You may think the title of this post is incomplete -- "most" among
which sample? However, I was explicitly requested by a forum
participant (I forget who) to shorten my subject titles.

The problem is to give the legal position which maximises the
total pipcount of both players.

Actually, I thought I had a solution but now I see that it's not so
obviously legal. And maybe the problem has more merit than I
thought. (I wonder if it's in Backgammon Funfair)
Perhaps we should put 13 checkers of each player on the bar and
make each player's 6 point for a total pipcount of 674.
I'd be surprised if we can beat that, but I suspect now that this
solution doesn't work.

Paul

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On 3/28/2023 8:36 AM, peps...@gmail.com wrote:

The problem is to give the legal position which maximises the
total pipcount of both players.

Actually, I thought I had a solution but now I see that it's not so
obviously legal. And maybe the problem has more merit than I
thought. (I wonder if it's in Backgammon Funfair)

Backgammon Funfair has a chapter on "Most checkers on the bar."
It makes the following claim: "It is impossible to have more than
fifteen checkers on the bar." This claim sounds plausible, but the
proof in the book strikes me as sketchy. Here's what it says:

"As shown in chapter 12, it is possible to have fifteen White checkers
on the bar. Then a Red checker can come *to* the bar only by being
hit by a White checker entering *from* the bar. Hence the total on
the bar cannot exceed fifteen."

To make this proof more rigorous, one has to show that it doesn't help
to start from some position other than all 15 of one side's checkers
on the bar. But I think this shouldn't be too hard to prove.

By the way, the Funfair book doesn't seem to address your question
explicitly.

---
Tim Chow

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On Tuesday, March 28, 2023 at 2:19:46 PM UTC+1, Timothy Chow wrote:

On 3/28/2023 8:36 AM, peps...@gmail.com wrote:

The problem is to give the legal position which maximises the
total pipcount of both players.

Actually, I thought I had a solution but now I see that it's not so obviously legal. And maybe the problem has more merit than I
thought. (I wonder if it's in Backgammon Funfair)

Backgammon Funfair has a chapter on "Most checkers on the bar."
It makes the following claim: "It is impossible to have more than
fifteen checkers on the bar." This claim sounds plausible, but the
proof in the book strikes me as sketchy. Here's what it says:

"As shown in chapter 12, it is possible to have fifteen White checkers
on the bar. Then a Red checker can come *to* the bar only by being
hit by a White checker entering *from* the bar. Hence the total on
the bar cannot exceed fifteen."

To make this proof more rigorous, one has to show that it doesn't help
to start from some position other than all 15 of one side's checkers
on the bar. But I think this shouldn't be too hard to prove.

By the way, the Funfair book doesn't seem to address your question explicitly.

I might be wrong but I think it was Donald Trump who claimed that no more
than 15 checkers can be on the bar.
Suppose you have your ace point made and I have my ace point made (after
all, copying Tim's backgammon would be unlikely to do me any harm).
You have a stack of three checkers on your 7 point (primality relevant).
I have a stack of two checkers on my 7 point and one on the bar.
Other than this, we each have ten blots on our outer and inner boards.
It seems to me that the great majority of twenty blots can be picked up
by doubles.
Maybe the 11 point checkers can't be profitably picked up because when we
roll a 65 we come in from the bar ourselves and don't gain anything.

Aha. But even if I roll doubles, I still bring in one checker off the bar that you may struggle
to hit. The issue is tricky, I think

After having googled it, it seems that the 15 argument is totally bogus, ignoring the
fact that people are actually allowed to roll doubles.

I might have better luck asking Joe Biden about it.

Paul

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On Tuesday, March 28, 2023 at 2:19:46 PM UTC+1, Timothy Chow wrote:

On 3/28/2023 8:36 AM, peps...@gmail.com wrote:

The problem is to give the legal position which maximises the
total pipcount of both players.

