The following question occurred to me while I was solving exercise 2.7(b) of the first edition of Mitzenmacher-Upfal. Let $X, Y$ be independent geometric random variables. Which of the following claims correctly gives the value of $\Pr(\max(X, Y) = i \cap X = Y)$, and why?
Claim 1:
By the definition of conditional probability, we have
$$\Pr(\max(X, Y) = i \cap X = Y) = \Pr(\max(X, Y) = i | X = Y) \Pr(X = Y)$$
If $X = Y$, then the event "$\max(X, Y) = i$" is equivalent to the event "$X = i \cap Y = i$". Therefore,
$$= \Pr(X = i \cap Y = i)\Pr(X = Y)$$
Claim 2:
The event "$\max(X, Y) = i \cap X = Y$" is equivalent to the event "$X = i \cap Y = i$", thus
$$\Pr(\max(X, Y) = i \cap X = Y) = \Pr(X = i \cap Y = i)$$
The first claim is wrong. $P(\max(X,Y) = i \mid X=Y)$ is actually $$\frac{P(\max(X,Y) = i \cap X=Y)}{P(X=Y)} = \frac{P(X=i \cap Y=i)}{P(X=Y)}.$$ Your mistake was dropping the conditioning on the event $X=Y$.
After this correction, both approaches give the same answer.