- Poker Hand Probability Problems Worksheets
- Poker Hand Probability Problems
- Poker Hand Probability Problems Worksheet
Intro
The following table lists, for each hand, the number and probability of a given hand. Five-Card Stud (Natural) Probabilities Hand Number Probability Straight Flush 2 40 0.00002 Four-of-a-Kind 624 0.00024 Full House 3744 0.00144 Flush 5108 0.00197 Straight 10,200 0.00393 Three-of-a-Kind 54,912 0.02113 Two Pair 123,552 0.04754. Often, the key to determining probability is selecting the best approach for a given problem. In Texas Hold'em, a player is dealt two down card (or pocket cards). The first card can be any one of 52 playing cards in the deck and the second card can be any one of the 51 remaining cards. The number of ways to arrange 5 cards out of 52 is (52:5) = 2,598,960. The odds of drawing any given hand are the number of ways it can be arranged divided by the total number of ways to arrange five cards above. Below are the number of ways to arrange each hand. The number of different royal flushes are four (one for each suit).
This is a problem concerning basic probability calculations from the text: 'A First Course in Probability Theory' by Sheldon Ross (8th addition).
Sample Problem
Chapter 2 #16. Poker dice is played by simultaneously rolling 5 dice. Show that:
a) P[no two alike] = .0926, b) P[one pair] = .4630, c) P[two pair] = .2315, d) P[three alike] = .1543, e) P[full house] = .0386, f) P[four alike] = .0193, g) P[5 alike]=.0008.
Solution
Notation, I will let ^ designate the power function. For example, 6^5 is 6 to the fifth power. 6! is 6 factorial, 6! = 6 * 5 * 4 * 3 * 2 * 1.
To calculate the Probabilities here, we will divide the number of occurrences for a particular event by the total possibilities in rolling 5 dice.
N[total] = total possibilities in rolling five dice = 6^5 = 7776
Note: N[total] is the number of ordered rolls. For example, if we rolled the dice one by one and rolled in order 3,4,5,6,2, it would be considered as different from if we rolled 2,3,4,5,6 in order.
a) P[no two alike]
There are 6 choices of numbers for the first dice. The 2nd dice must be one of the remaining 5 unchosen numbers, and the 3rd dice one of the 4 remaining unchosen numbers, and so on… This gives an ordered count, so:
N[no two alike] = 6 * 5 * 4 * 3 * 2 = 720
P[no two alike] = N[no two alike]/N[total] = 720/7776 = 0.09259259
b) P[one pair]
First we count the possible sets (unordered) of numbers. Here we have one number that is the pair, 6 choices. Then we must choose 3 different numbers from the remaining 5 values, as these three are equivalent, we have choose(5,3) = 5!/(3! * 2!) possibilities. So 6 * choose(5,3) is the total combinations of numbers. Since we are dealing with ordered counts, we must consider the orderings for each set of numbers. We have 5 die with 2 the same so the number of orderings is 5!/2!. Multiplying these together gives us the number of ordered samples:
N[one pair] = 6 * (5!/(3! * 2!)) * (5!/2) = 3600
P[one pair] = N[one pair]/N[total] = 0.462963
c) P[two pair]
Here we have 2 pairs which are equivalent, so we must choose 2 values from the 6 possible values, choose(6,2)= 6!/(2! * 4!). Then we must choose 1 value from the remaining 4 for the single value, 4 ways. Now we must consider the orderings, 5 die with 2 sets of 2 the same so the number of orderings is 5!/(2! * 2!).
N[two pairs] = (6!/(2! * 4!)) * 4 * (5!/(2! * 2!)) = 1800
P[two pairs] = N[two pairs]/N[total] = 0.2314815
d) P[three alike] (the remaining two cards are different)
There are 6 choices for the three of a kind. Then we must choose 2 different values from the remaining 5 choices, choose(5,2) = 5!/(2! * 3!). The number of orderings is 5 items with 3 being identical, which is 5!/3!.
N[three alike] = 6 * (5!/(3! * 2!)) * (5!/3!) = 1200
P[three alike] = N[three alike]/N[total] = 0.154321
e) P[full house] 3 alike with a pair.