Actually, I thought I had a solution but now I see that it's not so obviously legal. And maybe the problem has more merit than I
thought. (I wonder if it's in Backgammon Funfair)

Backgammon Funfair has a chapter on "Most checkers on the bar."
It makes the following claim: "It is impossible to have more than
fifteen checkers on the bar." This claim sounds plausible, but the
proof in the book strikes me as sketchy. Here's what it says:

"As shown in chapter 12, it is possible to have fifteen White checkers
on the bar. Then a Red checker can come *to* the bar only by being
hit by a White checker entering *from* the bar. Hence the total on
the bar cannot exceed fifteen."

To make this proof more rigorous, one has to show that it doesn't help
to start from some position other than all 15 of one side's checkers
on the bar. But I think this shouldn't be too hard to prove.

By the way, the Funfair book doesn't seem to address your question explicitly.

---
Tim Chow

Wait a minute. The 15 claim is total nonsense!
It's clearly possible for you to put all of my 15 checkers on the bar.
After all, backgammon is a competitive game so you're entitled to
be brutal in this context.
Clearly I can enter with doubles and hit four of your blots in the process. This introduces 4 more blots and enters 1 to leave 18 checkers on the bar.

No wonder, the argument can't be made rigorous. The argument might work
in a bg variant where doubles follow the same rules as non-doubles.

Paul

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On Tuesday, March 28, 2023 at 6:36:18 PM UTC+1, peps...@gmail.com wrote:

On Tuesday, March 28, 2023 at 2:19:46 PM UTC+1, Timothy Chow wrote:

On 3/28/2023 8:36 AM, peps...@gmail.com wrote:

The problem is to give the legal position which maximises the
total pipcount of both players.

Actually, I thought I had a solution but now I see that it's not so obviously legal. And maybe the problem has more merit than I
thought. (I wonder if it's in Backgammon Funfair)

Backgammon Funfair has a chapter on "Most checkers on the bar."
It makes the following claim: "It is impossible to have more than
fifteen checkers on the bar." This claim sounds plausible, but the
proof in the book strikes me as sketchy. Here's what it says:

"As shown in chapter 12, it is possible to have fifteen White checkers
on the bar. Then a Red checker can come *to* the bar only by being
hit by a White checker entering *from* the bar. Hence the total on
the bar cannot exceed fifteen."

To make this proof more rigorous, one has to show that it doesn't help
to start from some position other than all 15 of one side's checkers
on the bar. But I think this shouldn't be too hard to prove.

By the way, the Funfair book doesn't seem to address your question explicitly.

---
Tim Chow

Wait a minute. The 15 claim is total nonsense!
It's clearly possible for you to put all of my 15 checkers on the bar.
After all, backgammon is a competitive game so you're entitled to
be brutal in this context.
Clearly I can enter with doubles and hit four of your blots in the process. This introduces 4 more blots and enters 1 to leave 18 checkers on the bar.

No wonder, the argument can't be made rigorous. The argument might work
in a bg variant where doubles follow the same rules as non-doubles.

Paul

Backgammon Funfair shows that (a) the most checkers that can be on the bar is 15 and (b) the largest pip count that Red can have while still being smaller than White's is 362 and (c) when Red has 362, the most pips that White can have is 372, so that the
aggregate of pips of 734 is the meximum that can be reached from the starting position.

If you have 15 of your checkers on the bar and roll a double you hit either 0 or 1 of your opponent's checkers - not 4.

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On Tuesday, March 28, 2023 at 9:21:28 PM UTC+1, Raymond Kershaw wrote:

On Tuesday, March 28, 2023 at 6:36:18 PM UTC+1, peps...@gmail.com wrote:

On Tuesday, March 28, 2023 at 2:19:46 PM UTC+1, Timothy Chow wrote:

On 3/28/2023 8:36 AM, peps...@gmail.com wrote:

The problem is to give the legal position which maximises the
total pipcount of both players.

