#### Gib Z

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Basically, I have two die's , each with 20 faces, numbered 1 to 20. What is the chance that when I roll both of them, their sum will be less or equal to 15?

Basically what I did was:

Say I rolled the first dice, and got 1 - then there are 14 things on the next die i can get to fulfil the condition (1, 2 , 3, 4, ... 14). The chance of rolling 1 on the first die is 1/20. Then the chance of fulfilling the condition on the second die is 14/20. So the chance for this case is (1/20 * 14/20)

Next case, I rolled the first dice to get 2 - then i have 13/20 to fulfill conditions. So the chance for this case is (1/20 * 13/20)

Continuing in this fashion until the last case:

I roll the first dice, get 14, then i only have 1 more case to fulfill the condition, so the chance is (1/20 * 1/20)

Adding up all the cases probabilities:

[tex]\frac{1}{20} \left( \frac{1+2+3+4+5+...+14}{20} \right) = \frac{105}{400} = \frac{21}{80}[/tex]

Since we are putting an unnessicarily order on the results of the die's, we get each case of rolled numbers repeated twice, so we must divide our result by 2, so the final answer is 21/160.

I don't know, that just doesn't feel right to me. :(