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A 10-hand card is dealt from a well shuffled deck of 52 cards. What is the probability that the hand contains at least two cards from each of the four suits?

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  • $\begingroup$ What are your thoughts? What suit distributions satisfy the requirement? Can you calculate the chance of each one? $\endgroup$ – Ross Millikan Mar 26 '14 at 19:11
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    $\begingroup$ Hint: There are two possible suit length patterns that fulfil the conditions: 4-2-2-2 and 3-3-2-2. Count the number of hands for each of these, and divide by the total number of hands. $\endgroup$ – TonyK Mar 26 '14 at 19:26
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You either need the suits distributed $4222$ or $3322$. The chance of $4222$ is $$\frac{{4 \choose 1} (\text{suit with four cards}) {13 \choose 4}{13 \choose 2}^3}{52 \choose 10}$$ The chance of $3322$ is $$\frac{{4 \choose 2} (\text{suits with three cards}) {13 \choose 3}^2{13 \choose 2}^2}{52 \choose 10}$$ for a total of $\displaystyle\frac {7592832}{27657385} \approx 0.2745$

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  • $\begingroup$ I think we can choose cards with sequence 4222,2422,2242,2224 and 3322,3232 and so on, therefore finding answer by this way is impossible $\endgroup$ – user138232 Mar 26 '14 at 19:35
  • $\begingroup$ @user138232: Ross's answer is correct (although a little hard to read). Try to work out why. $\endgroup$ – TonyK Mar 26 '14 at 19:37
  • $\begingroup$ I had not seen completely Ross's answer, that's why i couldn't understand.now i understand. Thanks $\endgroup$ – user138232 Mar 26 '14 at 19:48

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