07-27-2013, 03:08 PM

###############################################

Double Deck Pinochle Card Distribution Probabilities - Singles Around

###############################################

Double Deck Pinochle Card Distribution Probabilities - Singles Around

###############################################

A knowledge of Double Deck Pinochle and combinatorics are useful for this discussion. Readers unfamiliar with Double Deck Pinochle may find the Definitions section at the end helpful. Please examine the rest of this (Great!) website for more information about Double Deck Pinochle. Many websites contain explanations on Combinations and Permutations.

Introduction

The question of card distribution probabilities presents itself in Pinochle all of the time.

This post answers two questions and provides the mathematics to show the solution. It points out an interesting observation from examining the solution to the second question.

Please note this is one solution approach, there are other ways to arrive at the same answer.

The two questions are:

- What is the Probability of getting exactly 1 Ace in each suit?

- What is the probability of getting Aces Around (at least 1 Ace in each suit)?

Discussion

Question #1: What is the Probability of getting exactly 1 Ace in each suit?

This can be answered using the combinations formula ( C(n,r) where one chooses r items from n things, or more commonly n choose r) and the following process.

Step 1

There are 4 identical Aces in 1 suit and any 1 can be selected, C(4,1) = 4 gives the possible ways to select 1 Ace from the 4 Aces in 1 suit.

Since the question asks for 1 Ace from each of the 4 suits, then multiplication of the number of possible combinations together for all four suits is required to give the number of ways to obtain exactly 1 Ace from each of the 4 suits:

A = C(4,1) • C(4,1) • C(4,1) • C(4,1) = C(4,1)^4 = 256 = Number of ways to get exactly 1 Ace from each of the 4 suits

Step 2

Four cards of the players 20 card hand has been selected leaving 16 cards to be selected from the remaining deck of 64 cards. The Pinochle Double Deck of 80 cards has had all 16 Aces removed leaving 64 to choose from thereby ensuring that no other Ace can appear in the player's hand. The number of ways that 16 cards can be chosen to complete the player's hand can be found from:

B = C(64,16) = 488,526,937,079,580

Step 3

Now multiplying A and B together gives the total number of possible hands containing exactly 1 Ace in each of the 4 suits:

C = A • B = 256 • 488,526,937,079,580 = 125,062,895,892,372,000 total possible hands containing exactly 1 Ace in each of the 4 suits

Please note the "C" used here is a variable and not the Combinations formula function.

Step 4

The total number of hands that can be dealt is found from choosing 20 from the 80 card deck.

D = C(80,20) = 3,535,316,142,212,170,000 total possible hands that can be dealt

Step 5

The probability of occurrence of exactly 1 Ace in each of the 4 suits can be found by dividing C by the Total possible hands that can be dealt, D.

Probability of exactly 1 Ace in each suit = C / D = 0.0353753075712533 = 3.53%

This means the Probability of exactly 1 Ace in each suit occurs on average 3.53% of the time.

Caution!

This is NOT the probability of having Single Aces Around.

This is NOT the probability of having Single Aces Around.

Step 6

By taking the inverse of E one gets the on average Frequency of Occurrence:

On average Frequency of Exactly 1 Ace in Each Suit = 1 / E = 28.27 Hands

This means that on average every 28.27 hands has exactly 1 Ace in each of the 4 suits. Given that 4 hands occur with each deal, then on average every 6 to 7 deals one Player at the Table will have a hand containing exactly 4 Aces Around.

Question #2: What is the probability of getting Single Aces Around (at least 1 Ace in each suit)?

This is a more complicated question because as long as there is at least 1 Ace in each suit, this is Aces Around. Consideration must be given to all of the possible combinations of Aces configurations that contain at least one Ace in each suit.

Step 1 Configuration (Config) Elaboration

There are 175 possible combinations of Aces configurations that give at least 1 Ace in each suit. They can be summarized into 20 configurations. Table 1 shows each of the 20 configurations where each of the four digits represents one of the four suits and the values represent the number of Aces in the suit. The order of suits is not represented in Table 1 and is not important here.

