How is calculated the frequency of a pair, a flush of 2 cards, a straight of 2 cards and a straight flush of 2 cards (52 cards, hand of 2, 3, 4, etc)?

Question

Introduction

Before starting, I would like to specify a few things:

1. The deck has 52 cards. From 1 (Ace) to 10, J, Q, K, all in four different suits: spades, diamonds, club and hearts. I don't add jokers because that is far beyond my possibilities of calculations.
2. For better precision, I won't use chance, instead I would use combinations or frequency / total possible combinations.
3. This is based on poker, but it isn't exactly poker. I mean, for example, the frequency on a pair in 5-cards poker is $2860={13 \choose 1}{4 \choose 2}{12 \choose 3}{4 \choose 1}^3$, this formula that can be found on Wikipedia already excludes the chance of having not two but three or four cards with the same rank (three or four of a kind) and also excludes the chance of a full house (three of a kind + pair). What I want to discover, is not the frequency of a pair, instead, is the frequency of having at least two cards of the same rank. I'm not using the rules of poker, so I don't want to remove from the frequency the combinations that are "better". For example, the hand Q Q Q J J, which normally would be a full house, for this calculation I want it to be included in the "pair" calculation, because if you have three Qs, that means that you have at least a pair of Q, same with the Js.

What do I want?

What I want to calculate is the frequency of drawing:

1. At least two cards of the same rank (a pair).
2. At least two cards of the same suit (a flush of 2).
3. At least two consecutive cards (a straight of 2).
4. At least two consecutive cards of the same suit (a straight of 2 flushed).

All the previous frequencies I wanted to calculate them with:

1. A 2-card hand.
2. A 3-card hand.
3. A 4-card hand.
4. A 5-card hand.
5. A 6-card hand.
6. A 7-card hand.

My Calculations

Before starting to show my math I will show you how I got it. At wikipedia, the chance of a four of a kind if a 5-card deck is $642={13 \choose 1}{4 \choose 4}{12 \choose 1}{4 \choose 1}$. I understand this as choose one of 13 ranks, then choose all four suits of that rank, for the last card choose one other rank and then one suit for that card. Then I tried to recreate the formula with different reasoning. $642={13 \choose 1}{4 \choose 4}{52-4 \choose 1}$ which I understand as choose one of 13 ranks, then choose all four suits of that rank, finally choose one card other than the four previously used.

Below I will show you a table with my math.

Hand	2 cards	3 cards	4 cards	5 cards	6 cards	7 cards
Total combinations	$1.326={52 \choose 2}$	$22.100={52 \choose 3}$	$270.725={52 \choose 4}$	$2.598.960={52 \choose 5}$	$20.358.520={52 \choose 6}$	$133.784.560={52 \choose 7}$
Pair	$78={13 \choose 1}{4 \choose 2}$	$3.900={13 \choose 1}{4 \choose 2}{52-2 \choose 1}$	$95.550={13 \choose 1}{4 \choose 2}{52-2 \choose 2}$	$1.528.800={13 \choose 1}{4 \choose 2}{52-2 \choose 3}$	$17.963.400={13 \choose 1}{4 \choose 2}{52-2 \choose 4}$	$165.263.280={13 \choose 1}{4 \choose 2}{52-2 \choose 5}$
Flush of 2	$312={13 \choose 2}{4 \choose 1}$	$15.600={13 \choose 2}{4 \choose 1}{52-2 \choose 1}$	$382.200={13 \choose 2}{4 \choose 1}{52-2 \choose 2}$	$6.115.200={13 \choose 2}{4 \choose 1}{52-2 \choose 3}$	$71.856.600={13 \choose 2}{4 \choose 1}{52-2 \choose 4}$	$661.053.120={13 \choose 2}{4 \choose 1}{52-2 \choose 5}$
Straight of 2	$208={13 \choose 1}{4 \choose 1}^2$	$10.400={13 \choose 1}{4 \choose 1}^2{52-2 \choose 1}$	$254.800={13 \choose 1}{4 \choose 1}^2{52-2 \choose 2}$	$4.076.800={13 \choose 1}{4 \choose 1}^2{52-2 \choose 3}$	$47.902.400={13 \choose 1}{4 \choose 1}^2{52-2 \choose 4}$	$440.702.080={13 \choose 1}{4 \choose 1}^2{52-2 \choose 5}$
Straight of 2 flushed*	$52={13 \choose 1}{4 \choose 1}$	$2.600={13 \choose 1}{4 \choose 1}{52-2 \choose 1}$	$63.700={13 \choose 1}{4 \choose 1}{52-2 \choose 2}$	$1.019.200={13 \choose 1}{4 \choose 1}{52-2 \choose 3}$	$11.975.600={13 \choose 1}{4 \choose 1}{52-2 \choose 4}$	$110.175.520={13 \choose 1}{4 \choose 1}{52-2 \choose 5}$

