Probability and Statistics

Forum Archive : Probability and Statistics

Chance of rolling x doubles in y rolls

From:   Raccoon
Date:   15 July 2007
Subject:   Chance of rolling X doublets in Y rolls
Forum:   GammOnLine

Chance of rolling exactly X doublets in Y rolls =

    Y!      5^(Y-X)
--------- * -------
X!*(Y-X)!     6^Y

Example: chance of rolling exactly 1 doublet in 3 rolls =

    3!      5^(3-1)   3*2   5^2    75
--------- * ------- = --- * --- = --- = 34.72%
1!*(3-1)!     6^3      2    6^3   216

Example: chance of rolling exactly 2 doublets in 4 rolls =

    5!      5^(4-2)    4*3*2*1    5^2   24    25     150
--------- * ------- = --------- * --- = -- * ---- = ---- = 11.57%
2!*(4-2)!     6^4     2*1 * 2*1   6^4    4   1296   1296

Example: chance of rolling 8 doubles in 12 rolls simplifies to

495 * 5^4th * 6^12th = 0.0142%

Example: if X = O, then the chance of rolling 0 doubles in Y rolls is
simply 5^Y/6^Y.

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
translated to

(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
> opponent?

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.

Douglas Zare
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Probability and Statistics

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Average luck of each roll  (Jørn Thyssen+, Feb 2004)  [GammOnLine forum]
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Chance of rolling x doubles in y rolls  (Raccoon+, July 2007)  [GammOnLine forum]
Chance of rolling x or more pips in y rolls  (Tom Keith, Feb 2004)  [GammOnLine forum] [Long message]
Clumping of random numbers  (Gary Wong, Sept 1998) 
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Distribution of points per game  (Roland Sutter, June 1999) 
Distribution of points per game  (Stig Eide+, Sept 1995)  [Recommended reading]
Expected variation in points after a series of games  (Achim Müller+, Feb 1999) 
How many games to decide who's better?  (Stephen Turner, Mar 1997) 
How often is too often?  (Gary Wong, Oct 1998) 
Losing after bearing off 14 checkers  (Daniel Murphy, July 1999) 
Number of games per match  (Jason Lee+, Jan 2005) 
Number of rolls to enter x checkers from bar  (Michael Depreli+, Mar 2011) 
Visualizing odds  (Daithi, Mar 2011) 

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