Distinguishing lottery probability cases from each other A lottery has two cases which my common sense has problems distinguishing from each other, when the combinatorics say their probabilities are different.
" A lottery has 100 tickets. One of them gives prize A, another gives prize B. John has bought 4 tickets."

*

*John tells you that he won prize A. What is the probability that he also won B?


*John tells you that he won at least one prize. What the probability that he won both prizes?
For 1) I think that you can simply remove 1 ticket from the pool. Then you have 3 tickets each giving $\frac{1}{99}$ chance of winning, totaling in a $\frac{3}{99}$ = $\frac{1}{33}$ = P(B|A).
For 2) I would think the same. Since I know he won either A or B, I can remove 1 ticket from the pool, and so the answer would also be $\frac{1}{33}$. But the answer is $\frac{1}{65}$:
"C" : winning at least one.
P(C) = 1 - "winning nothing" = 1 - $\frac{{98}\choose{4}}{{100}\choose{4}}$ = 1 - $\frac{152}{165}$ = $\frac{13}{165}$
P((A$\cap$B|C) = $\frac{P(A\cap B\cap C)}{P(C)}$ = $\frac{P(A\cap B)}{P(C)}$ = $\frac{P(B|A)\cdot P(A)}{P(C)}$ = $\frac{\frac{3}{99}\cdot \frac{4}{100}}{\frac{13}{165}}$ = $\frac{1}{65}$.
I know the combinatorics are right, but I would still say $\frac{1}{33}$ first thing if someone asked me. It still "feels" right. Could anyone point out what is wrong with my thinking here?
 A: Even though the labels A and B are arbitrary, they can still hold useful information about the different events we are considering. Winning one in particular (say A) is more specific than winning at least one (which could be A or B).
In each case we are looking for the probability of winning both prizes conditioning on different events:

*

*John won at least one prize and we know he won prize A.

*John won at least one prize.

Looking at it this way, you can see the first event is really a strict subset of the second one. This only works because we agreed on the labels before John drew the lottery tickets. Your intuition for the second case would work if we could assign the labels after we saw the result (we would just let that prize be "A").
This idea could be expressed equationally:
$$\frac{1}{33} 
= P(\mathrm{B} \mid \mathrm{A})
= P(\mathrm{A} \cap \mathrm{B} \mid \mathrm{A})
\neq P(\mathrm{A} \cap \mathrm{B} \mid \mathrm{A} \cup \mathrm{B})
=\frac{1}{65} 
$$
but I personally find Venn diagrams to be a bit more enlightening:
Venn diagram
A: This problem is a variant of the so-called Boy or Girl Paradox, see also this article. The translation is "win prize A = older child is a girl," "won at least one prize = at least one child is a girl" and "won both prizes = both children are girls." The specific probabilities are different, but the structure is the same.
