Poker Hand Combinations Probability
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*Poker Hand Combinations Probability Formula
*Poker Hand Combinations Probability Chart
The main underpinning of poker is math – it is essential. For every decision you make, while factors such as psychology have a part to play, math is the key element.
*The binomial coefficient can be used to calculate certain combinations of cards. Then, the counting principles of rule of sum and rule of product can be used to compute the frequency of each poker hand classification. Then, the probability of each poker hand classification is simply its frequency divided by 2,598,960.
*With such a hand one can have reasonable results, especially if you get a three-of-a-kind with flop. This hand is worse than it seems and raising will be improper. 6: 10.8%: Raise. A hand full of high cards must raise to reduce opponents. Side cards are lousy but two kings are enough for the start.
*The answer is 52.51, or 2,652. Carry this out to three-card poker: 52.51.50=132,600. With four cards, you could see 6,497,400 potential hands. Finally, we get to five-card poker.
In this lesson we’re going to give an overview of probability and how it relates to poker. This will include the probability of being dealt certain hands and how often they’re likely to win. We’ll also cover how to calculating your odds and outs, in addition to introducing you to the concept of pot odds. And finally we’ll take a look at how an understanding of the math will help you to remain emotional stable at the poker table and why you should focus on decisions, not results.
Hand Combinations in Poker. The term “combination” (or “combo” for short) refers to the different ways we can make a specific type of poker hand. Example: We are dealt AK preflop in Hold’em. How many different combinations of AK are there? If we were so inclined, we could list every possible way of making AK. 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.What is Probability?
Probability is the branch of mathematics that deals with the likelihood that one outcome or another will occur. For instance, a coin flip has two possible outcomes: heads or tails. The probability that a flipped coin will land heads is 50% (one outcome out of the two); the same goes for tails.Probability and Cards
When dealing with a deck of cards the number of possible outcomes is clearly much greater than the coin example. Each poker deck has fifty-two cards, each designated by one of four suits (clubs, diamonds, hearts and spades) and one of thirteen ranks (the numbers two through ten, Jack, Queen, King, and Ace). Therefore, the odds of getting any Ace as your first card are 1 in 13 (7.7%), while the odds of getting any spade as your first card are 1 in 4 (25%).
Unlike coins, cards are said to have “memory”: every card dealt changes the makeup of the deck. For example, if you receive an Ace as your first card, only three other Aces are left among the remaining fifty-one cards. Therefore, the odds of receiving another Ace are 3 in 51 (5.9%), much less than the odds were before you received the first Ace.
Want to see how poker math intertwines with psychology and strategy to give you a MASSIVE EDGE at the tables? Check out CORE and learn poker in the quickest and most systematic way:Pre-flop Probabilities: Pocket Pairs
In order to find the odds of getting dealt a pair of Aces, we multiply the probabilities of receiving each card:
(4/52) x (3/51) = (12/2652) = (1/221) ≈ 0.45%.
To put this in perspective, if you’re playing poker at your local casino and are dealt 30 hands per hour, you can expect to receive pocket Aces an average of once every 7.5 hours.
Housekeeping Shift Supervisor at Casino Arizona. Scottsdale, AZ: 9/24/2020 Floor Technician at Casino Arizona. Scottsdale, AZ: 9/27/2020 Security Officer at. Apply Online or Visit Us At: Casino Arizona Human Resources Center. Chaparral Road, Suite 1100 Scottsdale, AZ 85250. Experience Casino Arizona, the valley’s hottest destination for gaming, dining, and entertainment. Play our wide variety of multi-denominational slot machines, live blackjack, and fast-action Keno, or kick back and enjoy televised sporting events and live music. Casino arizona employment application.
The odds of receiving any of the thirteen possible pocket pairs (twos up to Aces) is:
(13/221) = (1/17) ≈ 5.9%.
In contrast, you can expect to receive any pocket pair once every 35 minutes on average.Pre-Flop Probabilities: Hand vs. Hand
Players don’t play poker in a vacuum; each player’s hand must measure up against his opponent’s, especially if a player goes all-in before the flop.
