25 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

- $3$
- $8$
- $9$
- $12$

0

Please re tag this to pigeon hole. I was looking in the GO book under pigeon hole.Its not probability

33 votes

Best answer

0

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

then A (3) is best choice

then A (3) is best choice

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)

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)

18 votes

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

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

5 votes

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

Hence **9 **is the answer.

Is it correct?

5 votes

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

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

0 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**.