5 digit poker test various probabilities calculations

[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

 

[nycmathguy]

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

What level of probability is this question?

 

[shivajikobardan]

What level of probability is this question?

High school level. But most likely this wasn't taught to most people in high school.(high school=year 11 or 12 starting from the grade nursery, ukg, lkg, 1,2,3....10,11,12. Then university starts. 1st year, 2nd year,...4th year)

 

[nycmathguy]

High school level. But most likely this wasn't taught to most people in high school.(high school=year 11 or 12 starting from the grade nursery, ukg, lkg, 1,2,3....10,11,12. Then university starts. 1st year, 2nd year,...4th year)

Thanks. Goto freemathhelp.com for help.