Finding expected probability in random number testing-poker test

[shivajikobardan]

An example of how probabilities are calculated in poker hand.

Probability and Statistics with Applications: A Problem Solving Text By Leonard Asimow, Ph.D., ASA, Mark Maxwell, Ph.D., ASA
You can ask me for more details about question, I won't paste them here, as it'd make the question too lengthy to view.

What problem I'm trying to do?

I am trying to find expected probability for random number independence testing aka poker test.

We've 10,000 random numbers of five digit each. They're assumed to be independent.

My calculations-:

1) Full house
10C1*9C1/10,000
=0.009

I'm correct. My only confusion here would be the denominator. Why is it 10,000?
According to the above example, should not it be 10C5?

Explanation of my thought process-:

First pick 1 digit out of 10 digits. Then next, pick another digit(only 1 digit as we need a pair), out of remaining 9 digits.

2) 1 pair:

Again I looked at that highlighted figure.
For one pair, from 10 digits, choose 1 digit. That 1 digit makes a pair. Now you've remaining 3 choices. But none of those choices can be same to each other. So,

10C1*9C1*8C1*7C1/10,000
=0.504
I'm correct here as well.

3) 3 of a kind:
Here, I need to pick only single digit for 3 places, then 2 different digits for the remaining 2 places.
So,
10C1*9C1*8C1/10,000
=0.072

Here, also I'm correct. But not anymore.

4) Four of a kind:

So from 10 digits, I need to pick 1 digit and out remaining 9 digits, I need to pick another 1 digit.
So, it should be 10C1*9C1/10,000
But it becomes similar to full house. This is wrong. I don't get why this became wrong.

5) 5 different digits:

This should've been simple, I got the answer but I got the answer greater than 1.

10C1*9C1*8C1*7C1*6C1/10,000
=3.024

I'm not sure why I got this. I am skeptical about the denominator since the start as I feel that's randomly chosen here unlike above where we did 52C5. If I increase 1 "zero" in denominator, the answer would be correct. (I've seen techniques like 10/10*9*10*8/10*7/10*6/10, but i prefer to do it as per the first poker example figure I showed so that it becomes simple for understanding).

6) Five of a kind:

It should be 10C1/10,000
=0.001
but it is instead 0.0001, so it's asking for another "zero" in the denominator for correct answer. I don't know why.
We have just 10,000 random numbers.

This is the reason for studying this-:

https://genuinenotes.com/wp-content/uploads/2020/03/Random-Numbers.pdf

 

[JeffM]

Unlike poker hands, strings of digits are sensitive to order. A poker hand cannot have two kings of spade, but a string of five digits can have two instances of 3. Random numbers cannot be modeled on poker hands.

46666 is a different string than 64666. 9KKKK is identical in value to K9KKK

Let’s start with how many different strings of 5 digits are possible.

There are 10 possibilities for choosing the first digit, 10 possibilities for choosing the second digit, 10 for the third digit, 10 for the fourth, and 10 for the fifth. So how many possibilities are there? Is that number = 10000?

Let’s go back to the strings that have four of one digit and one of a different digit.

How many possibilities are there for the digit with four mentions? How many possibilities for the digit mentioned once? How many possibilities for where in the sequence of five digits the one mentioned only once may fall? So there are how many distinct sequences of five decimal digits such that exactly four digits are the same?

Although it is not a guarantee, making sure that all your probabilities add up to 1 is a good way to catch errors.

 

[shivajikobardan]

Unlike poker hands, strings of digits are sensitive to order. A poker hand cannot have two kings of spade, but a string of five digits can have two instances of 3. Random numbers cannot be modeled on poker hands.

46666 is a different string than 64666. 9KKKK is identical in value to K9KKK

Let’s start with how many different strings of 5 digits are possible.

There are 10 possibilities for choosing the first digit, 10 possibilities for choosing the second digit, 10 for the third digit, 10 for the fourth, and 10 for the fifth. So how many possibilities are there? Is that number = 10000?

Let’s go back to the strings that have four of one digit and one of a different digit.

How many possibilities are there for the digit with four mentions? How many possibilities for the digit mentioned once? How many possibilities for where in the sequence of five digits the one mentioned only once may fall? So there are how many distinct sequences of five decimal digits such that exactly four digits are the same?

Although it is not a guarantee, making sure that all your probabilities add up to 1 is a good way to catch errors.

order isn't sensitive in this case? or is it? I'm not sure? because we are seeking how random the numbers are, i don't think order matters.

 

[JeffM]

It absolutely matters. You want know how many distinct strings consist of four 6’s and one 4.

46666
64666
66466
66646
66664

are distinct strings. You cannot say they count as only one.

In a poker hand, it makes no difference the order in which the cards are dealt. In numbers, the order of the digits does make a difference. Thinking about poker hands as a way to understand numbers is a snare and delusion.

 

[shivajikobardan]

It absolutely matters. You want know how many distinct strings consist of four 6’s and one 4.

46666
64666
66466
66646
66664

are distinct strings. You cannot say they count as only one.

In a poker hand, it makes no difference the order in which the cards are dealt. In numbers, the order of the digits does make a difference. Thinking about poker hands as a way to understand numbers is a snare and delusion.

no it's different thing . we're not seeing if they're exactly similar. we're testing if they have some repeating digits. i'm confident i've seen somewhere that it doesn't matter. i'll bring this proof later on. you can check free pdf of narsingh deo computer simulation.

 

[JeffM]

no it's different thing . we're not seeing if they're exactly similar. we're testing if they have some repeating digits. i'm confident i've seen somewhere that it doesn't matter. i'll bring this proof later on. you can check free pdf of narsingh deo computer simulation.

I understand that you are interested in the probabilities of certain classes and what defines the class does not depend on order. The number of distinct instances within each class does depend on order.