*At 12 cards $27.917.689.800={13 \choose 1}{4 \choose 1}{52-2 \choose 10}$ while the total number is $15.820.024.220={52 \choose 12}$

My problem is that as you can see, when you draw multiple hands, the combinations of for example a pair becomes higher than the total possible combination, a thing that is absurd and can be easily demonstrated with an example hand: 1, 2, 3, 4, 5, 6, 7, there are 7 different cards, none is a pair. Of course, a flush of 2 cards is allowed to get 100% if you draw 5 cards (it is impossible to get 5 different suits because there only exist 4 different); however, it should not reach 100% in a 4-card hand (an easy example would be drawing a spade, a club, a diamond and a heart, none is repeated).

Question: What am I missing with my calculations?

As my method of calculation of the extra cards after forming what I want (for example, the three cards remaining after having a pair using 5 cards) I used ${52-2 \choose 3}$ instead of what appears in Wikipedia, I made an experiment trying to get the same results of Wikipedia with my method.

The frequency of three of a kind in Wikipedia is $54.912={13 \choose 1}{4 \choose 3}{12 \choose 2}{4 \choose 1}^2$ which I understand as choose one card of the 13 ranks, then choose three of the four suits for that card, finally choose 2 cards that aren't from the previous rank, these cards each can be of any suit.

If I use ${13 \choose 1}{4 \choose 3}{52-4 \choose 2}=58.656$ I have to eliminate the frequency in which that two extra cards are a pair, because that would fall into Full House from normal poker (following Wikipedia), so $58.656-3.744=54.912$ which is fine. Now I go one step further.

If I use ${13 \choose 1}{4 \choose 3}{52-3 \choose 2}=61.152$ I have to eliminate the frequency in which that two extra cards are a pair (because of full house), and the frequency in that the remaining card of the suit appears and transform my three of a kind into a four of a kind $61.152-3.744-624=56.784$ which isn't 54.912. I discovered that if I multiply (in the previous formula) the frequency of a full house by 1.5, or I multiply the frequency of a four of a kind by 4 I get exactly 54.912, but I'm not sure why, and that might be what makes my math surpass the 100% (when comparing the frequency/total possible combinations).

Do you have any idea what I'm doing wrong?

What would be considered an answer?

An explanation of which mathematical expression I am understanding wrong, a correct expression to replace that part, an explanation of what means that suggested new expression and one example using any of the combinations that appear in my table above that surpass the 100% using that new suggested expression.

For example: "When you say this expression ... you are wrong because that means instead ... What you have to use is ... because that means ... For example, if we use this formula to get the frequency of a pair with a 7-cards hand (in which your math surpassed 100%), we get ... which is correct because ..."

Note: I'm not a math teacher, and I'm not studying for a math career, so please try to not be too much complicated in the explanation. I just know combinations because at university we had 3 classes of hypergeometric distributions (I'm studying business administration). This is not for the university, work or gambling, actually, it is for Dungeons & Dragons 5e.

Answer 1

At least two cards of the same rank:

The reason your numbers for at least two cards of the same rank are too large is that your method counts hands with more than one pair or more than two cards of the same rank multiple times.

The probability that there are at least two cards of the same rank among seven cards can be found by subtracting the probability that each card is of a different rank from $1$. If all seven cards are of different ranks, we must select $7$ of the $13$ ranks from which to select a card and one of the four cards from each of those ranks. \begin{align*} \Pr(\text{at least two cards of the same rank}) & = 1 - \Pr(\text{all cards are of different ranks})\\ & = 1 - \frac{\dbinom{13}{7}4^7}{\dbinom{52}{7}} \end{align*}

To do a direct count, we would have numerous cases, including

• exactly one pair
• exactly two pairs
• exactly three pairs
• exactly one three of a kind
• exactly two three of a kinds
• a three of a kind and a pair
• a three of a kind and two pairs
• a four of a kind
• a four of a kind and a pair
• a four of a kind and a three of a kind

where I am using the definitions from poker for a pair, three of a kind, and a four of a kind.

Your method counts each hand with exactly two pairs twice, once for each way of designating one of those pairs as the pair in the hand. Your method counts a hand with exactly one three of a kind three times, once for each of the $\binom{3}{2}$ ways of counting two of the three cards of that rank as the pair of that rank.

At least three cards of the same rank:

In your example of selecting at least three cards of the same rank from a hand with five cards, your method includes not only three of a kinds and full houses, it also includes four of a kinds.

The number of ways of selecting three of a kind is $$\binom{13}{1}\binom{4}{3}\binom{12}{2}4^2$$ for the reasons you stated.

