# Probability in card games “The opening hand”

I'm convinced this is a fairly simple problem, but I'm just having trouble determining which formulas I should be using.

I have a 44 card deck that includes 5 "trump" cards.

The game starts with me drawing an 8 card opening hand--I'm simply trying to figure out the probability of drawing at least 1 of the "trump" cards in my opening hand.

I just need to know the formulas so that if I were to adjust the deck (adding more non-trump cards, adding more trump cards, etc) or if I wanted to change the opening hand to 11 (since the player who takes the second turn draws 3 on his first turn).

Thank you,

JJ

• probability of drawing at least 1 of the "trump" cards=1-probability of drawing none of the "trump" cards. Now it should be easy – Qwerty Jul 6 '16 at 19:24
• @Qwerty not really "at least none" rather "at most none" (aka "none"). – quid Jul 6 '16 at 19:27

$$Pr(\text{at least one trump}) = 1-Pr(\text{no trump}).$$
Here $$Pr(\text{no trump}) = \frac{39\cdot 38\cdot 37\cdot 36\cdot 35 \cdot 34\cdot 33 \cdot 32}{44 \cdot 43 \cdot 42\cdot 41\cdot 40\cdot39\cdot38\cdot37}$$