Teaching MiniBridge - for the Mathematicians!

Only for those with an arithmetic bent, you can explain how, as declarer, you can calculate your likely number of tricks in a suit headed by the top honour cards.
This is how it is done:
+	Add together the number of cards in the suit held by the partnership
-	Take this total from 13 to give the number of cards held by the opponents. If this remainder is even, divide it by 2 to give a number x. If the remainder is odd, add 1 then divide by 2 to get x.
*	If there is an even break, as is reasonably likely, the remaining cards will split x : x between the two hands, or x : x-1 if they have an odd number.
/	If you have at least x top honours in the suit, you should be able to win all the available tricks in this suit.
For example: You hold AQ975 opposite K63.
	You have 8 cards in this suit, so the remainder when subtracted from 13 is 5. Add one and divide by two to give x = 3. So with a good break the suit will break 3 : 2. You have the three top honours in the suit, so should be able to take all the available tricks – in this case 5, as that is the length of the longer holding. Of course suits don’t always break evenly, but that is part of the fun of bridge – you don’t know for sure what will happen.

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This is how it is done: