Rolling One Die Twice

Date: 03/25/2002 at 17:40:25
From: Zack
Subject: Dice probablities

Hi,

I looked at the archives and found a couple of dice problems but still 
had problems.

My question is, If one die is rolled twice, what is the probability of 
getting a 10?

I came up with 1/6 for a 4, 1/6 for a 5, 1/6 for a 6, on each roll, 
for 3/16. Am I on the right track? Because of the number of 10 
combinations with the individual 4, 5, and 6, I got confused

Thank you.
Zack

Date: 03/25/2002 at 23:48:43
From: Doctor Twe
Subject: Re: Dice probablities

Hi Zack - thanks for writing to Dr. Math.

You're almost right; the numerator should be 3, because there are 3 
ways of getting a 10: (4,6), (5,5), or (6,4). But the denominator 
should be the total number of possible outcomes of the two dice, which 
is 6*6 = 36. [You could roll (1,1), (1,2), ..., (6,6).]

Another way to look at this is to make a chart. List the possible 
rolls of one die across the top, and the possible rolls of the other 
die along the left. Fill in the chart with the sums of the rows and 
columns. Then count the number of times each sum appears in the chart, 
like so:

               First Die
        | 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

Note that there are 36 outcomes on the chart, so all probabilities 
would be X/36, where X is the number of outcomes that you are 
counting.

I hope this helps. If you have any more questions, write back.

- Doctor TWE, The Math Forum
  http://mathforum.org/dr.math/