I am self-studying from the fourth edition of Statistics by Freedman, Pisani, and Purves, and I am having difficulty rationalizing the answer to one of the problems. The question is on page 232 of Chapter 13, shown below Screenshot of the question and answer
Question
Two draws are made at random without replacement from the box $\fbox{$\fbox{1}$ $\fbox{2}$ $\fbox{3}$ $\fbox{4}$}$. The first ticket is lost, and nobody knows what was written on it. True or false, and explain: the two draws are independent.
Answer
This is false. It's like saying someone doesn't have a temperature because you can't find the thermometer. To figure out whether two things are independent or not, you pretend to know how the first one turned out, then see if the chances for the second change. The emphasis is on the word "pretend."
When I first answered this question, I said the answer was True. I thought it was true because of the result of the following question.
Two draws are made at random without replacement from the box $\fbox{$\fbox{1}$ $\fbox{2}$ $\fbox{3}$ $\fbox{4}$}$. You win 1 dollar if the second ticket is a $\fbox{2}$. What is your chance of winning the dollar?
Let's say that $A$ is the event where the first draw is $\fbox{2}$ and $B$ is the event where the second draw is $\fbox{2}$. The $P(B)$ is then equal to
$$P(B) = P(B|A)P(A) + P(B|\neg A)P(\neg A)$$
$$P(B) = (0)(1/4) + (1/3)(3/4)$$
$$P(B) = 1/4$$
I believe that you can take the same approach shown above to the extreme and demonstrate that even when you sample without replacement to exhaust the population of tickets, the chances of drawing a $\fbox{2}$ as the last ticket without revealing any previous ticket is $1/4$. So when you draw two tickets randomly without replacement and lose the first ticket, losing the first ticket without knowing what was written on the ticket doesn't give you any information about the second ticket.
Yet the answer to the original question of sampling two tickets without replacement is dependent. Using the same definition for $A$ and $B$ and the definition of independence for conditional probability $P(B|A)=P(B)$ we can show:
$$P(B|A) = 0$$
$$P(B) = 1/4$$
$$P(B|A) \neq P(B)$$
Therefore the two events are dependent.
I am still trying to understand how I square both results. The results still seem contradictory because losing the ticket and not knowing what was written gives you no information to know the second ticket, yet the two events are dependent.