Here are some sample probabilities for most pre-flop situations:Post-Flop Probabilities: Improving Your Hand
Now let’s look at the chances of certain events occurring when playing certain starting hands. The following table lists some interesting and valuable hold’em math:
Many beginners to poker overvalue certain starting hands, such as suited cards. As you can see, suited cards don’t make flushes very often. Likewise, pairs only make a set on the flop 12% of the time, which is why small pairs are not always profitable.PDF Chart
We have created a poker math and probability PDF chart (link opens in a new window) which lists a variety of probabilities and odds for many of the common events in Texas hold ‘em. This chart includes the two tables above in addition to various starting hand probabilities and common pre-flop match-ups. You’ll need to have Adobe Acrobat installed to be able to view the chart, but this is freely installed on most computers by default. We recommend you print the chart and use it as a source of reference.Odds and Outs
If you do see a flop, you will also need to know what the odds are of either you or your opponent improving a hand. In poker terminology, an “out” is any card that will improve a player’s hand after the flop.
One common occurrence is when a player holds two suited cards and two cards of the same suit appear on the flop. The player has four cards to a flush and needs one of the remaining nine cards of that suit to complete the hand. In the case of a “four-flush”, the player has nine “outs” to make his flush.
A useful shortcut to calculating the odds of completing a hand from a number of outs is the “rule of four and two”. The player counts the number of cards that will improve his hand, and then multiplies that number by four to calculate his probability of catching that card on either the turn or the river. If the player misses his draw on the turn, he multiplies his outs by two to find his probability of filling his hand on the river.
In the example of the four-flush, the player’s probability of filling the flush is approximately 36% after the flop (9 outs x 4) and 18% after the turn (9 outs x 2).Pot Odds
Another important concept in calculating odds and probabilities is pot odds. Pot odds are the proportion of the next bet in relation to the size of the pot.
For instance, if the pot is $90 and the player must call a $10 bet to continue playing the hand, he is getting 9 to 1 (90 to 10) pot odds. If he calls, the new pot is now $100 and his $10 call makes up 10% of the new pot.
Experienced players compare the pot odds to the odds of improving their hand. If the pot odds are higher than the odds of improving the hand, the expert player will call the bet; if not, the player will fold. This calculation ties into the concept of expected value, which we will explore in a later lesson.Bad Beats
A “bad beat” happens when a player completes a hand that started out with a very low probability of success. Experts in probability understand the idea that, just because an event is highly unlikely, the low likelihood does not make it completely impossible.
A measure of a player’s experience and maturity is how he handles bad beats. In fact, many experienced poker players subscribe to the idea that bad beats are the reason that many inferior players stay in the game. Bad poker players often mistake their good fortune for skill and continue to make the same mistakes, which the more capable players use against them.Decisions, Not Results
One of the most important reasons that novice players should understand how probability functions at the poker table is so that they can make the best decisions during a hand. 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.
The good news is that there is a simple system, with powerful shortcuts & rules, that you can begin using this week. Rooted in GTO, but simplified so that you can implement it at the tables, The One Percent gives you the ultimate gameplan.
This 7+ hour course gives you applicable rules for continuation betting, barreling, raising, and easy ratios so that you ALWAYS have the right number of bluffing combos. Take the guesswork out of your strategy, and begin playing like the top-1%.Conclusion
A strong knowledge of poker math and probabilities will help you adjust your strategies and tactics during the game, as well as giving you reasonable expectations of potential outcomes and the emotional stability to keep playing intelligent, aggressive poker.
Remember that the foundation upon which to build an imposing knowledge of hold’em starts and ends with the math. I’ll end this lesson by simply saying…. the math is essential.Related Lessons
By Gerald Hanks
Gerald Hanks is from Houston Texas, and has been playing poker since 2002. He has played cash games and no-limit hold’em tournaments at live venues all over the United States.Related LessonsRelated LessonsShare:
We can use permutations and combinations to help us answer more complex probability questionsExample 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]
Hollywood casino amphitheater st. louis missouri. 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]Poker Hand Combinations Probability Formula
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 counte
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*Poker Hand Combinations Probability Formula
*Poker Hand Combinations Probability Chart
The main underpinning of poker is math – it is essential. For every decision you make, while factors such as psychology have a part to play, math is the key element.
