Calculating Probabilities in Poker Hand: An Example

In summary, the problem is trying to find the probability of a poker hand being full house, three of a kind, or five of a kind. The calculations show that for one pair, from 10 digits, choose 1 digit. That 1 digit makes a pair. For three of a kind, from 10 digits, pick 1 digit and 2 different digits for the remaining 2 places. For four of a kind, from 10 digits, pick 1 digit and 3 different digits. For five of a kind, from 10 digits, pick 1 digit and 4 different digits. However, for five of a kind, from 10 digits, pick 1 digit and 4 different digits. The problem becomes incorrect when the denominator is increased to 10,000. The
  • #1
shivajikobardan
674
54
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
 
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  • #2
For the denominator, think about thr total number of possibilities.

How are picking from 52 cards different from choosing digits?
 

Related to Calculating Probabilities in Poker Hand: An Example

What is the purpose of calculating probabilities in poker hand?

Calculating probabilities in poker hand allows players to make informed decisions about their gameplay, such as whether to bet, raise, or fold. It also helps players understand the likelihood of certain outcomes and adjust their strategy accordingly.

What factors should be considered when calculating probabilities in poker hand?

Factors that should be considered when calculating probabilities in poker hand include the number of players, the cards on the table, and the cards in your hand. Other factors such as player tendencies and betting patterns can also impact probabilities.

How is the probability of obtaining a specific hand in poker calculated?

The probability of obtaining a specific hand in poker is calculated by dividing the number of possible outcomes that result in that hand by the total number of possible outcomes. For example, the probability of obtaining a flush is calculated by dividing the number of ways to get a flush (5148) by the total number of possible hands (2,598,960), resulting in a probability of approximately 0.0019 or 0.19%.

Why is it important to understand probabilities in poker hand?

Understanding probabilities in poker hand can give players an edge in gameplay, as it allows them to make more informed decisions and minimize their losses. It also helps players identify when they have a strong hand and when their opponents may have a better hand.

How can probabilities in poker hand be used to improve gameplay?

Probabilities in poker hand can be used to improve gameplay by helping players make more strategic decisions. For example, if the probability of obtaining a certain hand is low, it may be more beneficial to fold rather than continue betting. It can also help players calculate their expected value and determine the best course of action based on the potential risks and rewards.

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