Re: Construct-a-position problem

On Jun 12, 11:41 am, David C. Ullrich <ullr...@xxxxxxxxxxxxxxxx>
wrote:
On Mon, 11 Jun 2007 23:27:23 -0400, Walter Trice



<wtrice6...@xxxxxxxxxxx> wrote:
pauldepst...@xxxxxxx wrote:
...
I doubt that a powerful watertight argument is possible without
resorting to doing the actual (probably tedious) arithmetic.

...

It isn't really that hard. You can ignore the chance of bearing off in
one roll with 4-4 or better, and you can also ignore the possibility
that the other guy rolls doubles on either his first or second turn,
because then the play doesn't matter. All you need to consider is the
sequences that lead up to a last-roll situation where you may or may not
have a double. To adjust for the cube, use the "double the difference
from 18" method, so that 26 winning rolls becomes 34, 23 becomes 28, 27
(or more) becomes 36, etc.

I'll give roll grouping, number of rolls in the group, number of rolls
that win next turn, and adjusted number of rolls for each play:

4/3

[33,22] .............2 36 36
[65,64,63,54,53,43] 12 26 34
[62,52,42].......... 6 25 32
[32]................ 2 23 28
[11]................ 1 17 17
[61,51,41,31,21].... 10 5 5

2/1

[65,64,54,22,33].... 8 36 36
[61,51,41] ..........6 23 28
[62,52,42] ..........6 29 36
[21]................ 2 11 11
[31,32]............. 4 5 5
[63,53,43].......... 6 6 6
[11]................ 1 17 17

So for 4/3, you have 2*36+12*34+6*32+2*28+1*17+10*5=795.

For 2/1, it's 8*36+6*28+6*36+2*11+4*5+6*6+1*17=767.

That's with the cube. For the cubeless comparison, just plug in the
cubeless numbers.

4/3: 2*36+12*26+6*25+2*23+1*17+10*5=647.
2/1: 8*36+6*23+6*29+2*11+4*5+6*6+1*17=695.

So we see that 4/3 comes out on top with a live cube, while 2/1 is
better cubeless.

Now I shall say a little prayer that Mozilla and Usenet do not destroy
my formatting so badly that what I have written becomes unintelligible...

Came out just fine here, thanks. I thought maybe there was an
explanation somewhat simpler than just working out all four
numbers, but I'm glad you posted this, because your calculations
are a lot simpler than mine would have been. (Took me a second to
see why that "double the difference from 18" made sense - I hope
I remember that next time.)

-- Walter Trice

************************

David C. Ullrich

For money, I would have played the correct ace. However, at DMP, I
would have erred and played the worse 4/3.
Assuming that there isn't time to do these computations over the
board, how does a strong player (I don't consider myself that strong)
arrive at the correct play at DMP? Walter's computations show that
4/3 is quite a serious mistake at DMP.

Thanks,

Paul

.

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