Poker Hand Combinations Probability

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Poker Hand Combinations Probability Formula
Poker Hand Combinations Probability Chart

While fluctuations in probability (luck) will happen from hand to hand, the best poker players understand that skill, discipline and patience are the keys to success at the tables. A big part of strong decision making is understanding how often you should be betting, raising, and applying pressure. Find the number of poker hands of each type. For the purposes of this problem, a poker hand consists of 5 cards chosen from a standard pack of 52 (no jokers). Also for the purposes of this problem, the ace can only be a high card. In other words, the card sequence A ;2;3;4;5 is not a straight, since the ace is a high card only. In this video we will go over the number of ways of getting the most common poker or sought poker hands using combinations and the multiplication principle.

We can use permutations and combinations to help us answer more complex probability questions

Probability of being dealt a certain starting hand There are a total of exactly 1,326 different starting hand combinations in Texas Hold’em poker. However, many of them are practically identical, e.g. A♥K♣ is exactly the same hand as A♦K♠ before the flop. The probability of being dealt a royal flush is the number of royal flushes divided by the total number of poker hands. We now carry out the division and see that a royal flush is rare indeed. There is only a probability of 4/2,598,960 = 1/649,740 = 0.00015% of being dealt this hand.

Example 1

A 4 digit PIN is selected. What is the probability that there are no repeated digits?

There are 10 possible values for each digit of the PIN (namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), so there are 10 × 10 × 10 × 10 = 10
⁴ = 10000 total possible PINs.

To have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 × 9 × 8 × 7, or notice that this is the same as the permutation
₁₀P₄ = 5040.

The probability of no repeated digits is the number of 4 digit PINs with no repeated digits divided by the total number of 4 digit PINs. This probability is

[latex]displaystylefrac{{{}_{{10}}{P}_{{4}}}}{{{10}^{{4}}}}=frac{{5040}}{{10000}}={0.504}[/latex]

Example 2

In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. In this lottery, the order the numbers are drawn in doesn’t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket.

In order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player’s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is
₄₈C₆ = 12,271,512. Of these possible outcomes, only one would match all six numbers on the player’s ticket, so the probability of winning the grand prize is:

[latex]displaystylefrac{{{}_{{6}}{C}_{{6}}}}{{{}_{{48}}{C}_{{6}}}}=frac{{1}}{{12271512}}approx={0.0000000815}[/latex]

Example 3

In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.

As above, the number of possible outcomes of the lottery drawing is
₄₈C₆ = 12,271,512. In order to win the second prize, five of the six numbers on the ticket must match five of the six winning numbers; in other words, we must have chosen five of the six winning numbers and one of the 42 losing numbers. The number of ways to choose 5 out of the 6 winning numbers is given by ₆C₅ = 6 and the number of ways to choose 1 out of the 42 losing numbers is given by ₄₂C₁ = 42. Thus the number of favorable outcomes is then given by the Basic Counting Rule: ₆C₅ × ₄₂C₁ = 6 × 42 = 252. So the probability of winning the second prize is

[latex]displaystylefrac{{{left({}_{{6}}{C}_{{5}}right)}{left({}_{{42}}{C}_{{1}}right)}}}{{{}_{{48}}{C}_{{6}}}}=frac{{252}}{{12271512}}approx{0.0000205}[/latex]

Try it Now 1

A multiple-choice question on an economics quiz contains 10 questions with five possible answers each. Compute the probability of randomly guessing the answers and getting exactly 9 questions correct.

Example 4

Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace.

In many card games (such as poker) the order in which the cards are drawn is not important (since the player may rearrange the cards in his hand any way he chooses); in the problems that follow, we will assume that this is the case unless otherwise stated. Thus we use combinations to compute the possible number of 5-card hands,
₅₂C₅. This number will go in the denominator of our probability formula, since it is the number of possible outcomes.

For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck. Since there are four Aces and we want exactly one of them, there will be
₄C₁ ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be ₄₈C₄ ways to select the four non-Aces. Now we use the Basic Counting Rule to calculate that there will be ₄C₁ × ₄₈C₄ ways to choose one ace and four non-Aces.

Putting this all together, we have

[latex]displaystyle{P}{left(text{one Ace}right)}=frac{{{left({}_{{4}}{C}_{{1}}right)}{left({}_{{48}}{C}_{{4}}right)}}}{{{}_{{52}}{C}_{{5}}}}=frac{{778320}}{{2598960}}approx{0.299}[/latex]

Example 5

Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces.

The solution is similar to the previous example, except now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same:

It is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier.

Try it Now 2

Compute the probability of randomly drawing five cards from a deck of cards and getting three Aces and two Kings.

Birthday Problem

Let’s take a pause to consider a famous problem in probability theory:

Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?

Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30/365, perhaps?). Let’s see if we should listen to our intuition. Let’s start with a simpler problem, however.

Example 6

Suppose three people are in a room. What is the probability that there is at least one shared birthday among these three people?

