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How to calculate expected probability for poker tests for random numbers independence?

shivajikobardan

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An example of how probabilities are calculated in poker hand.

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-:
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:
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.
 

wwfeldman

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An example of how probabilities are calculated

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?

i think you answered your own question.
"We've 10,000 random numbers of five digit each."

which means you have 10,000 number sets to choose from,
hence the total number of denominator choices is 10,000, not 10C5
 

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