We need one value for the three that are alike, 6 ways, and then we must choose from the remaining 5 values for the pair, 5 ways. The orderings are given by 5!/(3! * 2!).
N[full house] = 6 * 5 * (5!/(3! * 2!)) = 300
P[full house] = N[full house]/N[total] = 0.03858025
f) P[four alike]
We need one value for the four of a kind, and then one value from the remaining 5 for the last die. The number of orderings is given by 5!/4!.
N[four alike] = 6 * 5 * (5!/4!) = 150
P[four alike] = N[four alike]/N[total] = 0.01929012
g) P[five alike]
Here we need 1 value for the five alike, 6 ways. There is just 1 possible ordering as all five die are the same.
N[five alike] = 6
P[five alike] = N[five alike]/N[total] = 0.0007716049
Check: the numbers of each type must sum to N[total] = 7776:
720 + 3600 + 1800 + 1200 + 300 + 150 + 6 = 7776
Extra:
Straight
We can have two possible straights: one composed of (6,5,4,3,2) and one composed of (5,4,3,2,1). Each of these straights can be permuted 5! ways.
N[straight] = 2 * 5! = 240
P[straight] = N[straight]/N[total] = 0.0308642
Now let's run a simulation in R:
Poker Hand Probability Problems Worksheets
Results of simulation:
notwoalike: 0.09194
onepair: 0.464
twopair: 0.23187
threealike: 0.1554
fullhouse: 0.03727
fouralike: 0.01889
fivealike: 0.00063
straight: 0.03072
Ever wondered where some of those odds in the odds charts came from? In this article, I will teach you how to work out the probability of being dealt different types of preflop hands in Texas Holdem.
It's all pretty simple and you don't need to be a mathematician to work out the probabilities. I'll keep the math part as straightforward as I can to help keep this an easy-going article for the both of us.
- Probability calculations quick links.
A few probability basics.
When working out hand probabilities, the main probabilities we will work with are the number of cards in the deck and the number of cards we want to be dealt. So for example, if we were going to deal out 1 card:
- The probability of dealing a 7 would be 1/52 - There is one 7 in a deck of 52 cards.
- The probability of dealing any Ace would be 4/52 - There four Aces in a deck of 52 cards.
- The probability of dealing any would be 13/52 - There are 13 s in a deck of 52 cards.
In fact, the probability of being dealt any random card (not just the 7) would be 1/52. This also applies to the probability being dealt any random value of card like Kings, tens, fours, whatever (4/52) and the probability of being dealt any random suit (13/52).
Each card is just as likely to be dealt as any other - no special priorities in this game!
The numbers change for future cards.
A quick example... let's say we want to work out the probability of being dealt a pair of sevens.
- The probability of being dealt a 7 for the first card will be 4/52.
- The probability of being dealt a 7 for the second card will be 3/51.
Notice how the probability changes for the second card? After we have been dealt the first card, there is now 1 less card in the deck making it 51 cards in total. Also, after already being dealt a 7, there are now only three 7s left in the deck.
Always try and take care with the numbers for future cards. The numbers will change slightly as you go along.
Working out probabilities.
- Whenever the word 'and' is used, it will usually mean multiply.
- Whenever the word 'or' is used, it will usually mean add.
This won't make much sense for now, but it will make a lot of sense a little further on in the article. Trust me.
Probability of being dealt two exact cards.
Multiply the two probabilities together.
So, we want to find the probability of being dealt the A and K. (See the 'and' there?)
- Probability of being dealt A - 1/52.
- Probability of being dealt K - 1/51.
Now let's just multiply these bad boys together.
P = (1/52) * (1/51)
P = 1/2652
So the probability of being dealt the A and then K is 1/2652. As you might be able to work out, this is the same probability for any two exact cards, as the likelihood of being dealt A K is the same as being dealt a hand like 7 3 in that order.
But wait, we do not care about the order of the cards we are dealt!
When we are dealt a hand in Texas Hold'em, we don't care whether we get the A first or the K first (which is what we just worked out), just as long as we get them in our hand it's all the same. There are two possible combinations of being dealt this hand (A K and K A), so we simply multiply the probability by 2 to get a more useful probability.
P = 1/2652 * 2
P = 1/1326
You might notice that because of this, we have also worked out that there are 1,326 possible combinations of starting hands in Texas Holdem. Cool huh?
