Probability Theory Poker
2021年11月17日Register here: http://gg.gg/wx6m7
*Probability Theory Problems
*Probability Theory Poker Game
*Probability Theory Poker Rules
*Probability Theory Pierre De Fermat
*Probability Theory Practice
Poker Probability – Part 1: Basics of Probability. The market for uncertified physicians was not very big in 16th-century Italy. So when a young Girolamo Cardano found himself without a College membership or cash, he did what any reasonable person would do: he became a professional gambler. The classical definition of probability If there are m outcomes in a sample space, and all are equally likely of being the result of an experimental measurement, then the probability of observing an event that contains s outcomes is given by e.g. Probability of drawing an ace from a deck of 52 cards. Sample space consists of 52 outcomes. Poker-theory probability statistics. Asked Dec 5 at 19:32 Ftam 1. Calculating Odds for 3 of a kind. Odds probability equity.Intro
This is a problem concerning basic probability calculations from the text: “A First Course in Probability Theory” by Sheldon Ross (8th addition). National casino marketing jobs new york.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.SolutionProbability Theory Problems
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] = 0.0308642Probability Theory Poker Game
Now let’s run a simulation in R:
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
Intro
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 = 7776Probability Theory Poker Rules
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:Probability Theory Pierre De Fermat
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: 0.03072Probability Theory Practice
Register here: http://gg.gg/wx6m7
https://diarynote-jp.indered.space
*Probability Theory Problems
*Probability Theory Poker Game
*Probability Theory Poker Rules
*Probability Theory Pierre De Fermat
*Probability Theory Practice
Poker Probability – Part 1: Basics of Probability. The market for uncertified physicians was not very big in 16th-century Italy. So when a young Girolamo Cardano found himself without a College membership or cash, he did what any reasonable person would do: he became a professional gambler. The classical definition of probability If there are m outcomes in a sample space, and all are equally likely of being the result of an experimental measurement, then the probability of observing an event that contains s outcomes is given by e.g. Probability of drawing an ace from a deck of 52 cards. Sample space consists of 52 outcomes. Poker-theory probability statistics. Asked Dec 5 at 19:32 Ftam 1. Calculating Odds for 3 of a kind. Odds probability equity.Intro
This is a problem concerning basic probability calculations from the text: “A First Course in Probability Theory” by Sheldon Ross (8th addition). National casino marketing jobs new york.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.SolutionProbability Theory Problems
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] = 0.0308642Probability Theory Poker Game
Now let’s run a simulation in R:
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
Intro
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 = 7776Probability Theory Poker Rules
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:Probability Theory Pierre De Fermat
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: 0.03072Probability Theory Practice
Register here: http://gg.gg/wx6m7
https://diarynote-jp.indered.space
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