markdcasino Joined: Jan 27, 2018
• Posts: 8
July 31st, 2020 at 8:06:43 PM permalink
Taking a standard 52-card deck, what is are the combinations of drawing all 13 cards of the same suit if you randomly pull 14 cards? What is the probability of pulling 12 cards of the same suit if you randomly pull 14 cards? To figure out the combinations I am using the following formulas (I'm not sure which one is correct):

For 13 of the same suit out of 14 cards:
A: =combin(14,13)*4)*combin(38,0) or is it
B: =combin(14,13)*4)*combin(38,1)

For 12 of the same suit out of 14 cards:
A: =combin(14,12)*4)*combin(38,1) or is it
B: =combin(14,12)*4)*combin(38,2)

Total combinations should be =combin(52,14) = 1,768,966,344,600. Then of course to figure the probability, this figure would be the denominator with the specific combinations as the numerator.

Am I on the right track?
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5748
Thanks for this post from: August 1st, 2020 at 7:25:50 AM permalink
The first one is neither A nor B.
There are only four ways to get all 13 cards of a suit; for each one, you then draw one of the remaining 39 cards.
The answer is combin(13,13) * 4 * combin(39,1)

(a) How many ways are there to get 12 out of the 13 cards of a particular suit?
(b) How many different suits are there?
(c) How many cards do you still need to draw, and how many cards in the deck can they be? (Hint: you can't draw the 13th card of the same suit as the first 12)
markdcasino Joined: Jan 27, 2018
• Posts: 8
August 1st, 2020 at 7:52:26 AM permalink
Thank you. So for 12 would it be the following?

=combin(13,12)*4*combin(39,2)

11 would be
=combin(13,11)*4*combin(39,3)

Is that the right formula? And total combinations would still be =combin(52,14)?
Last edited by: markdcasino on Aug 1, 2020
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5748
Thanks for this post from: August 1st, 2020 at 8:19:30 AM permalink
Quote: markdcasino

Thank you. So for 12 would it be the following?

=combin(13,12)*4*combin(38,2)

11 would be
=combin(13,11)*4*combin(37,3)

Is that the right formula? And total combinations would still be =combin(52,14)?

Close. There are always 39 cards not in the suit.

12 is combin(13,12) * 4 *combin(39,2)
11 is combin(13,11) * 4 * combin(39,3)

Note that, once you get below 8, you have to allow for the possibility that you can have two suits with that many cards.
For example, with 14 cards, you can have 7 spades and 7 hearts.

Yes, the total number of combinations of 14 cards drawn from 52 is combin(52,14).
markdcasino Joined: Jan 27, 2018
• Posts: 8
August 3rd, 2020 at 10:32:01 AM permalink
Thank you for all your help. So when you get to seven or below, how does the formula change? It gets tricky because we would be double counting, right? Let's take seven for example, it's not as easy as

combin(13,7)*4*combin(39,7)

Six would not be as simple as:

combin(13,6)*4*combin(39,8)

I'm missing that extra layer of combinations so we aren't double counting.
gordonm888 Joined: Feb 18, 2015
• Posts: 4402
August 3rd, 2020 at 11:39:45 AM permalink
Quote: markdcasino

Thank you for all your help. So when you get to seven or below, how does the formula change? It gets tricky because we would be double counting, right? Let's take seven for example, it's not as easy as

combin(13,7)*4*combin(39,7)

Six would not be as simple as:

combin(13,6)*4*combin(39,8)

I'm missing that extra layer of combinations so we aren't double counting.

Excellent question! You are getting advanced, now. Let's take the example of six suited cards when selecting 14 cards from a deck of 52 cards,

You must realize that the suit distributions of your 14 cards can be:

6-8-0-0 *12 permutations
6-7-1-0 * 18
6-6-2-0 * 12
6-6-1-1 * 6
6-X-Y-Z where X,Y,Z are all <6 | *4 permutations

So
6-8-0-0 = 12* combin(13,8)* combin(13,6)
6-7-1-0 = 18*combin(13,7)* combin(13,6)* combin(13,1)
6-6-2-0 = 12*combin(13,6)* combin(13,6)* combin(13,2)
6-6-1-1 = 6*combin(13,6)* combin(13,6)* combin(13,2)

Then
6-X-Y-Z = 4*combin(13,6)*combin(39,8) - {6-8-0-0} - {6-7-1-0} - {6-6-2-0} - {6-6-1-1}

when I am doing complex calculations like this on a spreadsheet, I always make sure that I explicitly list all of the (double counting) configurations and calculate them explicitly so I can keep track of what I am doing.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
markdcasino Joined: Jan 27, 2018
• Posts: 8
August 4th, 2020 at 12:09:09 PM permalink
Thank you Gordon. So for seven, here is my breakdown following your formulas:

7-7-0-0 (12 permutations)
7-6-1-0 (18 permutations)

I'm not completely sure how we are calculating 12 and 18

7-7-0-0 = 12*combin(13,7)*combin(13,7)
7-6-1-0 = 18*combin(13,7)*combin(13,6)*combin(13,1)

So for seven it would be (combin(13,7)*4*combin(39,7)) - {7-7-0-0) - {7-6-1-0}
gordonm888 Joined: Feb 18, 2015
• Posts: 4402
August 4th, 2020 at 4:22:53 PM permalink
Quote: markdcasino

Thank you Gordon. So for seven, here is my breakdown following your formulas:

7-7-0-0 (12 permutations)
7-6-1-0 (18 permutations)

I'm not completely sure how we are calculating 12 and 18

7-7-0-0 = 12*combin(13,7)*combin(13,7)
7-6-1-0 = 18*combin(13,7)*combin(13,6)*combin(13,1)

So for seven it would be (combin(13,7)*4*combin(39,7)) - {7-7-0-0) - {7-6-1-0}

7-7-0-0 (6 permutations)
7-6-1-0 (24 permutations)

For example, the six permutations of 7-7-0-0 are:

7-7-0-0; 7-0-7-0; 7-0-0-7; 0-7-7-0; 0-7-0-7; and 0-0-7-7;

And you don't need 7-6-1-0 because it has only one way to make a seven card flush.

So: 7-7-0-0 = 6*combin(4,2)*combin(13,7)*combin(13,7)

My formula is different because the factor combin(4,2) accounts for selecting two of the four suits

and for seven it would be (combin(13,7)*4*combin(39,7)) - {7-7-0-0) which is the formula for making selecting 14 cards and making a single seven card flush.

The formula (combin(13,7)*4*combin(39,7)) + {7-7-0-0) gives you the total number of hands that will make one or more seven card flushes.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.