# Probability of drawing 3 cards in a 16 card deck where one of those 3 cards a duplicate in the deck?

Good day, I was playing Munchkin: Loot Letter (A "spinoff" of the card game Love Letter) with a friend of mine. There are 16 cards in the deck, and in 2 player games, you burn 3 cards face up. My friend managed to burn 2 unique cards and 1 card that had a duplicate card in the deck.

Let's say that the cards are A, B, and C where A and B are unique and Card C has 2 copies in the deck. The order does not matter as well.

I was thinking of the probability of this happening (burning 2 unique cards and 1 card that had a duplicate). At first, I was thinking along the lines of:

(1/16)(1/15)(1/14)

My general idea was that the chance of drawing 1st card was 1 out of the deck count, and the chance of drawing the 2nd card was 1 out of the deck count - 1, and the chance of drawing the 3rd card was 1 out of the deck count - 2 -- then those probabilities are multiplied.

However, this would only work assuming that all 16 cards are unique. However, one of the cards drawn/burned have 2 copies in the deck. Now I'm not sure how to take into account the probability of drawing a card that had 2 copies in the deck.

How does one proceed with calculating probability for this situation? I'm not sure how to proceed with this problem. Any help is very much appreciated. Thank you.

You have three disjoint cases:

1. You first draw A and B (in any order) and then C.
The probability of drawing A or B is $\frac{2}{16}$. The probability of drawing the other one is $\frac{1}{15}$. Then, the probability of drawing C is $\frac{2}{14}$.
2. You first draw C and than A and B (in any order).
The probability of drawing C is $\frac{2}{16}$. The probability of drawing A or B $\frac{2}{15}$. Then, the probability of drawing the other one is $\frac{1}{14}$.
3. You first draw A or B and than C and then the other one.
The probability of drawing A or B is $\frac{2}{16}$. The probability of drawing C $\frac{2}{15}$. Then, the probability of drawing the other one is $\frac{1}{14}$.
Then, the total probability is $$\dfrac{2}{16}\cdot \dfrac{1}{15} \cdot \dfrac{2}{14} + \dfrac{2}{16} \cdot \dfrac{2}{15} \cdot \dfrac{1}{14} +\dfrac{2}{16} \cdot \dfrac{2}{15} \cdot \dfrac{1}{14} = \dfrac{1}{280}$$
• Er no, the third card isn't equal to A or B. It's a Card C that has 2 copies in the deck. – Razgriz Mar 5 '16 at 14:04
• Oh, I misunderstood quite everything... How may copies of each card are there in the deck? – Τίμων Mar 5 '16 at 14:10
• Actually the one I'm interested is A and B (only 1 copy) and C (2 copies). – Razgriz Mar 5 '16 at 14:11
• So, you don't want to pick any three cards with that property, but exactly A, B and C. Ok, I'll edit my answer. – Τίμων Mar 5 '16 at 14:15
• One last. :)) The orders do not matter. :)) Sorry for failing to point that out previously. However, it seems that you took into account the "No specific order" already it seems. – Razgriz Mar 5 '16 at 14:38