Problem on selecting group of card from a well shuffled pack of card I have a problem I'm working on:

The minimum number of cards to be dealt from an arbitrarily shuffled deck of 52 card to guarantee that three cards are from some same suit is which amount?

I got a few solutions using the pigeonhole principle, but I don't have a clear idea about this how to solve this problem correctly. 
 A: Hint : What is the maximum number of cards that can be dealt such that it is possible that there are not three cards from the same suit ?
If you have this answer, the answer to your question is the same, plus $1$.
A: Your comment

pigeon hole principle states that "if n pigeons are assigned to m pigeon holes where n>>m than one pigeon hole must have ((n-1)/m)+1 pigeons. so m=4 and number of pigeon 3 so (n-1)/4 +1= 3 and you get n=9

gives the correct answer.
There are $4$ suits and $2 \lt 3$ so it is possible to have $4 \times 2=8$ cards without guaranteeing that three cards are from some same suit.  So $8$ cards are not enough.
However $8$ cards means all the pigeonholes are full and a $9$th card does guarantee that three cards are from some same suit. 
A: Although my math professor would hate me answering the question like this but, anyways... 
Assume a standard $52$ card deck with no jokers. Then, we have four suits. And $13$ of each suit. Pick a card and let's  arbitrarily assume it's a club. So, now our "hand" contains $\lbrace club_1 \rbrace$. 
Now, I may not choose from that suit again. Then, pick another card from a different suit. So, arbitrarily assume that I select a heart. And now our hand contains $\lbrace club_1, heart_1 \rbrace$. 
Continue this process until we have one kind of each suit. Our hand contains $\lbrace club_1, heart_1, spade_1, diamond_1 \rbrace$. I can then pick four more cards, in the same fashion and have I still have less than three cards of the same suit. That is, my hand has  $\lbrace club_1, heart_1, spade_1, diamond_1, club_2, heart_2, spade_2, diamond_2 \rbrace$. 
The next card I select must be from one of the four suits, and then I have made a hand which contains three suits of one kind. 
On this note, the other answers are much more proper in their logic, and make excellent use of mathematical principles. But I hope this answer gives you a little more intuition on the reasoning behind the pigeon hole principle.
A: Let's think about this systematically, using a smaller example: suppose we want the minimum number of cards drawn such that there must be a repeat in one suit, and suppose we're talking about a regular deck with no jokers. 
If you drew two cards, you could get repeat clubs- but it doesn't have to happen: you could draw a club and a heart, for example. Same with three cards. 
Now, if you drew four cards, there doesn't have to be a repeat: you could have drawn a club, heart, spade, and diamond. What happens when we draw one more card, bringing your total count to 5? That set of a club, heart, spade, and diamond will be joined by one more of a club, heart, space, or diamond. We don't know which one, but by drawing 5, we guarantee that at least one of the suits will have a repeat. 
So, the answer to that example is 5. Can you try the same for 3 cards being of the same suit, instead of 2? Try using the same logic I gave you. 
