MHB Calculating Probabilities in Poker Hand: An Example

[shivajikobardan]

An example of how probabilities are calculated in poker hand.

https://lh5.googleusercontent.com/pCynfBFNqfR00y8rEKWoXYkbSCGR310FpejMJ_iGWlwD7ttkCZjunp-TLKFMmU0A94CDsR4Bb-X8i6ai8RxiLLPdWlf1j9g6BZdjq1ppPZzp0JZOBjCVqwvKCK9XmGfg7Ks7VnN4IoWZIY3gqWvmKw
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-:
https://lh4.googleusercontent.com/xDjxqD8_wg0IXSRCdB51bWMOn-mwptbgDut1uDOC22EdDHlom1Dmi6yo7n2TLlEJsnT3xqAa1Ifo4JJIIh8cnnVoKSCnNaIQioCy6fPP5rKNF53jgsvoCCDJ_X32-CEyu4w5z3A0FqUEla037Us-7Q
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:
https://lh5.googleusercontent.com/bO1wsBA0d8FQty9ydQpGTtl3Zzlma8Z0qfeeABkzVg4UVBr2hM268mbUritJur8e0D5gn79KKItkM8TgMhfzEzLpVUT4C5Yvif--9JAA2wiAQYX9YST0uL8GLVPfZ2MvAKZ8VSnh5SLoQWDnx26RqA
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

 

[scottdave]

For the denominator, think about thr total number of possibilities.

How are picking from 52 cards different from choosing digits?