MHB Just need a little explanation. Dice Problem

[Enzipino]

I know this should be easy to understand but I just need a little clarification on the last part of my answer for this problem:

If three distinct dice are rolled, what is the probability that the highest value is twice the smallest value

I started this problem with the understanding that there were 3 possibilities in which the highest value would be twice the smallest: [1|2], [2|4], and [3|6]

Then my possible dice rolls would be:
[1|1|2], [1|2|2], [2|2|4], [2|3|4], [2|4|4], [3|3|6], [3|4|6], [3|5|6], [3|6|6]

From here I can see that there are 3 possible rearrangements of the form [x|x|y] and [x|y|y]. This is where I get stuck. My book has the final answer as: $\frac{(6*3)+(3*6)}{216}$ I understand where the 216 comes from. I just need clarification on the numerator as to where the 6 is coming from and why it's $(6*3) + (3*6)$

 

[Jameson]

I know this should be easy to understand but I just need a little clarification on the last part of my answer for this problem:

If three distinct dice are rolled, what is the probability that the highest value is twice the smallest value

I started this problem with the understanding that there were 3 possibilities in which the highest value would be twice the smallest: [1|2], [2|4], and [3|6]

Then my possible dice rolls would be:
[1|1|2], [1|2|2], [2|2|4], [2|3|4], [2|4|4], [3|3|6], [3|4|6], [3|5|6], [3|6|6]

From here I can see that there are 3 possible rearrangements of the form [x|x|y] and [x|y|y]. This is where I get stuck. My book has the final answer as: $\frac{(6*3)+(3*6)}{216}$ I understand where the 216 comes from. I just need clarification on the numerator as to where the 6 is coming from and why it's $(6*3) + (3*6)$

Hi Enzipino,

I'm also trying to see why they wrote it this way... maybe it will come to me. In the mean time we can get the answer the way you listed it.

You are correct there are 3 ways to arrange a list xxy, where order matters (and we are counting like it does). What about if you have xyz though? It's not 3 ways. How many is it?

You have 9 situations to count but they don't all have the some number of unique arrangements.

 

[Enzipino]

Hi Enzipino,

I'm also trying to see why they wrote it this way... maybe it will come to me. In the mean time we can get the answer the way you listed it.

You are correct there are 3 ways to arrange a list xxy, where order matters (and we are counting like it does). What about if you have xyz though? It's not 3 ways. How many is it?

You have 9 situations to count but they don't all have the some number of unique arrangements.

I believe the xyz part is where I'm getting messed up at. I can see 3 but do I multiply it by 2 because we have two different outcomes? That is we have the first situation being [2|4] and the other in [3|6]? If that's the case then we'd have 6.

 

[Jameson]

I believe the xyz part is where I'm getting messed up at. I can see 3 but do I multiply it by 2 because we have two different outcomes? That is we have the first situation being [2|4] and the other in [3|6]? If that's the case then we'd have 6.

When we have three distinct values there are 6 permutations of them. I'll list them - xyz, xzy, yxz, yzx, zyx, zxy.

For the xyy kind, there are just three. When you apply this to the list you made how many do you end up with?

 

[Enzipino]

When we have three distinct values there are 6 permutations of them. I'll list them - xyz, xzy, yxz, yzx, zyx, zxy.

For the xyy kind, there are just three. When you apply this to the list you made how many do you end up with?

Listing out all the forms of xyz, I have 18. (Kinda seeing where the $6*3$ part is starting to play in).

 

[Jameson]

I was talking about something like this:

[1|1|2] (3 perms) , [1|2|2] (3 perms) , [2|2|4] (3 perms), [2|3|4] (6 perms), [2|4|4] (3 perms), [3|3|6] (3 perms) , [3|4|6] (6 perms), [3|5|6] (6 perms) , [3|6|6] (3 perms)

Ah, I think I see where their notation comes from now. We have 6 sets where there are 3 permutations and 3 sets where there are 6 permutations.

 

[Enzipino]

I was talking about something like this:

[1|1|2] (3 perms) , [1|2|2] (3 perms) , [2|2|4] (3 perms), [2|3|4] (6 perms), [2|4|4] (3 perms), [3|3|6] (3 perms) , [3|4|6] (6 perms), [3|5|6] (6 perms) , [3|6|6] (3 perms)

Ah, I think I see where their notation comes from now. We have 6 sets where there are 3 permutations and 3 sets where there are 6 permutations.

Oh, I apologize. So then that would be why they did $(6*3)+(3*6)$. I see it now. Well could we have done the numerator in another way?

EDIT: OH They have another way to do the numerator which is $(3*3!)+(3*3!)$ which comes out to the same result.