Actually, I thought I had a solution but now I see that it's not so obviously legal. And maybe the problem has more merit than I
thought. (I wonder if it's in Backgammon Funfair)

Backgammon Funfair has a chapter on "Most checkers on the bar."
It makes the following claim: "It is impossible to have more than fifteen checkers on the bar." This claim sounds plausible, but the
proof in the book strikes me as sketchy. Here's what it says:

"As shown in chapter 12, it is possible to have fifteen White checkers on the bar. Then a Red checker can come *to* the bar only by being
hit by a White checker entering *from* the bar. Hence the total on
the bar cannot exceed fifteen."

To make this proof more rigorous, one has to show that it doesn't help to start from some position other than all 15 of one side's checkers
on the bar. But I think this shouldn't be too hard to prove.

By the way, the Funfair book doesn't seem to address your question explicitly.

---
Tim Chow

Wait a minute. The 15 claim is total nonsense!
It's clearly possible for you to put all of my 15 checkers on the bar. After all, backgammon is a competitive game so you're entitled to
be brutal in this context.
Clearly I can enter with doubles and hit four of your blots in the process.
This introduces 4 more blots and enters 1 to leave 18 checkers on the bar.

No wonder, the argument can't be made rigorous. The argument might work
in a bg variant where doubles follow the same rules as non-doubles.

Paul

Backgammon Funfair shows that (a) the most checkers that can be on the bar is 15 and (b) the largest pip count that Red can have while still being smaller than White's is 362 and (c) when Red has 362, the most pips that White can have is 372, so that

the aggregate of pips of 734 is the meximum that can be reached from the starting position.

If you have 15 of your checkers on the bar and roll a double you hit either 0 or 1 of your opponent's checkers - not 4.

Yes, I stand corrected. I was visualizing the scenario completely wrongly. It makes sense now.

Paul

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On 3/28/2023 4:21 PM, Raymond Kershaw wrote:

Backgammon Funfair shows that (a) the most checkers that can be on the bar is 15 and (b) the largest pip count that Red can have while still being smaller than White's is 362 and (c) when Red has 362, the most pips that White can have is 372, so that

the aggregate of pips of 734 is the meximum that can be reached from the starting position.

The final number of 734 might be correct, but it doesn't follow from
the other things you say here. For example, maybe the maximum total
occurs when Red has 350 pips and White has 385 pips.

---
Tim Chow

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On 3/28/2023 11:59 AM, peps...@gmail.com wrote:

Suppose you have your ace point made and I have my ace point made (after
all, copying Tim's backgammon would be unlikely to do me any harm).
You have a stack of three checkers on your 7 point (primality relevant).
I have a stack of two checkers on my 7 point and one on the bar.
Other than this, we each have ten blots on our outer and inner boards.
It seems to me that the great majority of twenty blots can be picked up
by doubles.

I assume you've abandoned this claim as well? It sounds like you're
imagining that we can move our checkers around the board and hit
blots when we're on the bar, which of course is illegal.

---
Tim Chow

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On Wednesday, March 29, 2023 at 1:12:31 AM UTC+1, Timothy Chow wrote:

On 3/28/2023 4:21 PM, Raymond Kershaw wrote:

Backgammon Funfair shows that (a) the most checkers that can be on the bar is 15 and (b) the largest pip count that Red can have while still being smaller than White's is 362 and (c) when Red has 362, the most pips that White can have is 372, so that

the aggregate of pips of 734 is the meximum that can be reached from the starting position.

The final number of 734 might be correct, but it doesn't follow from
the other things you say here. For example, maybe the maximum total
occurs when Red has 350 pips and White has 385 pips.

