Probability of getting a Four-of-a-kind?

[musicgold]

Homework Statement

Why is my method not getting the correct answer? What am I missing?

Homework Equations

The Attempt at a Solution

I am trying to find the probabilities of different hands in a 5-card poker (Texas hold' em). While the document below shows the answer, I like to use a method that is intuitive to me. My method is getting me the correct probability figures for the royal flush and straight flush hands, but not for a four-of-a-kind (FOK).

Here is how I like to think about the problem: Consider a FOK hand A1 A2 A3 A4, 9. That is 4 aces and one 9, where 1 means spades, 2 hearts, 3 diamonds and 4 clubs.

The probability of getting this particular hand is 4/52 x 3/51 x 2/50 x 1/49 x 1 = 0.00037%.

Since there are 13 possible hands, the total probability is = 0.00037% x 13 = 0.0048% or 1 in 20,825.

According to the article below the probability is 1 in 4,165. What am I missing? https://www.ece.utah.edu/eceCTools/Probability/Combinatorics/ProbCombEx15.pdf

 

[.Scott]

The problem is that you are looking for A1, A2, A3, A4, 9.
But 9, A1, A2, A3, A4, also works. And so do 3 other arrangements.

Here is your logic in more explicit terms. I will combine your "13" with the first pick:
On the first pick, any card will do: Hence p=1
On the second pick, any of three cards will do: Hence p=3/51
On the third pick, any of three cards will do: Hence p=2/50
On the second pick, any of three cards will do: Hence p=1/49
On the last pick, any card will do: Hence p=1

So you multiply all of them together and you get 20,825.
But by doing this, you are saying it is only acceptable for the non-matching card to appear as the last card. In fact, it can be in any of the 5 positions and you still have 4 of a kind.

So: 20,825/5 = 5165

 

[musicgold]

But 9, A1, A2, A3, A4, also works. And so do 3 other arrangements.

Ah! That's what I was missing! Thank you!

Here is your logic in more explicit terms. I will combine your "13" with the first pick:
On the first pick, any card will do: Hence p=1
On the second pick, any of three cards will do: Hence p=3/51
On the third pick, any of three cards will do: Hence p=2/50
On the second pick, any of three cards will do: Hence p=1/49
On the last pick, any card will do: Hence p=1

Yes, your approach is better than mine.

Also, I think you meant 4165 instead of 5165.

 

[musicgold]

A follow up question, if you don't mind.

I tried to use your method for a full house hand ( for example Q1 Q2 Q3 J1 J2). There has to be a triplet and a pair. So I went like this:
1 x 3/51 x 2/50 x 48/49 x 3/48 = 0.0144% or 1 in 6941.7

The 48/49 in the fourth term excludes the remaining Queen.

However as per the article the answer is 694.17. What am I missing? Looks like I am off by an order of 10!

 

[PeroK]

A follow up question, if you don't mind.

I tried to use your method for a full house hand ( for example Q1 Q2 Q3 J1 J2). There has to be a triplet and a pair. So I went like this:
1 x 3/51 x 2/50 x 48/49 x 3/48 = 0.0144% or 1 in 6941.7

The 48/49 in the fourth term excludes the remaining Queen.

However as per the article the answer is 694.17. What am I missing? Looks like I am off by an order of 10!

It's the same mistake as before. You could be dealt a Jack on the second card for example. The second card doesn't have to be the same as the first.

 

[PeroK]

$\binom{5}{2}=10$

Which is why you are out by a factor of ten..

 

[musicgold]

It's the same mistake as before. You could be dealt a Jack on the second card for example. The second card doesn't have to be the same as the first.

Thanks. But if you look at Scott's method, it doesn't need to worry about other possible arrangements. In my original method, I had to worry about other arrangements. For the full house calculation, I am using Scott's method.

 

[PeroK]

Thanks. But if you look at Scott's method, it doesn't need to worry about other possible arrangements. In my original method, I had to worry about other arrangements. For the full house calculation, I am using Scott's method.

If you want to calculate the probability of a full house using a card-by-card method, then the first card can be anything and so can the second card. The calculation then splits depending on whether the first two cards are a pair or not.

This is essentially the probability tree technique.

Note that if the first two cards are a pair, then the third card can be anything.