Probability of roll X or greater on Y Dice.

**takatok** · November 14th 2011, 04:20 PM

I want to make sure my analysis of the problem thus far is correct. I'm just learning about integer partitions so I am not sure exactly how to calculate the function I define as $G(x)$ .

Assume that $P(x,y)$ is the probality of rolling exactly X on Y Dice.
This should be:
$P(x,y)=\frac{\binom{x-1}{y}-G(x)}{6^Y}$
where:
$G(x)$ is the number of partitions of X that include Y members and at least 1 member over 6.

The probability of rolling X or greater on Y dice is then:
$\sum_{i=x}^{6^y} P(x,y)$ .

Calculating $G(x)$ :
Define $part(n,k)$ to be the number of partitions of n each of whose member is $\geq k$ . This is equivalent to the partitions of n which have at least k members.

$part(n,k) = 0\ for\ k>n$
$part(n,k) = 1\ for\ k=n$
$part(n,k) = part(n,k+1)+part(n-k,k)$
$part(n,1) =$ all the possible partitions of n

So
$part(x,y) -part(x,y-1)$ should be the all the ways I can partition X into exactly Y parts. However this could include members whose vlaue > 6.
(Also is the above just equal to $\binom{x-1}{Y}$
Or does the above imply that every member is also at least Y? (which is not at all what i want.)

I'm not sure at this point how to weed out the partitions that include invalid dice rolls?

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