Prime vs. prime cube action
tc...@lsa.umich.edu
-13--14--15--16--17--18--BAR--19--20--21--22--23--24-
| X . . . O O | | O O O . . X |
| O O | | O O O |
| | | O |
| | | |
| | | |
| | O | |
| | O | HOME | [1]
| | | |
| | | |
| | | |
| | | |
| O | | X X X X X |
| O . . . X X | | X X X X X X |
-12--11--10---9---8---7--BAR--6---5---4---3---2---1--
X is on roll and doubled. O took, and then X rolled 66. O held the cube,
and then X rolled 65. O then redoubled and X dropped.
Was this sequence of cube actions correct? I'm thinking that perhaps O
should have redoubled immediately after X's anti-joker 66, despite being
closed out. But I freely admit I don't understand this position very well.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
segl...@googlemail.com
Hi Tim,
is this a money game or a match (to n points)? It probably doesn't
matter much in this specific situation - but generally, cube decisions
depend also on the match score and the match length (i.e. how many
points both players are away from winning the match).
Now let's assume it was a mtch to 5 points, the score is 0 - 0.
In the given situation, X clearly has a cube, as he's a favourite to
close O out and also jump O's prime, unless he rolls all too big
numbers (thus having to break his home board) of course. O just has an
initial take, but NO redouble after X's double 6 (where it would be a
huge mistake), and most probably not even after X's 65. If O
redoubles, X has an easy take. Even after 6-5, when X had to leave a
blot on his 6 point, O still has to enter both his chequers from the
bar. This give X an opportunity to roll a 2 and escape his only back
man. And he has a huge lead in the race. No redouble so far for O.
Chris
Peter Schneider
<tc...@lsa.umich.edu> wrote
> The following position is taken from a game I played today.
>
> -13--14--15--16--17--18--BAR--19--20--21--22--23--24-
> | X . . . O O | | O O O . . X |
> | O O | | O O O |
> | | | O |
> | | | |
> | | | |
> | | O | |
> | | O | HOME | [1]
> | | | |
> | | | |
> | | | |
> | | | |
> | O | | X X X X X |
> | O . . . X X | | X X X X X X |
> -12--11--10---9---8---7--BAR--6---5---4---3---2---1--
>
>
> X is on roll and doubled. O took, and then
> X rolled 66. O held the cube,
> and then X rolled 65. O then redoubled and X dropped.
>
> Was this sequence of cube actions correct?
> I'm thinking that perhaps O
> should have redoubled immediately
> after X's anti-joker 66, despite being
> closed out.
Being closed out here is rather an advantage, isn't it? I'd think that if
anybody had an initial double it was not X but O (in all reality, probably
not just quite yet, but close). X will have trouble jumping the 5-prime;
after 3 or 4 rolls or so X will crash, and that's on average much fewer
rolls than needed to jump. Consequently yes, I'd have redoubled after X's
66. That O has 2 on the bar complicates matters somewhat because it will
take him long to enter both; consequently X has more time to jump in case
he isn't forced to expose checkers in his home board which get hit.
But still: Wrong double, redoubled too late.
Best,
Peter aka the juggler
tc...@lsa.umich.edu
It was a money game.
Your analysis makes sense to me. What I'm thinking now is that there might
be some small table of numbers worth calculating and memorizing to understand
these sorts of positions. Something like, if I need to roll a 2 followed by
a 6 to escape, what is the probability that I'll eat up at least 10 pips
(or 20 pips, or 30 pips, etc.) while waiting?
For a simpler problem, say I need to roll a 1, and I want to know the
probability p(n) that I will eat up at least n pips before I roll my first 1
(I'm not going to include the pips appearing on the other die when I roll
my first 1). Then unless I screwed up my math, here are some of the values
of p(n):
n p(n)
------------
58 .10
39 .20
27 .30
20 .40
13 .50
9 .58
1 .69
The table for "need to roll a 2" is not exactly the same, but ignoring that
distinction for the moment, we see that in the diagrammed position, the
probability would be about 30% that all of X's spare pips would be used up
before getting the 2 needed to escape the back checker. (This assumes of
course that O dances the whole time.)
