VP Combinatorics

Hello everyone,

I'm trying to understand the combinatorics for video poker. Wizard of Odds says 19,933,230,517,200 total combinations for Jacks or Better. This PDF states there are 1,661,102,543,100 combinations, which you can see on the probability table on page 9. However, from my messing around, I found that there are (52 choose 2)^2, or 6,754,593,081,600 combinations. My reasoning for that is because for every hand of the 2,598,960 pre-discard combinations, you can end up with every possible pre-card combination at post-discard. I did some small sample testing with drawing 3 cards from a 9 card deck and found the previous finding to be true. So can someone explain to me why the first or second total is correct while mine is not? I've seen the (52 choose 5)*(47 choose 5)*5 = 19,933,230,517,200 formula, but I don't understand why that is the answer.

An often asked question, that I once asked the wizard too. (and have unfortunately forgotten the answer to)

Somewhere on the WoO site, he explains it. You'll have to do a little searching under the VP section or in the Ask the Wiz section.

Quote: rsactuary

An often asked question, that I once asked the wizard too. (and have unfortunately forgotten the answer to)

Somewhere on the WoO site, he explains it. You'll have to do a little searching under the VP section or in the Ask the Wiz section.

Using the search engine at the top of this page for 19,933,230,517,200 finds this thread with a post by the Wizard.

I've read through this topic but didn't answer my question as to why you have to do these specific calculations to find the number of combinations. I understand the math, but don't understand its application.

See this:
http://www.math.utah.edu/~ethier/sample.pdf
Specifically p. 552.

Quote: ddr4lyfe

I've seen the (52 choose 5)*(47 choose 5)*5 = 19,933,230,517,200 formula, but I don't understand why that is the answer.

I think I can explain this.
Your (52 choose 5)^2 idea misses the fact that the initial cards are not returned to the deck when they are discarded so there are only 47 cards to be dealt after the discard. Leaving (52 choose 5)*(47 choose 5) combinations if you discard all 5 original cards.
However, you don't have to discard all 5 initial card, you can choose to keep 1 or more. The number of combinations for each number of discards is.
0 (52 choose 5)
1 (52 choose 5)*(47 choose 1)*(5 choose 1)
2 (52 choose 5)*(47 choose 2)*(5 choose 2)
3 (52 choose 5)*(47 choose 3)*(5 choose 3)
4 (52 choose 5)*(47 choose 4)*(5 choose 4)
5 (52 choose 5)*(47 choose 5)

To make the maths work you need to have a common multiple between each of the choices so we multiply the discard all 5 line by 5 giving (52 choose 5)*(47 choose 5)*5.
I hope this helps.

Quote: someone

I think I can explain this.
Your (52 choose 5)^2 idea misses the fact that the initial cards are not returned to the deck when they are discarded so there are only 47 cards to be dealt after the discard. Leaving (52 choose 5)*(47 choose 5) combinations if you discard all 5 original cards.
However, you don't have to discard all 5 initial card, you can choose to keep 1 or more. The number of combinations for each number of discards is.
0 (52 choose 5)
1 (52 choose 5)*(47 choose 1)*(5 choose 1)
2 (52 choose 5)*(47 choose 2)*(5 choose 2)
3 (52 choose 5)*(47 choose 3)*(5 choose 3)
4 (52 choose 5)*(47 choose 4)*(5 choose 4)
5 (52 choose 5)*(47 choose 5)

To make the maths work you need to have a common multiple between each of the choices so we multiply the discard all 5 line by 5 giving (52 choose 5)*(47 choose 5)*5.
I hope this helps.

All of this makes sense except for the last part. What's the point of the least common multiple? It makes sense to me to just sum up all of the 6 lines you had above. I don't see where the LCM comes in...

Quote: malgorium

All of this makes sense except for the last part. What's the point of the least common multiple? It makes sense to me to just sum up all of the 6 lines you had above. I don't see where the LCM comes in...

The problem with just adding the 6 lines is that they are not all equally likely to occur. To do any calculation of the correct play, you convert to have each result being equally likely. to do this you need an LCM.

Quote: malgorium

All of this makes sense except for the last part. What's the point of the least common multiple? It makes sense to me to just sum up all of the 6 lines you had above. I don't see where the LCM comes in...

Read the Ethier paper I cited above, he walks through the formulas that explain all of it.

Quote: MathExtremist

Read the Ethier paper I cited above, he walks through the formulas that explain all of it.

I've never seen this paper before. I will be reading it cover to cover at some point. What book is this in, by the way? It says Chapter 17

Quote: MathExtremist

Read the Ethier paper I cited above, he walks through the formulas that explain all of it.

Oh, ok I think this makes sense. Does this mean that the total number of different deal/draw combinations is 1,661,102,543,100, but the 19,933,230,517,200 number only comes in when calculating the probabilities?

Quote: rsactuary

I've never seen this paper before. I will be reading it cover to cover at some point. What book is this in, by the way? It says Chapter 17

"The Doctrine of Chances"
Video Poker: Ch 17 is the sample chapter on his academic website.

http://www.springer.com/gp/book/9783540787822

Quote: someone

I think I can explain this.
Your (52 choose 5)^2 idea misses the fact that the initial cards are not returned to the deck when they are discarded so there are only 47 cards to be dealt after the discard. Leaving (52 choose 5)*(47 choose 5) combinations if you discard all 5 original cards.
However, you don't have to discard all 5 initial card, you can choose to keep 1 or more. The number of combinations for each number of discards is.
0 (52 choose 5)
1 (52 choose 5)*(47 choose 1)*(5 choose 1)
2 (52 choose 5)*(47 choose 2)*(5 choose 2)
3 (52 choose 5)*(47 choose 3)*(5 choose 3)
4 (52 choose 5)*(47 choose 4)*(5 choose 4)
5 (52 choose 5)*(47 choose 5)

To make the maths work you need to have a common multiple between each of the choices so we multiply the discard all 5 line by 5 giving (52 choose 5)*(47 choose 5)*5.
I hope this helps.

But the initial deal counts as a possibility if we choose to keep the whole hand, no? For instance, if we had a 3 card deck and we deal 1 card:
For each initial deal, we can draw a 1, 2, or 3. For each of those, we can choose to keep it or discard it; if we discard it, we can get one of the two remaining.

E.g., if we are dealt a 1, we can keep it (final hand being "1"), or we can discard it, which could end up being a 2 or 3 (final hand being "2" or "3"). So given one of the initial possible hands, we were able to end up with every possible hand ("1" if we didn't discard, or "2" or "3" if we did discard the 1).

The idea applies to a 5 card, 2 dealt deck:

Initial Hand	Cards to Discard	Final Hand
1-2	None	1-2
	First	3-2
		4-2
		5-2
	Second	1-3
		1-4
		1-5
	Both	3-4
		3-5
		4-5

We end up with 10 (5 choose 2) possibilities for the first possible initial hand (1-2). We can have 10 (5 choose 2) initial hands for this deck (1-2, 1-3, 1-4, 1-5, 2-3, 2-4, 2-5, 3-4, 3-5, 4-5), and for each of these initial hands we can get every possible hand (5 choose 2).

So (5 choose 2)*(5 choose 2). Is this an incorrect assumption?

(52c5)^2 is the total number of combinations, but that number doesn't divide evenly into each type of draw. If you want to use integers in your tallying, you need to use a denominator that is a LCM of each type.