Probability of being dealt a certain hand.
Two exact cards is all well and good, but what if we want to work out the chances of being dealt AK, regardless of specific suits and whatnot? Well, we just do the same again...
Multiply the two probabilities together.
So, we want to find the probability of being dealt any Ace andany King.
- Probability of being dealt any Ace - 4/52.
- Probability of being dealt any King - 4/51 (after we've been dealt our Ace, there are now 51 cards left).
P = (4/52) * (4/51)
P = 16/2652 = 1/166
However, again with the 2652 number we are working out the probability of being deal an Ace and then a King. If we want the probability of being dealt either in any order, there are two possible ways to make this AK combination so we multiply the probability by 2.
Poker Hand Probability Problems
P = 16/2652 * 2
P = 32/2652
P = 1/83
The probability of being dealt any AK as opposed to an AK with exact suits is more probable as we would expect. A lot more probable in fact. Also, as you might guess, this probability of 1/83 will be the same for any two value of cards like; AQ, JT, 34, J2 and so on regardless of whether they are suited or not.
Probability of being dealt a range of hands.
Work out each individual hand probability and add them together.
What's the probability of being dealt AA or KK? (Spot the 'or' there? - Time to add.)
- Probability of being dealt AA - 1/221 (4/52 * 3/51 = 1/221).
- Probability of being dealt KK - 1/221 (4/52 * 3/51 = 1/221).
P = (1/221) + (1/221)
P = 2/221 = 1/110
Easy enough. If you want to add more possible hands in to the range, just work out their individual probability and add them in. So if we wanted to work out the odds of being dealt AA, KK or 7 3...
- Probability of being dealt AA - 1/221 (4/52 * 3/51 = 1/221).
- Probability of being dealt KK - 1/221 (4/52 * 3/51 = 1/221).
- Probability of being dealt 7 3 - 1/1326 ([1/52 * 1/51] * 2 = 1/1326).
P = (1/221) + (1/221) + (1/1326)
P = 359/36465 = 1/102
This one definitely takes more skill with adding fractions because of the different denominators, but you get the idea. I'm just teaching hand probabilities here, so I'm not going to go in to adding fractions in this article for now! This fractions calculator is really handy for adding those trickier probabilities quickly though.
Sample Problem
Chapter 2 #16. Poker dice is played by simultaneously rolling 5 dice. Show that:
a) P[no two alike] = .0926, b) P[one pair] = .4630, c) P[two pair] = .2315, d) P[three alike] = .1543, e) P[full house] = .0386, f) P[four alike] = .0193, g) P[5 alike]=.0008.
Solution
Notation, I will let ^ designate the power function. For example, 6^5 is 6 to the fifth power. 6! is 6 factorial, 6! = 6 * 5 * 4 * 3 * 2 * 1.
To calculate the Probabilities here, we will divide the number of occurrences for a particular event by the total possibilities in rolling 5 dice.
N[total] = total possibilities in rolling five dice = 6^5 = 7776
Note: N[total] is the number of ordered rolls. For example, if we rolled the dice one by one and rolled in order 3,4,5,6,2, it would be considered as different from if we rolled 2,3,4,5,6 in order.
a) P[no two alike]
There are 6 choices of numbers for the first dice. The 2nd dice must be one of the remaining 5 unchosen numbers, and the 3rd dice one of the 4 remaining unchosen numbers, and so on… This gives an ordered count, so:
N[no two alike] = 6 * 5 * 4 * 3 * 2 = 720
P[no two alike] = N[no two alike]/N[total] = 720/7776 = 0.09259259
b) P[one pair]
First we count the possible sets (unordered) of numbers. Here we have one number that is the pair, 6 choices. Then we must choose 3 different numbers from the remaining 5 values, as these three are equivalent, we have choose(5,3) = 5!/(3! * 2!) possibilities. So 6 * choose(5,3) is the total combinations of numbers. Since we are dealing with ordered counts, we must consider the orderings for each set of numbers. We have 5 die with 2 the same so the number of orderings is 5!/2!. Multiplying these together gives us the number of ordered samples:
N[one pair] = 6 * (5!/(3! * 2!)) * (5!/2) = 3600
P[one pair] = N[one pair]/N[total] = 0.462963
c) P[two pair]
Here we have 2 pairs which are equivalent, so we must choose 2 values from the 6 possible values, choose(6,2)= 6!/(2! * 4!). Then we must choose 1 value from the remaining 4 for the single value, 4 ways. Now we must consider the orderings, 5 die with 2 sets of 2 the same so the number of orderings is 5!/(2! * 2!).