I might carry out the more complicated calculation of a 2 followed by a 6
later, if someone else doesn't beat me to it.
segl...@googlemail.com
> In article <cee2acb0-4916-4c4c-930f-6fb513067...@r33g2000yqn.googlegroups.com>,
Hi Tim,
I agree, putting up such a table is a very good idea to get closer to
understanding the real odds in these positions.
As my math is probably not half as good as yours, I tried another
approach:
There are 25 rolls without a 1. The average roll including doublets
(again without 1s) counts 280/25 = 11.2 pips.
Any 1 (as any single number) occurs 11 times in 36 rolls (in about 30
% of the rolls), thus we need 3.27 rolls until it occurs.
3.27 rolls x 11.2 pips/roll = 36.6 pips in average 'used up' before
the first 1 appears.
Is it ok to calculate such an average (or do we need to take 4 rolls)?
> I might carry out the more complicated calculation of a 2 followed by a 6
> later, if someone else doesn't beat me to it.
>
Sure, such a calculation for the "need to roll a 2" (as required in
your position), is significantly more complex
- as the 1s also help to advance your back chequer.
Let's just assume the average 'pips usage' was about the same (maybe a
bit less) while waiting for the '2'. In the game you played,
interestingly, X rolled 2 of his most 'unfortunate' numbers when it
comes to minimizing unwanted pips 'usage' (6-6 and 6-5 = 35 pips)
after he had turned the cube. That was about the overall expectation
in only 2 (instead 3.3) moves, and in addition this forced him to
leave a blot on his 6-point. Couldn't be much worse, lol.
Plus, the 6 needed to jump O's 5-prime has to appear as well of
course. But with just a little 'luck', X should have managed to close
his board perfectly with just the average "pips usage", before he
rolls the 2 for the 24/22. Then we have O waiting for X to open up his
board to enter both chequers, opposed to X waiting for the one
required '6' to come ...
Now back to your game, O was actually a small favourite to win the
game after X's 6-5, but probably still not enough for him to recube
with still 2 on the bar - and if offered the cube, X should probably
still take.
I don't know if that is plausible. Admittedly, I rarely manage to do
the calculation over the board. I'm just not quick enough at this, so
I'm left to approximative evaluations, which is more of a
'feeling' :-) ... - but I definitely agree that it is a very good idea
to do it regularly - so we can actually see if our 'feeling' comes
close to the expectable reality.
Cheers
Chris
paulde...@att.net
Peter is right that you shouldn't double because you're likely to
crash before escaping. I think your opponent should have beavered
your double. After 66, your opponent can double you out. Big mistake
for your opponent to hold.
Paul Epstein
tc...@lsa.umich.edu
<paulde...@att.net> wrote:
>Peter is right that you shouldn't double because you're likely to
>crash before escaping. I think your opponent should have beavered
>your double. After 66, your opponent can double you out. Big mistake
>for your opponent to hold.
As it happens, I was the one on the bar.
But anyway, I'm getting confused again as to what I believe about the
position. Could someone perhaps post some rollout results? I don't
own Snowie or Jellyfish.
segl...@googlemail.com
> As it happens, I was the one on the bar.
>
> But anyway, I'm getting confused again as to what I believe about the
> position. Could someone perhaps post some rollout results? I don't
> own Snowie or Jellyfish.
Sure, I am running it right now on GNUbackgammon (which is free to use
for everyone).
As I want the results to be reliable, I'm running a overall 2-ply
rollout at 972 trials for every crucial position, so it might take a
few hours :-)
Cheers
Chris
tc...@lsa.umich.edu
>I might carry out the more complicated calculation of a 2 followed by a 6
>later, if someone else doesn't beat me to it.
O.K., I did this, and I believe that the probability of rolling a 2 and then
a 6 while using up at most 25 pips in the process is about 47%. This is not,
strictly speaking, the same as escaping the back checker before having to
break the home board, but it should be pretty close.
