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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. $12$

use pigeonhole principle Answer is 2*4+1=9
Please re tag this to pigeon hole. I was looking in the GO book under pigeon hole.Its not probability
N=(R-1)*k+1

k = no of pigeon holes

R=expected output here 3

N = minimum number

here we assign an R-1 number of items to k holes and to then add Rth item to any of the holes to get the minimum number

This is a clear picture of 52 cards. The minimum # of pigeons which assures atleast K+1 pigeons in some pigeon hole=Kn+1

$Here\ K+1=3\ \&\ n=4(suits)$

$\therefore 2\times 4+1=9$

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There are $4$ sets of cards. So, up till $8$ cards there is a chance that no more than $2$ cards are from a given set. But, once we pick the $9$$^{th}$one, it should make $3$ cards from any one of the sets. So, $(C)$ is the answer.
by

may be in first 3 attempt we get same(suits) card and question is asking about minimum.

then A (3) is best choice
minimum in worst case.
But it will not guarantee  that three cards are from same suit. Only Picking 9 or more ( 9 or10 or ......or 52) will guarantee that three cards are from same suit.

Minimum of( 9 ,10, 11...52)= 9, So, 9 is the answer(minimum number of cards to be dealt to guarantee that three cards are from same suit)
As suggested above also

apply pigeon hole, 4 holes (suits)

n pigeons(no of cards to be drawn)

floor [(n-1)/p] +1=3

floor[(n-1)/4]  =2

(n-1)/4  >= 2

n>=9

minimum 9 cards must be picked
by Any corrections or questions  to my solution are welcome

### 1 comment

I could not understandem logic behind to apply pigeonhole principle. But this technique help me lot.

Can this question be solved by calculating expectation:
Expectation of getting three cards from same deck->

E(x) = 13*(13/52) + 12*(12/51) + 11*(11/50)
=  8.49 => 9 cards

Is it correct?

by

### 1 comment

@shikhar May you please explain your approach more? I find it's correct use of "Expectation".
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

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.