The cost of a single game

What is the cost of a single game? During the game, each player will have to constantly choose between the game and raspasovki, and if the game is already assigned, the choice has between whist and pass. In this situation, the player must have some experience, or landmark. In our situation, this guide is the cost of a single game.
Thus, the cost of a single game will mean a change in the overall outcome (win or loss) of any player as a result of the game. This cost can easily determine if the paint bullet twice - before and after the end of the game in question. But this is not necessary, as there are common formula for calculating the cost of a single game for all players.

To Play: C (s) = 3 / 4P - 3 / 4SH (s) + 1 / 4SH (in) - B - P / 7 /.

C (i) - is the cost of playing the game for 3/4 - three-quarters of the prize, 3/4 - three-quarters of a fine play, 1 / 4SH (c) - one-fourth penalty Vista B - Vista Ave - bribes widow.

For VISTA: C (s) = -1 / 1 + 4P / 4SH (s) - 3 / 4SH (c) + 1 / 4SH (B2) + B

C (c) - the cost of games for Vista, -1 / 4P - one quarter of the prize, 1 / 4SH (s) - a fine player, 3 / 4SH (in) - three-quarters of its own fine if shortfall bribes 1 / 4SH (c2) - one-fourth penalty of the second whist at his failure, The - on the playing whist.

For dealer: C (s) = -1 / 1 + 4P / 4SH (s) + 1 / 4SH (2c) + Pr + K

C (c) - the cost of the game dealer, -1 / 4P - one quarter of the prize, 1 / 4SH (s) - one-quarter of fine plays, 1 / 4SH (2c) - one-fourth penalty Vista Ave - additional whist for taking bribes from a widow K - Consolation.
Knowing the formula calculation of P, W (i), W (c), B, and C can be, depending on the "k" - the number of gleaning or replayed bribes playing and "c" - the amount of bribes taken concrete Vista, determine the value of any single games for any player.
Let's take for an example, consider a situation where the player took bribes as much as assigned (k = 0).
The result of the game will be:
F (u) = 3 / 4x20 (n-5) - 2 (n-5) (10-n) - 2 (n-5) = Pr (n-5) (2N-5-2np)
Balance VISTA will:
B (s) = -1 / 4x20 (n-5) + 2 (n-5) = (n-5) (2c-5)
Balance deliverer will:
B (s) = -1 / 4x20 (n-5) + 2 (n-5) = Pr (n-5) (5-2np)
When Misery:
P (i) = 75, B (s) = D (c) = -25

Summarizes the results of playing, whist balance deliverer and minuscule balance in the table, below.

From the table it can be seen that the gain depends on the playing of the game and the amount of bribes from the widow (Pr), for which the deliverer wrote whist. The smallest gain (1 whist) obtained by sixfold the game, when the community cards were two aces. The greatest gain (75 whist) obtained by desyaternoy game and there is no explicit bribes in community cards, as well as minuscule.
When VISTA two players, the value of the loss depends on the amount of bribes taken (with gears and semernoy game can also have a positive balance). Let us assume that at semernoy play one of two VISTA took bribes. Based on the table, he lost 2 whist. It took one second vist bribe and lost 6 whist. Setter loses as much as whist not having taken bribes. Deliverers like playing loses or wins depending on the number of games and whist for bribes, which are determined to buy.
The calculations for cases where "k" is not equal to 0 will not lead, and the corresponding calculations in the tables show that below.

k <0 (remise)

The table shows that the value of losing the player at Remizov depends on the rank of the game and the number of gleaning bribes. Deliverer Winning is determined by the same factors. The presence of bribes in the Widow worsens in favor of the deliverer. Additional winnings VISTA determined by the number of bribes taken by them. For example, bribery was not to buy. Playing appointed sixfold game and remained without the two. Total 6 he gave bribes. The first vist was two bribes, the second 4 bribes were whist. The table shows that the loss was 54 playing whist. First whist whist won 18, and the second whist whist won 22, at deliverer gain estimated at 14 whist.

"a"> 0 (over-fulfillment of the order)

One of the players whist, and other passes

VISTA two

Consider the table, we see that when playing the over-fulfillment of the order basic loss falls on VISTA does not recruit the necessary number of bribes. For example, talon contains marriage playing appointed sixfold game and took bribes anymore. Vistovali two players, which respectively have two and a bribe. The player won 9.5 whist, whist won the first 1.5 whist, and the second lost 10.5 whist, whist deliverer lost 0.5.

Care "for their"

From the table we can see that the care "for their" at sixfold game is not very much and a bad decision. Losing accepting it contains one whist.

Two pass

If raspasovki balance will be playing:
B (i) = -3 / 4h10v + 1 / 4x10 (10-c) - 1/25 = 4h10ch - 10c - 2,5ch
Not who took bribes:
B (0) = 1 / 4h100-1 / 4x10 (h-1) + 7.5 = 35 - 2,5ch
h - the total number of players who did not take bribes.
The results of calculations of balance playing and who took bribes are given in the tables, below. In (h) - means the amount of bribes taken and the number of players who took no bribes.

(h) - means the number of bribes taken and the number of players who took no bribes.

Consider an example.

When raspasovki deliverer did not take any bribes, and according to the table above that won 32.5 whist in connection with which the balance and the bribes were distributed as follows:
1 hand (3 bribes) lost 7.5 whist;
2 arm (5 bribes) lost 27.5 whist;
3 arm (2 bribes) lost 2.5 vista.
Advantage to count the cost of the games that we have cited as an example on the page Example entries define.
1. Player A (playing) as a result of this game has won 22 of whist, whist and both lost 6 whist, whist deliverer lost 10.
2. The player won 30.5 whist, whist and setter respectively lost 37.5 and 7.5 whist, deliverer won by 4.5 vista widow.
3. The game was "not two" and lost 54 whist, whist whist won 20 each, deliverer won 14 whist.
4. Player G remained without bribes and lost 31 whist. Since deliverer back game (this row we did not), he owns a large part of the winnings - 21 whist. VISTA obtained only at the expense of its share of fine plays - every 5 whist.

Consider two examples that thanks to calculations that are shown in the tables above that, to find the best options for the player facing problems, especially at the stage of trade.

example 1

The first and second arm declare "pass". A third card hand contains 5 tricks on the game and no more than two bribes raspasovki. What to do in this case?
In this scenario, you need to pass. Buy more than one bribe is likely to fail, and this gives a gain no more than seven whist. There is a risk that did not manage to buy a bribe, then you can go without one, which could lead to the loss of 31 vista. If you assign raspasovku the prize 5 whist will be guaranteed (two bribes), and if the bribe is only one, the prize will increase to 15 whist.
In this situation (on the last hand) is more favorable to make a pass, having at hand the faithful 6 or 7 of bribes, as spending raspasovku without any bribes, a gain equal to the gain of the 8th game. Such raspasovku called artificial.
In the case where trade is already open and there is a very good whist, it is necessary to continue trading until his suit. Similarly, it is necessary to proceed, with five tricks in the first or the second arm.

example 2

Let's say one of the trading card and a good vistovaya 6 bribes when trump the worm, but the other player insists on not inferior, and has already announced "7 peak" what to do in this case?
If we continue to trade "7 of hearts," and in community cards are good cards, the prize will be equal to 18, and Vista (having good cards in the community cards it is unlikely), and failure will lose 58 and whist. If pasanut, and further open the game, it is very likely to put the opponent and without risk to win 22 vista. Even in the worst case (the game will be played) prize will be equal to two whist. Therefore, in this case it is better to refuse from further trading.

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