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 table. The mean is y/6. The standard deviation is sqrt(y * 5/36) or about sqrt(y) *3/8. 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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