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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

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    $\begingroup$ probability of drawing at least 1 of the "trump" cards=1-probability of drawing none of the "trump" cards. Now it should be easy $\endgroup$ – Qwerty Jul 6 '16 at 19:24
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    $\begingroup$ @Qwerty not really "at least none" rather "at most none" (aka "none"). $\endgroup$ – quid Jul 6 '16 at 19:27
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$$Pr(\text{at least one trump}) = 1-Pr(\text{no trump}).$$

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

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  • $\begingroup$ Thank you, that's exactly what I was looking for. $\endgroup$ – JJ Tyler Jul 6 '16 at 19:36
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    $\begingroup$ @JJTyler, if you found the answer helpful please consider marking it as 'accepted'. Glad I could help :) $\endgroup$ – fosho Jul 6 '16 at 19:37
  • $\begingroup$ @Dman: perhaps you could also write it a bit more formally, with binomial coefficients $\endgroup$ – Alex Jul 6 '16 at 20:48
  • $\begingroup$ @Alex, you are right, but I am not sure what level the OP is at. Perhaps he will not recognize them. $\endgroup$ – fosho Jul 6 '16 at 20:53

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