Dice question

[Zzm6899]

Hey all, could someone tell me where I went wrong? Doesn't make sense ?
Throwing three dice, what is the probability of the total sum of all three dices being ten, if one of the dice is a 1?
Answer would be 1/18
6,3,1 = 6/216 chance plus
5,4,1 = 6/216 chance
= 12/216

 

[tkhunny]

So, two dice, what is the probability of the total sum of both being nine?

	1​	2​	3​	4​	5​	6​
1​	2​	3​	4​	5​	6​	7​
2​	3​	4​	5​	6​	7​	8​
3​	4​	5​	6​	7​	8​	9​
4​	5​	6​	7​	8​	9​	10​
5​	6​	7​	8​	9​	10​	11​
6​	7​	8​	9​	10​	11​	12​

 

[JeffM]

Hey all, could someone tell me where I went wrong? Doesn't make sense ?
Throwing three dice, what is the probability of the total sum of all three dices being ten, if one of the dice is a 1?
Answer would be 1/18
6,3,1 = 6/216 chance plus
5,4,1 = 6/216 chance
= 12/216

Are you saying that the book's answer is 1/18, but your answer is 12/216?

Did you try reducing your answer to least terms?

By the way, your method is more intuitive to me than tkhunny's perfectly valid way of approaching it.

 

[tkhunny]

Are you saying that the book's answer is 1/18, but your answer is 12/216?

Did you try reducing your answer to least terms?

By the way, your method is more intuitive to me than tkhunny's perfectly valid way of approaching it.

haha Won't be the last time that happens!

 

[Zzm6899]

Are you saying that the book's answer is 1/18, but your answer is 12/216?

Did you try reducing your answer to least terms?

By the way, your method is more intuitive to me than tkhunny's perfectly valid way of approaching it.

That was my working out.. to ask what wrong wrong lol

So, two dice, what is the probability of the total sum of both being nine?

	1​	2​	3​	4​	5​	6​
1​	2​	3​	4​	5​	6​	7​
2​	3​	4​	5​	6​	7​	8​
3​	4​	5​	6​	7​	8​	9​
4​	5​	6​	7​	8​	9​	10​
5​	6​	7​	8​	9​	10​	11​
6​	7​	8​	9​	10​	11​	12​

Would the answer then be 4/36? As the probability

 

[pka]

Hey all, could someone tell me where I went wrong? Doesn't make sense
Throwing three dice, what is the probability of the total sum of all three dices being ten, if one of the dice is a 1?
Answer would be 1/18
6,3,1 = 6/216 chance plus
5,4,1 = 6/216 chance
= 12/216

That was my working out.. to ask what wrong wrong lol
Would the answer then be 4/36? As the probability

Your work in the original post is completely correct. But your arithmetic is wanting.
\(\dfrac{12}{216}=\dfrac{6}{108}=\dfrac{3}{54}=\dfrac{1}{18}\)

 

[Cubist]

Would the answer then be 4/36? As the probability

I think so. The bold word below means that you don't need to calculate the probability of throwing the die that shows 1, it's given that this has already happened...

Throwing three dice, what is the probability of the total sum of all three dices being ten, if one of the dice is a 1?

Therefore I agree with the comments in @tkhunny post#2, and you're correct that the probability is 4/36 (which can also be reduced!)

Or using your method...
chances of throwing 6,3 in some order is 2!/6²=2/36
chances of throwing 5,4 in some order is 2!/6²=2/36
sum of the above = 4/36 (then reduce)

 

[Steven G]

I'll repeat what has been said twice already. 12/216 = 1/18. What does this mean? It means that you and the textbook both have the same solution.

 

[Steven G]

I think so. The bold word below means that you don't need to calculate the probability of throwing the die that shows 1, it's given that this has already happened...


Therefore I agree with the comments in @tkhunny post#2, and you're correct that the probability is 4/36 (which can also be reduced!)

Or using your method...
chances of throwing 6,3 in some order is 2!/6²=2/36
chances of throwing 5,4 in some order is 2!/6²=2/36
sum of the above = 4/36 (then reduce)

I do not think the answer is 4/36. I am sure that you would not disagree that there are 216 possible outcomes if you roll three dice. In rolling two dice there are 4 outcomes that you get a sum of 9. Now for each of the 4 ways of getting a 9 you can insert the 1 in three different positions yielding 12 ways of obtaining a sum of 10. So the answer is 12/216. tkhunny's work up to a point is correct but seems to be assuming the 1 must be the result of the 1st die (or something like that).

 

[Cubist]

I do not think the answer is 4/36. I am sure that you would not disagree that there are 216 possible outcomes if you roll three dice. In rolling two dice there are 4 outcomes that you get a sum of 9. Now for each of the 4 ways of getting a 9 you can insert the 1 in three different positions yielding 12 ways of obtaining a sum of 10. So the answer is 12/216. tkhunny's work up to a point is correct but seems to be assuming the 1 must be the result of the 1st die (or something like that).

Your argument is very convincing! I'll have to think more about this tomorrow before I fully agree, it's getting very late here

EDIT: Actually I agree, I was assuming the 1 die was the first thrown. Very clever. I'll sleep in the corner tonight for my mistake

 

[Cubist]

...or could the probability be 12/75 because there are 75 ways of throwing 3 dice such that one of them shows a 1 (or does the question imply one or more shows a one?). Sorry, I really am going to sleep now...

