Problem of the Week #39: November 29


Cash game. Black owns the cube. Black on move.




(a) Black to play 5-1.




(b) Black to play 5-1.

I'm a complete amateur, but I'll give it a try:

I would go for 5/off 1/off and try and get as much off the board as possible. Protecting the single blot on 5 shouldn't be a factor because if white gets off the bar next roll, be it by 5 or 6, we're a dog to win the race. With a bit of luck here, we can still pull off a win.

edit: Just noticed there were two problems. Though, my (probably incorrect) play would be the same for both. 5/off 5/4 is a bit more tempting for (a).

in these spots i usually count the ( hypothethical ) moves left:
a) white has 7 rolls left , as black. in these case i dont let the blot
b) here the count is different cause white have 6 rolls left to clear, while black have 7 . In this case is necessary leave the blot and bearoff 2 chekers.

What formula are you using to count the hypothetical rolls?
I guess it is fairly easy for black but what are you doing for white?
I didn't think the positions were that close and it looks like white will win if he gets off the bar unless black can get hit and hit white again.

how often do we gain a full roll by leaving blot?
i figure ballperk 20% in A and 80% in B.
thus i`d take two off in B only

i take an average of 7/8 points per roll outside bearoff; this is obv not accurate, but could help in such a situations. in bearoff i count 2 chekers per roll .

The average number of pips you roll is larger than 8 because of doubles. You are probably overestimating the number of rolls required as a result of this.

in these situations the probability of black e white rolling a double is the same, and affects the game in the same way , being a race.
Thats why i make the count in that way.

Well, not really. White being on the bar and Black bearing off, doubles affect them quite differently I would say. For instance, any double below 5 would be very useful for Black but not for White.

I think you're making a mistake thinking about it this way. When it's a long race, I guess I can sort of buy it (but why use a smaller number when you know it's wrong?). But a bearoff is a completely different type of race.

Why not just count crossovers?

I think I pull two off each time. You're behind, you're already forced to break the point, and you're never getting gammoned. What do you gain by playing safe?

There is an even/odd thing going on, but I think you're too far behind right now to be worried about that. If white had a couple checkers left, you would probably have to take a closer look at it and maybe play safe on 1.

yes, the count is pratically the same.
i was answering the Bill' s problem: when i have to choose how to move my 5-1 , i have to made a statement and decide if i'm ahead or not. my decision is made considering :
1)white comin in in the next shot , which is likely but not sure
2) trying to figure out the rolls he needs to restart his bearoff ( i think it's 2 , again likely but not sure ). the average is fair enough to establish a rule ( and a move ). The race starts when white comes back on the board , and in that moment i figure doubles have the same value for both players ( accelerating bearoff restart for white, anf gaining rolls for black ). beacause of that i can easily don't consider the double as an option since it have the same probability for both players and affects the result the same way.

No real reason to get odd when every roll will take off 2 for every roll to come. You'll be unhappy if you throw 22 or 33 but you'll also be unhappy if you get hit.

Seems clear that 5/off 1/off is, relatively, much better in (b) than in (a) because you're further behind and it lets you get off in 6 more basic rolls when you don't get hit. I royally suck at the game though, so I can't say whether it's objectively the better move in either or both positions.

It seems like if you play safe in (a) and he doesn't get out, you have a good double because it's about 7 rolls each (once he gets out) you go first, and he stays stuck almost half the time. And if he does get out with a 6, it doesn't really hurt you, and if he gets out with a 5, it's better that you didn't get hit.

In (b) if you play safe, you're on 7 rolls instead of 6, and white is a turn faster a decent amount of the time because he has one less guy to bear off. You don't have a double if he stays stuck- it's not even clear to me that you're ahead. And if he does get out, and you still need 7 rolls to his 5.5 or so, you'd better crank up your luckbox. If he gets out with a 6 and you only need 6 rolls, you're not in bad shape going first. You might even have a chance to double if his next roll sucks.

a) play safe
b) gambool

I strongly disagree.

When white is making his way around the board, 11 is an above average roll for black, but a (very much) below average roll for white. They affect the positions in very different ways. 22 is in a similar spot (though 22 is closer to the average roll for white). So a third of the doubles favors black more than white. It's not a whole lot, but in this type of spot, it doesn't take a whole lot to turn the tables.

Sorry, not familiar with this terminology.
What is a crossover and how do you count it.

Thanks!

