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39 votes

The minimum number of cards to be dealt from an arbitrarily shuffled deck of $52$ cards to guarantee that three cards are from same suit is

  1. $3$
  2. $8$
  3. $9$
  4. $12$
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6 Answers

3 votes
3 votes
An easier way to think it as of worst case outcomes that is each time we take the card it is from different suit.

We know that there are 4 suits.

Let,

Heart= $H$

Spade= $S$

Diamond= $D$

Club= $C$

So the possible withdrawal sequence  (alternating/worst case)

 $HSDCHSDC\displaystyle$__ (this 9th card will make 3 cards of same suit.)

Therefore option C
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1 votes
1 votes

Let Min Number of cards be N.

 

Given,

No of holes (holes) = 4 (since there are 4 suits in pack of 52 cards)

Required no of cards (R) = 3 (since 3 cards of same suit required)

Using Generalized Pigeon Hole,

$\left \lceil \frac{N}{holes}\right\rceil$ = R

$\left \lceil \frac{N}{4}\right\rceil$ = 3

So minimum N satisfying above equation is 9 ( since $\left \lceil \frac{9}{4}\right\rceil$ = 3).

Hence Option C is correct.

Answer:

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