Probability Question, need help with a logial way. (2 Viewers)

[Fus Ro Dah]

The total possibilites is actually 42. This is because we have 6x6 ways of rolling dice, but we aren't actually counting the extra 6 rolls where we roll the same number on each dice.

How does having 6x6 not count having the same number on each side? It does.

Just consider the Cartesian Product: http://en.wikipedia.org/wiki/Cartesian_product

{1,2,3,4,5,6} X {1,2,3,4,5,6} ==> 36 elements in the set of pairs.

 

[barbernator]

for an elementary lesson in multiplication, 6x6=36

 

[uniquee]

for an elementary lesson in multiplication, 6x6=36

^

 

[RealiseNothing]

Lol, if you guys really need me to show you the combinations I will:

Let's not include repetition since it won't make a difference, that is, let's say (1,2) = (2,1) and is one outcome not two.

We have:

(1,1) = 2
(1,2) = 3
(1,3) = 4
(1,4) = 5
(1,5) = 6
(1,6) = 7

(2,2) = 4
(2,3) = 5
(2,4) = 6
(2,5) = 7
(2,6) = 8

(3,3) = 6
(3,4) = 7
(3,5) = 8
(3,6) = 9

(4,4) = 8
(4,5) = 9
(4,6) = 10

(5,5) = 10
(5,6) = 11

(6,6) = 12

There are 21 possibilities and 15 outcomes where it is even or >7.

Hence



Your welcome.

 

[RealiseNothing]

How does having 6x6 not count having the same number on each side? It does.

Just consider the Cartesian Product: http://en.wikipedia.org/wiki/Cartesian_product

{1,2,3,4,5,6} X {1,2,3,4,5,6} ==> 36 elements in the set of pairs.

It counts all non-identical pairs twice, but does not count identical pairs.

It does count (1,2) and (2,1) as being different, but does not count (1,1) and (1,1) as being two outcomes.

If you let one die be die A, and the other die B. We can see they are different by (1A,2B) and (1B,2A) compared to (1A,1B) and (1B,1A).

 

[Timske]

nice one

 

[RealiseNothing]

it's 2/3 lol..
what maths do you do?

1st in accelerated 4U.

And you?

 

[Fus Ro Dah]

It counts all non-identical pairs twice, but does not count identical pairs.

It does count (1,2) and (2,1) as being different, but does not count (1,1) and (1,1) as being two outcomes.

If you let one die be die A, and the other die B. We can see they are different by (1A,2B) and (1B,2A) compared to (1A,1B) and (1B,1A).

Why would {1A,1B} and {1B,1A} be counted as two separate things?

 

[barbernator]

Lol, if you guys really need me to show you the combinations I will:

Let's not include repetition since it won't make a difference, that is, let's say (1,2) = (2,1) and is one outcome not two.

We have:

(1,1) = 2
(1,2) = 3
(1,3) = 4
(1,4) = 5
(1,5) = 6
(1,6) = 7

(2,2) = 4
(2,3) = 5
(2,4) = 6
(2,5) = 7
(2,6) = 8

(3,3) = 6
(3,4) = 7
(3,5) = 8
(3,6) = 9

(4,4) = 8
(4,5) = 9
(4,6) = 10

(5,5) = 10
(5,6) = 11

(6,6) = 12

There are 21 possibilities and 15 outcomes where it is even or >7.

Hence



Your welcome.

the answer is 2/3.

 

[RealiseNothing]

Why would {1A,1B} and {1B,1A} be counted as two separate things?

That beckons the question as to why (1A,2B) and (1B,2A) be considered two different things?

A die doesn't know what numbers are on each face, so it doesn't discriminate between a 1 or 2.

 

[barbernator]

That beckons the question as to why (1A,2B) and (1B,2A) be considered two different things?

A die doesn't know what numbers are on each face, so it doesn't discriminate between a 1 or 2.

this is not a question about what something knows and doesnt know. It is about what can occur within a sample space. There are 2 possibilities within 36 that 1,2 are paired

 

[Fus Ro Dah]

That beckons the question as to why (1A,2B) and (1B,2A) be considered two different things?

They are two different things because we don't care about the output purely. We care about the dice from which the number came.

What you simply did when you counted the {1A,1B} and {1B,1A} case is that you implied 'non-commutativity' in a sense when counting the sets, which is incorrect in this case.

 

[uniquee]

1st in accelerated 4U.

And you?

congratulations buddy you're so smart.
but you're WRONG
you got beaten by a 2u
HAHA
pretty embarrassing I have to say.
I'll repeat my Method again, so you can get it into your head.
18/36 + 15/36 - 9/36
which eventually leads on to be 2/3
I'm correct you're wrong.

 

[uniquee]

well I repeated the method 4 -5 times.
so you stfu.
but he was arrogant and he thought he was correct.

 

[SpiralFlex]

You are all incorrect! Maths is about sharing ideas not fighting!

 

[RealiseNothing]

I still think the answer should be 5/7 though.

I don't see as to why (1,1) and it's reverse aren't counted as two seperate outcomes whilst (1,2) and (2,1) are.

What makes one situation different from the other?

 

[barbernator]

I still think the answer should be 5/7 though.

I don't see as to why (1,1) and it's reverse aren't counted as two seperate outcomes whilst (1,2) and (2,1) are.

What makes one situation different from the other?

pm me brah, ill talk it over

 

[deterministic]

I still think the answer should be 5/7 though.

I don't see as to why (1,1) and it's reverse aren't counted as two seperate outcomes whilst (1,2) and (2,1) are.

What makes one situation different from the other?

What's the probability of rolling a 2 and a 1? Compare this with the probability of rolling two 1s. That should answer your questions.

 

[Absolutezero]

Seriously guys, chill out. Don't make me bring out the banhammer.

 

[RealiseNothing]

I just ran a dice rolling program.

After 99 rolls, an even or >7 sum came up 70 times. Which is:



If we compare our possible answers:





Also after 84 rolls, it came up even or >7, 63 times, which is 75%.

It's really an interesting result. I might do more rolls and see what it starts to converge to.