Douglas Zare writes:
> What if the question is x OR MORE doubles in y rolls?
The simplest method (with a computer) is to add up the possibilities.
That's what I usually do.
A good(*) approximation when the number of rolls is large is to translate
the result of "x or more doubles" into "z or more standard deviations above
the mean," and then use a normal approximation, looking up the value z in a
The mean is y/6. The standard deviation is sqrt(y * 5/36) or about sqrt(y)
x is x-(y/6) above the mean, which is x-(y/6) / sqrt(y * 5/36) above the
mean. However, the number of doubles is a count, but the normal
distribution is continuous, so you should replace "x" with "from x-1/2 to
x+1/2," and "x or more" by "x-1/2 or more." So, "x or more" should be
(x-1/2 - y/6)/sqrt(y * 5/36) standard deviations above the mean.
For example, lets look at the statement Kit Woolsey made a while back that
200 or more doubles in 1000 rolls would seem suspicious. That is 199.5 -
166.7 / sqrt(5000/36) = 32.8 / 11.8 ~ 2.79 standard deviations above the
mean, which happens about 1 time in 375. Good intuition, Kit.
The actual value is about 1/314.
* For extreme values, the absolute error may be small, while the
proportional error is large. For example, you will get a small positive
estimate for the probability of at least y+1 doubles in y rolls, when the
actual probability is 0. Unfortunately, we are often interested in such
cases where the probability is small, such as in the above example.
> What are the odds (%) that one will roll x# doubles more than his
If it were a fixed number of rolls, you can just add up the possibilities,
or use a normal approximation to each, and then the difference between
normally distributed random variables is also normally distributed. If it
is from a backgammon game, then it depends greatly on the styles of the
players, since it is much easier to roll more doubles when the game is
longer, or in live play, when your opponent doesn't roll for a long time.