If X escapes before cracking then X will probably win a gammon. Even if X
cracks before escaping, X is not necessarily lost. So the initial position
is clearly a double for X. I'm still not sure if it's a take for O; if so,
it seems like a borderline take at best.
After the anti-joker 66, X is very likely to crack before escaping, but
again, it's quite complicated to assess X's chances of escaping before
O enters twice, since we also have to factor in the probability of O
sending another X checker back with a hit. Looks like a rollout may be
needed here.
For those who are interested in the mathematical details, here is how I
got the 47% figure. I assumed that double-1's counts as a 2, but that
other 1's count against us. The trick is to use a mathematical gadget
called a "generating function," which is a polynomial that keeps track
of probabilities and pip counts at the same time. The generating function
for "not rolling a 2" is the polynomial
a(x) = (1/18)*x^(1+3) + (1/18)*x^(1+4) + (1/18)*x^(1+5) + (1/18)*x^(1+6)
+ (1/36)*x^(3+3+3+3) + (1/18)*x^(3+4) + (1/18)*x^(3+5)
+ (1/18)*x^(3+6) + (1/36)*x^(4+4+4+4) + (1/18)*x^(4+5)
+ (1/18)*x^(4+6) + (1/36)*x^(5+5+5+5) + (1/18)*x^(5+6)
+ (1/36)*x^(6+6+6+6)
Each term here represents a roll that doesn't have a 2; the coefficient
is the probability of that roll and the exponent of x is the number of pips.
Similarly we can write down the generating function b(x) for "getting a 2
but not a 6," where now the exponent of x is the pip count of the roll
*minus* the 2 that we use profitably:
b(x) = (1/36)*x^(1+1+1+1-2) + (1/18)*x^(1+2-2) + (1/36)*x^(2+2+2+2-2)
+ (1/18)*x^(2+3-2) + (1/18)*x^(2+4-2) + (1/18)*x^(2+5-2)
Finally we can write down the generating function c(x) for "not getting a 6."
This is similar to a(x) and I won't bother writing c(x) down explicitly.
If we continue to roll until we get a 2 and then a 6, then the process can
be broken down into several stages:
1. failing to roll a 2;
2. then either rolling a 26, whereupon we're done, or
a. rolling a 2 but not a 6 (double-1's counts here);
b. continuing to fail to roll a 6;
c. finally rolling a 6.
The generating function for failing to roll a 2 is
1 + a(x) + a(x)^2 + a(x)^3 + ... = 1/(1 - a(x))
where we use the high-school formula for summing a geometric series.
Similarly the generating function for failing to roll a 6 is 1/(1 - c(x)).
The generating function for the full process of rolling a 2 and then a 6 is
(1/(1 - a(x))) * (1/18 + b(x) * (1/(1 - c(x))) * 11/36)
If we multiply this expression out then we get a function whose Taylor series
starts off
1/18 + (11/648)*x + (11/1296)*x^2 + (11/648)*x^3 + (245/11664)*x^4 + ...
The coefficient of x^n here is the probability that we will use up exactly
n pips before getting a 2 and then a 6. Summing up the coefficients up to
and including the term x^25 gives approximately 0.47277, which is where I
got my 47% figure.
segl...@googlemail.com
> In article <4a16c229$0$510$b45e6...@senator-bedfellow.mit.edu>, I wrote:
> >I might carry out the more complicated calculation of a 2 followed by a 6
> >later, if someone else doesn't beat me to it.
>
> O.K., I did this, and I believe that the probability of rolling a 2 and then
> a 6 while using up at most 25 pips in the process is about 47%. This is not,
> strictly speaking, the same as escaping the back checker before having to
> break the home board, but it should be pretty close.
>
> If X escapes before cracking then X will probably win a gammon. Even if X
> cracks before escaping, X is not necessarily lost. So the initial position
> is clearly a double for X. I'm still not sure if it's a take for O; if so,
> it seems like a borderline take at best.
>
> After the anti-joker 66, X is very likely to crack before escaping, but
> again, it's quite complicated to assess X's chances of escaping before
> O enters twice, since we also have to factor in the probability of O
> sending another X checker back with a hit. Looks like a rollout may be
> needed here.
Hello again everybody!
Wowww Tim - I don't get it. I am truly impressed by your stunning
mathematical skills - of which I can only dream of. Furthermore,
everything else you assumed (questionned) about the rollout outcomes
when you weren't exactly sure based on your own calculations, is true
as well! - Pheww, you just have to be excellent not only in backgammon
but also in any other game to be mastered on calculus of probability
or other mathematical analysis - Respect!! :-)
OK, here are the GNUbackgammon rollouts for the previously discussed
cube decisions.
Please note: Had O erronously redoubled X to 4 after move 1 (the
forced 66 played by X), that would have been a beaver for X. I didn't
manage to get that into the rollout output - but I have a screenshot
og this. If you're interested, I'm happy to run it again. Or please
just insert the match/position-IDs to GNUBg and roll it out for
yourself.
O's error in redoubling to 4 after X's move 3: 65 (played 6/1) lies in
handing over the cube too early in an unclear position. X can still
move 24/22 before O enters (and perhaps hits on the 6 point). If so,
he in return would have a recube to 8.
Finally, in the game text, I let X take O's final cube (to 4), but
this doesn't matter at all for the following rollout results:
The score (after 0 games) is: O 0, X 0
Match Information:
Date: 23.05.2009
GNU Backgammon Position ID: 2G6DAWBt2woCIA
Match ID : cAkFAAAAAAAA
+13-14-15-16-17-18------19-20-21-22-23-24-+ O: O
| X O O | O | O O O X | 0 points
| O O | O | O O O |
| | | O |
| | | |
| | | |
v| |BAR| | (Cube: 1)
| | | |
| | | |
| | | |
| O | | X X X X X |
| O X X | | X X X X X X | 0 points
+12-11-10--9--8--7-------6--5--4--3--2--1-+ X: X
Pip counts: O 142, X 93
* X doubles
Rollout details:
Centered 1-cube:
0,616 0,449 0,004 - 0,384 0,051 0,001 CL +0,631 CF +0,775
[0,002 0,002 0,000 - 0,002 0,001 0,000 CL 0,005 CF 0,012]
Player O owns 2-cube:
0,633 0,467 0,003 - 0,367 0,054 0,001 CL +1,364 CF +1,029
[0,002 0,002 0,000 - 0,002 0,001 0,000 CL 0,009 CF 0,013]
Full cubeful rollout with var.redn.
972 games, Mersenne Twister dice gen. with seed 859633756 and quasi-
random dice
Stop when std.errs. are small enough: ratio 0,1 (min. 144 games)
Play: world class 2-ply cubeful prune [world class]
keep the first 0 0-ply moves and up to 8 more moves within equity 0,16
Skip pruning for 1-ply moves.
Cube: 2-ply cubeful prune [world class]
GNU Backgammon Position ID: 2G6DAWBt2woCIA
Match ID : cBEFAAAAAAAA
+13-14-15-16-17-18------19-20-21-22-23-24-+ O: O
| X O O | O | O O O X | 0 points
| O O | O | O O O | Cube offered at 2
| | | O |
| | | |
| | | |
v| |BAR| |
| | | |
| | | |
| | | |
| O | | X X X X X |
| O X X | | X X X X X X | 0 points
+12-11-10--9--8--7-------6--5--4--3--2--1-+ X: X
Pip counts: O 142, X 93
* O accepts
Alert: wrong take ( -0,029)!Rollout details:
Centered 1-cube:
0,616 0,449 0,004 - 0,384 0,051 0,001 CL +0,631 CF +0,775
[0,002 0,002 0,000 - 0,002 0,001 0,000 CL 0,005 CF 0,012]
Player O owns 2-cube:
0,633 0,467 0,003 - 0,367 0,054 0,001 CL +1,364 CF +1,029
[0,002 0,002 0,000 - 0,002 0,001 0,000 CL 0,009 CF 0,013]
Full cubeful rollout with var.redn.
