Double Deck Pinochle Card Distribution Probabilities - Singles Around

[FLACKprb]

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Double Deck Pinochle Card Distribution Probabilities - Singles Around
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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:

1. What is the Probability of getting exactly 1 Ace in each suit?
   
   
2. 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.

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.

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RTFQ/RTFA

[richardpaulhall]

Informative, and terse!

[ToreadorElder]

You're doing the Aces Around the hard way.

Exactly 1 ace in each suit is indeed 4C1 ^ 4, * 64C16...because you're requiring that all the 16 other cards not be aces.

For aces around, you eliminate that. So it's 4C1^4 * 76C16. Yes, this includes the isolated double aces around...but that's easy to correct, if one really thinks it's necessary. Double aces around is 4C2^4 * 72C12. Just subtract that...but it should be a small enough number to not be terribly pertinent.

[FLACKprb]

(07-29-2013, 01:51 PM)ToreadorElder Wrote:  You're doing the Aces Around the hard way.

Exactly 1 ace in each suit is indeed 4C1 ^ 4, * 64C16...because you're requiring that all the 16 other cards not be aces.

For aces around, you eliminate that. So it's 4C1^4 * 76C16. Yes, this includes the isolated double aces around...but that's easy to correct, if one really thinks it's necessary. Double aces around is 4C2^4 * 72C12. Just subtract that...but it should be a small enough number to not be terribly pertinent.

It may be the hard way. I believe it is a correct way. I am very interested in learning other ways. I've been searching the web looking for Double Deck Probabilities. I have yet to find any site with a comprehensive set of data supported with backup presentations showing how the data was derived.

My detailed analysis shows:

Pr(Single Aces Around) = 0.2068 = 20.68%
Occurs on average 1 in every 4.84 hands.
Or to put it another way 5 out of 24 hands on average.
Therefore, on average in 6 deals there will be 5 hands at the Table (any of the four players) that have had Aces Around.

Record your next games and see how many times Singles Around of each rank occur. The analysis applies to any rank.

I've only presented Single Aces Around on this website as of this date. I've got the detailed analysis for the following:

Pr(Double Aces Around) = 0.0021 = 0.21%.
Occurs on average 1 in every 467 hands.

Pr(Triple Aces Around) = 0.000000367 = 0.0000367%.
Occurs on average every 1 out of 2,720,627 hands.

Pr(Quadruple Aces Around) = 0.0000000000001797 = 0.00000000001797%.
Occurs on average every 1 out of 5,564,132,328,278 hands.

(07-29-2013, 01:51 PM)ToreadorElder Wrote:  For aces around, you eliminate that. So it's 4C1^4 * 76C16. Yes, this includes the isolated double aces around...but that's easy to correct, if one really thinks it's necessary. Double aces around is 4C2^4 * 72C12. Just subtract that...but it should be a small enough number to not be terribly pertinent.

By your analysis and recommendation ignoring the 0.56% from 4C2^4 * 72C12:

Single Aces Around is 4C1^4 * 76C16 = 0.7842 = 78.42%

This would make Single Aces occurring on average 1 out of every 1.275 hands.

Do you get Single Aces Around approximately every 4 out of 5 hands on average?

Do you see Single Aces Around at the Table by any player every deal?

(07-29-2013, 01:51 PM)ToreadorElder Wrote:  ... Double aces around is 4C2^4 * 72C12.

By your analysis:

Double Aces Around is 4C2^4 * 72C12 = 0.0056 = 0.56%

This would make Double Aces occurring on average 1 out of every 178 hands.

You may not consider this significant but it is the difference between seeing on average Double Aces every 1 out of 467 hands (0.21%, my number) versus 1 out every 178 hands (0.56%, your number).

If you take an average game at 8 deals per game (something I'd like to gather data on) then with 4 hands per deal then someone at the table will:

By my number, 0.21%, you'd see Double Aces on average every 14 to 15 games.

By your number, 0.56%, you'd see Double Aces on average every 5 to 6 games.

Do you see Double Aces every 5 to 6 games?

If my detailed analysis is in error, please show me specifically where the error of my ways occurred so that I may learn. I'm willing to admit mistakes. I will correct it. I would like to see information on this website supported by understandable backup analysis data that an interested person can examine and decide accuracy for themselves.

RTFQ/RTFA

[ToreadorElder]

Then we have a problem.

My argument is this: we're specifying aces around. That requires 1 ace in each suit...the 4C1 factors. But the other 16 cards *don't matter* in the sense that you have your aces around. They can be anything, except for the 4 cards that have been specified before. That's why it feels like 76C16 is correct. Yes, if we're saying "no double aces around (or better)" there's the slight error when that does happen, but that's a small error. It doesn't account for the extreme difference. So why isn't it 76C16? It's not a rhetorical question, or being obstinate. SOMETHING is wrong, so what is it? You're making case-by-case counting arguments, I'm making a general one. I cover cases you don't, but RARE ones (if you add all your cases that have 8 or more total aces, it's less than 1%). If the numbers don't relate, then something's wrong somewhere. The fact that there's multiples is likely the problematic factor for the 76C16; it may be double-counting cases. If so there *may not be* a simple closed-form expression for this probability, and you actually *have* to do it on a case by case basis.

Hmm...yes, I'm double counting with 76C12. Another way to (incorrectly) express the 5-ace aces around case is this:

4C1^4 (one of each) * 12C1 (any one ace) * 64C15 (any 15 non-aces)
The 1112 case count is 4C1^3 (1 ace in 3 suits) * 4C2 (2 aces in 1 suit) * 4C1 (the 2-ace suit can be any of 4 suits) * 64C15.

Drop 4C1^3 from each, and the 64C15. We have 4C1 * 12C1 versus 4C2 * 4, or 4*12 versus 6*4. A factor of 2...oh. Gee. 4C1^4 * 12C1, double counts. Say we've got 2 aces of diamonds. 'Number' them...1,2,3,4. The diamond ace in 4C1 is 1...in 12C1 is 3. Then the diamond ace in 4C1 is 3...in 12C1 it's 1. Oh. Double count. Oops. Gets nastier with the 6-ace case, insofar as trying to create a closed-form expression. Bummer.