# Probability of specific card being in the deck vs your players hand?

If there are 7 cards left in a deck of cards, and 5 cards in the other players hand and you've seen all the cards except for the following:

Hearts:♥ 10 Q K A

Clubs: ♣ 10 J Q K A

Diamonds: ♦ A

1. What is the probability that the Ace of Diamonds is in the deck?
2. What is the probability that the other player's hand contains at most 1 spade?
3. What is the probability that the other player's hand contains at most 1 spade, given that the ace of diamonds is in the deck?

My first thought for number 1 is simply P(A) = 7/12, but I have a feeling that is wrong.

For number 2, wouldn't the answer be 762/792?

Since $$\frac{\binom{2}{2}\binom{10}{3}}{\binom{12}{5}}$$ represents the number of hands where both spades are drawn,

P(B) = 1- $$\frac{\binom{2}{2}\binom{10}{3}}{\binom{12}{5}}$$ = $$\frac{762}{792}$$ where A is the event that at most one spade is drawn.

For number 3, I know in order to compute P(B|A), you need to know the intersection of A $$\cap$$ B, which I am confused about how to compute because they are aren't disjoint.

How do I go about solving these?

• For problem $1$, you are perfectly correct.
– lulu
Jan 24 '19 at 0:05
• For $2$, your formula and method are correct, but the computation is wrong. $\binom {10}3=120$.
– lulu
Jan 24 '19 at 0:08
• For 3, the same method you used for $2$ works here, though you only have $6$ cards left to place in the deck and only $11$ total cards to choose from.
– lulu
Jan 24 '19 at 0:10
• Of course, this is assuming that the twelve cards are randomly distributed for some reason. Jan 24 '19 at 0:18
• OP: By "randomly distributed," I think Graham Kemp means that the cards are identically distributed (e.g., the diamond ace is no more likely to be in the deck a priori than any of the other cards). Jan 24 '19 at 0:47

To justify your answer for Question 1: Let's count the number of ways the Ace of Diamonds could be in the deck. There would have to be $$6$$ other cards in the deck, so there are $$\binom{11}{6}$$ ways this can happen (and the remaining $$5$$ cards must be in your opponent's hand. There are $$\binom{12}{7}$$ total ways the cards could be distributed between the deck and the opponent's hand if we don't care where the Ace of Diamonds is. So the probability the Ace of Diamonds is in the deck is: $$\frac{\binom{11}{6}}{\binom{12}{7}}=\frac{\frac{11!}{6!5!}}{\frac{12!}{7!5!}}=\frac{11!7!}{12!6!}= \frac{7}{12}.$$

As others have stated, your formula for Question 2 is correct, but you made an arithmetic error.

For Question 3, let $$B=\{\text{other player's hand has at most one spade}\}$$ and $$A=\{\text{Ace of Diamonds is in the deck}\}$$. As you said, you need to compute $$P(B\mid A)$$. Since the event $$B$$ is somewhat complicated - the other player's hand could have zero or one spade - but $$B^c$$ is simple, because it means the other player's hand has both spades. Now you can use the fact that $$P(B\mid A) = 1-P(B^c \mid A)$$. Now you just need to compute $$P(B^c \mid A)=P(B^c \cap A)/P(A)$$. The numerator involves counting the number of ways that both spades are in the opponent's hand and the Ace of Diamonds is in the deck, which I believe you should be able to do.

• Thank you! So I computed T's complement and got 2/11, is this correct? Since 2C2 * 9C3 / 11C5... but I feel like I am missing something. Jan 24 '19 at 0:29
• What is $T$? The calculation you just wrote shows that $P(B^c \mid A)=2/11$, so now what is $P(B\mid A)$?
– kccu
Jan 24 '19 at 0:34
• Sorry I meant B :P Jan 24 '19 at 0:39
• So P(B |A) = 9/11? Jan 24 '19 at 0:41
• Yes, that is correct.
– kccu
Jan 24 '19 at 0:41