972 games, Mersenne Twister dice gen. with seed 859633756 and quasi-
random dice
Stop when std.errs. are small enough: ratio 0,1 (min. 144 games)
Play: world class 2-ply cubeful prune [world class]
keep the first 0 0-ply moves and up to 8 more moves within equity 0,16
Skip pruning for 1-ply moves.
Cube: 2-ply cubeful prune [world class]
_____________
Move number 1: X to play 66
GNU Backgammon Position ID: 2G6DAWBt2woCIA
Match ID : QQkbAAAAAAAA
+13-14-15-16-17-18------19-20-21-22-23-24-+ O: O (Cube: 2)
| X O O | O | O O O X | 0 points
| O O | O | O O O |
| | | O |
| | | |
| | | |
v| |BAR| |
| | | |
| | | |
| | | |
| O | | X X X X X | Rolled 66
| O X X | | X X X X X X | 0 points
+12-11-10--9--8--7-------6--5--4--3--2--1-+ X: X
Pip counts: O 142, X 93
* X moves 13/1 8/2 7/1
_____________
Move number 2: O to play 33
GNU Backgammon Position ID: d9sGACDYboMBYA
Match ID : AYENAAAAAAAA
+12-11-10--9--8--7-------6--5--4--3--2--1-+ O: O (Cube: 2)
| O O | O | O O O X | 0 points
| O O | O | O O O | Rolled 33
| | | O |
| | | |
| | | |
^| |BAR| |
| | | |
| | | |
| | | X X |
| O | | X X X X X X |
| O | | X X X X X X | 0 points
+13-14-15-16-17-18------19-20-21-22-23-24-+ X: X
Pip counts: O 142, X 69
* O cannot move
_____________
Move number 3: X to play 65
GNU Backgammon Position ID: 2G6DAWB32wYAIA
Match ID : QQkXAAAAAAAA
+13-14-15-16-17-18------19-20-21-22-23-24-+ O: O (Cube: 2)
| O O | O | O O O X | 0 points
| O O | O | O O O |
| | | O |
| | | |
| | | |
v| |BAR| |
| | | |
| | | |
| | | X X |
| O | | X X X X X X | Rolled 65
| O | | X X X X X X | 0 points
+12-11-10--9--8--7-------6--5--4--3--2--1-+ X: X
Pip counts: O 142, X 69
* X moves 6/1
_____________
Move number 4: O on roll, cube decision? GNU Backgammon Position ID:
77YFACDYboMBYA
Match ID : AQEAAAAAAAAA
+12-11-10--9--8--7-------6--5--4--3--2--1-+ O: O (Cube: 2)
| O O | O | O O O X | 0 points
| O O | O | O O O | On roll
| | | O |
| | | |
| | | |
^| |BAR| |
| | | |
| | | X |
| | | X X |
| O | | X X X X X |
| O | | X X X X X X | 0 points
+13-14-15-16-17-18------19-20-21-22-23-24-+ X: X
Pip counts: O 142, X 64
* O doubles
Alert: wrong double ( -0,043)!Rollout details:
Player O owns 2-cube:
0,678 0,131 0,002 - 0,322 0,168 0,000 CL +0,321 CF +0,596
[0,001 0,002 0,000 - 0,001 0,002 0,000 CL 0,004 CF 0,007]
Player X owns 4-cube:
0,682 0,132 0,005 - 0,318 0,168 0,000 CL +0,667 CF +0,553
[0,002 0,003 0,000 - 0,002 0,002 0,000 CL 0,010 CF 0,014]
Full cubeful rollout with var.redn.
691 games, Mersenne Twister dice gen. with seed 859633756 and quasi-
random dice
Stop when std.errs. are small enough: ratio 0,1 (min. 144 games)
Play: world class 2-ply cubeful prune [world class]
keep the first 0 0-ply moves and up to 8 more moves within equity 0,16
Skip pruning for 1-ply moves.
