TITLE: The case of the missing ninth of a $2$€ coin
QUESTION [12 upvotes]: In answering Expected value of the number of bills, I came across a phenomenon the likes of which I don't think I've encountered before, and I'd like to know more about it.
You draw coins, each coin independently being a $1$€ coin or a $2$€ coin with equal probability. Obviously you'd expect to draw as many $2$€ coins as $1$€ coins. In particular, the expectation of $A-B$, where $A$ is the number of $1$€ coins drawn and $B$ is the number of $2$€ coins drawn, is $0$ after any given number of draws.
However, conditional on reaching a total value of $n$ euros, the expectation of $A-B$ tends to $\frac13$ for $n\to\infty$ (in fact, it's positive for all $n\gt2$): The probability to reach $n$ euros with $k$ $2$€ coins and $n-2k$ $1$€ coins is $\binom{n-k}k2^{k-n}$, so the expectation of $B$ is
\begin{eqnarray*}
&&\frac{\sum_{k=0}^n\binom{n-k}k2^{k-n}k}{\sum_{k=0}^n\binom{n-k}k2^{k-n}}=\frac{\frac2{27}(3n-1)+O\left(2^{-n}\right)}{\frac23+O\left(2^{-n}\right)}=\frac n3-\frac19+O\left(2^{-n}\right)\;,\\
\end{eqnarray*}
the expectation of $A=n-2B$ is $\frac n3+\frac29+O\left(2^{-n}\right)$, and  the expectation of $A-B$ is $\frac13+O\left(2^{-n}\right)$.
This is rather counterintuitive (to me): For any given number of coins the expectation is $0$, but for any given value of the coins it's positive. This is the stuff that paradoxes are made of if you're not careful how you talk about it, e.g.: “Someone is playing this game. What value do you expect for $A-B$?” – “$0$.” – “So far they drew $137$€. Now what do you expect?” – “$\frac13$.”
The resolution here is (as it often is) that the conditions aren't properly defined – we don't know why and when the person is telling us this amount. If they fixed a number of draws to wait for and then told us the total value at that point, the correct answer would still be $0$; if they fixed a total value to wait for and then told us when it was reached, the correct answer would be $\frac13$, but then the paradox of changing our mind just because we were told some number, no matter which one, wouldn't arise, because it's the stopping protocol that makes the difference.
Still, a certain uneasy sense of paradox remains, even if it temporarily retreats under the glare of careful analysis.
I don't have any concrete questions about this, but I'd be interested to hear about any other cases where such a phenomenon occurs, or names by which it's known, or approaches to deal with it, and perhaps also to ease that lingering sense of paradox.

REPLY [2 votes]: The many interesting answers have provided insight into the problem from various angles. I feel the most important insight I took from them is that the precise stopping criterion is crucial and that the bias towards the smaller coins occurs only when we condition on exactly reaching a certain total, but is $0$ when we condition on reaching at least that total. We can make intuitive sense of this: If you take smaller steps, you’re more likely to hit exactly a certain total than if you take larger steps. Thus you shouldn't be surprised if conditioning on hitting a certain total makes it more likely that you took small steps.
In fact, in an analysis somewhat similar to that in antkam’s answer, but focusing on the coin that completed the total $n$, we can attribute the entire bias of $\frac13$ to that one coin. Since we get $\frac32$ euros on average per coin, we hit $\frac23$ of all totals for large $n$. Thus, we have probability $\frac23$ to reach $n-2$ and probability $\frac23$ to reach $n-1$. Since the probability that we reach neither is $0$, the probability that we reach both must be $\frac13$. So we have three equiprobable cases: Reaching only $n-2$, reaching only $n-1$, or reaching both. Each of these leaves just one way to reach exactly $n$, so they remain equiprobable when we condition on reaching exactly $n$. In two of them we need a $1$€ coin to reach $n$, and only in one we need a $2$€ coin. Thus the entire expected bias $E[A-B]=\frac13$ is due to this last coin, which is more likely to be $1$€ because that makes it more likely to hit $n$. We can also make sense of the value $\frac13$ of the bias in this intuitive way: If we drew only $1$€ coins, we'd have probability $1$ to hit $n$. If we drew only $2$€ coins, we'd have probability $\frac12$ to hit $n$. Thus, if we were doing either of these with equal probability $\frac12$ and didn’t know which, and we learned that we hit a certain target, the conditional probabilities that we were drawing $1$€ or $2$€ coins would be $\frac1{1+\frac12}=\frac23$ and $\frac{\frac12}{1+\frac12}=\frac13$, respectively.