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

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Homework Help Overview

The discussion revolves around calculating the odds of achieving a specific sum when rolling a weighted six-sided die twice. The die has a unique probability distribution where rolling a 1 is three times more likely than rolling any of the other numbers (2 through 6), which are equally probable.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the probability of rolling a sum of 4 by considering different outcomes from two rolls. They express confusion over their calculations and seek a faster method.
  • Some participants question the necessity of separating the rolls into first and second, suggesting that considering all possible pairs of outcomes might be more efficient.
  • There is a mention of a chart created by the original poster that successfully finds the answer, but they are looking for a more straightforward approach.
  • One participant points out inconsistencies in the original poster's reasoning regarding the probabilities associated with different outcomes.

Discussion Status

The discussion is active, with participants exploring various methods of calculating the probabilities. Some guidance has been offered regarding the approach to consider all pairs of rolls, and there is an acknowledgment of the original poster's chart as a valid method, though they seek a less meticulous alternative.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the methods they can use or the depth of their exploration. The unique weighting of the die is a critical factor in their calculations.

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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? O.o 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
 
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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?
 
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.
 
pugfug90 said:
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.
 
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.
 
Just wondering if there was an equational way?
 

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