There are a lot of ways there could be at least one shared birthday. Fortunately there is an easier way. We ask ourselves “What is the alternative to having at least one shared birthday?” In this case, the alternative is that there are
no shared birthdays. In other words, the alternative to “at least one” is having none. In other words, since this is a complementary event,

P(at least one) = 1 – P(none)

We will start, then, by computing the probability that there is no shared birthday. Let’s imagine that you are one of these three people. Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that the second person does not share your birthday? There are 365 days in the year (let’s ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is 364/365. Now we move to the third person. What is the probability that this third person does not have the same birthday as either you or the second person? There are 363 days that will not duplicate your birthday or the second person’s, so the probability that the third person does not share a birthday with the first two is 363/365.

We want the second person not to share a birthday with you
and the third person not to share a birthday with the first two people, so we use the multiplication rule:

[latex]displaystyle{P}{left(text{no shared birthday}right)}=frac{{365}}{{365}}cdotfrac{{364}}{{365}}cdotfrac{{363}}{{365}}approx{0.9918}[/latex]

and then subtract from 1 to get

P(shared birthday) = 1 – P(no shared birthday) = 1 – 0.9918 = 0.0082.

This is a pretty small number, so maybe it makes sense that the answer to our original problem will be small. Let’s make our group a bit bigger.

Example 7

Suppose five people are in a room. What is the probability that there is at least one shared birthday among these five people?

Continuing the pattern of the previous example, the answer should be

[latex]displaystyle{P}{left(text{shared birthday}right)}={1}-frac{{365}}{{365}}cdotfrac{{364}}{{365}}cdotfrac{{363}}{{365}}cdotfrac{{362}}{{365}}cdotfrac{{361}}{{365}}approx{0.0271}[/latex]

Note that we could rewrite this more compactly as

[latex]displaystyle{P}{left(text{shared birthday}right)}={1}-frac{{{}_{{365}}{P}_{{5}}}}{{365}^{{5}}}approx{0.0271}[/latex]

which makes it a bit easier to type into a calculator or computer, and which suggests a nice formula as we continue to expand the population of our group.

Example 8

Suppose 30 people are in a room. What is the probability that there is at least one shared birthday among these 30 people?

Here we can calculate

[latex]displaystyle{P}{left(text{shared birthday}right)}={1}-frac{{{}_{{365}}{P}_{{30}}}}{{365}^{{30}}}approx{0.706}[/latex]

which gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least one shared birthday!

If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn’t studied probability!) You wouldn’t be guaranteed to win, but you should win more than half the time.

This is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts. If you still don’t believe the math, you can carry out a simulation. Just so you won’t have to go around rounding up groups of 30 people, someone has kindly developed a Java applet so that you can conduct a computer simulation. Go to this web page:
http://www-stat.stanford.edu/~susan/surprise/Birthday.html, and once the applet has loaded, select 30 birthdays and then keep clicking Start and Reset. If you keep track of the number of times that there is a repeated birthday, you should get a repeated birthday about 7 out of every 10 times you run the simulation.

Try it Now 3

Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?

[latex]displaystyle{P}{left({9} text{ answers correct}right)}=frac{9cdot4}{(5^{10})}approx0.0000037[/latex] chance
[latex]displaystyle{P}{left(text{three Aces and two Kings}right)}=frac{{{left({}_{{4}}{C}_{{3}}right)}{left({}_{{4}}{C}_{{2}}right)}}}{{{}_{{52}}{C}_{{5}}}}=frac{{24}}{{2598960}}approx{0.0000092}[/latex]
[latex]displaystyle{P}{left(text{shared birthday}right)}={1}-frac{{{}_{{365}}{P}_{{10}}}}{{365}^{{10}}}approx{0.117}[/latex]

David Lippman, Math in Society, “Probability,” licensed under a CC BY-SA 3.0 license.

Introduction

Derivations for Five Card Stud

I have been asked so many times how I derived the probabilities of drawing each poker hand that I have created this section to explain the calculation. This assumes some level mathematical proficiency; anyone comfortable with high school math should be able to work through this explanation. The skills used here can be applied to a wide range of probability problems.

The Factorial Function

If you already know about the factorial function you can skip ahead. If you think 5! means to yell the number five then keep reading.

The instructions for your living room couch will probably recommend that you rearrange the cushions on a regular basis. Let's assume your couch has four cushions. How many combinations can you arrange them in? The answer is 4!, or 24. There are obviously 4 positions to put the first cushion, then there will be 3 positions left to put the second, 2 positions for the third, and only 1 for the last one, or 4*3*2*1 = 24. If you had n cushions there would be n*(n-1)*(n-2)* ... * 1 = n! ways to arrange them. Any scientific calculator should have a factorial button, usually denoted as x!, and the fact(x) function in Excel will give the factorial of x. The total number of ways to arrange 52 cards would be 52! = 8.065818 * 10⁶⁷.

