poker test probability confusion

[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

[youngdan]

The denominator of 10,000 in your calculations is the total number of possible five-digit combinations that can be made from 10 distinct digits, which is 10^5 (10 raised to the power of 5) or 10,000.
For example, in the case of a full house, you're looking for 1 pair and 3 of a kind. There are 10 possible digits for the pair, and 9 possible digits for the 3 of a kind. The probability of this outcome happening is (10C1 * 9C1) / 10,000.
For 5 different digits, the probability of getting 5 different digits out of 10 possible digits is (10C1 * 9C1 * 8C1 * 7C1 * 6C1) / 10,000.
For five of a kind, the probability of getting 5 of the same digit out of 10 possible digits is (10C1) / 10,000.
If you're looking for a more detailed explanation I recommend checking out commercial link removed by moderator, they have a variety of resources and tutorials available that may be helpful.

Edited January 20, 2023 by Phi for All
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[Genady]

5 minutes ago, youngdan said:

10^5 (10 raised to the power of 5) or 10,000.

Regardless of the topic, 10^5 (10 raised to the power of 5) is 100,000 rather than 10,000.