Five 6-sided dice are rolled. What is the probability that exactly one of the dice shows 1, and one of the dice shows 2?

wiseowl Apr 12, 2022

#1**+1 **

We have a binomial probability

We want to choose any 2 of the 5 to show a one and a two

Each has a probability of (1/6) of being rolled

And the other 3 die can each show any value except one or two (4 outcomes for each of the other three die)

The probability is

C (5,2) * (1/6)^2 * (4/6)^3 = 20 / 243

CPhill Apr 12, 2022

#2**0 **

CPhill: A minor typo in this bit: 5 nCr 2*(1/6)^2***(5/6)^3 ==1250 / 6^5==625 / 3888==16.075%**

Guest Apr 13, 2022

#3**+1 **

It's up to interpretation. The problem isn't specific on what exactly it's asking for. You could interpret it as Chris did, with exactly one 1 and exactly one 2, or you could interpret it the way you did, where you need at least one 2.

BuilderBoi Apr 13, 2022

#4**0 **

The question is quite clear. There is no ambiguity; it should not be subjected to second-guessing.

*Five 6-sided dice are rolled. What is the probability that exactly one of the dice shows 1, and one of the dice shows 2? *(The comma, hanging there like a limp dick in a love-making session, is superfluous. It neither enhances nor alters the meaning of the statement or question.)

**In any case, neither of the above answers gives the correct solution for this question.**

CPhill’s solution is not a binomial probability. The probabilities (1/6 and 4/6) do not sum to one (1).

Mr. BB’s ascii slop and atrocious math presentation is a binomial probability, but it is not a solution for this problem. **It sure as hell is NOT a solution for a single “1” and at least one “2”. **

See below for solutions and explanations.

GA

--. .-

GingerAle
Apr 18, 2022

#5**0 **

Solutions:

The easiest method to solve this (IMHO) is to note there are (6^5) unique arrangements of five (5) dice, and each with 1 to 6 pips. Reserve two of these dice and populate them with (1, 2) and (2, 1). Populate the remaining three spaces with dice populated with 3 to six pips. Divide this by the total population sample.

\(\large nPr(5,2) \normalsize* \dfrac{4^3}{6^5} = 0.164609\)

Note this is a permutation of a set of 2 from 5. (A combination will only count the sets). This counts both the sets and their order; that is there are 10 set of (1, 2) and 10 sets of (2, 1) for a total of 20 unique sets. With 2 spaces for two of the dice counted and populated with (1, 2) and (2, 1), the 1’s and 2’s are removed from the remaining sample space giving (4^3). The product of (nPr(5,2) * 4^3 = 1280) gives the number of sets of five with a single “1”and a single “2” in each set. Dividing this by 6^5 gives the probability that a roll of five dice will produce one of these sets.

**This method works for any selection of fixed dice with any population size.**

----------------

Other solutions:

Changing CPhill’s equation (C (5,2) * (1/6)^2 * (4/6)^3) to a permutation, presents the correct solution probability. nPr (5,2) * (1/6)^2 * (4/6)^3 = 0.164609. This equation resembles a binomial probability with a missing fraction. Note that the exponent (2) corresponds to the (r) and the sum of the exponents (5) corresponds to the (n). Why does this work? IDK... This is a *Nancy Drew mystery*, and Gingerlock Holms has yet to solve it.

Best guess: The asymmetry of probability is corrected in the permutations. This equation will work for any (n) when (r=2), with corresponding adjustments to the exponents. It does not work when (r>2).

GA

--. .-

GingerAle
Apr 18, 2022