This topic was previously discussed in another thread, but I don't think
it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
and this point seems unresolved in that thread. I aim to resolve this
point here, completely rigorously.

Let us call a position "crowded" if it contains strictly more than 15 checkers
on the bar.
Suppose, by contradiction, that we have reached the first occurrence
of a crowded position in a game.
Let k be the number of checkers that were brought in from the bar on the
previous roll.
If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
hit myself so none of my checkers get sent to the bar, leaving a max of 15.
Suppose k is 1. If I bring that checker in, we again get a max of 15, as
before, because I can't hit myself.
If the previous roll is a dance, then the position repeats contradicting the
claim of first occurrence.
Suppose k >= 2. If I dance completely, the previous argument applies.
If I dance with one man, we either end up with fewer checkers on the
bar or the same number of checkers on the bar depending on whether
the entering number hits. If I don't dance at all, then I take at least two
checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.

Paul

On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
> This topic was previously discussed in another thread, but I don't think
> it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
> and this point seems unresolved in that thread. I aim to resolve this
> point here, completely rigorously.
>
> Let us call a position "crowded" if it contains strictly more than 15 checkers
> on the bar.
> Suppose, by contradiction, that we have reached the first occurrence
> of a crowded position in a game.
> Let k be the number of checkers that were brought in from the bar on the
> previous roll.
> If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
> hit myself so none of my checkers get sent to the bar, leaving a max of 15.
> Suppose k is 1. If I bring that checker in, we again get a max of 15, as
> before, because I can't hit myself.
> If the previous roll is a dance, then the position repeats contradicting the
> claim of first occurrence.
> Suppose k >= 2. If I dance completely, the previous argument applies.
> If I dance with one man, we either end up with fewer checkers on the
> bar or the same number of checkers on the bar depending on whether
> the entering number hits. If I don't dance at all, then I take at least two
> checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
>
> Paul

Sorry, k needs to be the number of the last-roller's checkers that were on the bar before the
most recent shake.

Paul

On 3/30/2023 8:15 AM, peps...@gmail.com wrote:
> On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
>> This topic was previously discussed in another thread, but I don't think
>> it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
>> and this point seems unresolved in that thread. I aim to resolve this
>> point here, completely rigorously.
>>
>> Let us call a position "crowded" if it contains strictly more than 15 checkers
>> on the bar.
>> Suppose, by contradiction, that we have reached the first occurrence
>> of a crowded position in a game.
>> Let k be the number of checkers that were brought in from the bar on the
>> previous roll.
>> If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
>> hit myself so none of my checkers get sent to the bar, leaving a max of 15.
>> Suppose k is 1. If I bring that checker in, we again get a max of 15, as
>> before, because I can't hit myself.
>> If the previous roll is a dance, then the position repeats contradicting the
>> claim of first occurrence.
>> Suppose k >= 2. If I dance completely, the previous argument applies.
>> If I dance with one man, we either end up with fewer checkers on the
>> bar or the same number of checkers on the bar depending on whether
>> the entering number hits. If I don't dance at all, then I take at least two
>> checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
>>
>> Paul
>
> Sorry, k needs to be the number of the last-roller's checkers that were on the bar before the
> most recent shake.

Thanks for clarifying...I was having trouble following your argument.

So when you say, "we have reached the first occurrence of a crowded
position," you mean for example that White has just completed her
checker play, leaving a crowded position with Black on roll? And
k is the number of White's checkers on the bar? And in your narrative,
is "I" White or Black?

---
Tim Chow

On Thursday, March 30, 2023 at 1:49:02 PM UTC+1, Timothy Chow wrote:
> On 3/30/2023 8:15 AM, peps...@gmail.com wrote:
> > On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
> >> This topic was previously discussed in another thread, but I don't think
> >> it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
> >> and this point seems unresolved in that thread. I aim to resolve this
> >> point here, completely rigorously.
> >>
> >> Let us call a position "crowded" if it contains strictly more than 15 checkers
> >> on the bar.
> >> Suppose, by contradiction, that we have reached the first occurrence
> >> of a crowded position in a game.
> >> Let k be the number of checkers that were brought in from the bar on the
> >> previous roll.
> >> If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
> >> hit myself so none of my checkers get sent to the bar, leaving a max of 15.
> >> Suppose k is 1. If I bring that checker in, we again get a max of 15, as
> >> before, because I can't hit myself.
> >> If the previous roll is a dance, then the position repeats contradicting the
> >> claim of first occurrence.
> >> Suppose k >= 2. If I dance completely, the previous argument applies..
> >> If I dance with one man, we either end up with fewer checkers on the
> >> bar or the same number of checkers on the bar depending on whether
> >> the entering number hits. If I don't dance at all, then I take at least two
> >> checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
> >>
> >> Paul
> >
> > Sorry, k needs to be the number of the last-roller's checkers that were on the bar before the
> > most recent shake.
> Thanks for clarifying...I was having trouble following your argument.
>
> So when you say, "we have reached the first occurrence of a crowded
> position," you mean for example that White has just completed her
> checker play, leaving a crowded position with Black on roll?