*The binomial coefficient can be used to calculate certain combinations of cards. Then, the counting principles of rule of sum and rule of product can be used to compute the frequency of each poker hand classification. Then, the probability of each poker hand classification is simply its frequency divided by 2,598,960.
*With such a hand one can have reasonable results, especially if you get a three-of-a-kind with flop. This hand is worse than it seems and raising will be improper. 6: 10.8%: Raise. A hand full of high cards must raise to reduce opponents. Side cards are lousy but two kings are enough for the start.
*The answer is 52.51, or 2,652. Carry this out to three-card poker: 52.51.50=132,600. With four cards, you could see 6,497,400 potential hands. Finally, we get to five-card poker.
In this lesson we’re going to give an overview of probability and how it relates to poker. This will include the probability of being dealt certain hands and how often they’re likely to win. We’ll also cover how to calculating your odds and outs, in addition to introducing you to the concept of pot odds. And finally we’ll take a look at how an understanding of the math will help you to remain emotional stable at the poker table and why you should focus on decisions, not results.
Hand Combinations in Poker. The term “combination” (or “combo” for short) refers to the different ways we can make a specific type of poker hand. Example: We are dealt AK preflop in Hold’em. How many different combinations of AK are there? If we were so inclined, we could list every possible way of making AK. 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.What is Probability?
Probability is the branch of mathematics that deals with the likelihood that one outcome or another will occur. For instance, a coin flip has two possible outcomes: heads or tails. The probability that a flipped coin will land heads is 50% (one outcome out of the two); the same goes for tails.Probability and Cards
When dealing with a deck of cards the number of possible outcomes is clearly much greater than the coin example. Each poker deck has fifty-two cards, each designated by one of four suits (clubs, diamonds, hearts and spades) and one of thirteen ranks (the numbers two through ten, Jack, Queen, King, and Ace). Therefore, the odds of getting any Ace as your first card are 1 in 13 (7.7%), while the odds of getting any spade as your first card are 1 in 4 (25%).
Unlike coins, cards are said to have “memory”: every card dealt changes the makeup of the deck. For example, if you receive an Ace as your first card, only three other Aces are left among the remaining fifty-one cards. Therefore, the odds of receiving another Ace are 3 in 51 (5.9%), much less than the odds were before you received the first Ace.
Want to see how poker math intertwines with psychology and strategy to give you a MASSIVE EDGE at the tables? Check out CORE and learn poker in the quickest and most systematic way:Pre-flop Probabilities: Pocket Pairs
In order to find the odds of getting dealt a pair of Aces, we multiply the probabilities of receiving each card:
(4/52) x (3/51) = (12/2652) = (1/221) ≈ 0.45%.
To put this in perspective, if you’re playing poker at your local casino and are dealt 30 hands per hour, you can expect to receive pocket Aces an average of once every 7.5 hours.
Housekeeping Shift Supervisor at Casino Arizona. Scottsdale, AZ: 9/24/2020 Floor Technician at Casino Arizona. Scottsdale, AZ: 9/27/2020 Security Officer at. Apply Online or Visit Us At: Casino Arizona Human Resources Center. Chaparral Road, Suite 1100 Scottsdale, AZ 85250. Experience Casino Arizona, the valley’s hottest destination for gaming, dining, and entertainment. Play our wide variety of multi-denominational slot machines, live blackjack, and fast-action Keno, or kick back and enjoy televised sporting events and live music. Casino arizona employment application.
The odds of receiving any of the thirteen possible pocket pairs (twos up to Aces) is:
(13/221) = (1/17) ≈ 5.9%.
In contrast, you can expect to receive any pocket pair once every 35 minutes on average.Pre-Flop Probabilities: Hand vs. Hand
Players don’t play poker in a vacuum; each player’s hand must measure up against his opponent’s, especially if a player goes all-in before the flop.
