Ten million people enter a certain lottery. For each person, the chance of winning is one in ten million, independently I am solving the following question in Blitzstein & Hwang:
"Ten million people enter a certain lottery. For each person, the chance of winning is
one in ten million, independently.
(b) Congratulations! You won the lottery. However, there may be other winners. Assume
now that the number of winners other than you is W ~ Pois(1), and that if there is
more than one winner, then the prize is awarded to one randomly chosen winner. Given
this information, findnd the probability that you win the prize (simplify)."
My solution is to use Poisson approximation to first approximate the PMF and then calculate the desired probability.
Calculate the PMF: Let X be a r.v of the number of people who win the lottery. Let $N=10^7$. Then
$$
X=I_1+I_2+...+I_N \\
\lambda=E[X]=P(I_1)+P(I_2)+...+P(I_N)=N 1/N=1
$$
By the Poisson approximation we have then
$$
P(X=k) \approx e^{-1}1^k/k!
$$
Calculate the probability that you win: Let $A$ be the event that you win.  Then, I know that we can calculate this probability by summing over all possible events $W=k$,
$$
\begin{aligned} 
P(A) &= \sum_{k=0}^\infty P(A|W=k)P(W=k) \\
&= \sum_{k=0}^\infty\frac{1}{k+1} \frac{e^{-1} \lambda }{k!} \\
&= e^{-1} (e^1-1) = 1-e^{-1}
\end{aligned} 
$$
Thus, $1-e^{-1}$ is the correct answer. But let us approach this problem from a slightly different angle. Let us seperate the case where I win with certainty $P(A|W=0)=1$ from the events where I do not win with certainty. And use that $P(W >0) =1- P(W =0)=1- \frac{e^{-1} 1 }{0!}=1-e^{-1}$.
$$
\begin{aligned} 
P(A) &= P(A|W=0)P(W=0) + P(A|W >0)P(W >0)  \\
&= 1 e^{-1} + P(A|W >0)(1-e^{-1}) \\
&= e^{-1} +  \sum_{k=1}^\infty\frac{1}{k+1} (1-e^{-1})  \\
 &=  ???
\end{aligned} 
$$
And now the problem is that $\sum_{k=1}^\infty \frac{1}{k+1}$  is a sum that diverges, as far as I can see. My question is then how can this second approach lead to a dead end? This second approach of viewing this problem is, for me, a more intuitive view of analyzing the problem. Yet this intuition is leading me to a dead end. Your guidance and comments are most welcomed.
My question is related to Probability that you win the prize, but mine question is different.
 A: Your mistake is the incorrect intuition that
$$ P(A\mid W>0) \stackrel?=
 P(A\mid W=1) + P(A\mid W=2) + \cdots + P(A\mid W=N). $$
Here is a correct formulation:
\begin{align}
P(A\mid W>0) &= \frac{P(A \cap W>0)}{P(W>0)} \\
&= \frac{P(A\cap W=1) + P(A\cap W=2) + \cdots + P(A\cap W=N)}{P(W>0)} \\
&= \frac{P(A\cap W=1)}{P(W>0)} + \frac{P(A\cap W=2)}{P(W>0)} + \cdots + \frac{P(A\cap W=N)}{P(W>0)} \\
\end{align}
Now since $P(W>0) > P(W=k)$ for any $k,$ clearly
\begin{align}
 \frac{P(A\cap W=1)}{P(W>0)} &< \frac{P(A\cap W=1)}{P(W=1)} = P(A\mid W=1),\\
 \frac{P(A\cap W=2)}{P(W>0)} &< \frac{P(A\cap W=1)}{P(W=2)} = P(A\mid W=2),\\
  &\qquad\vdots\\[1ex]
 \frac{P(A\cap W=N)}{P(W>0)} &< \frac{P(A\cap W=N)}{P(W=N)} = P(A\mid W=N),
\end{align}
and therefore
$$ P(A\mid W>0) < P(A\mid W=1) + P(A\mid W=2) + \cdots + P(A\mid W=N). $$
In summary, every term in your sum, as you wrote it, is too large by a factor of $P(W>0)/P(W=k).$
We could even say that your mistake is saying that
$$
\frac{a + c + e}{b + d + f} \stackrel?=
\frac ab + \frac cd + \frac ef. 
$$
