Welcome toChinese Net Bridge
Add to Favorites | 中文版
Position:Home>Entry bridge>
Computational card mound
From;    Author:Stand originally

Bright hand: K3
Banker: 62
In this design and color banker has 2 to be defeated by mound. Although bright hand has K, but it can win 1 mound to will rely on A of hold of which card hand, so, should go up in this design and color now plan be defeated by mound for 2.
Bright hand: 432
Banker: Q65
This design and color of banker should plan be defeated by mound for 3. Really, the Q of banker can win 1 cluster possibly, but the positional place that this must rely on AK of the other side again. This kind of case compared with on exemple come more win cluster not easily.

Not complete join piece

Sometimes, your one party lacks a piece in a design and color or card of a few magnify, hold does not join one set completely piece structure, right now, when computation is defeated by mound must scrupulous. Consider the following case:
Bright hand: 432
Banker: AQJ
You have a ——K is defeated by mound, the likelihood in the following card games can catch it, eliminate this to be defeated by mound thereby. But engrave here plan be defeated by mound for 1.
Bright hand: KJ10
Banker: 543
In A, K, Q, J, 10, you lacked A, Q in 5 magnify card two pieces, 2 are defeated by mound.

Bright hand: KJ4
Banker: 532
3 pieces of ——A, Q is lacked in 5 magnify card, 10, likelihood complete play away of 3 mound card, answer so plan be defeated by mound for 3.
Bright hand: AQ10
Banker: 432
A can get 1 mound for certain, but the other side has KJ, you just may want play away 2 mound, answer so plan be defeated by mound for 2.
Other combines a form
Sometimes, when computation is defeated by mound you need will big shop sign and length union to rise consideration. Look please below this example:

Bright hand: KQ76
Banker: 5432
In this design and color, you just lack A, J. 10, 9, 8, 5 pieces of cards, you have how many to be defeated by mound to depend on actually which hostile have A and this 5 pieces of cards the allocation in hand of hand of card of two the other side circumstance. If they are 3 - 2 allocate (a the other side holds 3 pieces, another the other side holds 2 pieces) , you are defeated by 2 mound at most, be defeated by 1 mound possibly also only. If they are 4-1 allocation; You are likely play away 2 mound, it is 3 mound even. So, when hold of your one party when afore-mentioned combination, should estimate commonly be defeated by mound for 2.
The person that just began to learn bridge as will tell, wanting the shop sign that composes to this kind to have a correct estimation is more difficult. Become gradually familiar to various combination when you after rising, you will discover estimation is defeated by mound is an easy thing.

Of computational pack of cards be defeated by mound

Hitting a pair to have before will entering into the agreement, should calculate first be defeated by mound. This pair of card is 4S enters into the agreement below, allow you to together 3 are defeated by mound only. Calculate how many is defeated by mound together?
上一页 1 2 3 4 56 7 下一页
About us | Legal Notices | Sitemap | Links | Partner