Unbalanced partnerships in Chattahoochee Rummy.
§1 Introduction. It is entirely possible to play partnership Chattahoochee with a number of players that is a mathematical prime, but to still achieve a balanced level of competition offering players equal opportunities to score. Three methods are given below, and each might be described as a "symmetrical distribution of asymmetries".
These methods can also be considered for other card games where a victory is achieved by earning a large number of points.
For the sake of providing concrete numbers, this report gives examples which generally assume that five players play five games. In clockwise order, those players are Amber, Betty, Cindy, Diana, and Ellie, as on the home page.
In table one appear their individual game scores, before any kind of partnership calculations are considered:
| table one | ||||||
| individual game scores |
game | |||||
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | ||
| player | Amber | 86 | 54 | 72 | 60 | 82 |
| Betty | 63 | 81 | 51 | 68 | 98 | |
| Cindy | 80 | 62 | 91 | 94 | 74 | |
| Diana | 71 | 53 | 95 | 78 | 58 | |
| Ellie | 59 | 93 | 84 | 66 | 88 | |
These numbers are not intended to be realistic representatives of actual Chattahoochee scoring. In reality, at least one player in every game will have a score of zero, and other players will often have scores of well over 100. Instead, the values here were selected to minimize the occurence of repeated numbers in the subsequent tables. This is so that the reader can more easily trace the calculations.
§2 First method. This involves five players playing five games, or any multiple of five.
Each game has one partnership with two players (a duo), and one partnership of three (a trio). Each player plays two games in different duos, and three games in different trios. The partnerships rotate.
At the end of the five games, their individual scores are be added as in an ordinary (balanced) partnership game. Here is an example of how the figures work out with the following partnerships:
| table two (using values from table one) | ||||
| game | duo's score | trio's score | ||
|---|---|---|---|---|
| 1 | Amber-Cindy | 86 + 80 = 166 | Ellie-Betty-Diana | 59 + 63 + 71 = 193 |
| 2 | Betty-Diana | 81 + 53 = 134 | Amber-Cindy-Ellie | 54 + 62 + 93 = 209 |
| 3 | Cindy-Ellie | 91 + 84 = 175 | Betty-Diana-Amber | 51 + 95 + 72 = 218 |
| 4 | Diana-Amber | 78 + 60 = 138 | Cindy-Ellie-Betty | 94 + 66 + 68 = 228 |
| 5 | Ellie-Betty | 88 + 98 = 186 | Diana-Amber-Cindy | 58 + 82 + 74 = 214 |
Each player's total score is then the sum of her partnership's scores from the five games as in table three:
| table three (using values from table two) | |
| player | player's total score |
|---|---|
| Amber | 166 + 209 + 218 + 138 + 214 = 945 |
| Betty | 134 + 218 + 228 + 186 + 193 = 959 |
| Cindy | 175 + 228 + 214 + 166 + 209 = 992 |
| Diana | 138 + 214 + 193 + 134 + 218 = 897 |
| Ellie | 186 + 193 + 209 + 175 + 228 = 991 |
A duo's score will typically be only about two-thirds of a trio's score. Some players might like to perform an adjustment on the scores to make them more compatible. This can be achieved without introducing fractions, by:
After that, the scores can be added as in table four. This step is essential if the number of games to be played is not a multiple of the number of players.
| table four (using values from table two) | |
| player | player's adjusted total score |
|---|---|
| Amber | 3 × 166 + 2 × 209 + 2 × 218 + 3 × 138 + 2 × 214 = 2194 |
| Betty | 3 × 134 + 2 × 218 + 2 × 228 + 3 × 186 + 2 × 193 = 2238 |
| Cindy | 3 × 175 + 2 × 228 + 2 × 214 + 3 × 166 + 2 × 209 = 2325 |
| Diana | 3 × 138 + 2 × 214 + 2 × 193 + 3 × 134 + 2 × 218 = 2066 |
| Ellie | 3 × 186 + 2 × 193 + 2 × 209 + 3 × 175 + 2 × 228 = 2343 |
The adjustment ratio of 3 to 2 is convenient, and it does provide a good first approximation to equivalence. However, once the scores from a large body of games have been recorded, statisticians might determine that the ratio needs to be fine-tuned.
§3 Second method. The following is a completely different method of achieving balance, and it does not require that the number of games played be a multiple of five. It relies on some familiar numbers:
| table five (excerpt of table one) | ||
| individual game score |
game | |
|---|---|---|
| 1 | ||
| player | Amber | 86 |
| Betty | 63 | |
| Cindy | 80 | |
| Diana | 71 | |
| Ellie | 59 | |
Each player supports the second player to her left. Support is one-way. For instance, Amber supports Cindy, but Cindy does not support Amber. Amber supports Cindy by adding to her melds, and by delegating a discard to her; but Cindy can do neither of these things for Amber.
To reflect this in the scoring, Amber's total score will be the sum of her own individual score and Cindy's individual score; but Cindy's total score will not include Amber's. Instead, Cindy's total will include Ellie's, who is the second player to Cindy's left, and whom Cindy supports. Here is a listing of all the supports:
Here are their scores:
| table six (using values from table five) | ||
| combined game score |
game | |
|---|---|---|
| 1 | ||
| player | Amber | 86 + 80 = 166 |
| Betty | 63 + 71 = 134 | |
| Cindy | 80 + 59 = 139 | |
| Diana | 71 + 86 = 157 | |
| Ellie | 59 + 63 = 122 | |
Rather than the second player from the left, equally well chosen could have been the first player on the right, et cetera. The idea is that there is one chain of support that touches every player. In particular, no two players ever support each other; but if they were to, their relationship would be that of an ordinary partnerhip.
Here is is another configuration demonstrating the possibilies of support. With six players, adding Flora, there can be two separate triangles of support, as follows:
This is not the same as two ordinary partnerships of three, because for instance Cindy does not support Amber.
Here is a further variation. Although eight is not a prime number, it is relatively prime to three. With two additional players named Greta and Helen, each player supports the third player to her left, in one chain of support:
§4 Third method. This system involves intransitive partnerships and is best explained with a seven-person game, using all the players from the previous section except Helen. Each player is partnered with the player second on her left; and partnered with the player second on her right. However, those two other players are not partners with each other.
For example, Amber and Cindy are partners, Cindy and Ellie are partners, but Amber and Ellie are not partners. Explained another way, Amber and Cindy support each other, Cindy and Ellie support each other, but neither Amber nor Ellie supports the other.
The complete list of partnerships:
Here are their individual scores:
| table seven (extension of table five) | ||
| individual game score |
game | |
|---|---|---|
| 1 | ||
| player | Amber | 86 |
| Betty | 63 | |
| Cindy | 80 | |
| Diana | 71 | |
| Ellie | 59 | |
| Flora | 73 | |
| Greta | 94 | |
Each player's total score is the sum of three values: her own score, her left partner's score, and her right partner's score:
| table eight (using values from table seven) | ||
| combined game score |
game | |
|---|---|---|
| 1 | ||
| player | Amber | 86 + 80 + 73 = 239 |
| Betty | 63 + 71 + 94 = 228 | |
| Cindy | 80 + 59 + 86 = 225 | |
| Diana | 71 + 73 + 63 = 207 | |
| Ellie | 59 + 94 + 80 = 233 | |
| Flora | 73 + 86 + 71 = 230 | |
| Greta | 94 + 63 + 59 = 216 | |
§5 Comments. Many arrangements of unbalanced partnerships are complicated enough that players will want to obtain markers to place on the card table to clearly indicate who is supporting whom, or who is partnered with whom; and to use electronic equipment to calculate scores.