Declare or pass

Always before the first announcement there is a question what to do? Start trading or pass? The need to choose between trade and the cross appears in three cases:
a) The player is on the first hand;
b) a player has a second hand, and on the first hand have said "pass";
c) on the first and second arm said "pass".
If your hand has 6 true bribes or good (2-3 bribes) raspasovka, something that is not of selected problems. Therefore, there is a case before us, when the hand is not six or even five bribes, but raspasovki them may be a sufficient number. Thus, there is the need to choose a less comfortable situation. A more pleasant case of choosing between good and raspasovki good game can be just the third hand.
In order to make a choice, you need to compare the average expected outcome (COP) and raspasovki games.
Take into account that the right or wrong of our choice is highly dependent on the two cards that are in the community cards. How many options can be a widow? In fact, a little bit. This is the number of combinations of the twenty-two cards remaining as 10 cards in our hand. It turns out 231 option. Assuming that any combination of the community cards equally probable, we have to do the calculation procedure for all possible options for a widow (231 times). Further, by dividing the sum of the results by the number of implementations, we determine the average expected outcome of the game at a particular scenario, the hands, not knowing widow. Now if you compare the result with the average expected outcome raspasovki, we will get the answer to this question.
Try to understand with a simple example. From the tables of page value separate games see that the gain in sixfold game of whist 7 6 (a) = 7c and loss 6 (-1) = - 31V. Balance plays raspasovku is: B (2) = 5V, B (3) = - 5V, B (4) = - 15V, B (5) = - 25V. Based on these values, we can determine at what probability of success depending on the GRA raspasovki likely to start trading.
For this purpose, the COP (6)> PDS (R).
Consider the case when you need to take two raspasovki bribes.
We have: p (-31) + (1-p) 7> 5
where p - is likely to remain in the game without a sixfold.
Hence, the probability will be equal to 0.95.
In practice, it is much easier to compare the likelihood of not found the GRA, and the number of bribes raspasovki. Therefore, in the table below represent the data with this in mind.

Determination of the average value of the game and raspasovki be complicated. By further added that some of the factors of the game, which significantly affect the result, almost defy detail and formalization. Therefore, to argue for a long time not give all the data in the table prepared №1, which provides guidance on selecting the ad, depending on the arrangement and the hands to play. The lack of numbers indicating hand means that the recommendation applies to any of them.

Table 1

Supplement table combinations of figures, in which you should always declare "6 peak."
1. T ... T ... T ... T ...
2. TC ... T ... T ... ...
3. TC ... T ... CD ... ...
4. TC ... T ... ... KV10, ...
5. TC ... T ... Dhhh ...
6. TDV ... T ... T ... ...
7. TC ... TC ..., ..., ...

In conclusion, let us define prikupnoy suit. Under prikupnoy suit refers to the suit, adherence to which at least one card from the talon, provides an increase in the hands of bribes. Prikupanya ointment usually contains not less than three cards.
What might be probable prikupnuyu suit at least one card of the widow? The answer of this question is in Table 2.

table 2

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