Ok, where are we?
We have a backgammon board with 2 players, 15 stones each, both taking turns rolling two 6-sided die that average 7 per turn, with a 2.5% chance of double 6’s and a 16.7% chance of any doubles at all. The game has three stages: starting, mid-game and bear off. And there are two primary strategies: play a back game, or break into a running game. Plus, there’s two kinds of risk: The risk that a blot can be sent to the bar, and the possibility your blot getting on the bar and not being able to get back into the game. It’s not always wise to hit every blot that you can (you could get hit back, or you could fill up your own home field with the enemy’s stones).
Let’s look more at the mid-game, and what happens if you hit too many blots
If the goal of the start game is to take advantage of the first 2-3 turns and set yourself up against your opponent using standardized responses (as described in part 2), then the mid-game is where the jockeying happens to get you to where you can start bearing off.
The kind of jockeying depends on whether the start stage put you into a better position for initiating a running game, or if you’d rather try for a back game. In a running game, you want to pull in your farthest two stones from the backfield (24-point for Red, 1-point for White) and try to avoid leaving blots that can be hit by your opponent (if White moves both stones from 1-point to 4-point, then it’s more-or-less safe to start leaving blots behind him on the 1-, 2- and 3-points). But, for a back game, you want to leave both stones on the 24-point as long as you can, and move everything on the 13- and 8-points into your home field to make up a big wall of blocks, two stones each on 6 consecutive points. Then, hit your opponent’s blot to put him on the bar and not let him back into the game until after you start bearing off.
So, that jockeying in the mid-game can vary from game to game, and the success of each strategy will depend largely on the dice you get. Just because your strategy is good doesn’t mean you’ll always get the right dice to continue carrying it out. That is, the 2.5% chance that something will go wrong repeats with every turn, and after 50 turns, it WILL happen. You can try to minimize risk, but you can’t eliminate it entirely.
Ok, the game starts, White goes first and rolls 6-5. He moves one stone from 1-point to 12-point. Red gets 1-2, moving a stone from 24 to 23, and one from 13 to 11. White gets 1-3, moving a stone from 17 to 20, and one from 19 to 20, making a block. Red gets 3-4. Generally, he’d want to use the 2 blots to make the block on 20-point, but that’s already covered. He could play it safe and move one stone from 13-point to 6-point, but that puts him in an inflexible position. He takes the risk and moves from 13 to 10 and 13 to 9. That leaves three blots, and a 33% chance any of them will get hit. White rolls 6-2, and moves one stone from 1-point to 9, hitting the blot there and putting it on the bar.
Red isn’t too worried. There’s a 32/36=89% chance of getting back on the board, with the option of making a block anywhere on the 21-, 22-, 23- and 24-points. But, he rolls 6-6 and is stuck on the bar. White rolls double 1’s, hitting Red’s blots on 10- and 11-points, putting them on the bar as well, then continues to 12-point, and moves one stone from 19- to 20-point.
Generally, this would look grim for Red. But, he rolls a 1-2, making blocks on the 23- and 24-points. White rolls 3-4, and moves one stone from 6-point to 19-point. Red rolls a 4-3, putting the last stone back on the board, and making a block by moving one stone from 24- to 21-point. It should be obvious now that White has a problem. Any blot on his side of the board is going to be vulnerable, and two of the points in his home field are now blocked so it’s going to be harder to initiate a running game.
White rolls 1-6, and continues to play it safe, again going from 13-point to 19-point. At this stage, there’s nothing stopping Red from leaving blots. He rolls 3-4, and moves the two stones on 13-point back to 10- and 9-points. White gets double 1’s, and moves the two stones on 17- to 19-point. Red gets 1-3, and moves stones from 8- and 6-points to make the block on 5-point. White gets 6-3, moving from 12- to 18-point, and 13- to 15-point, leaving 2 blots. White rolls 5-1, and moves from 12-point to make the block on 18. Red gets 5-2, and makes the block on 4-point. White gets 1-6, and moves one stone from 12 to 18, and one from 19 to 20. Red rolls 6-4, and makes the block on the 3-point. White rolls 5-2, and moves one more stone from 12 to 19.
(The tide turns.)
At this stage, Red has 4 consecutive blocks, and a (6-1, 6-2, 6-3, 6-4, 6-5, 6-6, 1-6, 2-6, 3-6, 4-6, 5-6, 4-2, 2-4, 5-1, 1-5 = 15/36) 42% chance of hitting that blot on point 15. He rolls 6-5, moving both stones from point-21 to points 16 and 15, moving them out of his back field and putting White’s blot on the bar. White needs a 1 or 2 to get back on the board, which is 1 out of 3 turns.
There’s not a lot of reason to play the full game out. White is all jammed up around points 18 through 20, while Red is in a much more flexible position. Even if White can get back on the board, it will be difficult to get out of Red’s home field, while Red is focused on putting blocks on the 1- and 2-points. When White does get that lone blot past that wall of Red blocks, Red will probably just send it back to the bar again. Red’s plan is to have blocks on points 1- through 5-, and then start bearing off. If White does get on the board, it will be on point-6, BEHIND Red, turning this into a running game that Red is more likely to win.
There is still some strategy at this point. With 3 stones on the 5 point, Red will have a problem if he rolls a 5-6 – he’ll have to bear off 2 stones from the 5-point, leaving a blot that White has a 1/6 chance of hitting. The ideal is to constantly move stones in from the highest point number without having an odd number of stones on that point. Another reason for avoiding an odd number of stones on the highest point is if you have 5 stones and roll high doubles – you bear off 4 stones and leave a blot behind. In this game, Red doesn’t need to worry too much if he is sent to the bar because he has an 80% of getting back on the board in one turn, and a 50% chance of getting past White’s current wall. Red still has the advantage in the running game, and could still survive White’s hitting him on the way back to his home field.
BUT… There’s a chance (maybe 5-10%) that Red couldn’t get off the bar while White builds up blocks in his own home field. This is why Red wants to play a conservative game while bearing off, avoiding odd numbers of stones on the highest point of the wall until White is on the board behind him and no longer a threat.
Another option is to use the doubling cube. When White is on the bar and Red is starting to bear off, Red challenges White to double the game. If White accepts and then loses, it’s at twice the points as before. If he declines, he automatically loses the game and Red doesn’t have to worry about how the dice would turn out. If the idea is to win money, Red would want to double the game a bit earlier than this, making White struggle over whether to accept. If Red is already bearing off, then White will probably decline the challenge and forfeit the game. Red wouldn’t get the extra money, but it does let both players go on to the next game and start over again.
To be continued.