The number of ways of selecting a full house is $$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}$$

The number of ways of selecting four of a kind is $$\binom{13}{1}\binom{4}{4}\binom{12}{1}\binom{4}{1}$$

Again, we need to be careful about choosing three representatives of a rank as being the three cards of that rank. If your hand actually contains a four of a kind, you count this hand four times, once for each of the $\binom{4}{3}$ ways of designating three of those four cards as the three cards of that rank.

At least two cards of the same suit:

You count each hand with more than two cards from the same suit multiple times and each hand in which there are at least two cards apiece from two or more suits multiple times.

The easiest way to calculate the probability that at least two cards are of the same suit is to subtract the probability that they are all of different suits from $1$. Since there are only four suits, the answer must be $1$ whenever the number of cards drawn is at least five since we would otherwise need more than four suits. For a hand with four cards, the probability that at least two cards are of the same suit is \begin{align*} \Pr(\text{at least two cards of the same suit}) & = 1 - \Pr(\text{all cards are of different suits})\\ & = 1 - \frac{52 \cdot 39 \cdot 26 \cdot 13}{52 \cdot 51 \cdot 50 \cdot 49}. \end{align*}

At least two cards of consecutive ranks:

In poker, an ace can be treated as the card of highest rank or lowest rank, but not both. The number of two card straights is $13 \cdot 4^2$, since there are $13$ ways to select the low card and four ways to select the suit of each rank.

Suppose we wish to calculate the probability that there are at least two cards of consecutive ranks in a hand with five cards (ignoring the possibility that an Ace could be considered the highest rank). Again, it is easier to subtract the probability that no two ranks are consecutive from $1$.

First, we will choose the ranks. To do so, we will first solve a simpler problem, namely arranging nine blue and five green balls so that no two green balls are consecutive.

Line up nine blue balls in a row. This creates ten spaces, seven between consecutive blue balls and two at the ends of the row. $$\square b \square b \square b \square b \square b \square b \square b \square b \square b \square$$ Choose five of these nine spaces for the green balls, which can be done in $\binom{10}{5}$ ways. Label the balls from left to right by $$A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A$$ The labels on the green balls represent the selected ranks.

However, we cannot count an Ace as both the lowest and highest card. Such configurations would have three green balls among the eight spaces between successive blue balls, so there are $$\binom{9}{3}$$ impermissible ways to select the ranks so that no two of them are consecutive, giving $$\binom{10}{5} - \binom{8}{3}$$ ways to select the ranks of the five cards so that no two of them are consecutive.

For each rank, there are four possible suits. Hence, the probability that at least two cards are of the same rank is \begin{align*} \Pr(\text{at least two consecutive ranks}) & = 1 - \Pr(\text{no consecutive ranks})\\ & = 1 - \frac{\left[\dbinom{10}{5} - \dbinom{8}{3}\right]4^5}{\dbinom{52}{5}} \end{align*}

At least two cards of consecutive ranks which are of the same suit: You correctly calculated the result for two cards.

For three cards, there are still $13$ ways to choose the lowest rank and four ways to choose the suit. There are $50$ ways to choose a third card. However, multiplying these numbers counts each hand with three consecutive cards of the same suit twice, once for each pair of consecutive cards of the same suit. There are $12$ ways to choose the lowest rank of three consecutive cards and they must be of the same suit as above. Hence, the number of three-card hands with at least two consecutive cards of the same suit is $$\binom{13}{1}\binom{4}{1}\binom{50}{1} - \binom{12}{1}\binom{4}{1} = \binom{4}{1}\left[\binom{13}{1}\binom{50}{1} - \binom{12}{1}\right]$$

For four cards, there are numerous cases:

• exactly two cards with consecutive ranks of the same suit
• exactly three cards with consecutive ranks of the same suit
• two disjoint pairs of two cards with consecutive ranks of the same suit*
• four cards of consecutive ranks of the same suit

*Note that it is possible for the pairs to have the same ranks ($5\clubsuit, 6\clubsuit, \color{red}{5\diamondsuit}, \color{red}{6\diamondsuit}$), have the same suit ($7\spadesuit, 8\spadesuit, J\spadesuit, Q\spadesuit$), or neither ($\color{red}{9\heartsuit}, \color{red}{10\heartsuit}, A\spadesuit, 2\spadesuit$).

This problem becomes increasingly complicated as the number of cards increases. Finding the complementary probability is also tricky since we have to eliminate all hands in which no two cards are of consecutive ranks and all hands in which cards with consecutive ranks are of different suits. We did the first task above. As the number of cards increases, eliminating the cases in which cards with consecutive ranks are of different suits also becomes complicated.