We lost two cards from a deck of $52$ cards. If we extract a card of this deck, what is the probability we get a diamond? We lost two cards from a deck of $52$ cards. If we extract a card of this deck, what is the probability we get a diamond?
I'm confused here.
Suppose we lost two diamonds, then
$$P(\text{we get a diamond})=\frac{11}{50}$$
but, if we lost two cards of the other suits, we have
$$P(\text{we get a diamond})=\frac{13}{50}$$
if we lost one of the diamonds and one card from the other suits, we have:
$$P(\text{we get a diamond})=\frac{12}{50}$$
but here I'm not sure, can someone help me?
 A: Another approach is to see that by symmetry, the answer must be the same for any of the four suits, since there was no information about the suit of either of the lost cards.  Since the probabilities associated with each of the four suits must still sum to one, the probability of drawing a diamond is just $1/4$.
Now, if you had to determine the probability that, for instance, at least one of the first two draws was a diamond, well, that might get a little harder...
A: Using the second method in my Comment, and using R as a calculator.
There are d = 52*51*50 ways to choose three cards without replacement.
The number of ways for the third card to be a Diamond is 
n = 39*38*13 + 39*13*12 + 13*39*12 + 13*12*11. The ratio n/d is the answer: 1/4.
d = 52*51*50
n = 39*38*13 + 39*13*12 + 13*39*12 + 13*12*11
n/d
## 0.25

A: Here is an equivalent way to understand Brian’s answer:
Suppose that you will draw three cards from the deck. What is the probability that the third card you draw from the deck will be a diamond?
A: Suppose you deal out all the cards, face down, in a line, and then turn over the leftmost card.  What's the probability that it is a diamond?  $\frac14$, of course.
Now deal out the cards as before, but this time, before you turn over the leftmost card, you take two cards off the right-hand end and, without looking at them, you throw them away.  What is the probability now that the leftmost card is a diamond?
