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Most checkers on the bar -- solution

Posted By: Nack Ballard
Date: Wednesday, 9 June 2010, at 2:04 a.m.

The posers were originally posted in this thread:

1) What is the largest number of checkers (both colors counted) that can legally be on the bar at the same time?

(2) What is the largest number of checkers with the two colors having the same number on the bar?

Bob Koca was the first to get the right answers of 15 for (1) and 12 for (2).

Ray offered a well articulated solution to (2):

In his home board, Black has 2 checkers on his 6pt and 1 checker on each of the other five points. In her home board, White has 3 checkers on her 6pt. Black sends 12 White checkers to the bar (made possible because White keeps dancing with 66). Then Black rolls a 1 and plays 6/5 so that Black has 1 checker on each of his home board points. Then White rolls 21 to send 2 Black checkers to the bar. Black fans. Then White rolls 43 to send 2 more Black checkers to the bar. Black fans. Finally White rolls 65 to send 2 more Black checkers to the bar. At this point, Black and White each have 6 checkers on the bar.

If White has only 2 checkers on her 6pt, then Black can send 13 White checkers to the bar. The above sequence then results in 7 White checkers and 6 Black checkers on the bar.

And Ray's recap of both (1) and (2) was:

In summary, the maximum number of checkers on the bar is 15 (all of one color); the maximum number of checkers on the bar is 13 if each color has to be on the bar; and the maximum number of checkers on the bar is 12 if there are an equal number of each color (Nack's constraint).

Steve corrected the second statement in the above paragraph: It seems to me that if you can get 15 checkers of one color on the bar, you can get 15 total checkers of both colors on the bar. Imagine I close out all 15 of your checkers. Then I leave a blot while bearing off, and you roll a number that enters and hits. Now the bar contains 14 of your checkers and 1 of mine.

Ray quickly acknowledged that correction, and added: As well as the 14 + 1 described by Steve, 13 + 2 is also possible by the same process. The full generalisation of Nack's questions is to list all the feasible combinations, whether or not each color has the same number of checkers on the bar. So 15 goes with only 0; 14 goes with only 0 or 1; 13 goes with 0, 1 or 2; ---- I strongly suspect that Nack knows all this.

Your suspicion is correct, Ray :) And kudos for such a thorough answer! You even correctly noted (or at least implied) that it is impossible when one player has 6 on the bar for the other player to have more than 7. Your example demonstrates you've grasped why, but I'll state the "proof" below for others:

Say that White is the player with the initial mass of checkers on the bar. When Black slots four inside points then breaks his final point thereby slotting all six points, White can hit at most 2 checkers per roll, and if Black himself faces a pointless board he will be entering at least as fast as White can hit. To get more than two Black checkers in the air, White must own a point. Hence when one player has 6 blots, the max-bar situation for the other player is reached by: White has 13 on the bar, then in three rolls hits 6 blots (with 7 still in the air), Black having fanned twice on White's one-point board.

Finally, you are right when you say...

So 15 goes with only 0; 14 goes with only 0 or 1; 13 goes with 0, 1 or 2; ----

However, the series does not continue with "12 goes with 0, 1, 2 or 3," etc. The total number of checkers on the bar cannot exceed 13 except in the temporary situations of 14+1, 13+2, 13+1 and 12+2, for the same reason that 7+6 is a cap when one player has 6 on the bar. A player must own an inside point in order for the number of his checkers on the bar to increase while the other player's number decreases on a 1 for 1 basis.

Nack