# Conditional probability questions

1. Three dice are rolled. If no two show the same face, what is the probability that one is an ace?
2. Given that a throw with ten dice produced at least one ace, what is the probability p of two or more aces?

Answers provided by the Author William Feller is

1. $$1-\frac{5*4*3}{6*5*4}=\frac12$$

2.$$p=1-\frac{10*5^9}{6^{10}-5^{10}}=0.6147724$$

I don't understand how did the author calculate these answers? If any member knows may answer these questions.

• Hint: 1. P(neither is ace) = (number of combinations where each dice has different face and it isn't ace) / (number of combinations where each dice has different face). P(at least one ace) = 1 - P(neither is ace).
– Tim
Jan 22, 2019 at 13:24
• Intuitive hints: (1) asks for the proportion of three-element subsets of $\{1,2,3,4,5,6\}$ that do not include $1$; that is, they are subsets of $\{2,3,4,5,6\}.$ (2) The value subtracted represents the chance that none of the other $9$ faces is an ace--that is, it's one of the $5$ values $\{2,3,4,5,6\}.$ Its numerator is the sum over all $10$ dice of the value $5^9.$ Its denominator must come from the calculation of conditional probability and can be recognized as the total number of outcomes with at least one ace.
– whuber
Jan 24, 2019 at 20:40

Explanation of Question 1

The desired probability is: $$P(\text{an ace} | \text{each face is different})$$

So, one of the first things that we should calculate is how many different ways we can roll three die with the faces all being different (the given in the question). This number becomes the size of our sample space (also, the question implicitly assumes that order does not matter. We then want to calculate how many of those possible rolls (in the sample space) do not have an ace. The difference is our desired probability.

Worked out solution:

Ways to roll three die with all different faces: 6*5*4

• First dice in the roll can be any of the 6 faces
• Second dice in roll can have any face other than that on the first (5)
• Third dice in roll can have any face other than first two die (4)

Ways to roll three die with all different faces and no ace: 5*4*3

• First dice in the roll can be any of the 5 non-ace faces
• Second dice in roll can have any face other than the first or an ace (4)
• Third dice in roll can have any face other than first two die or an ace (3)

Putting these together:

$$P(\text{no ace} | \text{each face is different}) = \frac{5*4*3}{6*5*4}$$

$$P(\text{an ace} | \text{each face is different}) = 1 - P(\text{no ace} | \text{each face is different})$$

$$P(\text{an ace} | \text{each face is different}) = 1 - \frac{5*4*3}{6*5*4} = \frac{1}{2}$$