Computer Dice

Forum Archive : Computer Dice

FIBS: Entering from the bar

From:   Tom Keith
Date:   3 April 1997
Subject:   FIBS dice -- MORE data

The following data is taken from the archive of matches
collected by Mark Damish's Big Brother program and stored
at "".  A total of 2553 matches were
analyzed (8463 games), and in those matches the player
on roll had a checker (or checkers) on the bar 91503 times.
The following table summarizes how often that player was
able to enter at least one of his checkers, classified
according to the number of points that were open in the
opponent's home board.  I leave it to more statistically-
inclined readers to interpret this data.

Points Open    Successes / Attempts     Rate*36  (Expect)

     0                 0 /  6486          0.00      (0)
     1              4973 / 16910         10.59     (11)
     2             10802 / 19529         19.91     (20)
     3             14170 / 18972         26.89     (27)
     4             14953 / 16847         31.95     (32)
     5             12288 / 12629         35.03     (35)
     6               130 /   130         36.00     (36)

   Total           57316 / 91503


Chuck Bower  writes:

Tom Keith wrote:
>                                                       --added by CRB---
> Points Open  Successes / Attempts   Rate*36  (Expect) expected variance
>                                                       entries
>      0               0 /  6486        0.00      (0)
>      1            4973 / 16910       10.59     (11)   5166.94  3588.66
>      2           10802 / 19529       19.91     (20)  10849.44  4821.98
>      3           14170 / 18972       26.89     (27)  14229.00  3557.25
>      4           14953 / 16847       31.95     (32)  14975.11  1663.90
>      5           12288 / 12629       35.03     (35)  12278.19   341.06
>      6             130 /   130       36.00     (36)
> TK Total         57316 / 91503
> CRB Total        57186 / 84887                       57498.60 13972.34

      The last two columns were added by me.  I don't count the first
and last rows, which are a good "sanity check" on the programming but
of doubtful use statistically.  (If you do the statistics with them,
I think you'd get the same answer...  Check me, please!)

     So the discrepancy (in standard deviations) using Rob' method is:

          (57186 - 57498.6)/sqrt(13972.34) = -2.64

For a 2-tailed experiment, this has a probability of 0.8% of being
consistent with random dice.  I BEG ALL OF YOU TO PLEASE CHECK MY

     So, the plot thickens.  I think there are a couple things worth

1) In this experiment, as in Rob's, there was a big discrepancy for
entering on a five point board, with not much on the other size
boards.  This may be a clue IF there is a problem.  It may also be
just a coincidence.

2) The RATIO's of expected/actual have gotten closer to 1, BUT, the
much larger number of trials (84,887 compared to 431) is more sensitive
and thus the discrepancy in terms of number of standard deviations is
larger than before.  This shouldn't be too surprising.

3) It would be interesting to see the case--ONLY one checker on the
bar--enumerated separately.  This might or might not tell something.
In particular, if there is a "programming" error in Tom's data
collection (no offense, PLEASE!) it might show up.  Also, if there
really is a random number generator problem in FIBS, this extra
piece of info might be another clue.

     Out of curiosity, did anyone take you up on your bet, Kit?


Kit Woolsey  writes:

Robert P. Smith ( wrote:
> We are now left with three active hyptoheses:
> A)  FIBS dice are non-random.
> B)  There was a programming error by Tom Keith.
> C)  Big_Brother happened to observe an extended set of unlikely events
> (correctly calculated by Chuck Bower to be <0.8% probability).
> Also, for the one-point open case the observation is 3.24 standard
> deviations away from the expected value, which has a two-tailed
> probability of 0.12%.
> Explanations anyone?
> Rob Smith

Yeah, I think I can give it a try.  There is a fourth possible
hypothesis:  That the Big_Brother data isn't complete!

Not complete, you ask?  In what way?  Doesn't Big_Brother record the full
games?  Maybe not.

Consider the following (all-too-familiar) scenario:  You are stuck on the
bar against a closed board, and your opponent starts peeling his
checkers off.  The situation is such that you MUST hit a shot to have a
chance, either because you are stone cold gammoned if you don't or
because your opponent doesn't need a gammon to win the match.
Eventually, you enter.  What do you do?  You don't play the move -- you
resign!  We would have to check it out, but my guess is that when this
happens Big_Brother doesn't record that last entering dice roll or the
entrance which was never made.  Note that if you had stayed on the bar,
the roll would have been recorded.  Consequently, some of the entering
numbers which should be in the data simply aren't there, while all of the
flunking numbers are.  Granted this only occurs occasionally, but it
doesn't have to happen very often to cause the apparently skewed results
Tom's survey got us.

Great programming job, Tom.  However, if I happen to be correct about the
missing data do you think you could tweak your program a bit to take
these last rolls which were not recorded into account (if you already did
so, my hypothesis goes down the drain, of course).

Tom Keith  writes:

Good thinking!  I checked the BB data.  In 500 games, the game ended when
a player resigned while he had a checker on the bar.  It seems likely in
the vast majority of those cases that the resigning player actually
rolled first and then resigned when he saw that he was forced to enter
his checker.  BB does not seem to record the roll if no play is made.
Here's a breakdown of those games:

    Points Open    Games Resigned With
                   a Checker on the Bar
         0                  5
         1                 37
         2                126
         3                155
         4                113
         5                 52
         6                 12
       Total              500

If we assume that all these games represent rolls that enter (probably
not quite true), then the previous totals can be revised as follows:

Points Open
     0           5/ 6491    =  0.03/36   <-(Obviously false assumpt here)
     1        5010/16947    = 10.64/36
     2       10928/19655    = 20.02/36
     3       14325/19127    = 26.96/36
     4       15066/16960    = 31.98/36
     5       12340/12681    = 35.03/36
     6         142/  142    = 36.00/36

E&OE :)


Stephen Turner  writes:

OK, I'll analyse Tom Keith's data with the Woolsey Correction:

> Points Open   observed      expected  variance
>      1        5010/16947     5178.3     3596.0
>      2       10928/19655    10919.4     4853.1
>      3       14325/19127    14345.3     3586.3
>      4       15066/16960    15075.6     1675.1
>      5       12340/12681    12328.8      342.5
               -----          -----      -----
               57669          57847      14053

(57847-57669)/sqrt(14053) = 178/118.5 = 1.5

This is well within normal amounts of randomness. The first row is a bit
out, but the other rows are now very close to their theoretical answers. It
looks as if Kit has hit the nail right on the head once again.
Did you find the information in this article useful?          

Do you have any comments you'd like to add?     


Computer Dice

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FIBS: Entering from the bar  (Tom Keith+, Apr 1997) 
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