A small formula to simplify the bear off process in backgammon
A small formula to simplify the bear off process in backgammon |
I thought the game was over, but my opponent said that I had to move my 4 piece to the 1 slot, then take the 2 off with the 6 roll. She said I had to do this because I had the opportunity to move my piece. I've got $... riding on this, please respond." We have received many messages as the above one and decided to provide a useful formula for beginners. The formula is quite simple: "If the number of the dice that you have rolled is equal or bigger than the number of the row* that your checker** is located you bear off. If the dice number is smaller than the row number you must play that number by moving your checker towards rows with smaller number." Row*: the space, the slot, the field... that you play or place your checkers. 24 rows exist on the backgammon board. Checker**: the stone, the piece... that you play on the board. Each player has 15 checkers. The formula may seem difficult but don't worry, it's not, let me explain! To start to bear off you should enter all your checkers into your home area and in your home area you have 6 rows. Let's put a number to each of these 6 rows. The first one is "r1" which is also the row that your opponent's two checkers are placed on the beginning of the game. Then you have "r2", "r3", "r4", "r5" lined side by side and the last one is "r6" in which you place five checkers on the beginning. Now you know that the row number is a number between 1 and 6. Same of the dice number! Now suppose that you have one checker in row 5 and one checker in row 2 (better is to use your real backgammon board and simulate those possibilities), and you have rolled a "6 + 3", 6 is bigger than 5 and 3 is bigger than 2, and you bear off your two checkers! Suppose you have 2 checkers in "r4" and one checker in "r1", you have rolled a "4 + 3", you bear off one checker from "r4" with your dice number 4, and for dice number 3 you must move one checker from "r4" to "r1". You have 3 checkers in "r6" and 2 checkers in "r3", 4 checkers in "r2" and 3 checkers in "r1". You have rolled a "5 + 4". Unfortunately you will move 2 checkers from "r6" to "r2" and "r1", and you can not bear any checker off. You have 3 checkers in "r6", 3 in "r5", 2 in "r4", 4 in "r2" and 3 in "r1", and you have rolled a "double3", because your "r3" is empty you must move 2 checkers from "r6" to "r3" and bear 2 checkers off from "r3". You have 3 checkers in "r3", 4 checkers in "r2" and 5 checkers in "r1", you have rolled a "5 + 4". You will bear 2 checkers off from "r3". You have 2 checkers in "r6", 4 checkers in "r5", 1 checker in "r4", 3 checkers in "r3", 3 checkers in "r2" and 2 checkers in "r1". You have rolled a "double4". You will bear 1 checker off from "r4", move 2 checkers from "r6" to "r2" and 1 checker from "r5" to "r1". Only one checker has been born off. Maybe the worst position is when you have only two checkers remaining in "r2" and you roll one "1", for example a "3 + 1". While 3 is bigger than "r2", 1 is smaller, and you will bear one checker off but move the last one to "r1" and wait for the next hand. Same for "2 + 1" or "4 + 1" We hope you will find this useful. For any help send your question by email. |