N[two pairs] = (6!/(2! * 4!)) * 4 * (5!/(2! * 2!)) = 1800
P[two pairs] = N[two pairs]/N[total] = 0.2314815
d) P[three alike] (the remaining two cards are different)
There are 6 choices for the three of a kind. Then we must choose 2 different values from the remaining 5 choices, choose(5,2) = 5!/(2! * 3!). The number of orderings is 5 items with 3 being identical, which is 5!/3!.
N[three alike] = 6 * (5!/(3! * 2!)) * (5!/3!) = 1200
P[three alike] = N[three alike]/N[total] = 0.154321
e) P[full house] 3 alike with a pair.
We need one value for the three that are alike, 6 ways, and then we must choose from the remaining 5 values for the pair, 5 ways. The orderings are given by 5!/(3! * 2!).
N[full house] = 6 * 5 * (5!/(3! * 2!)) = 300
P[full house] = N[full house]/N[total] = 0.03858025
f) P[four alike]
We need one value for the four of a kind, and then one value from the remaining 5 for the last die. The number of orderings is given by 5!/4!.
N[four alike] = 6 * 5 * (5!/4!) = 150
P[four alike] = N[four alike]/N[total] = 0.01929012
g) P[five alike]
Here we need 1 value for the five alike, 6 ways. There is just 1 possible ordering as all five die are the same.
N[five alike] = 6
P[five alike] = N[five alike]/N[total] = 0.0007716049
Check: the numbers of each type must sum to N[total] = 7776:
720 + 3600 + 1800 + 1200 + 300 + 150 + 6 = 7776
Extra:
Straight
We can have two possible straights: one composed of (6,5,4,3,2) and one composed of (5,4,3,2,1). Each of these straights can be permuted 5! ways.
N[straight] = 2 * 5! = 240
P[straight] = N[straight]/N[total] = 0.0308642
Now let's run a simulation in R:
Poker Hand Probability Problems Worksheets
Results of simulation:
notwoalike: 0.09194
onepair: 0.464
twopair: 0.23187
threealike: 0.1554
fullhouse: 0.03727
fouralike: 0.01889
fivealike: 0.00063
straight: 0.03072
Ever wondered where some of those odds in the odds charts came from? In this article, I will teach you how to work out the probability of being dealt different types of preflop hands in Texas Holdem.
It's all pretty simple and you don't need to be a mathematician to work out the probabilities. I'll keep the math part as straightforward as I can to help keep this an easy-going article for the both of us.
- Probability calculations quick links.
A few probability basics.
When working out hand probabilities, the main probabilities we will work with are the number of cards in the deck and the number of cards we want to be dealt. So for example, if we were going to deal out 1 card:
- The probability of dealing a 7 would be 1/52 - There is one 7 in a deck of 52 cards.
- The probability of dealing any Ace would be 4/52 - There four Aces in a deck of 52 cards.
- The probability of dealing any would be 13/52 - There are 13 s in a deck of 52 cards.
In fact, the probability of being dealt any random card (not just the 7) would be 1/52. This also applies to the probability being dealt any random value of card like Kings, tens, fours, whatever (4/52) and the probability of being dealt any random suit (13/52).
Each card is just as likely to be dealt as any other - no special priorities in this game!
The numbers change for future cards.
A quick example... let's say we want to work out the probability of being dealt a pair of sevens.
- The probability of being dealt a 7 for the first card will be 4/52.
- The probability of being dealt a 7 for the second card will be 3/51.
Notice how the probability changes for the second card? After we have been dealt the first card, there is now 1 less card in the deck making it 51 cards in total. Also, after already being dealt a 7, there are now only three 7s left in the deck.
Always try and take care with the numbers for future cards. The numbers will change slightly as you go along.
Working out probabilities.
