Solving Probability Question: Rolling 2 Die's w/ 20 Faces - Answer 21/160

[Gib Z]

I just thought of this during class and don't know if I'm right. If I'm wrong, please explain the error in my logic. Thank you!

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:

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

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. :(

 

[HallsofIvy]

I just thought of this during class and don't know if I'm right. If I'm wrong, please explain the error in my logic. Thank you!

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:

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

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. :(

Why are you adding the values? Those are not probabilities. All you need to do is count:
(1, 14), (2, 13), (3, 12), ..., (12, 3), (13, 2), (14, 1) all are different ways of getting "15". There are 14 such out of the 20²= 400 ways the two die can come up so the probability of rolling "15" with two 20 sided dice is 14/400= 7/200.

 

[Gib Z]

That is indeed the part that confused me the most!

Say we go to case 2, where we rolled a 2 with the first die, then out of the 13 possible things left, say we get a 3.

Then in case 3, where we roll a 3 first, then one of the cases we count again is the 2.

So we counted the case of getting a 2 and a 3 as separate to getting 3 and 2 didnt we! ?

 

[Office_Shredder]

You always treat the dice as being separate... otherwise when you count the number of possible rolls, you wouldn't have 20*20 possibilities. Just pretend one die is red and the other is white if it's bothering you

And Halls, those are probabilities, since you're not finding the probability of getting a 15, you're finding the probability of getting less than or equal to 15

 

[Gib Z]

But if I did pretend one die was red, and the other white, then doesn't my method count the case eg 5 on Red, 3 on White, as well as 3 on Red, 5 on White, ie 2 separate cases when really when we roll it, the dice are indistinguishable and they are both the same case? O god I am bad at this >.<