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Reservations will be accepted for NS or EW positions when received by email no later than midnight the night before the game. A confirming email will be sent the next morning. Any players making such a reservation but not arriving will have this privilege rescinded

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SITE ADDRESSES

MONDAY & WEDNESDAY

FRIDAY 12:30 Centerville Masonic Lodge

SATURDAY 12:15 MASHPEE COA Open Pairs and 299'er

Margie Sullivan was awarded the Bermuda National Tourism Award for 22 years of arranging tours to Bermuda for the ACBL Regional there!  Congrats, Margie!

UCBC SMOKING POLICY  

PLAYER MUST PLAY THEIR OWN DUMMY CARDS PRIOR TO EXITING TO SMOKE.  

IF THE ROUND IS CALLED AND 3 PLAYERS AWAIT THE RETURN OF THE FOURTH WHO IS OUTSIDE SMOKING THERE WILL BE A 3 MATCHPOINT PENALTY ASSESSED WHICH WILL DOUBLE ON EACH SUBSEQUENT INFRACTION

The Theory of Expected Holding

by Ken Camilleis

Many of the bids you make are a priori, which means without any knowledge of the contents of any of the other three hands at the table. Bridge is a game that involves taking calculated risks based on statistical probabilities of bad results. If a tactical action produces, say, fifteen good results and five poor ones, over the span of ten duplicate sessions, then that treatment is heavily favored. Some situations dictate a unilateral action where “you pays your money and you takes your chances.”  Such bidding is based on the theory of expected holding (TEH.)

Preempts are a classic example of applying the TEH. Say you're in first seat white vs. red and you hold QTxxxx, xx, KJx, xx. There's actually a specific holding that you can “expect” partner (or either of the opponents) to have!  Of course it will not be that exact holding, but the statistically most likely occurrence. Let's start with the spades. You hold six of them, which means there are seven you don't hold. These seven spades are divided some way among the three other players, so by dividing 7 by 3 you get 2-1/3, which means you “expect” partner to hold 2-1/3 spades. So you round this down to two, which means you likely have an 8-card spade fit.  Now count the high card points in the spade suit. You have only 2, which means there are 8 HCP in the A, K and J distributed among the other three hands.  Divide 8 by 3 and you get 2-2/3 so you “expect” partner to hold 2-2/3 HCP which rounds up to 3.  So guess what … partner's most likely holding in spades, assuming a nonbiased random shuffle, is Kx!  Now apply this reasoning to the other three suits and you map partner's “expected” hand to be the likes of Kx, QJxx, Qxx, Kxxx. As you are missing 34 HCP, diving 34 by 3 leaves 11-1/3 HCP for your partner which may include an ace and one less king and/or queen. So the opponents probably don't have game.  Do you pass as dealer?

If you pass, your LHO may be in position to open the bidding, and by the time it comes back to you both opponents may have exchanged a wealth of information and it may not be profitable to enter the auction at that point. However, if you open with 2S, three people have to start their communications at the three level or with a takeout double that will likely let you off the hook. You are taking a two-to-one (66.67%) chance that the one person who will be inconvenienced by your preempt isn't your partner. If you catch partner with void, AKJx, Axxxx, KQTx once and something in the neighborhood of KJxx, x, Qxxxx, xxx or KJ, Axx, AQxx, KQxx two other times, you get two matchpoints for every one you give. Even the first hand may not be so bad if partner holds his nose and passes as 2S may make if the spades break 4-3. The logic behind TEH also applies to the tough decision an opponent needs to make over a direct preempt when neither his LHO nor partner have made a call.

The TEH can also be extended to defense, especially with regard to making an opening lead when the opponents have exchanged little information (such as 1S-2S-4S or 1NT-3NT.) Say your long suit is QTxxxx … you know what to expect from pard – Kx!  If the lead goes into declarer's AKJ, that's life, but application of the TEH over the long term may produce many more good results than bad ones.

Top Percentage UCBC Games

The Midrank Method of Nonparametric Statistical Sampling

by Ken Camilleis

As duplicate bridge players, did you know that the process of “matchpointing” scores on a given board has a scientific name for it?  This process is referred to as the midrank method of statistical sampling. Here's how it works:

Let n = the number of times a board is played.  When ranking scores that are all distinct, the highest score receives n-1 matchpoints and the lowest score gets zero matchpoints, and every other score tabulated would be ranked according to its relative algebraic position. However, what happens when two (or more) scores are the same?  In that case, one half-matchpoint will be assigned for each score that the score in calculation ties.  Say a deal is played six times and the results are +140, +110, +110, +100, -50 and -100.  The +140 would get 5 matchpoints and the +100 would get 2 matchpoints but the two scores that tie at +110 would receive the midrank (the midpoint value) between 2 and 5 which is 3.5.  

If more than two scores tie, the calculated score gets 1/2 matchpoint for each score it ties. Suppose a board is played 13 times and 12 of the scores are +650 and the other is 680. The 680 gets 12 matchpoints and all the 650s get 5.5, which is 11/2, the midrank of the remaining 12 scores of 650.

Every now and then, the same result occurs across an entire board, in which case everyone who played it would be awarded (n-1)/2 matchpoints, which is the exact average score for the board. If a deal were played nine times, and everyone achieved the same score, the midrank value of 4 matchpoints would be assigned to all pairs.

Probability of getting dealt a hand with all of the same suit 

by Ken Camilleis

The number of combinations of n items taken r at at time is denoted nCr, calculated as follows:  nCr = n! / ((n-r)! r!)

A bridge deal consists of the process of taking combinations of 52 cards 13 at a time and distributing them around the table. So let n = 52 and r = 13. We have:

52C13 = 52!/39!13! = 52*51* ...*40 / 13! =

3954242643911239680000 / 6227020800 = 635013559600

13/52 * 12/51 * ... * 1/40 = the probability of all of any particular suit (spades, hearts, diamonds or clubs) = 1 in 635,013,559,600

Divide this by four to get the probabilty of this for any suit, which is 1 in

158,753,389,900

That's one hundred fifty-eight billion, seven hundred fifty-three million, three hundred eighty-nine thousand nine hundred