 

[Steven G]

...or could the probability be 12/75 because there are 75 ways of throwing 3 dice such that one of them shows a 1 (or does the question imply one or more shows a one?). Sorry, I really am going to sleep now...

The problem reads if one of the dice is a 1. How would you interrupt that? If all three dice were a 1 for example and someone asked me if one of the dice was a one I would say yes.
Good night

 

[Cubist]

The problem reads if one of the dice is a 1. How would you interrupt that? If all three dice were a 1 for example and someone asked me if one of the dice was a one I would say yes.
Good night

...good morning!

OK, then there are 91 ways of throwing 3 dice so that at least one die shows a 1 (see * below). Therefore the final probability could be 12/91 (this fraction can't be reduced). I use the words "could be" because I'm not sure. It would be nice to have verification from other helpers or the OP.

--

* - I got the number 91 via computer, going through all the 216 ways of throwing 3 dice and counting the number of ways that contained at least one 1.

Here's another way of doing it without a computer:- #outcomes containing exactly one "1" is choose(3,1)*1*5*5=75; #containing exactly two "1" is choose(3,2)*1*1*5=15; #containing exactly three "1" is choose(3,3)*1=1. Add them all up and we get 75+15+1=91.

 

[Steven G]

...good morning!

OK, then there are 91 ways of throwing 3 dice so that at least one die shows a 1 (see * below). Therefore the final probability could be 12/91 (this fraction can't be reduced). I use the words "could be" because I'm not sure. It would be nice to have verification from other helpers or the OP.

--

* - I got the number 91 via computer, going through all the 216 ways of throwing 3 dice and counting the number of ways that contained at least one 1.

Here's another way of doing it without a computer:- #outcomes containing exactly one "1" is choose(3,1)*1*5*5=75; #containing exactly two "1" is choose(3,2)*1*1*5=15; #containing exactly three "1" is choose(3,3)*1=1. Add them all up and we get 75+15+1=91.

You had me thinking that you were correct for a few minutes. I even confirmed that your work is correct. There is one major problem. You answered a different problem! You forgot that the sum must be 10!! I guess that you need to stay in the corner again.

 

[Cubist]

...You forgot that the sum must be 10!! I guess that you need to stay in the corner again.

Isn't the following correct?

12   <- number of permutations that sum to 10 AND have at least one 1
--
91   <- number of permutations that have at least one 1

 

[Steven G]

Isn't the following correct?

12   <- number of permutations that sum to 10 AND have at least one 1
--
91   <- number of permutations that have at least one 1

Ooops, I was thinking that you were saying that the answer was 91/216. Your work seems fine. Surely this is a conditional probability.

 

[Cubist]

Ooops, I was thinking that you were saying that the answer was 91/216. Your work seems fine. Surely this is a conditional probability.

Yes I think so. The original question:-

Throwing three dice, what is the probability of the total sum of all three dices being ten, if one of the dice is a 1?

is surely the same as:-

Throwing three dice, what is the probability of the total sum of all three dice being ten given that one (or more) of the dice is a 1?

OP - do you agree/ understand?

 

[JeffM]

No doubt the student is confused by now (if the student is even still paying attention).

I have three dice. Let's say one is red, one is blue, and one is green.

What is the probability that one die comes up 1 and the other two dice sum to 9.

The dice are fair, and the rolls are independent.

Red die 1, blue die 6, green die 3 Probability = (1/6)(1/6)(1/6) = 1/216.
Red die 1, blue die 3, green die 6 Probability 1/216.
Red die 1, blue die 5, green die 4 Probability 1/216.
Red die 1, blue die 4, green die 5 Probability 1/216.

Blue die 1, red die 6, green die 3 Probability 1/216.
Blue die 1, red die 3, green die 6 Probability 1/216.
Blue die 1, die 5, green die 4 Probability 1/216.
Blue die 1, red die 4, green die 5 Probability 1/216.

Green die 1, red die 6, blue die 3 Probability 1/216.
Green die 1, red die 3, blue die 6 Probability 1/216.
Green die 1, die 5, blue die 4 Probability 1/216.
Green die 1, red die 4, blue die 5 Probability 1/216.

That probability is 12/216 = 1/18.

The numbers of ways to roll 1 can be computed in two different ways

[MATH]\dbinom{3}{1} * 1 * 5^2 + \dbinom{3}{2} * 1^2 * 5 + \dbinom{3}{3} * 1^3 =[/MATH]
[MATH]3 * 25 + 3 * 15 + 1 * 1 = 75 + 15 + 1 = 91.[/MATH]
Or [MATH]216 - 5^3 = 216 - 125 = 91.[/MATH]
Thus, if we are looking for a conditional probability, the answer is 12/91.

I have no idea what the exact question posed to the student was.

 

[Zzm6899]

This is some um intresting stuff... The teacher hasn't given me a response but logically it would be many of these answers.... I would have went for the probability of getting a sum of 9 then adding the chance of getting a one..

 

[Steven G]

There is only one answer which depends on whether or not your teacher assumes exactly one 1 or at least one 1.
I personally think that if it does not state exactly one 1 then it means at least one 1.
As I stated earlier, if one, two or three 1s were flipped and someone asked me if a 1 was flipped I would say yes (otherwise I wouldn't be truthful).