Guide 4 (Word Translations) Chapter 5

[ankitp]

A gambler rolls three fair six-sided dice. What is the probability that two of the dice show the same number, but the thirds show a different number?

The answer in the book says 15/36 but I have a different one 5/36

6 ways of choosing the first dice * 1 way of choosing the 2nd dice * 5 ways of choosing the last dice = 6 * 1 * 5 = 30

Total number of ways = 6^3 = 96

30/96 = 5/36.

What am I doing wrong, I disagree with answer

[StaceyKoprince]

First, 6^3 does not equal 96. 6^3 = 216. Interestingly, this isn't your mistake, because later on, you reversed your mistake exactly by dividing 96 by 6 and getting 36. You got lucky with that part of it.

Second, you calculated the circumstance in which the first 2 match but not the third. What about the circumstance in which the first and 3rd match, but not the middle? Or the circumstance in which the 2nd and 3rd match but not the first?

Let's call a Match "M" and a mismatch "U." There are five ways to have MMU. But there are also five ways to have MUM and another five ways to have UMM. You actually need to multiply your 5 by 3 (because there are three different ways to arrange the letters M, M, and U). That gives you 15.