Well, yes. But how could I possibly have meant anything else?
Please could you explain the ambiguity here?

> And k is the number of White's checkers on the bar? And in your narrative,
> is "I" White or Black?

White.

> Tim Chow

On Thursday, March 30, 2023 at 5:01:37 PM UTC+1, peps...@gmail.com wrote:
> On Thursday, March 30, 2023 at 1:49:02 PM UTC+1, Timothy Chow wrote:
> > On 3/30/2023 8:15 AM, peps...@gmail.com wrote:
> > > On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
> > >> This topic was previously discussed in another thread, but I don't think
> > >> it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
> > >> and this point seems unresolved in that thread. I aim to resolve this
> > >> point here, completely rigorously.
> > >>
> > >> Let us call a position "crowded" if it contains strictly more than 15 checkers
> > >> on the bar.
> > >> Suppose, by contradiction, that we have reached the first occurrence
> > >> of a crowded position in a game.
> > >> Let k be the number of checkers that were brought in from the bar on the
> > >> previous roll.
> > >> If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
> > >> hit myself so none of my checkers get sent to the bar, leaving a max of 15.
> > >> Suppose k is 1. If I bring that checker in, we again get a max of 15, as
> > >> before, because I can't hit myself.
> > >> If the previous roll is a dance, then the position repeats contradicting the
> > >> claim of first occurrence.
> > >> Suppose k >= 2. If I dance completely, the previous argument applies.
> > >> If I dance with one man, we either end up with fewer checkers on the
> > >> bar or the same number of checkers on the bar depending on whether
> > >> the entering number hits. If I don't dance at all, then I take at least two
> > >> checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
> > >>
> > >> Paul
> > >
> > > Sorry, k needs to be the number of the last-roller's checkers that were on the bar before the
> > > most recent shake.
> > Thanks for clarifying...I was having trouble following your argument.
> >
> > So when you say, "we have reached the first occurrence of a crowded
> > position," you mean for example that White has just completed her
> > checker play, leaving a crowded position with Black on roll?
> Well, yes. But how could I possibly have meant anything else?
> Please could you explain the ambiguity here?
> > And k is the number of White's checkers on the bar? And in your narrative,
> > is "I" White or Black?
> White.
>
> > Tim Chow

My argument needs another minor correction.
It is possible to bring in two checkers and hit three checkers but that only happens if that
last roll brings in all that player's checkers so that checkers of only one colour remain on the bar.

Paul

On Thursday, March 30, 2023 at 1:49:02 PM UTC+1, Timothy Chow wrote:
> On 3/30/2023 8:15 AM, peps...@gmail.com wrote:
> > On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
> >> This topic was previously discussed in another thread, but I don't think
> >> it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
> >> and this point seems unresolved in that thread. I aim to resolve this
> >> point here, completely rigorously.
> >>
> >> Let us call a position "crowded" if it contains strictly more than 15 checkers
> >> on the bar.
> >> Suppose, by contradiction, that we have reached the first occurrence
> >> of a crowded position in a game.
> >> Let k be the number of checkers that were brought in from the bar on the
> >> previous roll.
> >> If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
> >> hit myself so none of my checkers get sent to the bar, leaving a max of 15.
> >> Suppose k is 1. If I bring that checker in, we again get a max of 15, as
> >> before, because I can't hit myself.
> >> If the previous roll is a dance, then the position repeats contradicting the
> >> claim of first occurrence.
> >> Suppose k >= 2. If I dance completely, the previous argument applies..
> >> If I dance with one man, we either end up with fewer checkers on the
> >> bar or the same number of checkers on the bar depending on whether
> >> the entering number hits. If I don't dance at all, then I take at least two
> >> checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
> >>
> >> Paul
> >
> > Sorry, k needs to be the number of the last-roller's checkers that were on the bar before the
> > most recent shake.
> Thanks for clarifying...I was having trouble following your argument.
>
> So when you say, "we have reached the first occurrence of a crowded
> position," you mean for example that White has just completed her
> checker play, leaving a crowded position with Black on roll? And
> k is the number of White's checkers on the bar? And in your narrative,
> is "I" White or Black?
>
> ---
> Tim Chow