Cube: 2-ply cubeful prune [world class]
_____________
Move number 5: O doubles to 4 GNU Backgammon Position ID:
77YFACDYboMBYA
Match ID : ARkAAAAAAAAA
+12-11-10--9--8--7-------6--5--4--3--2--1-+ O: O
| O O | O | O O O X | 0 points
| O O | O | O O O |
| | | O |
| | | |
| | | |
^| |BAR| |
| | | |
| | | X |
| | | X X |
| O | | X X X X X | Cube offered at 4
| O | | X X X X X X | 0 points
+13-14-15-16-17-18------19-20-21-22-23-24-+ X: X
Pip counts: O 142, X 64
* X accepts
Alert: wrong double ( -0,043)!Rollout details:
Player O owns 2-cube:
0,678 0,131 0,002 - 0,322 0,168 0,000 CL +0,321 CF +0,596
[0,001 0,002 0,000 - 0,001 0,002 0,000 CL 0,004 CF 0,007]
Player X owns 4-cube:
0,682 0,132 0,005 - 0,318 0,168 0,000 CL +0,667 CF +0,553
[0,002 0,003 0,000 - 0,002 0,002 0,000 CL 0,010 CF 0,014]
Full cubeful rollout with var.redn.
691 games, Mersenne Twister dice gen. with seed 859633756 and quasi-
random dice
Stop when std.errs. are small enough: ratio 0,1 (min. 144 games)
Play: world class 2-ply cubeful prune [world class]
keep the first 0 0-ply moves and up to 8 more moves within equity 0,16
Skip pruning for 1-ply moves.
Cube: 2-ply cubeful prune [world class]
__________________________
Cube statistics for game 1
Player O
X
Total cube decisions 2
2
Close or actual cube decisions 2
2
Doubles 1
1
Takes 1
1
Passes 0
0
Missed doubles below CP (EMG (Points)) 0
0
Missed doubles above CP (EMG (Points)) 0
0
Wrong doubles below DP (EMG (Points)) 1 (-0,043 ( -0,086))
0
Wrong doubles above TG (EMG (Points)) 0
0
Wrong takes (EMG (Points)) 1 (-0,029 ( -0,029))
0
Wrong passes (EMG (Points)) 0
0
Error total EMG (Points) -0,073 ( -0,116)
+0,000 ( +0,000)
Error rate mEMG (Points) -36,4 ( -0,058)
+0,0 ( +0,000)
Cube decision rating Awful!
Supernatural
Output generated Sat May 23 06:59:58 2009
by GNU Backgammon 0.90-mingw (Text Export version 1.91)
Cheers :-)
Chris
tc...@lsa.umich.edu
<segl...@googlemail.com> wrote:
>OK, here are the GNUbackgammon rollouts for the previously discussed
>cube decisions.
Very nice...thank you. So, it seems that both my opponent and I
overestimated the power of O's 5-prime.
David C. Ullrich
>In article <1d6015c4-ddb1-4ed6...@s12g2000yqi.googlegroups.com>,
> <paulde...@att.net> wrote:
>>Peter is right that you shouldn't double because you're likely to
>>crash before escaping. I think your opponent should have beavered
>>your double. After 66, your opponent can double you out. Big mistake
>>for your opponent to hold.
>
>As it happens, I was the one on the bar.
>
>But anyway, I'm getting confused again as to what I believe about the
>position. Could someone perhaps post some rollout results? I don't
>own Snowie or Jellyfish.
gnubg is free! Run, don't walk to www.gnubg.org .
(Note the '.org'. You don't want gnubg.com - they're evil people.)
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
segl...@googlemail.com
>Very nice...thank you.
You're very welcome.
It was a good exercise for me, this being the first GNUbg rollout I
exported/posted.
Next time, I will keep the rollout details info shorter and maybe
adding it once is enough.
>So, it seems that both my opponent and I
>overestimated the power of O's 5-prime.
Those decisions are hard to spot over the board.
There are many excellent players here, and nobody got it right at
first sight.
I thought O would have a take after X's double, which proved wrong as
well.
