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In the card game bridge, each of 4 players is dealt a hand of 13 of the 52 cards.

  1. What is the probability to get 13 cards that no one is Diamond?
  2. What is the probability to get 13 cards that one suit is not appeared at all? I need to show it is 0.051.

thank you!

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  • $\begingroup$ In 2, what do you mean by shape? Do you mean suit? $\endgroup$ – user264781 Oct 28 '15 at 17:46
  • $\begingroup$ yes! suit! @user264781 $\endgroup$ – user283468 Oct 28 '15 at 17:48
  • $\begingroup$ For $(1)$, follow the hints in gt6989b's answer. For $(2)$, you need to use Inclusion exclusion principle, $$\begin{align} \verb/Prob/(\text{one suit is missing}) = & \binom{4}{1} \verb/Prob/(\text{one specific suit is missing}) \color{blue}{\leftarrow\text{ answer from }(1)}\\ - & \binom{4}{2} \verb/Prob/(\text{two specific suits are missing})\\ + & \binom{4}{3} \verb/Prob/(\text{three specific suits are missing}) \end{align} $$ $\endgroup$ – achille hui Oct 28 '15 at 18:05
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hints

For (1), answer the following:

  1. How many possible bridge hands are there (52 total cards, drawing 13)?
  2. How many cards are there in the deck without diamonds?
  3. How many bridge hands are there without diamonds (drawing 13 cards from answer to 2 above)?
  4. What is the probability in question?

Update with your progress, and see if you can do (2) as well, happy to guide you.

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  • $\begingroup$ 1. $52\choose 13$ 2. $39\ \text{cards}$ 3. $39\choose 13$ 4. is for OP :) $\endgroup$ – user66407 Oct 28 '15 at 17:47
  • $\begingroup$ @user66407 why are you giving the OP answers without him working on this problem himself??? $\endgroup$ – gt6989b Oct 28 '15 at 17:52
  • $\begingroup$ the answer for 1 will be (13 from 39) divided to (13 from 52) ? thanks to both of you! $\endgroup$ – user283468 Oct 28 '15 at 17:57
  • $\begingroup$ @user283468 yes, but to do (2) you can make this is in how many ways if you don't know which suit will be missing? $\endgroup$ – gt6989b Oct 28 '15 at 17:59

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