I think we can be a lot simpler as follows:
Clearly the 15 figure can be obtained by making an inner board point and then strolling around,
picking up all 15 blots.
But can this be beaten?
To get more than 16 checkers on the bar, we need checkers on the bar of both colours.
Consider the position immediately before the maximum was first obtained, and consider the
next roll.
If all that roll did was bring checkers in from the bar, with no time to do extra non-entering plays.
Then clearly there were at least as many checkers entered as blots hit, and we didn't increase
the on-bar number.
Alternatively, suppose that this prior roll did make non-entering plays. That means that
the prior roller had all their checkers off the bar and we don't beat 15.

Paul

On Friday, March 31, 2023 at 2:07:45 PM UTC+1, peps...@gmail.com wrote:
> On Thursday, March 30, 2023 at 1:49:02 PM UTC+1, Timothy Chow wrote:
> > On 3/30/2023 8:15 AM, peps...@gmail.com wrote:
> > > On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
> > >> This topic was previously discussed in another thread, but I don't think
> > >> it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
> > >> and this point seems unresolved in that thread. I aim to resolve this
> > >> point here, completely rigorously.
> > >>
> > >> Let us call a position "crowded" if it contains strictly more than 15 checkers
> > >> on the bar.
> > >> Suppose, by contradiction, that we have reached the first occurrence
> > >> of a crowded position in a game.
> > >> Let k be the number of checkers that were brought in from the bar on the
> > >> previous roll.
> > >> If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
> > >> hit myself so none of my checkers get sent to the bar, leaving a max of 15.
> > >> Suppose k is 1. If I bring that checker in, we again get a max of 15, as
> > >> before, because I can't hit myself.
> > >> If the previous roll is a dance, then the position repeats contradicting the
> > >> claim of first occurrence.
> > >> Suppose k >= 2. If I dance completely, the previous argument applies.
> > >> If I dance with one man, we either end up with fewer checkers on the
> > >> bar or the same number of checkers on the bar depending on whether
> > >> the entering number hits. If I don't dance at all, then I take at least two
> > >> checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
> > >>
> > >> Paul
> > >
> > > Sorry, k needs to be the number of the last-roller's checkers that were on the bar before the
> > > most recent shake.
> > Thanks for clarifying...I was having trouble following your argument.
> >
> > So when you say, "we have reached the first occurrence of a crowded
> > position," you mean for example that White has just completed her
> > checker play, leaving a crowded position with Black on roll? And
> > k is the number of White's checkers on the bar? And in your narrative,
> > is "I" White or Black?
> >
> > ---
> > Tim Chow
> I think we can be a lot simpler as follows:
> Clearly the 15 figure can be obtained by making an inner board point and then strolling around,
> picking up all 15 blots.
> But can this be beaten?
> To get more than 16 checkers on the bar, we need checkers on the bar of both colours.
> Consider the position immediately before the maximum was first obtained, and consider the
> next roll.
> If all that roll did was bring checkers in from the bar, with no time to do extra non-entering plays.
> Then clearly there were at least as many checkers entered as blots hit, and we didn't increase
> the on-bar number.
> Alternatively, suppose that this prior roll did make non-entering plays. That means that
> the prior roller had all their checkers off the bar and we don't beat 15.
>
> Paul
These rather tortuous explanations are dealt with more succinctly in Backgammon Funfair: "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."
I don't see this argument as being in any way sletchy.

