# GATE2000-1.1

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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. $12$

edited
20
use pigeonhole principle Answer is 2*4+1=9
0
Please re tag this to pigeon hole. I was looking in the GO book under pigeon hole.Its not probability
2
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
1
1

This is a clear picture of 52 cards.

0

The minimum # of pigeons which assures atleast K+1 pigeons in some pigeon hole=Kn+1

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

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

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.

edited
0
may be in first 3 attempt we get same(suits) card and question is asking about minimum.

then A (3) is best choice
3
minimum in worst case.
3
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

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?

0
@shikhar May you please explain your approach more? I find it's correct use of "Expectation".

Any corrections or questions  to my solution are welcome

0
I could not understandem logic behind to apply pigeonhole principle. But this technique help me lot.
1 vote
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

edited

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.

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