1 club can be 3+ cards in club suit. In this particular case, we can assume East typically has no 5 card major and fill a figure in Spade or Heart. The important thing is that as long as the opponent has a bid, a corresponding number will appear in a position in the cross-number table image.

Rather than trying to do the mental math on 4 rows and 4 columns, most bridge teachers and experts recommend starting with the number cards of 'missing ' in each suit and mentally noting the most likey distribution of those cards; e.g. 5 unseen cards in a suit will be lying 3-2 (3 in one hidden hand and 2 in the other). Obviously adjustments are made based on bidding and leads. Then as tricks are played in that suit, if both hidden hands follow, you note that the remaining 3 are likely 2-1. With this method only 4 pairs of small numbers need be committed to memory and repetitive subtracting from 13 needs not be done. When a less likely distribution is revealed (one of the hidden hands does not follow suit) the location of all the remaining cards are revealed and the distribution of that suit across all 4 hands becomes known. If at any point in the play the shape of one hidden hand is determined by these methods, the fourth hand also becomes completely known. (note that this requires learning the exact count of only 3 suits in that hand)

The key point is the cross positioning, which determines the hand pattern and card missing in suits.

This is a good question. In this game, the declarer calculates the HCP of the opponents after LHO's opening lead. The opponents have 15 HCPs (40 - 25). Since LHO made 1 spade opening bid. LHO has 11 to 15 HCP. The HCPs possible distributions for RHO and LHO are listed below. ---------------------- LHO RHO ---------------------- 11 4 12 3 13 2 14 1 15 0 ---------------------- Ace is 4 HCP. So there is 20% probability for RHO has club Ace. In another word, the declarer has 80% probability to win the contract by using the plan.

Yes, you're right.