You need to guess the demolition
Consider 3 examples when you need to guess the demolition of the player.
Example 1
1st hand |
2nd hand |
3rd hand |
|
♠ |
DV |
7810T |
KB9 |
♣ |
AT 8 |
710D |
TV9 |
♦ |
108 |
79D |
TKB |
♥ |
ДВ108 |
78 |
TO |
In this situation, the player must clearly demolish the Awer and one of the ladies.
The first move must be made from clubs or diamonds, for example, the move is made from a tambourine.
The player plays a small hand, the third hand beats with an ace. Now the move is passed in the spades to the first hand and again the move to the tambourine. It becomes clear that if in a demolition the diamond lady, then the miner is no longer caught - there is not enough gear on the right hand.
So, you need to decide before the first move which lady to catch. If the demolition is not guessed, then the miner will be won.
Example 2
1st hand | 2nd hand | 3rd hand | |
♠ | TD | 78910BC | - |
♣ | K98 | 710D | TV |
♦ | 108 | 710 | TKDV8 |
♥ | KB98 | T | D107 |
In the demolition of a hearts ace and a lady's queen, or a diamond ten.
If the demolition is guessed, then the player will have two bribes, otherwise there will be one.
The first 2 moves are made from the peak, with the third hand put two clubs.
The third move with a small club.
The player puts the top ten and from what kind of demolition, gets one (in the demolition of the diamond ten) or two (in the demolition of a lady) bribes.
Naturally, the player, trying to involve players in a more gambling game, can skip the club by putting the seven.
But here you can do it cleverly - to repeat the course with a small club.
Then the player again receives one or two bribes on the same conditions.
Example 3
1st hand | 2nd hand | 3rd hand | |
♠ | D | 7810T | KB9 |
♣ | K8 | 710D | TV9 |
♦ | 108 | 79D | TKB |
♥ | TDV109 | 78 | TO |
In the demolition there are either two ladies, or a lady and a dozen dozen.
The first move is made from the treble eight.
Already here the game can end if the player puts the top ten.
Suppose that the player puts the trefoil seven and the third hand beats with an ace.
Then assuming that in the demolition of two clubs, we pass the move to the first hand in the peaks and make a move from the tambourine.
The player puts a small, and the third hand beats an ace.
Next move with the king of hearts.
The first hand takes a move on the ace and makes a move in the heart.
It would be a gross mistake not to repeat the move, since having a retreat in this suit, the player could pass without any problems and the second tambourine.
On the second move, on the hearts of the third hand, a peak comes down (to increase the number of bribes, if the player goes to the "collective").
Now the decisive move in the tambourine.
Playing or beating and receiving one bribe, or puts a seven.
In the latter case, the third hand beats the king, makes a move to the peak, which allows her to take the king's first hand from the first hand and make a move in the right suit.
If he suddenly guesses, he will take 3 bribes, otherwise he will be lucky.
The probability that you can guess from the two suits you need is 50%.
This means that one of two similar mizers passes, and the other can be caught in 3 bribes.
Hence from here you can find its average cost: (-300 + 100): 2 = -100.
Hence, from the point of view of probabilities, the variant of the "collective" and the variant with the voluntary taking of one bribe are equivalent, which gives the player to choose any continuation of the game.
Judging from practice, more often players agree to one bribe.
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