# of cards to draw to get five of one suit

[N5PN]

Problem: 52 card deck, how many cards must one draw to ensure five cards of the same suit?

It is possible to get the first five cards drawn all one suit, the least number of cards you could draw to get five of one suit.
To ENSURE five cards of one suit, I think it is required to think of the other end of the outcome possibilities,
which I think is when each card drawn is in sequence, a different suit.
That way if the questions was changed to how many to get 'two cards of the same suit' the fifth card drawn would have to match one of the first four.
It looks like this can be extended to five cards of the same suit.

I solved it drawing a row-column matrix, with four columns representing the suits, and the rows representing the cards being drawn --
Each row has four cells, one for each column, and for representing the worst case, the drawn card number sequence is this,
with the first card drawn being a Spade, the second one a Club, the third a diamond, the fourth a Heart and in sequence each card drawn is the next suit.

Sp Club Dia. Hearts
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
17
It seems clear by this matrix that to ensure five cards of one suit, one must draw at least 17 cards.

It also appears that this matrix method can be extended or modified in either the number of rows or columns to make a generic tool.
Could we use this to obtain the number of cards necessary to be drawn for an imaginary deck of cards with 6 suits to get 3 of the same suit?
I think so, it seems to be the first card draw number in the third row of that six-column matrix, or the thirteenth card drawn, meaning one needs draw thirteen cards for that case.

Now my questions:
1. Is 17 the right answer?

2, I want to find the formula (or really prove the equation I think gives the formula) for the general case --

3. is this a combination problem in statistics that is explained by a standard formula? I could not seem to find this specific problem in the examples in my math book on combinations, and it does not look like permutation.

4. Does a 52 card deck have anything to do with solving the problem? Or is there a different method/formula to get the same answer that does use the fact of 52 cards in the deck?
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I think the formula for the problem is: 'N' represents the number of suits (or columns in the matrix) - 'C' represents the number of cards of the same suit needed - (the rows)
(C x N) - (N - 1) --> CN -N +1

For 5 cards, four suits this equals 20 - 4 + 1 or 17

3 cards, six suits = 18 - 6 + 1 = 13

2 cards four suits = 8 - 4 +1 = 5

So it seems to work, but is there a more formal way to prove it is the general formula?

 

[Deleted member 4993]

you can draw 16 cards and have 4 cards each from 4 different suites - worst case.

The 17 th card must be the 5 th card of one of the suits.

This is called Pigeon hole principle - the formula would be N*(C-1) + 1

 

[Denis]

Ok N5PN: there's a drawer containing over 100 black socks and over 100 white socks.
If you pull socks out at random (it's dark!), how many do you need to pull to get a pair the same color?

 

[galactus]

Think about it. If you draw out 3 socks and there are only two colors, the third is going to match one of them. Correct?.

Suppose on the first draw you get a white sock. The second draw is either going to be black or white. If it's white, there's a match. If it's black, then there is one of each color. On the third draw, you either have a white or a black.

We have either WWB, WBB, WBW. A match regardless. 3 draws are needed to assure a match. Note, the 100 white and 100 black. These numbers really do not matter. It could be 5 white and 5 black and it would still be 3 draws to assure a match.