On 23 Mai, 11:23, David C. Ullrich <dullr...@sprynet.com> wrote:
>
> gnubg is free! Run, don't walk towww.gnubg.org.
I agree, you should get it. There is a detailed documentation coming
with the program, but here's also another more user-friendly
tutorial: http://www.bkgm.com/gnu/AllAboutGNU.html
Have a good weekend.
Chris
tc...@lsa.umich.edu
tc...@lsa.umich.edu
to answer the question, "How many spare pips do I need if I want a
25%/50%/75% chance of leaping a 5-prime, if I am x pips away from the edge?"
I've continued to make the simplifying assumptions I mentioned before; in
particular, if x = 5, say, then I don't use rolls like 21 to creep up closer
to the edge of the prime (not an insane assumption since the closer you are
to the prime, the harder it is to roll what you need to get to the edge).
value of x 25% 50% 75%
--------------------------------------------
1 13 30 54
2 12 28 51
3 11 25 46
4 10 23 43
5 9 23 41
It makes sense that the numbers decrease down each column, both because
larger values of x can be rolled in more ways, and because if x is small
then your bad rolls will on average have more pips than if x is large.
What's slightly surprising to me is how spread apart the 25%/50%/75%
columns are. If I'm trying to roll a 1 then a 6, then I have a 50%
chance of success if I have 30 pips to spare, but if I want a 75%
chance of success then I need 54 pips to spare. And even if I have
only 13 pips to spare then I still have a 25% chance of success.
David C. Ullrich
>In article <n2gf15l3cd6dp6dj8...@4ax.com>,
>David C. Ullrich <dull...@sprynet.com> wrote:
>>gnubg is free! Run, don't walk to www.gnubg.org .
>
>Actually, I don't own a computer either.
Hmm. Surely you can convince the administrator of whatever
system you use that gnubg is vitally important for your vitally
important research?
Never mind...
tc...@lsa.umich.edu
muratk <mu...@compuplus.net> wrote:
>whenever i feel like i need some entertainment, i come to
>read RGB... never fails :))) thank you clowns :)))
Why not entertain yourself further by using the figures I calculated to win
some bets? Forgetting about backgammon, challenge someone to roll two dice
until they roll a 1, and then a 6. Chances are their intuitions about the
odds will be wrong.
muratk
> muratk <mu...@compuplus.net> wrote:
>> whenever i feel like i need some entertainment, i come to
>> read RGB... never fails :))) thank you clowns :)))
> Why not entertain yourself further by using the figures I
> calculated to win some bets?
I tried many times but nobody would bet against me... :(
> Forgetting about backgammon, challenge someone to roll two
> dice until they roll a 1, and then a 6. Chances are their
> intuitions about the odds will be wrong.
Why forget about backgammon?? Remember when I asked in RGB
about SW tools to calculate complex odds based on several
rolls in a row? Well, that's exactly what I was intending
to do (i.e. betting money on predicting that gnudung would
roll certain dice combinations during the next N rolls)...
MK
Peter Schneider
<tc...@lsa.umich.edu> wrote
> In article <n2gf15l3cd6dp6dj8...@4ax.com>,
> David C. Ullrich <dull...@sprynet.com> wrote:
>>gnubg is free! Run, don't walk to www.gnubg.org .
>
> Actually, I don't own a computer either.
But you have an account on an MIT machine. Just compile and install gnubg
for your account only, no need for an admin. It's worth it, no it's
indispensible, actually. I appreciate logical and high level "symbolic"
analysis (a lot!) but it gets you only so far without a reality check in
the shape of a rollout. Computer help has readjusted the backgammon
concepts that we humans hold quite a bit.
Best
Peter Schneider
<segl...@googlemail.com> wrote
> As I want the results to be reliable, I'm running a overall 2-ply
> rollout at 972 trials for every crucial position, so it might take a
> few hours :-)
Not sure how much I would trust a bot here in this prime vs. prime
position. If the decisive events like crashing/jumping are too far away the
bot may not evaluate this position properly. Probably a good idea to check
with a cubeless rollout, in order to avoid bad drops (or cube action in
general) which would distort the picture. What do the bot experts think?
paulde...@att.net
...
