色子的概率问题第一个色盅里有3个色子,第二个色盅里有4个色子（每个色子可以显示1到6）,问第二个色盅里色子点数之和大于第

色子的概率问题
第一个色盅里有3个色子,第二个色盅里有4个色子（每个色子可以显示1到6）,问第二个色盅里色子点数之和大于第一个色盅里色子点数之和的概率?
最好有模型

funwo 说的很对很经典!
我“狗尾续貂”一下：
概率问题有两条法则：一个是加法,一个是乘法
为了方便说明,a[i]表示第一个碗是i点的概率,b[j]表示第二个碗是j点的概率,那么所求的就是:
a[3]*b[4]+a[3]*b[5]+...+a[3]*b[24]
+a[4]*b[5]+a[4]*b[6]+...+a[4]*b[24]
+.
+a[18]*b[19]+a[18]*b[20]+...+a[18]*b[24]
第一个碗的点数在3-18之间 先分别算出3点-18点的概率各是多少
a[3] is 1/216 0.46296296%
a[4] is 3/216 1.388889%
a[5] is 6/216 2.777778%
a[6] is 10/216 4.6296296%
a[7] is 15/216 6.9444447%
a[8] is 21/216 9.722222%
a[9] is 25/216 11.574074%
a[10] is 27/216 12.5%
a[11] is 27/216 12.5%
a[12] is 25/216 11.574074%
a[13] is 21/216 9.722222%
a[14] is 15/216 6.9444447%
a[15] is 10/216 4.6296296%
a[16] is 6/216 2.777778%
a[17] is 3/216 1.388889%
a[18] is 1/216 0.46296296%
第二个碗的点数在4-24之间 算出4点-24点的概率各是
b[4] is 1/1296 0.07716049%
b[5] is 4/1296 0.30864197%
b[6] is 10/1296 0.77160496%
b[7] is 20/1296 1.5432099%
b[8] is 35/1296 2.7006173%
b[9] is 56/1296 4.3209877%
b[10] is 80/1296 6.1728396%
b[11] is 104/1296 8.024691%
b[12] is 125/1296 9.6450615%
b[13] is 140/1296 10.802469%
b[14] is 146/1296 11.265432%
b[15] is 140/1296 10.802469%
b[16] is 125/1296 9.6450615%
b[17] is 104/1296 8.024691%
b[18] is 80/1296 6.1728396%
b[19] is 56/1296 4.3209877%
b[20] is 35/1296 2.7006173%
b[21] is 20/1296 1.5432099%
b[22] is 10/1296 0.77160496%
b[23] is 4/1296 0.30864197%
b[24] is 1/1296 0.07716049%
计算
a[3]*b[4]+a[3]*b[5]+...+a[3]*b[24]
+a[4]*b[5]+a[4]*b[6]+...+a[4]*b[24]
+.
+a[18]*b[19]+a[18]*b[20]+...+a[18]*b[24]
=74.28305041152264%
附上程序:
public class Sezi {
public static void main(String[] args) {
// TODO Auto-generated method stub
int i,j,k,l,sum;
int ttt=216;
int tttt=1296;
int[] a=new int[19];
int[] b=new int[25];
double s=0;
for(i=0;i