On Friday, March 31, 2023 at 6:29:26 PM UTC+1, Raymond Kershaw wrote:
> On Friday, March 31, 2023 at 2:07:45 PM UTC+1, peps...@gmail.com wrote:
> > On Thursday, March 30, 2023 at 1:49:02 PM UTC+1, Timothy Chow wrote:
> > > On 3/30/2023 8:15 AM, peps...@gmail.com wrote:
> > > > On Thursday, March 30, 2023 at 1:12:52 PM UTC+1, peps...@gmail.com wrote:
> > > >> This topic was previously discussed in another thread, but I don't think
> > > >> it was fully resolved. Tim said "The proof in the book strikes me as sketchy"
> > > >> and this point seems unresolved in that thread. I aim to resolve this
> > > >> point here, completely rigorously.
> > > >>
> > > >> Let us call a position "crowded" if it contains strictly more than 15 checkers
> > > >> on the bar.
> > > >> Suppose, by contradiction, that we have reached the first occurrence
> > > >> of a crowded position in a game.
> > > >> Let k be the number of checkers that were brought in from the bar on the
> > > >> previous roll.
> > > >> If k is 0, then we can't end up with more than 15 checkers on the bar. I can't
> > > >> hit myself so none of my checkers get sent to the bar, leaving a max of 15.
> > > >> Suppose k is 1. If I bring that checker in, we again get a max of 15, as
> > > >> before, because I can't hit myself.
> > > >> If the previous roll is a dance, then the position repeats contradicting the
> > > >> claim of first occurrence.
> > > >> Suppose k >= 2. If I dance completely, the previous argument applies.
> > > >> If I dance with one man, we either end up with fewer checkers on the
> > > >> bar or the same number of checkers on the bar depending on whether
> > > >> the entering number hits. If I don't dance at all, then I take at least two
> > > >> checkers off the bar while hitting <= 2 checkers. In all cases, the claim of first occurrence is contradicted.
> > > >>
> > > >> Paul
> > > >
> > > > Sorry, k needs to be the number of the last-roller's checkers that were on the bar before the
> > > > most recent shake.
> > > Thanks for clarifying...I was having trouble following your argument.
> > >
> > > So when you say, "we have reached the first occurrence of a crowded
> > > position," you mean for example that White has just completed her
> > > checker play, leaving a crowded position with Black on roll? And
> > > k is the number of White's checkers on the bar? And in your narrative,
> > > is "I" White or Black?
> > >
> > > ---
> > > Tim Chow
> > I think we can be a lot simpler as follows:
> > Clearly the 15 figure can be obtained by making an inner board point and then strolling around,
> > picking up all 15 blots.
> > But can this be beaten?
> > To get more than 16 checkers on the bar, we need checkers on the bar of both colours.
> > Consider the position immediately before the maximum was first obtained, and consider the
> > next roll.
> > If all that roll did was bring checkers in from the bar, with no time to do extra non-entering plays.
> > Then clearly there were at least as many checkers entered as blots hit, and we didn't increase
> > the on-bar number.
> > Alternatively, suppose that this prior roll did make non-entering plays.. That means that
> > the prior roller had all their checkers off the bar and we don't beat 15.
> >
> > Paul
> These rather tortuous explanations are dealt with more succinctly in Backgammon Funfair: "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."
> I don't see this argument as being in any way sketchy.

Of course it's sketchy. The argument shows only that, if 15 checkers of the same colour are on the bar, then no other checkers are on the bar.
It does nothing to show why we can't have (for example) 14 white checkers and two black checkers on the bar.

My argument, in the post you're replying to, is much clearer and if you do produce a next edition, you're welcome to use it.

Paul Epstein

On 3/31/2023 1:29 PM, Raymond Kershaw wrote:
> These rather tortuous explanations are dealt with more succinctly in Backgammon Funfair: "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."
> I don't see this argument as being in any way sletchy.

The trouble is that this argument only says what can happen if you
start with 15 White checkers on the bar. It doesn't prove that
there isn't some other way of getting to (for example) 8 White
checkers and 8 Red checkers on the bar without first getting 15
checkers of one color on the bar.

---
Tim Chow

On Saturday, April 1, 2023 at 12:48:29 AM UTC+1, Timothy Chow wrote:
> On 3/31/2023 1:29 PM, Raymond Kershaw wrote:
> > These rather tortuous explanations are dealt with more succinctly in Backgammon Funfair: "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."
> > I don't see this argument as being in any way sletchy.
> The trouble is that this argument only says what can happen if you
> start with 15 White checkers on the bar. It doesn't prove that
> there isn't some other way of getting to (for example) 8 White
> checkers and 8 Red checkers on the bar without first getting 15
> checkers of one color on the bar.

From a content point of view, this reads like a duplicate of what I posted.
However, I had a (justifiably) irritated tone in my response, which is not
present in yours.

Paul