> Why not entertain yourself further by using the figures I calculated to win
> some bets? Forgetting about backgammon, challenge someone to roll two dice
> until they roll a 1, and then a 6. Chances are their intuitions about the
> odds will be wrong.
> --
Actually, a similar bet was a common way for hustlers to make money in
the 19th century.
If I remember rightly, you need to shake a pair of dice at least 24 or
25 times to have a >=50% probability of hitting a 66. However, many
people assume that because 18 = 0.5 * 36, a person is 50-50 to hit 66
in a series of 18 shakes. So hustlers have made money by offering an
even-money bet where the bet-taker bets on being able to hit at least
one 66 in 18 rolls.
Paul Epstein
tc...@lsa.umich.edu
I'm obviously not a bot expert, but the hand calculation I did makes me
confident that the initial position was a clear double for X.
Whether it was a take for O seems more up for grabs. The gnubg rollout
suggested a close drop, and like you I'm not sure I trust that result.
The crashing/jumping could be 5 or 6 plies in the future.
After the 66 and the 65, the position has clarified more, with O's hit
being the most important threat. Since that threat is immediate, the
bot will certainly see it. It may not see that X can jump, but if anything
that should mean that the bot *underestimates* X's chances, so if it says
that X has a take then it would seem we should believe that X has a take.
Whether O has a double while closed out is trickier. I think not, again
based on my hand calculations for rolling a 2 then a 6---the gammon chances
are still quite high. I agree, though, that it's risky just to take the
bot's word for it in that position.
tc...@lsa.umich.edu
<paulde...@att.net> wrote:
>If I remember rightly, you need to shake a pair of dice at least 24 or
>25 times to have a >=50% probability of hitting a 66. However, many
>people assume that because 18 = 0.5 * 36, a person is 50-50 to hit 66
>in a series of 18 shakes. So hustlers have made money by offering an
>even-money bet where the bet-taker bets on being able to hit at least
>one 66 in 18 rolls.
Ah yes, I remember reading about this in a Martin Gardner article.
Another amusing problem along these lines is a roulette question that I heard
from Joe Buhler (presented here with some modifications of my own). Say I
bet a dollar on the number 1 at the roulette wheel. Win or lose, I then bet
a dollar on the number 2. Win or lose, I then bet a dollar on the number 3,
and so on around the wheel. I go around the wheel three full times, betting
a dollar each time. (I skip 0, of course; I'm also assuming there's no 00,
although I don't think it matters much for the question I'm asking.)
Question: What is the probability that at the end I'll be behind?
A. greater than 99%;
B. 90% to 99%;
C. 75% to 90%;
D. 50% to 75%;
E. less than 50%.
The problem is more fun if you try to guess the answer first before
calculating it in detail.
paulde...@att.net
> In article <c7479e32-0158-4ccd-8563-6a0baa51f...@v4g2000vba.googlegroups.com>,
Most casinos do have a 00. I assume that this is a normal casino
except for the fact that the casino's rake comes only from the
additional 0, and that no 00 exists.
The gambler has 108 attempts at a 1/37 event. The gambler will be
behind if this 1/37 event occurs less often than 3 times out of 108.
Using informal reasoning, the gambler can be considered lucky if the
event happens 4 or more times and unlucky if the event happens 2 or
fewer times. How about the case that the event happens 3 times?
Well, then the gambler is lucky because the expected number is 2.92
rather than 3 but only very slightly lucky. So a score of 3 is pretty
much a neutral event.
So the problem asks how likely it is that the gambler is particularly
unlucky. So the answer seems to be E, less than 50%. A person is not
usually unlucky (by definition -- for example, failing to win the
lottery is not usually considered "unlucky").
Careful attention must be paid here to whether the relevant inequality
is strict or not.
The point is that the most typical event -- a score of exactly 3 -- is
ruled out by asking if you'll be "behind". If "behind" is replaced by
"not ahead", the answer would be very different.
Paul Epstein