# Uniform and Independent Probability (Simple Problem)

#### blackrobe89

##### New Member
Hey there,

I have a quick question.

The problem:
3 numbers, say X, Y, and Z are chosen from a set of numbers (1 to n) where n is greater or equal to 1, in a uniform and independent manner.

Question 1
What's the chance that all these three numbers take on the same value, i.e. X=Y=Z?
My intuition
Let v equal to some value in the set (1 to n)
Let P (X = v) be the probability X equals v
Let P (Y = v) be the probability Y equals v
Let P (Z = v) be the probability Z equals v
P (X = Y = Z) = P (X = v) * P (Y = v) * P (Z = v)
P (X = Y = Z) = (1/n) * (1/n) * (1/n)
P (X = Y = Z) = (1/n)^3

Question 2
What's the chance that all these three numbers take on different values?
My intuition
Let v1, v2, v3 be values in the set (1 to n), such that v1 != v2 != v3
Let P (X = v1) be the probability X equals v1
Let P (Y = v2) be the probability Y equals v2
Let P (Z = v3) be the probability Z equals v3
P (X != Y != Z) = P (X = v1) * P (Y = v2) * P (Z = v3)
P (X != Y != Z) = (1/n) * (1 - (1/n)) * (1 - (1/n)^2)

Question 3
What's the expected number of different values?
My intuition
Let N be the number of different values X, Y, Z can take
Range of N = (1, 2,..., n)
E[N] = [N * P (X = v1)] + [(N - 1) * P (Y = v2)] + [(N - 2) * P (Z = v3)]

I'm not quite sure of the 2nd and 3rd, but I'm hoping that I'm not that far off.
Any help would be really appreciated.
Thanks a lot.

#### Dason

For question 1 you wrote:
P (X = Y = Z) = P (X = v) * P (Y = v) * P (Z = v)

which isn't true. The right side implies that X = Y = Z = v but the left hand side has nothing about v. You'll need to correct for this.

#### BGM

##### TS Contributor
P (X = Y = Z) = P (X = v) * P (Y = v) * P (Z = v)
This is not correct. Note that
$$\Pr\{X = Y = Z = v\} = \Pr\{X = v, Y = v, Z = v\}$$ $$= \Pr\{X = v\}\Pr\{Y = v\}\Pr\{Z = v\}$$
To evaluate the required probability, use law of total probabilities:
$$\Pr\{X = Y = Z\} = \sum_{v=1}^n \Pr\{X = Y = Z|X = v\}\Pr\{X = v\}$$
Intuitively speaking, you can imagine you drawing one of the $$X, Y, Z$$ first. Then require the latter two taking the same value. Since it is uniform it is equivalent to two of them taking a specific value.

Question 2 can use the similar technique if you understand question 1. So let you think about it first.

For question 3, note that the number of different values that $$X, Y, Z$$ can take can only be 1, 2, 3 respectively, i.e. the support of $$N$$
In question 1,2 you have calculated $$\Pr\{N = 1\}$$ and $$\Pr\{N = 3\}$$ respectively.

#### blackrobe89

##### New Member
Intuitively speaking, you can imagine you drawing one of the $$X, Y, Z$$ first. Then require the latter two taking the same value. Since it is uniform it is equivalent to two of them taking a specific value.
I see. I'm not sure if I understand this correctly..

Lets say I have:

Set $$S = \{1,2,3\}$$
$$Pr\{X=1\} = \frac{1}{3}$$
$$\frac{3!}{[(3-3)! * 3!]}$$ combinations of selecting 3 numbers from S
$$Pr\{X = Y = Z = 1\} = \frac{3!}{[(3-3)! * 3!]} * (\frac{1}{3})^3 * (1 - \frac{1}{3} )^{(3-3)}$$

The will equal 11.12% I think does that seem right?

#### BGM

##### TS Contributor
Yes but that will complicate the matter. You consider the numbers of random variables getting a specific number in which you can use the Binomial distribution. Thats correct. But more simply is just use the independent assumption which shown above.

#### blackrobe89

##### New Member
If I change the notation to:

Let $$A, B, C$$ be the three variables chosen independently and uniformly from the set of integers $$S = \{1,..,n\}$$

Let $$X$$ be a random variable of the number of different values $$A, B, C$$ could take

so
$$X = 1$$ if all variables take the same value
$$X = 2$$ if only 2 variables take the same value
$$X = 3$$ if none of the variables take the same value

$$\Pr\{$$any value$$\} = \frac{1}{n}$$ meaning any value is equally likely.

$$\sum_{i=1}^{n}\frac{1}{n} * \Pr\{X_i = 1\}$$

$$\sum_{i=1}^{n}\frac{1}{n} *\frac{1}{3!}$$

But I can't really work out the probability that all three variables are the equal i.e.$$\Pr\{X_i = 1\}$$ but I know that there are $$3!=6$$ permutations

#### BGM

##### TS Contributor
I assuming you are focus on part c) now.

The first half is good. Very clear formulation.

However,

$$\sum_{i=1}^{n}\frac{1}{n} * \Pr\{X_i = 1\}$$
is wrong (actually I do not know why you think like this). The expectation is

$$1 \times \Pr\{X = 1\} + 2 \times \Pr\{X = 2\} + 3 \times \Pr\{X = 3\}$$

where $$\Pr\{X = 1\}$$ is calculated in the 1st question, and

$$\Pr\{X = 3\}$$ is calculated in the 2nd question.

And both of these calculations can be very simple without using the permutation coefficients.

#### blackrobe89

##### New Member
Oh oops I think I was counting A, B, C as a set of 3 numbers itself and then computing their permutations.

Then I think
$$\Pr\{X = 1\} = 3 \times \Pr\{$$any value$$\} = 3 \times \frac{1}{n}$$

and

$$\Pr\{X = 3\} = \frac{1}{n}\times\frac{1}{n-1}\times\frac{1}{n-2}\times...\times1 = \frac{1}{n!}$$