Poker Hand Probability Problems

• Poker Hand Probability Problems Solving
• Poker Hand Probability Problems
• Poker Hand Probability Problems Involving
• Poker Hand Probability Problems Worksheets

Intro

This is a problem concerning basic probability calculations from the text: “A First Course in Probability Theory” by Sheldon Ross (8th addition).

Poker starting hands by position, poker starting hand winning probabilityBring Vanna White and fun puzzles to your home and get ready to play, poker starting hands by position. Wheel of Fortune is a video game adapation of the famous TV quiz game show that was released for multiple gaming consoles including the SEGA Genesis. The probability of collecting royal flush in poker is 1 to 649 740. The odd to catch this combination on the flop with pocket broadway cards is equal to 0.0008%. If there is a potential royal flush on the board, the probability that it will be collected on the turn is 2%, and till the river – 4%. What is the probability that a 5-card poker hand has exactly three spades? What is the probability that it has at least three spades? Problem 10) To fulfill the requirements for a certain degree. A student can choose to take any 7 out of 20 courses. With the constraint that at least 1 of the 7 courses must be a statistics. The Probability of drawing a given hand is calculated by dividing the number of ways of drawing the hand (Frequency) by the total number of 5-card hands (the sample space; =, ). For example, there are 4 different ways to draw a royal flush (one for each suit), so the probability is 4 / 2,598,960, or one in 649,740. The Probability of drawing a given hand is calculated by dividing the number of ways of drawing the hand (Frequency) by the total number of 5-card hands (the sample space; =,). For example, there are 4 different ways to draw a royal flush (one for each suit), so the probability is 4 / 2,598,960, or one in 649,740.

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.

Poker Hand Probability Problems Solving

Solution

Poker Hand Probability 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] = 0.0308642

Now let’s run a simulation in R:

Poker Hand Probability Problems Involving

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

Poker Hand Probability Problems Worksheets