Rolling One Die TwiceDate: 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/ |
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