*much*more complex than the four table movement. Edwin would have had to be taking opium or something for this dream to materialize.

As always, there's an odd number of (total) tables (because after counting the stationary pair, we have 2n – 1 "teams" of a real pair and a phantom pair – where

*n*is the number of tables). In this case,

*n*is 3 so we need five total tables. E/W pairs still move down 2 tables and N/S pairs 3 tables. The difference from the four-board movement is in the boards. The board sets still move down a table each round. But in the three-table Howell, there are only four board sets to begin with (if you recall, in the last round all three tables relay the remaining board set).

So the new rules applying to the boards are as follows: the two (logical) tables on either side of "Table 1" (the table at which the stationary pair resides) viz. tables 2 and 5, are twinned; and the remaining two tables, viz. tables 3 and 4, are also twinned. When a pair arrives at one of these twinned tables, they look to see which boards are in place. If there are none (table 5 starts with no boards – the 5th set doesn't appear until the final round) or if they have already played the boards, they "borrow" the boards from the twin table.

Otherwise, it works pretty much exactly the same was as does the four-table Howell movement. Let's see how things work out for the three table movement that corresponds to this movement on the Bridge Guys site:

logical # | actual # | Rnd 1: board set | Rnd 1: NS pair | Rnd 1: EW pair | Rnd 2: board set | Rnd 2: NS pair | Rnd 2: EW pair |
---|---|---|---|---|---|---|---|

1 | 1 | 1 | 6 (1) | 1 | 2 | 6 (2) | 2 |

2 | twinned with 5 | 2 | 3 | 4 | 3 | 4 | 5 |

3 | twinned with 4 | 3 | 5 | 2 | 4 | 1 | 3 |

4 | 3 | 4 | 2 | 5 | none (4) | 3 | 1 |

5 | 2 | none (2) | 4 | 3 | 1 | 5 | 4 |

Other movements (more than four tables) are actually simpler even than the four-table Howell. In these other movements, the bye-stand tables (the tables that the phantom pairs "play" at) are at consecutive tables. It is logical therefore that the tables on either side of the bye-stand should be numbered 1 and

*n*. Note that the position of "table 1" in the three and four-table Howell movements is completely arbitrary.

In practice, as the number of rounds for a full Howell must be 2n – 1 (everyone plays everyone else), some Howells could become rather lengthy affairs. n=7 for instance, gives us the practical limit: 13 rounds of 2 boards. Beyond that come the "three-quarter" Howells where the number of rounds is curtailed.

A seven-table Howell fits very nicely as the final session of a "complete" movement wherein 26 pairs meet in a regular 13-round Mitchell followed by two-interlaced 7-table Howells. In this latter session, each "team" is made up of one pair from the N-S pairs from the first session and one pair from the E-W pairs from the first session. The teams are of course not real (their scores are not summed) and are only teams in the sense of Howell's dream. But in this movement,

*both*pairs actually play bridge (unlike in non-interlaced Howells) so there are no bye-stands – all tables are in actual use.

You can also run this elegant two-session movement with 20 pairs, 14 pairs and 8 pairs, although in practice the latter is not done.

How is the initial table assignment determined? The movement and everything else makes perfect sense, but there is no mention of why you start the teams at the tables that they start at.

ReplyDeleteThe initial table assignment is arbitrary and often done differently (that's why Howell movement guide cards don't always agree). But typically, the first round is set up as if it were a Mitchell with the appropriate phantom pairs.

ReplyDelete