Crossovers count the number of times a checker must "cross over" into the next quadrant of the board (including on/off the board). So if a checker on the bar has 5 crossovers (bar -> 4th -> 3rd -> 2nd -> 1st -> home) and a checker on the 15 point has 3 crossovers.

a) 5/off 5/4

b) 5/off 1/off

In a) you have 14 checkers off after 5/off 5/4 and 13 checkers off after 5/off 1/off. Assuming you bear off 2 checkers with each roll (yes you might get more with doubles and less with some rolls) you need 7 rolls after this play to get off. Taking the chance on getting hit does not gain you anything.

In b) if you play 2 off you have 12 checkers left which means you need 6 rolls to get off while if you only play 1 off you need 7 rolls, so you gain a roll at the risk of being hit.

Better late than never...

In both Parts (a) and (b), Black has rolled 51, and is forced to bear off one man with his five. The only question is whether he should bear off a second checker, 5/off, 1/off, leaving a blot, or play safe, 5/off, 5/4. Either way, White can enter on his next turn with any five or six.

The key number is the five. So long as White doesn't roll it, Black's best move is to pull two checkers. Even if White dances, bearing off two is no worse than playing safe, and, in some cases, will save Black half a roll. Of course, if White enters with a six, then the race is on, and Black will be glad to have borne off the extra man.

When White rolls the five, things are a bit tighter, but even then Black may be better off being hit than otherwise. If White enters on the next roll, his lead will be so large that Black's best chance may be to come around again and try for a hit of his own, particularly if White rolls his joker 55.

Part (a) - Assuming White Rolls a Five
White needs twenty-five pips to enter and come all the way around, plus three rolls to bear off the six men on his one and two points. With his home-board checkers crowded onto the lower points, White will be lucky to bear off his trailing checker in three rolls; he will probably need three and a half. White is favored to be off in six or seven rolls.

If Black plays safe, he will have fourteen checkers left after this turn. He will likely need seven or eight turns to clear his board. Unfortunately for him, he can't miss if he is to be off in seven, and White rolls first. Even when Black rolls optimally, and White doesn't, White is favored to win in seven rolls.

Alternatively, Black can leave his blot to be hit. From the bar, he might have a 20%-30% chance to re-hit, and get back in the game. Otherwise, he's probably a loser. Let's play for contact in Part (a).

My solution in Part (a): 5/off, 1/off.

Part (b) - Assuming White Rolls a Five
White has only five home board checkers, so he is favored to be off in six rolls, although seven is still a possibility. If Black plays safe, he will have thirteen checkers left after this turn. He will likely need seven more rolls, even if he misses once. Again, White gets the first roll.

Here, the race is closer, but once again, even when White rolls suboptimally, he's favored to win in seven rolls.

My solution in Part (b): 5/off, 1/off.

For the Record
I am so often wrong that I like to post my record in these messages. It's kind of a truth-in-advertising thing. I have been answering these problems without the use of a bot, and before checking the excellent solutions of others, since Problem 28. My record at this writing is 54% correct.
Correct: 28a, 29, 30, 32, 35, 36, 38
Incorrect: 28b, 31, 33a, 33b, 34, 37

An article in Roberties "Inside backgammon" - late 80s? described similar positions. Black would be about 75% favorite if white was on the bar and Black was 3 piecess off.
Black would then redouble and white still take . But if Black had 4 off it was a pass (more than 80% to black). If black was only 2 off, it is not a double especially where he has two open entry points.

Taking a man off instead of safetying may have two purposes: either to reach even number of checkers left or to have an additional chance to hit enemy checker if hit by the enemy checker on the bar. OR BOTH (as here)

I called these positions to myself BAREOFF positions. I saw a mistake in such position made in a published match of Grandell. When I did some research I got an idea that the BAREOFF was correct if the opponent was at least 9 men off (here only 8) and perhaps white has a blot in his home. In a real situation (no bots) I would play safe here.

I think the French giant (sixprime) sorry can't remember name now... had an article in gammonvillage on this theme later.

François Tardieu wrote this article on voluntary blot leaving in bear off against an enemy man on the bar.

Sorry did not see Robertie's solution before my first post . I am new here, you know ...

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François Tardieu wrote this article on voluntary blot leaving in bear off against an enemy man on the bar.

Sorry did not see Robertie's solution before my first post . I am new here, you know ...

Also my discussion applied to 39a, because I did not see 39b at first.

Robertie's solution explains all

Hi Svilo!
very nice to see u here ... still enjoying your table. Bye

This.