Table 1 Possible Single Aces Configurations

In this table each configuration (Config) digit represents a suit without regard to a specific suit.

Code:

`Please note this Table Scrolls within the code window.`

Config

1) 1111

2) 1112

3) 1113

4) 1114

5) 1122

6) 1123

7) 1124

8) 1133

9) 1134

10) 1144

11) 1222

12) 1223

13) 1224

14) 1233

15) 1234

16) 1244

17) 1333

18) 1334

19) 1344

20) 1444

Step 2 Configuration Count

Now consider each configuration and determine how many ways the each digit can be arranged among the four suits to get the count of the ways.

For example, given a configuration of XXYZ, the total number of ways the letters can be arranged is found from:

C(4,2) • C(2,1) • C(1,1) = 6 • 2 • 1 = 12.

Where

C(4,2) = 6 represents the number of ways the 2 Xs can be positioned in the 4 spaces.

C(2,1) = 2 represents the number of ways the 1 Y can be positioned in the remaining 2 spaces.

C(1,1) = 1 represents the number of ways the 1 Z can be positioned in the remaining 1 space.

Applying this process to each of the 20 summary configurations of Table 1 yields the count of possible configurations for each one as shown in Table 2.

For example, 1112 has a count of 4 when enumerated with each digit position representing a specific suit is 1112, 1121, 1211, and 2111.

Table 2 Configurations Count

In this table each configuration (Config) digit represents a suit without regard to a specific suit.

Code:

`Please note this Table Scrolls within the code window.`

Config Count Deriving Formula

1) 1111 1 C(4,4)

2) 1112 4 C(4,3) • C(1,1)

3) 1113 4 C(4,3) • C(1,1)

4) 1114 4 C(4,3) • C(1,1)

5) 1122 6 C(4,2) • C(2,2)

6) 1123 12 C(4,2) • C(2,1) • C(1,1)

7) 1124 12 C(4,2) • C(2,1) • C(1,1)

8) 1133 6 C(4,2) • C(2,2)

9) 1134 12 C(4,2) • C(2,1) • C(1,1)

10) 1144 6 C(4,2) • C(2,2)

11) 1222 4 C(4,1) • C(3,3)

12) 1223 12 C(4,1) • C(3,2) • C(1,1)

13) 1224 12 C(4,1) • C(3,2) • C(1,1)

14) 1233 12 C(4,1) • C(3,1) • C(2,2)

15) 1234 24 C(4,1) • C(3,1) • C(2,1) • C(1,1)

16) 1244 12 C(4,1) • C(3,1) • C(2,2)

17) 1333 4 C(4,1) • C(3,3)

18) 1334 12 C(4,1) • C(3,2) • C(1,1)

19) 1344 12 C(4,1) • C(3,1) • C(2,2)

20) 1444 4 C(4,1) • C(3,3)

Summing the Counts of Table 2 yields 175 possible configurations for Single Aces Around.

Step 3 Group Count

The Aces from each suit contributing to each configuration are selected from the set of 4 Aces in the suit. Thus, for each configuration the number of possible ways to select these Aces must be accounted.

For example, given the configuration 1234, the total Group Count is found from:

C(4,1) • C(4,2) • C(4,3) • C(4,4) = 4 • 6 • 4 • 1 = 96

The factor C(4,1) = 4 represents the number of ways 1 Ace can be selected from 4.

The factor C(4,2) = 6 represents the number of ways 2 Aces can be selected from 4.

The factor C(4,3) = 4 represents the number of ways 3 Aces can be selected from 4.

The factor C(4,4) = 1 represents the number of ways 4 Aces can be selected from 4.

Table 3 provides the elaborated Group Counts for each Single Aces Configuration with their deriving formula.

Table 3 Group Counts

In this table each configuration (Config) digit represents a suit without regard to a specific suit.