The Combinatorial Function

Assume you want to form a committee of 4 people out of a pool of 10 people in your office. How many different combinations of people are there to choose from? The answer is 10!/(4!*(10-4)!) = 210. The general case is if you have to form a committee of y people out of a pool of x then there are x!/(y!*(x-y)!) combinations to choose from. Why? For the example given there would be 10! = 3,628,800 ways to put the 10 people in your office in order. You could consider the first four as the committee and the other six as the lucky ones. However you don't have to establish an order of the people in the committee or those who aren't in the committee. There are 4! = 24 ways to arrange the people in the committee and 6! = 720 ways to arrange the others. By dividing 10! by the product of 4! and 6! you will divide out the order of people in an out of the committee and be left with only the number of combinations, specifically (1*2*3*4*5*6*7*8*9*10)/((1*2*3*4)*(1*2*3*4*5*6)) = 210. The combin(x,y) function in Excel will tell you the number of ways you can arrange a group of y out of x.

Now we can determine the number of possible five card hands out of a 52 card deck. The answer is combin(52,5), or 52!/(5!*47!) = 2,598,960. If you're doing this by hand because your calculator doesn't have a factorial button and you don't have a copy of Excel, then realize that all the factors of 47! cancel out those in 52! leaving (52*51*50*49*48)/(1*2*3*4*5). The probability of forming any given hand is the number of ways it can be arranged divided by the total number of combinations of 2,598.960. Below are the number of combinations for each hand. Just divide by 2,598,960 to get the probability.

Poker Math

The next section shows how to derive the number of combinations of each poker hand in five card stud.

Royal Flush

There are four different ways to draw a royal flush (one for each suit).

Straight Flush

The highest card in a straight flush can be 5,6,7,8,9,10,Jack,Queen, or King. Thus there are 9 possible high cards, and 4 possible suits, creating 9 * 4 = 36 different possible straight flushes.

Four of a Kind

There are 13 different possible ranks of the 4 of a kind. The fifth card could be anything of the remaining 48. Thus there are 13 * 48 = 624 different four of a kinds.

Full House

There are 13 different possible ranks for the three of a kind, and 12 left for the two of a kind. There are 4 ways to arrange three cards of one rank (4 different cards to leave out), and combin(4,2) = 6 ways to arrange two cards of one rank. Thus there are 13 * 12 * 4 * 6 = 3,744 ways to create a full house.

Flush

There are 4 suits to choose from and combin(13,5) = 1,287 ways to arrange five cards in the same suit. From 1,287 subtract 10 for the ten high cards that can lead a straight, resulting in a straight flush, leaving 1,277. Then multiply for 4 for the four suits, resulting in 5,108 ways to form a flush.

Straight

The highest card in a straight can be 5,6,7,8,9,10,Jack,Queen,King, or Ace. Thus there are 10 possible high cards. Each card may be of four different suits. The number of ways to arrange five cards of four different suits is 4⁵ = 1024. Next subtract 4 from 1024 for the four ways to form a flush, resulting in a straight flush, leaving 1020. The total number of ways to form a straight is 10*1020=10,200.

Three of a Kind

There are 13 ranks to choose from for the three of a kind and 4 ways to arrange 3 cards among the four to choose from. There are combin(12,2) = 66 ways to arrange the other two ranks to choose from for the other two cards. In each of the two ranks there are four cards to choose from. Thus the number of ways to arrange a three of a kind is 13 * 4 * 66 * 4² = 54,912.

Two Pair

There are (13:2) = 78 ways to arrange the two ranks represented. In both ranks there are (4:2) = 6 ways to arrange two cards. There are 44 cards left for the fifth card. Thus there are 78 * 6² * 44 = 123,552 ways to arrange a two pair.

One Pair

There are 13 ranks to choose from for the pair and combin(4,2) = 6 ways to arrange the two cards in the pair. There are combin(12,3) = 220 ways to arrange the other three ranks of the singletons, and four cards to choose from in each rank. Thus there are 13 * 6 * 220 * 4³ = 1,098,240 ways to arrange a pair.

Nothing

First find the number of ways to choose five different ranks out of 13, which is combin(13,5) = 1287. Then subtract 10 for the 10 different high cards that can lead a straight, leaving you with 1277. Each card can be of 1 of 4 suits so there are 4⁵=1024 different ways to arrange the suits in each of the 1277 combinations. However we must subtract 4 from the 1024 for the four ways to form a flush, leaving 1020. So the final number of ways to arrange a high card hand is 1277*1020=1,302,540.

Specific High Card

For example, let's find the probability of drawing a jack-high. There must be four different cards in the hand all less than a jack, of which there are 9 to choose from. The number of ways to arrange 4 ranks out of 9 is combin(9,4) = 126. We must then subtract 1 for the 10-9-8-7 combination which would form a straight, leaving 125. From above we know there are 1020 ways to arrange the suits. Multiplying 125 by 1020 yields 127,500 which the number of ways to form a jack-high hand. For ace-high remember to subtract 2 rather than 1 from the total number of ways to arrange the ranks since A-K-Q-J-10 and 5-4-3-2-A are both valid straights. Here is a good site that also explains how to calculate poker probabilities.