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Appendix A is a glossary. I've included all terminology used in this book. I've included none of the jargon you and your little friends use amongst yourselves.

Appendices B and C are sets of output from a bidding simulation program. (Appendix F contains larger sets available for download or browsing.) The primary goal of the program was to estimate the probabilities of overcoming deficits of varying magnitude. (In other words, how often a team could outscore the opposing team by 10 points, or by 20 points, etc.)

The first step was writing a single-hand analyzer. The analyzer is comprised of two algorithms: one algorithm estimates the number of tricks a hand is expected to take, and the other estimates the probability of the hand succeeding as a nil bid. These algorithms are similar to those described in Chapter 3.

The single-hand program bids the hands in a vacuum; it makes no adjustments for position, tournament score, bids of other players, etc. The four-hand analyzer program, however, deals the entire deck to four simulated players, each of which analyzes their hand the same way as the single-hand simulator, then makes adjustments in reaction to their position and the other bids, similarly to the adjustments recommended in Chapter 2.

When determining the probability of various outcomes, one large assumption is made: that non-nil bids are never broken. I agree this is unrealistic during normal play. Outcome differences of 110, 120 or 130 are expected when one team gets set.

However, the main function of the simulator is to estimate win/loss probabilities in the end game. I would argue that the frequency of a team overcoming a 30 point deficit to win the game should NOT include those times when they've set the other team, because the fact they set the other team is irrelevant. The only thing that's relevant is that they took 3 more tricks than their opponents, or otherwise outscored them by 30 points.

For example, let's say West-East needs to overcome a 30-point deficit. If West-East makes a team bid of 6, North-South is not going to bid more than 4, even if under normal circumstances they might bid more. In other words, the only way East-West is going to win the game is if they receive a hand that score 30 more points than their opponents. The output reveals approximately how often this will happen.

Here's an example from the single-player output:



0.6000   1.20  Q42s 
0.1200   1.00  AJ7h 
0.9300   1.70  AK732c 
0.6000   0.30  Q6d 
0.0402   4.20  Total

NilEV: -45.9824  Bid: 4

If this hand were bid as a nil:

its spades would be covered approximately 60% of the time
its hearts would be covered approximately 12% of the time
its clubs would be covered approximately 93% of the time
its diamonds would be covered approximately 60% of the time

The product of these probabilities, 4.02%, is the approximate probability of the entire hand being covered. You can expect to lose an average of 45.89 points by bidding nil.

If this hand were not bid as a nil:

its spades would take an average of 1.2 tricks
its hearts would take an average of 1 trick
its clubs would take an average of 1.7 tricks
its diamonds would take an average of 0.3 tricks

The sum of these estimates is 4.20. The program would choose to bid 4 with this hand. If the NilEV were higher than ten times the bid, then program would bid nil.


The first section of Appendix B contains details about each round of the simulation. The second section is a list of approximately how frequently each single-player bid is made. The third section is a list of how frequently a nil bid has a success percentage of 100, 95, etc. The fourth section is a list of how frequently each team bid is made. The fifth section is a list outcome frequencies. For example, the entry:


-5     0.0345     0.0939     0.9061

...means that there is a 3.26% chance the first team will be outscored by exactly 50 points on a given round (not counting bags); there is a 9.39% chance the first team will be outscored by 50 points or more; and there is a 90.61% chance the first team will be outscored by 40 points or fewer.

This entry:


  3     0.0689     0.8592     0.1408
...means that there is a 6.73% chance the first team will outscore the second team by exactly 30 points on a given round; there is a 85.92% chance the first team will outscore the second team by 30 points or fewer (or will be outscored by the second team); and there is a 14.08% chance the first team will outscore the second team by 40 points or more.


Part I of Appendix D contains tables of probabilities of another player having a certain number of a suit given that you have a certain number of that suit. By referring to this table...

3 of a suit
 
  exactly at most at least
       
0 0.0084 0.0084 1.0000
1 0.0639 0.0722 0.9916
2 0.1917 0.2639 0.9278
3 0.2959 0.5598 0.7361
4 0.2589 0.8188 0.4402
5 0.1332 0.9519 0.1812
6 0.0404 0.9923 0.0481
7 0.0070 0.9993 0.0077
8 0.0007 1.0000 0.0007
9 0.0000 1.0000 0.0000
10 0.0000 1.0000 0.0000
11 0.0000 1.0000 0.0000
12 0.0000 1.0000 0.0000

...you can see that if you have three of a particular suit, the probability of a particular other player (for example, your partner) having exactly four of the same suit is 0.2959. The probability of a particular player having AT LEAST four of the same suit is 0.4402. The probability of a particular player having AT MOST four of the same suit is 0.8188.

Note that this chart does not tell you, for example, the probability of ALL players having at least three suits, or of ANY player having at most three suits. It only gives probabilities with regards to a specific other player.

In order to determine the probability of ANY playing having at most three suits, you can rely on Part II of Appendix D. Part II contains tables of probabilities of a certain distribution among all players given that you have a certain number of that suit. By referring to this table...

If you have 8 of a suit
5 0 0 0.0022
4 1 0 0.0161
3 2 0 0.0387
3 1 1 0.0839
2 2 1 0.1374

...you can see that if you have eight of a suit, the probability that your LHO has three of that suit, your partner has one of that suit, and your RHO has one of that suit is 0.0839. (In other words, a distribution of 8-3-1-1 given that you have 8.) 0.0839 is also the probability of a distribution of 8-1-3-1, and also of 8-1-1-3.

Let's say you have 8 of a suit, and you want to know the probability of both of your opponents having at least two in that suit. You'll want to sum the probabilities of these distributions:


8-3-0-2		0.0387
8-2-0-3		0.0387
8-2-1-2		0.1374

       = 0.2148.
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