endgame

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spleischft

Sep 8, 1999, 3:00:00 AM9/8/99
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How to calculate you opponents chances of winning the game in situations
like this:
You are about to bear off, and you opponent has ONE checker on the bar and
maybe he has borne some of his checkers off too.

The End-game-spleischft formula:

conditions:
1) Your opponent is on the bar (only one checker)
2) You have a closed inner table with spare checkers on 4,5 and 6.
3) your opponents innertable-checkers are smoothly distributed.
5) Claus Thomsen is a killer on FIBS.

If your opponent has borne x checkers off, his chance of winning the game
is:

************************************************
if x is between 0 and 7:
opponent wins 2*x*(x+1)/3+5 out of 100 games.
if x is between 8 and 14:
opponent wins 6*(x+1) out of 100 games.
************************************************
examples:
+24-23-22-21-20-19-+---+18-17-16-15-14-13-+
| O O O O | | |
| O O O O | | |
| O O | | |
| | | |
| | | |
| | | |
| | | |
| | O | |
| X X X | | |
| X X X X X X | | |
| X X X X X X | | |
+-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+

the conditions is ok so: how many checkers has white borne off? -> 4
4*5=20 20*2=40 40/3=13 13+5=18
opponent wins 18% of all games

Example 2
+24-23-22-21-20-19-+---+18-17-16-15-14-13-+
| O O | | |
| O O | | |
| O O | | |
| O | | |
| O | | |
| | | |
| | | |
| | O | |
| X X X | | |
| X X X X X X | | |
| X X X X X X | | |
+-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+

White has borne 6 pieces off.
6*7=42 42*2=84 84/3=28 28+5=33
BUT whites checkers is not smoothly distributed! the two checkers on 24
should have been on 22 - but since white dont need any extra rolls because
of this we will not give him extra %'s.
opponent wins: 33%

The hard one :-)
+24-23-22-21-20-19-+---+18-17-16-15-14-13-+
| O O O | | |
| O O O | | |
| O | | |
| | | |
| | | |
| | | |
| | | |
| | O | |
| X X X | | |
| X X X X X X | | |
| X X X X X X | | |
+-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
white has borne 7 checkers off but white doesn't have a smooth distribution
of checkers :-(((
- We will have to create a equivalent-diagram. How? Lets say O would need
one or two more rolls now than if his checkers was smoothly distributed. So
I will add two checkers to his innertable and distribute his checkers.

new equivalent diagram
+24-23-22-21-20-19-+---+18-17-16-15-14-13-+
| O O O | | |
| O O O | | |
| O O O | | |
| | | |
| | | |
| | | |
| | | |
| | O | |
| X X X | | |
| X X X X X X | | |
| X X X X X X | | |
+-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
now: white has 5 checkers borne off.
5*6=30 30*2=60 60/3=20 20+5=25

but the players checkers is placed worse than in the standard condition
(spare checkers on 4-5-6) so we will give white 9%
opponent wins: 25% + 9% = 34%

In the last diagram I gave white 9% and here is why:

this *NEW* table tells us the how many %'s we should give the opponent if
our spare checkers is on 1-2-3 instead of 4-5-6:

opponent has again borne (X) checkers off

the BLUWDUCH table:
--------------------------
1, 1%
2, 2%
3, 3%
4, 5%
.
9, 9% (here is x=9 and we see that we need to add 9% :-D)
.
14,5%
interpolate between 4 and 9. ANd interpolate between 9 and 14

Final example to show the use of the formula for x between 9 and 14:
+24-23-22-21-20-19-+---+18-17-16-15-14-13-+
| O O | | |
| O O | | |
| | | |
| | | |
| | | |
| | | |
| | | |
| | O | |
| X X X | | |
| X X X X X X | | |
| X X X X X X | | |
+-1--2--3--4--5--6-+---+-7--8--9-10-11-12-+
white has borne 10 checkers off. x is now greater than 7, so we will use the
other formula
6*(10+1)=6*11=66 (6*(X+1))
opponent wins: 66%

What do you think?

__ : this thing sucks.
__ : power to the spleischft and bluwduch formula.