---
Tim Chow

Tim - White cannot have 385 pips. The maximum a player could have is 375 = 15*25

--- SoupGate-Win32 v1.05
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On 3/29/2023 5:43 AM, Raymond Kershaw wrote:

On Wednesday, March 29, 2023 at 1:12:31 AM UTC+1, Timothy Chow wrote:

On 3/28/2023 4:21 PM, Raymond Kershaw wrote:

Backgammon Funfair shows that (a) the most checkers that can be on the bar is 15 and (b) the largest pip count that Red can have while still being smaller than White's is 362 and (c) when Red has 362, the most pips that White can have is 372, so that

the aggregate of pips of 734 is the meximum that can be reached from the starting position.

The final number of 734 might be correct, but it doesn't follow from
the other things you say here. For example, maybe the maximum total
occurs when Red has 350 pips and White has 385 pips.

---
Tim Chow

Tim - White cannot have 385 pips. The maximum a player could have is 375 = 15*25

Good point. But what about this: Red has 360 pips and
White has 375 pips?

---
Tim Chow

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On Wednesday, March 29, 2023 at 2:18:45 PM UTC+1, Timothy Chow wrote:

On 3/29/2023 5:43 AM, Raymond Kershaw wrote:

On Wednesday, March 29, 2023 at 1:12:31 AM UTC+1, Timothy Chow wrote:

On 3/28/2023 4:21 PM, Raymond Kershaw wrote:

Backgammon Funfair shows that (a) the most checkers that can be on the bar is 15 and (b) the largest pip count that Red can have while still being smaller than White's is 362 and (c) when Red has 362, the most pips that White can have is 372, so

that the aggregate of pips of 734 is the meximum that can be reached from the starting position.

The final number of 734 might be correct, but it doesn't follow from
the other things you say here. For example, maybe the maximum total
occurs when Red has 350 pips and White has 385 pips.

---
Tim Chow

Tim - White cannot have 385 pips. The maximum a player could have is 375 = 15*25

Good point. But what about this: Red has 360 pips and
White has 375 pips?

Tim Chow

If you can show how to get a 360 and 375 position then I would accept that the aggregate pips are one more than 734.

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On Wednesday, March 29, 2023 at 4:00:45 PM UTC+1, Raymond Kershaw wrote:

On Wednesday, March 29, 2023 at 2:18:45 PM UTC+1, Timothy Chow wrote:

On 3/29/2023 5:43 AM, Raymond Kershaw wrote:

On Wednesday, March 29, 2023 at 1:12:31 AM UTC+1, Timothy Chow wrote:

On 3/28/2023 4:21 PM, Raymond Kershaw wrote:

Backgammon Funfair shows that (a) the most checkers that can be on the bar is 15 and (b) the largest pip count that Red can have while still being smaller than White's is 362 and (c) when Red has 362, the most pips that White can have is 372, so

that the aggregate of pips of 734 is the meximum that can be reached from the starting position.

The final number of 734 might be correct, but it doesn't follow from
the other things you say here. For example, maybe the maximum total
occurs when Red has 350 pips and White has 385 pips.

---
Tim Chow

Tim - White cannot have 385 pips. The maximum a player could have is 375 = 15*25

Good point. But what about this: Red has 360 pips and
White has 375 pips?

Tim Chow

If you can show how to get a 360 and 375 position then I would accept that the aggregate pips are one more than 734.

But this reverses the burden of proof!
The burden of proof is on you to show that a 360/375 position is impossible.

Paul

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On Wednesday, March 29, 2023 at 1:25:38 AM UTC+1, Timothy Chow wrote:
....It sounds like you're

imagining that we can move our checkers around the board and hit
blots when we're on the bar, which of course is illegal.

No, I'm not imagining that.
I was visualizing it that way, but no longer.
I understand the issue now, although (like you),
I'm not totally convinced by the claim.

The actual situation mirrors reality quite closely.
Suppose you're at a bar, and you leave the bar and playfully hit one person. The bystanders and bar staff might well just assume that you're just joshing with that other person, and it might well be no big deal.

But suppose you leave the bar and hit four people.
Then your actions may well be considered illegal and I suppose the cops
might be called and declare your behaviour illegal.

Paul

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