazing understanding of what happens in the “infinity.” I just asked another question, which builds on this one. I hope you can help me out with it: math.stackexchange.com/questions/90917/… In that question, I’m once again just asking for some intuition about what happens in the “infinity,” and how to properly measure it. – Topology Dec 12 '11 at 20:51

The problem is an imprecise use of concepts like "what happens at infinity". The following are correct:

• The expected value of your fortune eventually increases above any finite bound.
• The probability that you haven't gone broke eventually decreases below any positive bound.

This is usually expressed by

• The expected value goes to infinity.
• The probability goes to zero.

Note, however, that these latter statements derive their meaning solely from having been defined to mean precisely the former statements. Whatever you may associate with the second set of statements that goes beyond the first set of statement is not part of their mathematical definition.

Thus, there is no paradox because there is no point "at infinity" at which you are simultaneously expectedly wealthy and certainly broke.

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Thanks for your insights. Upvoted it! By the way, I just asked a new question that builds somewhat on this one. I hope you have the time to look at that too: math.stackexchange.com/questions/90917/… – Topology Dec 12 '11 at 20:53