# Can you explain this counterintuitive conditional expectation result intuitively?

Consider the following experiment.

We throw a three-sided die with sides $$1$$, $$2$$ and $$3$$ infinitely many times. Let $$T_i$$ denote the outcome of the $$i$$'th throw. Define $$N:=\min\{i:T_i\neq1\}$$. Let $$X$$ be the event that $$T_N=2$$ and let $$Y$$ be the event that $$T_N=3$$.

Some calculation (*) leads to the result that $$\mathbb{E}(N)=\mathbb{E}(N|X)=\mathbb{E}(N|Y)=3/2$$.

Let $$Z$$ be the event that $$T_i\neq3$$ for all $$i$$. Some calculation (**) leads to the result that $$\mathbb{E}(N|Z)=2$$.

I find it very unintuitive that $$\mathbb{E}(N|X)\neq\mathbb{E}(N|Z)$$. Obviously we have $$Z\subsetneq X$$. However, the information $$Z$$ gives, which $$X$$ does not give, intuitively only affects what comes after the $$N$$'th throw. So how is it possible that the probability distribution of $$N$$ is different when conditioning on $$X$$ or $$Z$$?

(*) We have $$\mathbb{P}(X)=\mathbb{P}(Y)$$ and $$\mathbb{E}(N|X)=\mathbb{E}(N|Y)$$ by symmetry. Also notice that $$X$$ and $$Y$$ partition the event space, so $$\mathbb{P}(X)+\mathbb{P}(Y)=1$$, so $$\mathbb{P}(X)=\mathbb{P}(Y)=\frac12$$. Since $$\mathbb{P}(T_i\neq1)=2/3$$, we have $$\mathbb{E}(N)=3/2$$. By the principle of divide and conquer, we have $$\mathbb{E}(N)=\mathbb{P}(X)\mathbb{E}(N|X)+\mathbb{P}(Y)\mathbb{E}(N|Y)$$, so we find $$\mathbb{E}(N)=\mathbb{E}(N|X)=\mathbb{E}(N|Y)=3/2$$.

(**) We have $$\mathbb{P}(T_i\neq1|Z)=1/2$$, so $$\mathbb{E}(N|Z)=2/1=2$$.

By the way, if you have a suggestion for a better, more specific title, be my guest. I could not come up with a good descriptive title for this very specific question.

• do you mean $Z \subsetneq X$? – mathworker21 Aug 31 at 3:47
• Correct, I fixed it – SmileyCraft Sep 1 at 15:45

In all honesty, I am confused why you are bringing $$X$$ into the situation. Why aren't you just saying "The information $$Z$$ gives only affects what comes after the $$N$$'th throw, so why isn't $$E(N) = E(N | Z)$$?" The issue here (as well as in your actual question) is that you have the quantifiers/chronology backwards; $$N$$ isn't determined yet.
You are of course correct that once we know $$N$$, then $$X$$ and $$Z$$ provide the same information up to the $$N^{th}$$ throw. However, being in the world of $$Z$$ significantly affects the (expected) value of $$N$$. View $$E(N | Z)$$ as saying "I guarantee you that 3 won't be rolled; how long will you have to wait on average to see a 2?", the (partial) answer to which is obviously "longer than I would have to wait to see a 2 or 3 with no other conditions". I.e., rolling a non-1 on a two-sided die takes longer than rolling a non-1 on a 3-sided die.
To see it more clearly, start from scratch. Consider only a 2-sided die with sides 1,2. Let $$T_i$$ be the outcome of the $$i^{th}$$ throw and $$N = \min\{i : T_i \not = 1\}$$. Let $$Z$$ be the event that $$T_i \not = 2$$ for all $$i$$ (or, if you want to make $$Z$$ have positive measure, define $$Z$$ to be the event that $$T_i \not = 2$$ for $$i = 1,2,\dots,10^{100}$$). Then $$E(N) = 2$$ while $$E(N | Z) = \infty$$ (or some very very large number).
I think the main problem is that $$P(Z)=0$$, that means $$Z$$ almost surely does not happen. Arguing about expected values under conditons that almost surely do not happen are bound to be counterintutive, they are roughly equivalent to $$\frac00$$ limit forms in calculus, as both $$P(N \cap Z)=0$$ and $$P(Z)=0$$.
• In this case $Z$ basically only reduces the three sided dice to a two sided dice. Not too bad imo. You can also define $Z_n$ as $T_i\neq1$ for all $i\leq n$. Then for any event $\omega$ we have $\lim\mathbb{P}(\omega|Z_n)=\mathbb{P}(\omega|Z)$. This way you can avoid conditioning on an event of zero probability. – SmileyCraft Jan 28 at 23:23