- Whenever the word 'and' is used, it will usually mean multiply.
- Whenever the word 'or' is used, it will usually mean add.
This won't make much sense for now, but it will make a lot of sense a little further on in the article. Trust me.
Probability of being dealt two exact cards.
Multiply the two probabilities together.
So, we want to find the probability of being dealt the A and K. (See the 'and' there?)
- Probability of being dealt A - 1/52.
- Probability of being dealt K - 1/51.
Now let's just multiply these bad boys together.
P = (1/52) * (1/51)
P = 1/2652
So the probability of being dealt the A and then K is 1/2652. As you might be able to work out, this is the same probability for any two exact cards, as the likelihood of being dealt A K is the same as being dealt a hand like 7 3 in that order.
But wait, we do not care about the order of the cards we are dealt!
When we are dealt a hand in Texas Hold'em, we don't care whether we get the A first or the K first (which is what we just worked out), just as long as we get them in our hand it's all the same. There are two possible combinations of being dealt this hand (A K and K A), so we simply multiply the probability by 2 to get a more useful probability.
P = 1/2652 * 2
P = 1/1326
You might notice that because of this, we have also worked out that there are 1,326 possible combinations of starting hands in Texas Holdem. Cool huh?
Probability of being dealt a certain hand.
Two exact cards is all well and good, but what if we want to work out the chances of being dealt AK, regardless of specific suits and whatnot? Well, we just do the same again...
Multiply the two probabilities together.
So, we want to find the probability of being dealt any Ace andany King.
- Probability of being dealt any Ace - 4/52.
- Probability of being dealt any King - 4/51 (after we've been dealt our Ace, there are now 51 cards left).
P = (4/52) * (4/51)
P = 16/2652 = 1/166
However, again with the 2652 number we are working out the probability of being deal an Ace and then a King. If we want the probability of being dealt either in any order, there are two possible ways to make this AK combination so we multiply the probability by 2.
Poker Hand Probability Problems
P = 16/2652 * 2
P = 32/2652
P = 1/83
The probability of being dealt any AK as opposed to an AK with exact suits is more probable as we would expect. A lot more probable in fact. Also, as you might guess, this probability of 1/83 will be the same for any two value of cards like; AQ, JT, 34, J2 and so on regardless of whether they are suited or not.
Probability of being dealt a range of hands.
Work out each individual hand probability and add them together.
What's the probability of being dealt AA or KK? (Spot the 'or' there? - Time to add.)
- Probability of being dealt AA - 1/221 (4/52 * 3/51 = 1/221).
- Probability of being dealt KK - 1/221 (4/52 * 3/51 = 1/221).
P = (1/221) + (1/221)
P = 2/221 = 1/110
Easy enough. If you want to add more possible hands in to the range, just work out their individual probability and add them in. So if we wanted to work out the odds of being dealt AA, KK or 7 3...
- Probability of being dealt AA - 1/221 (4/52 * 3/51 = 1/221).
- Probability of being dealt KK - 1/221 (4/52 * 3/51 = 1/221).
- Probability of being dealt 7 3 - 1/1326 ([1/52 * 1/51] * 2 = 1/1326).
P = (1/221) + (1/221) + (1/1326)
P = 359/36465 = 1/102
This one definitely takes more skill with adding fractions because of the different denominators, but you get the idea. I'm just teaching hand probabilities here, so I'm not going to go in to adding fractions in this article for now! This fractions calculator is really handy for adding those trickier probabilities quickly though.
Overview of working out hand probabilities.
Hopefully that's enough information and examples to allow you to go off and work out the probabilities of being dealt various hands and ranges of hands before the flop in Texas Holdem. The best way to learn how to work out probabilities is to actually try and work it out for yourself, otherwise the maths part will just go in one ear and out the other.
I guess this article isn't really going to do much for improving your game, but it's still pretty interesting to know the odds of being dealt different types of hands.
I'm sure that some of you reading this article were not aware that the probability of being dealt AA were exactly the same as the probability of being dealt 22! Well, now you know - it's 1/221.
Other useful articles.
- Poker mathematics.
- Pot odds.
- Equity in poker.
Go back to the poker odds charts.
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Poker Hand Probability Problems Worksheet
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