Code:

`Please note this Table Scrolls within the code window.`

Config Group Count Deriving Formula

1) 1111 256 C(4,1)^4

2) 1112 384 C(4,1)^3 • C(4,2)

3) 1113 256 C(4,1)^3 • C(4,3)

4) 1114 64 C(4,1)^3 • C(4,4)

5) 1122 576 C(4,1)^2 • C(4,2)^2

6) 1123 384 C(4,1)^2 • C(4,2) • C(4,3)

7) 1124 96 C(4,1)^2 • C(4,2) • C(4,4)

8) 1133 256 C(4,1)^2 • C(4,3)^2

9) 1134 64 C(4,1)^2 • C(4,3) • C(4,4)

10) 1144 16 C(4,1)^2 • C(4,4)^2

11) 1222 864 C(4,1) • C(4,2)^3

12) 1223 576 C(4,1) • C(4,2)^2 • C(4,3)

13) 1224 144 C(4,1) • C(4,2)^2 • C(4,4)

14) 1233 384 C(4,1) • C(4,2) • C(4,3)^2

15) 1234 96 C(4,1) • C(4,2) • C(4,3) • C(4,4)

16) 1244 24 C(4,1) • C(4,2) • C(4,4)^2

17) 1333 256 C(4,1) • C(4,3)^3

18) 1334 64 C(4,1) • C(4,3)^2 • C(4,4)

19) 1344 16 C(4,1) • C(4,3) • C(4,4)^2

20) 1444 4 C(4,1) • C(4,4)^3

Step 4 Filler Cards Number of Ways

The selection of Aces is complete. The remaining cards to fill out the Player's hand is found by counting the number of Aces in the specific configuration, subtracting this count from 20, then choosing the resulting number of filler cards from the 64 cards of the deck that has the Aces removed.

Given a configuration of WXYZ, let q = W + X + Y + Z represent the total number of Aces in the Player's hand for that configuration, then s = 20 - q number of cards must be chosen from the 64 cards to complete the hand.

The number of ways that s can be chosen from the 64 cards is given by

C(64,s) = 64 choose s

Table 4 provides the filler cards number of ways for each configuration and the deriving formula.

Table 4 Filler Cards Number of Ways

In this table each configuration (Config) digit represents a suit without regard to a specific suit.

Code:

`Please note this Table Scrolls within the code window.`

Config Filler Cards Number of Ways Deriving Formula

1) 1111 488,526,937,079,580 C(64,16)

2) 1112 159,518,999,862,720 C(64,15)

3) 1113 47,855,699,958,816 C(64,14)

4) 1114 13,136,858,812,224 C(64,13)

5) 1122 47,855,699,958,816 C(64,14)

6) 1123 13,136,858,812,224 C(64,13)

7) 1124 3,284,214,703,056 C(64,12)

8) 1133 3,284,214,703,056 C(64,12)

9) 1134 743,595,781,824 C(64,11)

10) 1144 151,473,214,816 C(64,10)

11) 1222 13,136,858,812,224 C(64,13)

12) 1223 3,284,214,703,056 C(64,12)

13) 1224 743,595,781,824 C(64,11)

14) 1233 743,595,781,824 C(64,11)

15) 1234 151,473,214,816 C(64,10)

16) 1244 27,540,584,512 C(64,9)

17) 1333 151,473,214,816 C(64,10)

18) 1334 27,540,584,512 C(64,9)

19) 1344 4,426,165,368 C(64,8)

20) 1444 621,216,192 C(64,7)

Step 5 Determining the Individual Configurations Probability

The Probability of Single Aces Around for each configuration is found by taking the product of the Configuration Count in Table 2 times the Group Count in Table 3 times the Filler Ways in Table 4 and dividing by the total possible hands that can be dealt, C(80,20) = 3,535,316,142,212,170,000.

Table 5 provides the total number of possible hands that can be dealt containing the specified Aces configuration and the Probability of Single Aces Around for each configuration.

