... and thinking about their implications for playing craps.

There were three sets of dice (A, B, C) to be analyzed. We saw in class that "C" was suspicious, with "A" and "B" looking normal. The plot (which I made in Excel; you can see the raw data in the .xls file in this directory or here as html) summarizes what I found for the "C" dice, using several different approaches to "rolling" them. Also on the plot on the far right are my results from last year.

What I got 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.

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,

Then p = 2.040/6 = 0.340.

To get 2 σ, the 95% confidence range of that p, the formula is

where (a) we divide by 6 dice to convert "how many 1's or 6's" to a probability fraction, and (b) we divide by sqrt(100) because we want the variation in the average of the 100 data points. (That's the 1/sqrt(N) I've been emphasizing.)

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.

(Warning what follows is my convincing myself that the difference is statistically significant. Unless you've taken a statistics course, it may not make all that much sense. That's OK; when/if you need to this sort of stuff for your own experiments, ask someone. All I'm really doing is looking at how far apart the data points are, and whether "outside the error bars" means that they really are different.)

Subtracting the means and combining the σ's gives

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. Here "=" at the front means "put this formula into this box", and the C21:C120 indicate a range of data cells where my numbers were. I actually clicked and dragged through them to put that into the formula.

The functions I used were

(By the way, if anyone wants any help making plots or whatever in Excel or with other tools, come find me.)

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 evaluate that properly.

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:

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.

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 even in a fairly simple (one might even say contrived) case like this isn't all that easy, either in terms of the repetitive dice rolling you need (thousands of dice rolls - not all that different from a lot of the tedious science I've done, actually) or in terms of the math you may well need to get your point across (ditto).

FYI, I spent about an hour rolling dice, and about three hours thinking about what I had, doing the calculations and plot in Excel, analysing the craps game, writing it all up, and putting it online. Your mileage may vary.

So there you are.

http://cs.marlboro.edu/ courses/ spring2006/empirical_science/ dice_and_statistics/ jims_results
last modified Sunday February 11 2007 11:05 pm EST