Here are some sample probabilities for most pre-flop situations:Post-Flop Probabilities: Improving Your Hand
Now let’s look at the chances of certain events occurring when playing certain starting hands. The following table lists some interesting and valuable hold’em math:
Many beginners to poker overvalue certain starting hands, such as suited cards. As you can see, suited cards don’t make flushes very often. Likewise, pairs only make a set on the flop 12% of the time, which is why small pairs are not always profitable.PDF Chart
We have created a poker math and probability PDF chart (link opens in a new window) which lists a variety of probabilities and odds for many of the common events in Texas hold ‘em. This chart includes the two tables above in addition to various starting hand probabilities and common pre-flop match-ups. You’ll need to have Adobe Acrobat installed to be able to view the chart, but this is freely installed on most computers by default. We recommend you print the chart and use it as a source of reference.Odds and Outs
If you do see a flop, you will also need to know what the odds are of either you or your opponent improving a hand. In poker terminology, an “out” is any card that will improve a player’s hand after the flop.
One common occurrence is when a player holds two suited cards and two cards of the same suit appear on the flop. The player has four cards to a flush and needs one of the remaining nine cards of that suit to complete the hand. In the case of a “four-flush”, the player has nine “outs” to make his flush.
A useful shortcut to calculating the odds of completing a hand from a number of outs is the “rule of four and two”. The player counts the number of cards that will improve his hand, and then multiplies that number by four to calculate his probability of catching that card on either the turn or the river. If the player misses his draw on the turn, he multiplies his outs by two to find his probability of filling his hand on the river.
In the example of the four-flush, the player’s probability of filling the flush is approximately 36% after the flop (9 outs x 4) and 18% after the turn (9 outs x 2).Pot Odds
Another important concept in calculating odds and probabilities is pot odds. Pot odds are the proportion of the next bet in relation to the size of the pot.
For instance, if the pot is $90 and the player must call a $10 bet to continue playing the hand, he is getting 9 to 1 (90 to 10) pot odds. If he calls, the new pot is now $100 and his $10 call makes up 10% of the new pot.
Experienced players compare the pot odds to the odds of improving their hand. If the pot odds are higher than the odds of improving the hand, the expert player will call the bet; if not, the player will fold. This calculation ties into the concept of expected value, which we will explore in a later lesson.Bad Beats
A “bad beat” happens when a player completes a hand that started out with a very low probability of success. Experts in probability understand the idea that, just because an event is highly unlikely, the low likelihood does not make it completely impossible.
A measure of a player’s experience and maturity is how he handles bad beats. In fact, many experienced poker players subscribe to the idea that bad beats are the reason that many inferior players stay in the game. Bad poker players often mistake their good fortune for skill and continue to make the same mistakes, which the more capable players use against them.Decisions, Not Results
One of the most important reasons that novice players should understand how probability functions at the poker table is so that they can make the best decisions during a hand. 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.
The good news is that there is a simple system, with powerful shortcuts & rules, that you can begin using this week. Rooted in GTO, but simplified so that you can implement it at the tables, The One Percent gives you the ultimate gameplan.
This 7+ hour course gives you applicable rules for continuation betting, barreling, raising, and easy ratios so that you ALWAYS have the right number of bluffing combos. Take the guesswork out of your strategy, and begin playing like the top-1%.Conclusion
A strong knowledge of poker math and probabilities will help you adjust your strategies and tactics during the game, as well as giving you reasonable expectations of potential outcomes and the emotional stability to keep playing intelligent, aggressive poker.
Remember that the foundation upon which to build an imposing knowledge of hold’em starts and ends with the math. I’ll end this lesson by simply saying…. the math is essential.Related Lessons
By Gerald Hanks
Gerald Hanks is from Houston Texas, and has been playing poker since 2002. He has played cash games and no-limit hold’em tournaments at live venues all over the United States.Related LessonsRelated LessonsShare:
We can use permutations and combinations to help us answer more complex probability questionsExample 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]
Hollywood casino amphitheater st. louis missouri. 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]Poker Hand Combinations Probability Formula
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 counte
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