Hi !

It may look simple but I've got doubts since my friend told me he had a different answer ^^

Given 3 throws of dices (6-sided), independently.. what is the probability that the sum of all the numbers displayed is superior or equal to 15 ?

Thanks

Edit : scuse me but I don't know how to say correctly that we throw 3 times a dice ^^

2. Originally Posted by Moo
Hi !

It may look simple but I've got doubts since my friend told me he had a different answer ^^

Given 3 throws of dices (6-sided), independently.. what is the probability that the sum of all the numbers displayed is superior or equal to 15 ?

Thanks

Edit : scuse me but I don't know how to say correctly that we throw 3 times a dice ^^
You have the following outcomes:

4;5;6
4;6;6
4;6;5

5;5;6
5;4;6
5;6;4
5;6;5
5;5;5

6;4;5
6;5;4
6;5;5
6;6;6

Are these all the possibilities? Just double check them.

The total number of possibilities you can throw is $6^3$ = 216

And I've found 12 possibilities there.

Although maybe something like 5;5;5 should be counted for all the possibilities in which the dice can be arranged? Which will be 6 different possibilities if I'm not mistaken.

EDIT: If this is wrong, forgive me. My brain is fried

3. Probability has never been my strong point, so here is a brute force approch.

See the tree diagrams below

From the above diagram we can see that there are 10 ways to roll a sum of 15 so $P(15)=10 \left( \frac{1}{6}\right)^3=\frac{10}{216}$

looking at the next we get

$P(16)=6\left( \frac{1}{6}\right)^3=\frac{6}{216}$

$P(17)=3 \left( \frac{1}{6}\right)^3=\frac{3}{216}$

And there is only one way to roll an 18 so $P(18)=\frac{1}{216}$

So finally we get $P(15)+P(16)+P(17)+P(18)=\frac{20}{216}=\frac{5}{54 }\approx 0.0926$

4. Originally Posted by janvdl
You have the following outcomes:

~~

Are these all the possibilities? Just double check them.

The total number of possibilities you can throw is $6^3$ = 216

And I've found 12 possibilities there.

Although maybe something like 5;5;5 should be counted for all the possibilities in which the dice can be arranged? Which will be 6 different possibilities if I'm not mistaken.

EDIT: If this is wrong, forgive me. My brain is fried

Ok, I did the same reasoning !
Though I beat you at being tired because I counted 17 possibilites xD (oh I see from Tessy the case 6-6-3)

-----------------------------
My friend told me you had to count the possibility (5,5,6) only once, for example... because it was redundant or something like that.

I don't know if the problem is clearly stated : when do you know if you have to count the permutations, and when don't you have to ?

-----------------------------

Thanks prof' !

Originally Posted by TheEmptySet
~~

So finally we get $P(15)+P(16)+P(17)+P(18)=\frac{20}{216}=\frac{5}{54 }\approx 0.0926$
Exhaustive... And nice handwriting !

Thanks to both of you, though it doesn't yield the same result. I was mostly interested by the reasoning

5. If you expand $\left( {\sum\limits_{k = 1}^6 {x^k } } \right)^3$ the last four terms are $10x^{15} + 6x^{16} + 3x^{17} + x^{18}$.
If we add the coefficients we get 20. That is the number of ways that we get a sum of at least 15 when tossing three dice (or rolling a die three times).

6. Originally Posted by Plato
If you expand $\left( {\sum\limits_{k = 1}^6 {x^k } } \right)^3$ the last four terms are $10x^{15} + 6x^{16} + 3x^{17} + x^{18}$.
If we add the coefficients we get 20. That is the number of ways that we get a sum of at least 15 when tossing three dice (or rolling a die three times).
But where does this formula come from ??

7. Originally Posted by Moo
But where does this formula come from ??
You can look up the topic Generating Functions. Given the power of today’s CAS’s and calculators they are really useful in counting problems.
Here is an example. Say you are in a country with money in one, five and ten units. You want to know how many ways you can make a sum of fifty units. Expand $\left( {\sum\limits_{k = 0}^{50} {x^k } } \right)\left( {\sum\limits_{k = 0}^{10} {x^{5k} } } \right)\left( {\sum\limits_{k = 0}^5 {x^{10k} } } \right)$ the coefficient of $x^{50}$ answers the question.

8. Originally Posted by Plato
You can look up the topic Generating Functions. Given the power of today’s CAS’s and calculators they are really useful in counting problems.
Here is an example. Say you are in a country with money in one, five and ten units. You want to know how many ways you can make a sum of fifty units. Expand $\left( {\sum\limits_{k = 0}^{50} {x^k } } \right)\left( {\sum\limits_{k = 0}^{10} {x^{5k} } } \right)\left( {\sum\limits_{k = 0}^5 {x^{10k} } } \right)$ the coefficient of $x^{50}$ answers the question.
I've never dealt with such a thing, but now I remember I already asked you ^-^
I'll have a look on it when I have time

Thanks

9. Originally Posted by Plato
You can look up the topic Generating Functions. Given the power of today’s CAS’s and calculators they are really useful in counting problems.
Here is an example. Say you are in a country with money in one, five and ten units. You want to know how many ways you can make a sum of fifty units. Expand $\left( {\sum\limits_{k = 0}^{50} {x^k } } \right)\left( {\sum\limits_{k = 0}^{10} {x^{5k} } } \right)\left( {\sum\limits_{k = 0}^5 {x^{10k} } } \right)$ the coefficient of $x^{50}$ answers the question.
Originally Posted by Moo
I've never dealt with such a thing, but now I remember I already asked you ^-^
I'll have a look on it when I have time

Thanks
Same here, I want to see how this works. Very very interesting.

Originally Posted by janvdl
Same here, I want to see how this works. Very very interesting.
I am also interested in-- well, OK, obsessed by-- generating functions, and have been for some time. Several books on combinatorial analysis have good introductions to the subject, for example "Applied Combinatorics" by Tucker.

To further whet your interest, here is a link to an article from "Poker Digest", of all places, with a very nicely explained example:

Beauty and the Beholder

And here is what the statistician Frederick Mosteller had to say about his first exposure to generating functions (from http://www.umass.edu/wsp/statistics/tales/mosteller.html):
A key moment in my life occurred in one of those classes during my sophomore year. We had the question: When three dice are rolled what is the chance that the sum of the faces will be 10? The students in this course were very good, but we all got the answer largely by counting on our fingers. When we came to class, I said to the teacher, "That's all very well - we got the answer - but if we had been asked about six dice and the probability of getting 18, we would still be home counting. How do you do problems like that?" He said, "I don't know, but I know a man who probably does and I'll ask him." One day I was in the library and Professor Edwin G Olds of the Mathematics Department came in. He shouted at me, "I hear you're interested in the three dice problem." He had a huge voice, and you know how libraries are. I was embarrassed. "Well, come and see me," he said, and I'll show you about it." "Sure, " I said. But I was saying to myself, "I'll never go." Then he said, "What are you doing?" I showed him. "That's nothing important," he said. "Let's go now."

So we went to his office, and he showed me a generating function. It was the most marvelous thing I had ever seen in mathematics. It used mathematics that, up to that time, in my heart of hearts, I had thought was something that mathematicians just did to create homework problems for innocent students in high school and college. I don't know where I had got ideas like that about various parts of mathematics. Anyway, I was stunned when I saw how Olds used this mathematics that I hadn't believed in. He used it in such an unusually outrageous way. It was a total retranslation of the meaning of the numbers. [Albers, More Mathematical People].