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Killing two birds with one stone

Here is a counting method I developed, which I have found especially useful and use all the time. It is probably recorded in some books too.

It counts the number of ways to kill two birds with one stone (well, in fact, with two dice), for instance enter and hit, enter and cover, hit and cover. It counts only moves played by two different checkers, e.g. it does not count as "enter and hit" a move that enters a checker and hits an indirect shot with the same checker.

Let n be the total number of dice that kill at least one bird. Let a be the number of dice that kill bird one, but not bird two. Let b be the number of dice that kill bird two, but not bird one. Then the number of rolls that kill both birds is

n^2 - a^2 - b^2

Here is an example from one of my games - it was not hard to find since such positions happen very often. How many hits for White (on the bar) ?

is nabla score: 14 pip: 146

15 point match

pip: 151 score: 12 is hamkachan

XGID=--B--aD-D-a-cBb--c-cBb---A:1:-1:1:55:12:14:0:15:10

to play 55

eXtreme Gammon Version: 2.00

White (my opponent) enters ("bird one") with a 1, 2, 3 or 5, and hits ("bird two") with a 1 or a 3, so we have n = 4 (dice 1,2,3,5) ; a = 2 (dice 2 and 5) ; b = 0 (no die that hits but does not enter). So the number of enter-and-hits is 4^2 - 2^2 - 0^2 = 16-4 = 12. Yes, it is that simple.

Err, and of course my opponent rolled a joker double 5 and hit both my blots by playing all the way with the entered checker. So the real number of hits was not 12, but 13. But no formula will take care of such long shots, you will still have to spot them by yourself.

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