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BACKGAMMON PROBLEM
OF THE WEEK - ARCHIVE
Problem #1, Black to play 5-2
24 23 22 21 20 19 18 17 16 15 14 13
1). 23-21-16
2). 13-8-6
3). 13-8, 23-21
4). 13-8, 13-11
1 2 3 4 5 6 7 8 9 10 11 1 2
Discussion
Problem of the Week #1, Black to play 5-2
Black has the choice of escaping a back man, advancing and attacking White from
his midpoint, or doing a little of each.
Playing the five from the midpoint (13) down to the 8 point gives Black a direct 4
shot if White gets a man in from the bar on his next roll. Black chances for a
gammon are excellent provided White does not secure an advanced anchor on
the 4 point. So 13-8 for the 5 is a must move. Now what about the 2?
Continuing from the 8 point to the 6 point with the deuce changes Black “hit”
shot roll from a “4” to a “2” if White comes in. The probability of getting a 2 is 12
rolls (all 2’s (11 different two dice combinations) plus double 1’s). The probability
of rolling a 4 is 15 rolls (all 4’s (11 combinations) plus 3-1, 1-3, double 2’s and
doubles 1’s). Fifteen ways to hit White is better than twelve ways to hit White so
we can eliminate Choice #2.
Choice 1 uses the entire roll to escape one back man from White’s 22 point. Black
wants to gammon White so he must use 5 to move a man to the 8 point. Choice 1
is now ruled out. The remaining problem is what to do with the 2 move?
Moving the deuce from the midpoint (13) to the 11 point only provides four
indirect rolls to hit White on the 4 point should he come in (6-2, 2-6, 5-3, 3-5).
That is not a powerful attacking play. On the other side, playing 23-21 with the
deuce is great for escaping. Now 6-6, 5-5, and 6-5 are great rolls for Black rather
than mini-disasters were Black to keep both his men on White’s 23 point.
The BEST play is Option 3. Attack White by moving from the midpoint to the 8
point. Start your escape by moving from the 23 point to the 21 point.
Congratulations if you chose this option. You are playing winning backgammon.
Problem #2, Black is on roll. What is the best play?
24 23 22 21 20 19 18 17 16 15 14 13
1 2 3 4 5 6 7 8 9 10 11 12
Problem of the Week #2, Black is on roll. What is the best play?
This end game position occurred in a $1.00/point money game recently played between two members of our club. It was a very exciting game with the advantage swinging back and forth with the rolls of the dice. When both players started bearing off, White was decisively ahead. Two moves prior to the position displayed , Black had eight men left to bear off while White was down to three men left to bear off to win the game.
For Black’s previous move, he rolled double 4’s and removed two men from his 4-point and two men from his 2-point. That double 4 roll left him with four men on his ace point. White’s previous roll was 5-2 which resulted in the removal of only one man and leaving two men remaining on his 3-point.
Incredibly, Black realized he had a possibility to win this game that seemed hopelessly lost just two moves before. Seeing White leaving two men on his 3- point, Black immediately redoubled the Cube to 16! White took some time to weigh his decision on whether to Take or Drop the Cube.
White knew that Black would win immediately with the roll of any double. There are exactly 6 rolls of the dice that would give Black an immediate win: 1-1, 2-2, 3-3, 4-4, 5-5, and 6-6. The other 30 possible rolls would enable Black to bear off two of the four men remaining. The probability was 83% (30/36) that Black would leave two men remaining. Consider that EVENT #1.
So now White calculated the number of rolls which would result in him leaving one man on his board. White knew if he rolled a single “1” , double 1’s or a single “2” that he would lose the game. How many “bad” rolls of the dice contain a “1” or a “2”? The answer is 19 rolls (1-1, 1-2, 2-1, 1-3, 3-1, 1-4, 4-1, 1-5, 5-1, 1-6, 6-1, 2-3, 3-2, 2-4, 4-2, 2-5, 5-2, 2-6, 6-2). Only 2-2 would get both men off! That means White has a probability of 47% (17/36) of making one of the 17 winning rolls. Consider that EVENT #2.
So for White to win, BOTH Event #1 and Event #2 must occur. That means White’s probability of winning Event # 1 x White’s probability of winning Event # 2 is 83% x 47% = 39%. White correctly accepted the double even though his odds of winning were less than 50%. Let’s analyze the monetary outcomes of a “Take” or a “Drop” of this redouble over 100 matches.
If White drops, he immediately loses $8 x 100 matches which is $800. If White accepts the Cube, he loses $16 x 100 matches x 61% = $976. However, White will win $16 in 39% of the 100 matches which equals $624. Over 100 matches, by accepting the redouble to $16, White will lose only $976 - $624 = $352 as opposed to $800 if White drops every time.
It is hard to accept a Cube redoubled to 16 in this position. However, the correct decision is clearly to take the redouble to 16 rather than drop the Cube. Black was correct to redouble to 16 and White was correct to Take the Cube. Congratulations if you got the right answers. You are well on your way to proper Cube management and understanding Cube equity.
You ask what happened in the actual game? Black redoubled to 16. White took the Cube. Black rolled double 5’s and ended the game to yells, screams and expletives from kibitzers who were rooting for both players. That game epitomized the ecstasy and heartbreak of this grand game of backgammon that all of us know and love all too well!
Discussion
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