What is 19,933,230,517,200 in Jacks or Better Video Poker?

The number 19,933,230,517,200 comes up as the total number of combinations in Jacks or Better Video Poker.
Would somebody please provide a formula for this number in terms of combinations like C(52,5) (which is 52!/47!5!)?
Please include underlying hypothesis such as number of decks or whatever. The number is the size of the complete
sample space of something. The number is so large I can't guess what that sample space is. Thanks and I'm very
grateful for your help. Also, if you could suggest a good book or a website on probability calculations underlying card games
like Five Card Draw, that would be great. Thanks again.

Quote: rjs357

The number 19,933,230,517,200 comes up as the total number of combinations in Jacks or Better Video Poker....

This is a little hard to explain, so read carefully.

First, there are C(52,5)=2,598,960 ways to deal the initial five cards.

The number of possible combinations on the draw depends on the number of cards the player discards. The following table shows the draw combinations in the second column.

Discard	Draw combinations	Weight	Product
0	1	7,669,695	7,669,695
1	47	163,185	7,669,695
2	1,081	7,095	7,669,695
3	16,215	473	7,669,695
4	178,365	43	7,669,695
5	1,533,939	5	7,669,695

However, the weight of hands on the draw need to be weighted according to how many cards the player discarded. For example, a royal after keeping four to a royal is much more likely than a royal after tossing everything. To keep the combinations integers we need to find the least common multiple of the draw combinations. The answer to that is 7,669,695, which equals 5*C(47,5)

So C(52,5)*C(47,5)*5 = 19,933,230,517,200.

Quote: rjs357

The number 19,933,230,517,200 comes up as the total number of combinations in Jacks or Better Video Poker.
Would somebody please provide a formula for this number in terms of combinations like C(52,5) (which is 52!/47!5!)?
Please include underlying hypothesis such as number of decks or whatever. The number is the size of the complete
sample space of something. The number is so large I can't guess what that sample space is. Thanks and I'm very
grateful for your help. Also, if you could suggest a good book or a website on probability calculations underlying card games
like Five Card Draw, that would be great. Thanks again.

What do you mean, it comes up? Where did you get it from?

At first I thought that it might be the total number of possible (deal, draw) pairs but it looks too big even for that (it's almost triple the the square of the number of deals)

One question related to this has to do with least optimal play. It's well known that video poker offers advantage play, and I often wonder what actual observed hold percentages are for some machines. But the absolute lowest payback percentage difference based on intentionally holding the worst possible cards every time is a number that I am not familiar with nor have I ever heard of anyone determining (least optimal play backback percentage based on a given game and pay table).

Quote: rjs357

Also, if you could suggest a good book or a website on probability calculations underlying card games
like Five Card Draw, that would be great. Thanks again.

For actual 5 card draw, this book shows calculations in significant detail.

http://www.amazon.com/DRAW-POKER-ODDS-Mathematics-Classical/dp/9738752051

For video poker, Stuart Ethier offers an analysis of "video poker" as the sample chapter of his book "The Doctrine of Chances" on his faculty website.

http://www.math.utah.edu/~ethier/sample.pdf

Quote: AxiomOfChoice

What do you mean, it comes up? Where did you get it from?

At first I thought that it might be the total number of possible (deal, draw) pairs but it looks too big even for that (it's almost triple the the square of the number of deals)

Oh, I somehow missed the wizard's reply before I wrote mine. Never mind..

(edit: I actually think that we were writing at the same time, so his reply was not there before I started replying but was there by the time I hit "post")

Quote: Ahigh

One question related to this has to do with least optimal play. It's well known that video poker offers advantage play, and I often wonder what actual observed hold percentages are for some machines. But the absolute lowest payback percentage difference based on intentionally holding the worst possible cards every time is a number that I am not familiar with nor have I ever heard of anyone determining (least optimal play backback percentage based on a given game and pay table).

We did it awhile back for 9/6 JoB (99.54% return w/optimal strategy). It's a 2.3717% return if you are allowed to throw away a dealt Royal. Most machines don't let you do that, so the return one would realistically expect is 2.495%.

https://wizardofvegas.com/forum/gambling/video-poker/14822-theoretical-payout-on-9-6-job-with-opposite-to-optimal-strategy/

As for actual "return", I've read that it's often 2 to 4% below optimal. It depends on the game and market though. JoB is played much more accurately than Ultimate X for example. I would also expect players on the Strip to play video poker worse because they are more often trying the game for the first time than some Vegas local.

Quote: tringlomane

We did it awhile back for 9/6 JoB (99.54% return w/optimal strategy). It's a 2.3717% return if you are allowed to throw away a dealt Royal. Most machines don't let you do that, so the return one would realistically expect is 2.495%.

https://wizardofvegas.com/forum/gambling/video-poker/14822-theoretical-payout-on-9-6-job-with-opposite-to-optimal-strategy/

As for actual "return", I've read that it's often 2 to 4% below optimal. It depends on the game and market though. JoB is played much more accurately than Ultimate X for example. I would also expect players on the Strip to play video poker worse because they are more often trying the game for the first time than some Vegas local.

so your saying if you make the worst possible holds you would only get back 2.3%. I thought it would be higher then that. Sometimes no matter what you do you end up with a pay.

Quote: AxelWolf

so your saying if you make the worst possible holds you would only get back 2.3%. I thought it would be higher then that. Sometimes no matter what you do you end up with a pay.

I think you're overlooking the multitude of losing deal hands, such as 2-3-6-10-Q, where you would hold everything.

Quote: AxiomOfChoice

Oh, I somehow missed the wizard's reply before I wrote mine. Never mind..

(edit: I actually think that we were writing at the same time, so his reply was not there before I started replying but was there by the time I hit "post")

Ninja'd. Makes us all look silly one time or another.

Quote: dwheatley

Ninja'd. Makes us all look silly one time or another.

Except that the Wizard's answer was the right one and mine was just asking wtf was going on :)

Thanks for your response.
I understand how you arrived at 19,933,230,517,200.
I see that 5*C(47,5) is the LCM of the set {C(47,n)| n=0,1,2,3,4,5} but I'm struggling to find a probabilistic justification to your approach,
"To keep the combinations integers we need to find the least common multiple of the draw combinations."

I want to respectfully submit the following alternative.
The video poker player starts with 5 cards, there are C(52,5) possibilities.
From there, Player draws n cards, where n = 0,1,2,3,4,5.
Player must choose n cards from his/her 5 card hand, there are C(5,n) ways of doing this.
Player is then dealt n cards from the remaining 47 card deck, there are C(47,n) possible outcomes to the draw.
From the sum rule for probabilities of nonintersecting events, summing over n should lead to the size of our sample space,
C(52,5)[C(5,0)*C(47,0) + C(5,1)*C(47,1) + C(5,2)*C(47,2) + C(5,3)*C(47,3) + C(5,4)*C(47,4) + C(5,5)*C(47,5)]
= 2,598,960 * 2,598,960 = C(52,5)^2.
The formula for the factor with the sum can be obtained by direct calculation of from Vandermonde's Identity.
This is the value mentioned by AxiomOfChoice in the post following the your post.
As was pointed out there by AxiomOfChoice, the ratio 19,933,230,517,200 / C(52,5)^2 is about 2.95.

Would you please explain in just a little more detail how your LCM approach relates to the size of our sample space?

Thanks

My apologies, it was careless of me not to include the following reference,
Please see my reply to the narrator's response.