Feb 2006
Jim Mahoney

 -- Measuring some dishonest dice probabilities
    and thinking about their implications for playing craps.

My results this year surprised me.  I tried to be thorough by rolling
'em several different ways, expecting that all would give me the same
results - but they didn't.

I also see now that I wasn't thorough enough in writing down last year
exactly what I did.  In particular, although I recorded how many 1's
and 6's I got from rolling 10 dice, and although I know that I rolled
them in batches, I'm not sure now *how many* dice I put into one of
the bottles at once.  I'm guessing that it makes a difference, and
that the "larger area is more likely" doesn't apply when the dice get
too crowded.

This year I rolled 6 dice at once in 3 different jars.  The idea was
to get a rolling technique that was quick, simple, and didn't require
picking up the dice each time - I left the lid on the jar, shook it,
and then turned it upside down, and counted how many of the dice were
1 or 6 as they lay in the cap.  Once complication was that fitting all
six in the cap was possible but not automatic; I often had to shuffle
the jar back and forth to get all six to lie flat in the cap.

Also, the jar with the yellow cap broke part way through - the force
of the dice rolling cracked the glass.

I also rolled the six in an more typical way, cupping them in my hands,
shaking them to mix 'em up, then tossing them into an empty plastic
toolbox (to keep them from rolling off the table) so they'd bounce off
a "back wall" like at a gaming table.

                                        mean p(1 or 6)    2 sigma         N dice
 glass jar (brown cap)               0.340             0.038           600
 glass jar (yellow cap)              0.322             0.054           258
 glass jar (blue cap)                0.336             0.054           342
 plastic toolbox                      0.403             0.049           300
 2005 data (blue cap -               0.437             0.032          1000
            five dice per roll ? )


To illustrate the procedure for generating these numbers, I'll use the
first "brown cap" data.

The raw data is "how many 1's or 6's with 6 dice" = (2, 2, 1, 1, ... )
for 100 total rolls.  Then for these numbers,
  mean = 2.040, s = 1.154 (using "(n-1) best estimate" standard dev)
Then p = 2.040/6 = 0.340.

To get 2 sigma, the 95% confidence range of that p, the formula is
  2 sigma = 2 * s / 6 / sqrt(100)
where (a) we divide by 6 to convert from a number to a fraction, and
(b) we divide by sqrt(100) because we want the variation in the
average of 100 numbers.

The surprise is that rolling six dice in the jars gave results
consistent with completely honest dice - not at all what I expected,
or what I got last year.  I *think* the difference is that last year I
rolled fewer dice in a single roll in the jars, giving them more room
to in the cap - but since I didn't record my procedure last year, and
since I haven't tested that idea, I guess for now it'll just remain a
guess.

In particular, the 1st set of "brown cap" data and the 2005 data are
significantly different.  Subtracting the means and combining the
sigma's :

 x = delta p = 0.437 - 0.340 = 0.097
 s = sigma delta p = sqrt( 0.019**2 + 0.016**2 ) = 0.025
 so z = x/s = 3.88 (z is normal with mean=0 and stdev=1)

A value that extreme happens by random chance only this fraction :
2*(1-NORMSDIST(3.88)) = 0.000105 of the time.  Not likely.

I did my numeric work in Excel, using expressions like
 "=AVERAGE(C21:C120)" as cell formulas.  
The functions I used were
 AVERAGE(x1,x2,...)
 STDEV(x1,x2,...)
 COUNT(x1,x2,...)
 NORMSDIST(z)  = 1 sided cumulative area under normal where z has sigma=1

So my working hypothesis is that six dice rolled in a glass jar do
*not* give a good measure of the probability of these crooked dice as
rolled in normal circumstances.  It'd take more time than I have now
to test that.

At least the data this year from rolling in the toolbox and last
year's data are consistent.  Combining the two with weighted averages
gives me a best estimate of p(1)=p(6)=0.203 with sigma=0.013 .

Now - craps.  The rules look like this:
 * 1st roll :
    * 7 or 11 wins
    * 2, 3, or 12 loses
    * 4, 5, 6, 8, 9, 10 sets the "point"
 * successive rolls (if point is set)
    * point wins
    * 7 loses
    * any other number means keep going.
I wrote a short computer program to report the odds of winning with
these dice.  The results are a bit counter-intuitive: you actually win
*less* often with the loaded dice.

 probability of winning with honest dice = 0.493
 probability of winning with p(1)=p(6)=0.203 dice = 0.471

The reason behind this is that while the dice increase the odds of
your winning on the first throw (because 7 and 11 are more likely),
most of the time you *don't* win on the first throw; instead you have
to make your point.  This is true both with the honest dice
(probability of continuing = 0.667) and the crooked dice (probability
of continuing = 0.626).  And once you have to make your point, the
first 7 loses - which is more likely with the crooked dice.
The details are at the end of the craps.pl file.

The upshot of all this is that getting the right answer isn't entirely
trivial, either in terms of some tedious dice rolling (not all the
different from a lot of the science I've done) or in terms of the math
you need (ditto).  But it can still be kind of fun, IMHO.