I’d really like to learn to play Go, even though I have no one to play with. I’d also like to learn Backgammon – even though I have no one to play with. Rather than hijack my own Go thread, I’ll start this Backgammon thread for when I start looking into that game.
I have no idea what septimus is talking about.
I have no idea what you have no idea about, so I’ll cover the basics :).
The ‘coded’ part is a representation of a backgammon board. The Xs and Os represent the different coloured checkers. The bottom of the diagram is X’s side, and the 24 points are numbered from X’s point of view. X is trying to bring all his checkers round from point 24 to point 1 (well, point 0 really, i.e. past point 1 and off the board - if you get a checker to point 0, it is known as “bearing off” and that checker takes no further part in the game. The goal of the game is to bear off all your checkers before your opponent does so with theirs). We can see from the diagram that X is miles ahead in this game, as he has already borne off 12 of his 15 checkers, only 3 remain - one on his “6 point” and 2 on his “5 point”. O still has all 15 of his checkers in play, most round on his side of the board but 3 of them are way behind on X’s side of the board.
septimus is saying that it is X’s turn and he has rolled a 3 and a 1 with his 2 dice. What X would like to be able to do is move the checker from point 6 to point 5 with his roll of 1, and then say that he has no way of playing the 3 - because O ‘owns’ point 2, 3 steps away from point 5, by virtue of having more than one of his checkers on it. This means X is not allowed to move his checkers on to point 2. X would like to make this move because it leaves all three of his checkers safe from being landed on (“hit”) by O on O’s next turn. HOWEVER, the rules of backgammon are that you must use the numbers on both dice if possible (even if it could be disadvantageous to do so). Incidentally, I prefer this way of expressing it as it avoids the issue septimus is explaining.
Going by this rule, X has to play his 3 first, which means moving the checker on point 6 to point 3 (i.e. “6/3”). Then he must play the 1 - he can’t do it with the checker now on point 3, as it is blocked by O sitting on point 2, so he must move one of the checkers on point 5 to point 4 (“5/4”). This leaves him with one checker each on points 5, 4, and 3. As they are all single checkers, they are known as “blots”. And if O is able to land on one or more of them with his next roll (which he is able to do unless he rolls double-six, double-five, or a five and a six - commonly noted as “6-6”, “5-5”, and “6-5” respectively), he ‘hits’ them, which means his checker takes the place of the X checker, and the X checker is put on the bar (in the middle), which means it must begin again from point 24. This would swing the advantage back to O, because X could then only come off the bar with a roll containing a 1, since all the other points in what is called O’s “home board” (i.e. points 24-19 from X’s point of view) are blocked by O’s checkers.
Moving on, if X were to roll a six and a one in the position shown, he is NOT obliged to bear off the checker on point 6 (“6/0”) followed by moving 5/4 with the one. Instead, he can use the one first, moving 6/5, followed by moving 5/0 (because once you have no checkers further away from 0 than the number you have rolled, you can use that number to get your furthest checker to 0). So although X has not used all the spots on his dice in that scenario, he has legally used both dice - which is why I would express it that way rather than “using the full roll”.
Does that help at all? Any other questions?
OK, so X rolls a 6 and a 1. So X uses the 1 to move the checker from 6 to 5, so there are now three checkers on 5. Then X moves one of the 3 checkers on 5 off of the board, even though he has a 6 and it only takes 5 to get off. (Other games require the exact number.) This leaves X with 2 checkers on 5. Right?
So you’re not allowed to land on an opponent’s point if the point has 2 or more checkers on it. Are you allowed to cross a point that has 2 or more checkers on it?
Yes. And it’s probably clear, but you still have to consider each die separately. Say you have a checker on 6, and the opponent has 2 checkers each on 3 and 4.
| |
| o o |
|x o o |
|6--5--4--3--2--1|
If you roll a 2-3, you’re stuck. You can’t total it as 5 and move to the clear space on 1, because you’re blocked with both the 2 roll and 3 roll. A 4-1 roll, on the other hand, would allow you to move.
Weird.
Doesn’t seem weird to me, but I guess I’ve been playing backgammon for a long time. It’s just that when you roll a 2 and a 3 on two separate dice that means you have to make two separate moves: one of two, and then another of 3. You can’t just add them together. I feel like similar rules exist in other games using dice or similar, but I’m blanking.
No one ever plays frontgammon…
Excellent explanation, Dead Cat ! But one nitpick: X doesn’t have to play his 3 first; he can play the 1 first and discover a different way to leave three blots: 5/4/1*.
Exactly right. In this specific situation, you do not need an exact number. Any number higher than the highest point you occupy can be used to bear a checker off that point (but not off lower points you hold).
Yes, you can jump over points owned by your opponent, subject to the restriction mentioned by TroutMan. Of course, you can’t jump over a series of six points all held by your opponent, because there is no number you can roll (on one die) that is greater than six. Such a position is known as a ‘prime’, or more precisely in this case a ‘full prime’ or six-prime’. This is almost always a very powerful situation for the player holding the prime. A ‘five-prime’ can also be useful as it can only be jumped over by a roll containing a six.
Good point, quite so. Though I suspect this is a worse play, as it means (as far as I can make out) that there are no rolls at all by O that do not hit at least one blot on the next turn! Which is unusual to say the least.