Table 5 Number of Possible Single Aces Configurations Hands & Their Probability of Occurrence

In this table each configuration (Config) digit represents a suit without regard to a specific suit.

Code:

`Please note this Table Scrolls within the code window.`

Config Number of Hands Probability of Occurrence

1) 1111 125,062,895,892,372,000 0.0353753075712533

2) 1112 245,021,183,789,138,000 0.0693067250375576

3) 1113 49,004,236,757,827,600 0.0138613450075115

4) 1114 3,363,035,855,929,340 0.0009512687750253

5) 1122 165,389,299,057,668,000 0.0467820394003514

6) 1123 60,534,645,406,728,200 0.0171228379504554

7) 1124 3,783,415,337,920,510 0.00107017737190346

8) 1133 5,044,553,783,894,020 0.00142690316253795

9) 1134 571,081,560,440,832 0.000161536207079768

10) 1144 14,541,428,622,336 0.00000411319045804965

11) 1222 45,400,984,055,046,100 0.0128421284628416

12) 1223 22,700,492,027,523,100 0.00642106423142078

13) 1224 1,284,933,510,991,870 0.000363456465929478

14) 1233 3,426,489,362,644,990 0.000969217242478608

15) 1234 348,994,286,936,064 0.0000987165709931915

16) 1244 7,931,688,339,456 0.00000224355843166344

17) 1333 155,108,571,971,584 0.0000438740315525296

18) 1334 21,151,168,905,216 0.00000598282248443585

19) 1344 849,823,750,656 0.000000240381260535369

20) 1444 9,939,459,072 0.00000000281147673140782

Step 6 Determining the Single Aces Around Probability

Finally, the Probability of having Single Aces Around is found by summing the Probabilities for each configuration.

This is the Probability of having Single Aces Around.

Probability of Single Aces Around = 0.206809180253003 = 20.68%

Step 7 Frequency of Occurrence

Taking the reciprocal of the Probability provides the on average frequency of occurrence.

On average Frequency of occurrence = 1 / Probability = 1 / 0.206809180253003 = 4.84 Hands

Observations

On average every 4 to 5 hands a player will see Single Aces Around.

Since this same analysis applies to any rank. The following may be stated.

A player on average will see 5 Single Arounds meld of a particular rank in 24 hands. Since four hands are dealt with each deal, then in 6 deals, 5 players will on average have had say Aces around. The same can be said for Kings, Queens, and Jacks (Tens as well, but no meld for Tens Around!).

Interesting Note

Examining the data used to compute these probabilities (see Table 5 individual configurations probabilities) shows, it is more likely to have 5 (6.9%) or 6 (6.1%) cards of the same rank Around than it is to have just 4 (3.5%) or more than 6 (4.1%, 7 through 16 cards of the same rank in a Player's hand add up to the 4.1%). This means when Single Aces are declared Around one can reasonable expect that player to have 5 or 6 Aces! This is a manifestation of Pascal's Triangle.

Definitions

This section contains background information to help understand this post. Please examine the rest of the website for more information on Double Deck Pinochle.

The presentation here applies to Double Deck Pinochle consisting of an 80 card deck:

- Four Suits: Spades, Hearts, Clubs, and Diamonds

- Each Suit has four ordered Ranks: Ace, Ten, King, Queen, and Jack with Ace high

- Each Rank in each Suit has four identical cards

Double Deck Four Player Partnership Pinochle has two Two-Player teams that sit opposite each other at the Table. For discussion, going clockwise around the Table label the players as follows. The player to your left is Left, L, the player opposite you is your Partner, P, the player on your right is Right, R, and You, Y. L and R are partners. Y and P are partners.

All 80 cards are dealt, 20 to each player, for each hand played.

Combinations formula:

C(n,r) = n! / ( r! • (n-r)! ) where you are choosing r items from n items (or more commonly n choose r) without regard to order and without replacement. When order matters then the Permutations formula should be used.

###################################

RTFQ/RTFA

Ta!

--FLACKprb

--FLACKprb