# Probability of rolling a "1" on a die conditioned on when all rolls are different.

So I have this problem:

I am rolling a six sided die 3 times. Conditioned on the rolls all being different, what's the probability at least one die is a "1"

So I worked it out like this:

probability of not getting a 1 on the first roll is 5/6

probability of not getting a 1 on the second roll is 4/6

probability of not getting a 1 on the second roll is 3/6

I then just did (5/6) * (4/6) * (3/6) to get 60/216 possible conditions where you would not roll a 1.

Doing 216-60 you get that 158/216 possible solutions (or 13/18) possible solutions for rolling a "1" when all numbers are different. Does this make sense? The number seems a bit large and I am not sure how to check it.

• Welcome to Math.SE! Congratulations on asking a homework question that follows the guidelines!
– Dahn
Feb 2, 2014 at 19:28

You're almost there, but you made a small error:

Hint: How many possible outcomes are there on the second roll?

• should it be 4/5? I thought about that but was not sure. Feb 2, 2014 at 19:32
• So it would be 4/5 and 3/6 which makes more sense because there are 120 possible combinations when no numbers can be the same Feb 2, 2014 at 19:36
• 4/5, yes, 3/6, no!
– Dahn
Feb 2, 2014 at 19:37
• I meant 3/4 sorry Feb 2, 2014 at 19:41
• Yes, that's right. However, make sure you check drhab's and Austin's answers, too, they provide a different (and possibly even more insightful) way of thinking about the question :)
– Dahn
Feb 2, 2014 at 19:43

This question is probably simplest if we disregard the order in which the dice are rolled. Imagine instead of rolling the same die three times, you have three identical dice in your hand and throw them all at once.

A successful outcome is one in which one of the dice shows a "1" and the other two dice are distinct members of the set $\{2, 3, 4, 5, 6\}$. There are $\binom{5}{2}$ successful outcomes.

The number of outcomes in which all faces are distinct correspond to any selection of three distinct members of the set $\{1, 2, 3, 4, 5, 6\}$. There are $\binom{6}{3}$ total outcomes.

The probability of a success is therefore $$\frac{\binom{5}{2}}{\binom{6}{3}} = \frac{1}{2}.$$

• A computer simulation I ran gave a result of $0.4999$, which confirms the answer of $1/2$. Feb 2, 2014 at 19:43

You can interpret this as electing $3$ distinct elements of set $\left\{ 1,2,3,4,5,6\right\}$. The probability for a specific number to be elected is $\frac{3}{6}=\frac{1}{2}$. This is also true for number $1$.

Alternatively the number of subsets $\left\{ i,j,k\right\}$ with $\left|\left\{ i,j,k\right\} \right|=3$ is $\binom{6}{3}$, and the number of subsets $\left\{ 1,j,k\right\}$ with $\left|\left\{ 1,j,k\right\} \right|=3$ is $\binom{5}{2}$ leading to probability: $$\binom{5}{2}\binom{6}{3}^{-1}=\frac{1}{2}$$

• Ahh, the first way is so perfectly clear and efforetlessly true once I read it, any tips on how to see that in a question like this?
– Dahn
Feb 2, 2014 at 19:50
• @DahnJahn I think that unfortunately it is quite seldom that an approach like this can be used. I would like to give you tips, but honestly speaking I don't have any. Thanks for your compliment, though :-). Feb 2, 2014 at 19:55