Dice with weighted sides, looking for odds of a sum:faster way

[pugfug90]

Dice with weighted sides, looking for odds of a sum:faster way please

Homework Statement

6 side die is weighted so prob of rolling 2,3,4,5,6 is same and rolling 1 is 3x rolling a 2. If die is thrown twice, what's odds of getting a sum of 4?

Homework Equations

The Attempt at a Solution

x+x+x+x+x+3x represents the "chances" of rolling a 2,3,4,5,6, and 1. So 2,3,4,5,6 individually have a 1/8 chance of rolling each time. 1 has a 3/8 chance of hitting each time.

a)For sum of 4 from 2 rolls, I need to get 1,2,3 on either rolls. Chance of getting that on first roll is 1/8 + 1/8 + 3/8 = 5/8.

If I get a 1 or 2 on first roll, then my 2nd roll can only be one result, so 5/8*1/8=5/64. Then now it's weird. If I get a 3, I multiply 5/8 by 3/8 right? Answer is 7/64 though and for 3, it'd be 15/64.

Also, 7/64 doesn't match the 5/64 I got for getting a 1 or 2 on 2nd roll.

b)I made a chart. I got the correct answer this way, but can someone show me a faster way?

http://s000.tinyupload.com/index.php?file_id=15919977748640328011

 

[Hurkyl]

If I get a 1 or 2 on first roll, then my 2nd roll can only be one result, so 5/8*1/8=5/64.

That can't be right. 5/8 is the odds of getting a 1, 2, or 3 on the first roll... but you said you were considering the case where you got a 1 or 2 on the first roll.


Incidentally, why bother breaking the problem into "first roll" and "second roll"? Why not just consider each possible pair of rolls as outcomes?

 

[pugfug90]

Incidentally, why bother breaking the problem into "first roll" and "second roll"? Why not just consider each possible pair of rolls as outcomes?

You mean 5/8 * 5/8? That wouldn't equal 7/64. My chart works out to 7/64, but I want to find a less meticulous way.

 

[Hurkyl]

Incidentally, why bother breaking the problem into "first roll" and "second roll"? Why not just consider each possible pair of rolls as outcomes?

You mean 5/8 * 5/8? That wouldn't equal 7/64. My chart works out to 7/64, but I want to find a less meticulous way.

Well, most of the outcomes in {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} do not have a sum of 4. So clearly the answer is not the probability of getting anyone of these outcomes.

 

[Hurkyl]

I can finally see your chart. I'm not sure what the problem with that approach is -- unless you mean writing the entire chart. Of course you don't need to write down all 36 outcomes; you only need to write down the 3 relevant ones.

 

[pugfug90]

Just wondering if there was an equational way?