2 Tough VP Questions

I have been thinking about some odds lately while playing JoB on the WoO site. I have some question I think are very difficult to calculate. Let's see if anyone is up to the challenge.

1) What percentage of the time, if the "best choice" is made, will the best outcome occur? In other words, how often would the player get a better outcome if he knew the cards that were going to come up and made a different choice based on this knowledge?

2/3) What percentage of all hands dealt could a Royal occur if the cards to come were known? This actually may not be that difficult to figure out. However as a related question, how often would this happen if the "best choice" is made? Or how often are Royals missed because the "best choice" wasn't the one leading to the Royal?

By "best choice" I mean the correct selection of holds based on the WoO perfect rules. At first I figured that anyone would need to know what cards were going to come up next to figure out the answer these questions. But now I'm not even sure that is true. Just thinking about how to answer these makes my head hurt.

1) 100%. If you make the best choice, the end result will be the highest possible return given the cards held. Yes, I know this is not what you meant by your question, but you shouldn't focus on final outcome. If you focus on holding the cards that give you the best return, long run, you will do better.

For a perfect peeker, I get a probability of a royal of 1 in 23,080.75.

I've thought about this same question before. Payback percentage of say 9/6 JoB if you knew the next five cards. You never know when a machine will F up!

Quote: downtowner

I have been thinking about some odds lately while playing JoB on the WoO site. I have some question I think are very difficult to calculate. Let's see if anyone is up to the challenge.

1) What percentage of the time, if the "best choice" is made, will the best outcome occur? In other words, how often would the player get a better outcome if he knew the cards that were going to come up and made a different choice based on this knowledge?

2/3) What percentage of all hands dealt could a Royal occur if the cards to come were known? This actually may not be that difficult to figure out. However as a related question, how often would this happen if the "best choice" is made? Or how often are Royals missed because the "best choice" wasn't the one leading to the Royal?

By "best choice" I mean the correct selection of holds based on the WoO perfect rules. At first I figured that anyone would need to know what cards were going to come up next to figure out the answer these questions. But now I'm not even sure that is true. Just thinking about how to answer these makes my head hurt.

Excellent question, downtowner! I, too, have considered this idea. Furthermore, what would be the value of knowing the next one, two, three,... cards after the deal. It comes to mind more often on Loose Deuces, when I choose one of two pairs and get the wrong card(s) on the deal.

By the way, in the field of Decision Analysis, this idea is referred to as the "value of perfect information." What is VP analysis if not a huge decision tree?

Wish I had more time to analyze further - would be an interesting idea for a VP variant.

I think there is an IGT video poker variation where you pay a fee and can see the next card. As far as I know, it never caught on.

Quote: Wizard

For a perfect peeker, I get a probability of a royal of 1 in 23,080.75.

That's what I get as well:

TDG: "622,200 deals with at least one 2/Royal; you have a 1/16,215 chance that the next 3 cards complete the Royal"

On a hand like As Ks Qh Jh 5c, you are calculating a 1/16,215 chance. I understand how you got that. You assumed you held the hearts and drew three cards.

But isn't it 2/16,215 because BOTH the three royal cards in spades AND the three royal cards in hearts would qualify?

Similarly for a hand like Ah Ks Qd Jc 5x, your formula implies there is one way to draw four perfect royal cards --- but actually there are four because the royal could be in any of the four suits.

This isn't going to affect the final number by much --- but it does make it more likely than you indicate.

If I'm wrong, please correct me.

Quote: BobDancer

TDG: "622,200 deals with at least one 2/Royal; you have a 1/16,215 chance that the next 3 cards complete the Royal"

On a hand like As Ks Qh Jh 5c, you are calculating a 1/16,215 chance. I understand how you got that. You assumed you held the hearts and drew three cards.

But isn't it 2/16,215 because BOTH the three royal cards in spades AND the three royal cards in hearts would qualify?

No, because in that case, you have to choose to keep either the two spades or the two hearts. In either case, only one set of three drawn cards will complete the Royal Flush.

The same with your second example - you can keep only one of the four cards, and the four cards dealt have to be in that suit. The probability of making the Royal will be the same regardless of which of the four cards you choose to keep.

The only exception to this "rule" is if you do not have any 10 or higher cards in your initial five; in that case, the next five cards can be any of the four possible Royal Flushes.

Quote: ThatDonGuy

No, because in that case, you have to choose to keep either the two spades or the two hearts. In either case, only one set of three drawn cards will complete the Royal Flush.

The same with your second example - you can keep only one of the four cards, and the four cards dealt have to be in that suit. The probability of making the Royal will be the same regardless of which of the four cards you choose to keep.

The only exception to this "rule" is if you do not have any 10 or higher cards in your initial five; in that case, the next five cards can be any of the four possible Royal Flushes.

I understood the original premise was that a perfect peeker would look first and THEN decide what to play. So on the As Ks Qh Jh 5c hand, if he peeked and saw three royal spade cards he would hold AK. If he saw three royal heart cards he would hold QJ.

You tell me I have to pick the hearts or spades first and then peek. That can't be right because the peeking wouldn't give me any value because I already made my choice. If peeking is going to give me value, I should be able to choose EITHER AK or QJ after I peek.

What am I missing?

Quote: BobDancer

I understood the original premise was that a perfect peeker would look first and THEN decide what to play. So on the As Ks Qh Jh 5c hand, if he peeked and saw three royal spade cards he would hold AK. If he saw three royal heart cards he would hold QJ.

You tell me I have to pick the hearts or spades first and then peek. That can't be right because the peeking wouldn't give me any value because I already made my choice. If peeking is going to give me value, I should be able to choose EITHER AK or QJ after I peek.

What am I missing?

You're not missing anything; I was.

My answer would have been correct if the question was what is the probability of a royal for a "royal or nothing" player. In other words, a perfect player against a pay table where only a royal paid anything. The strategy would be simple -- keep the cards to a royal in the suit where you have the most. In a tie, pick randomly.

A perfect peeker would have a better chance. Consider the hand on the deal: Ah Kc Qc Jc 10c and the next cards on the draw were, in order, Kh Qh Jh 10h 2s. The royal or nothing player would keep the four clubs -- and get a pair of kings. The perfect peeker would keep the Ah only and get the royal.

That said, my probability for the perfect peeker is 1 in 20304.37624 (edited). Can I get anyone to second that? It is a tricky problem. Be careful to not double-count situations like getting a royal on the deal, with a royal in a different suit right behind it on the draw.

I assumed that the five draw cards were randomly drawn from the 47 not used on the deal, in a specific order, and they would come out in that order according to how many cards the player discarded.

Quote: BobDancer

I understood the original premise was that a perfect peeker would look first and THEN decide what to play. So on the As Ks Qh Jh 5c hand, if he peeked and saw three royal spade cards he would hold AK. If he saw three royal heart cards he would hold QJ.

You tell me I have to pick the hearts or spades first and then peek. That can't be right because the peeking wouldn't give me any value because I already made my choice. If peeking is going to give me value, I should be able to choose EITHER AK or QJ after I peek.

What am I missing?

Doing it that way, I get 1 / 20,858.57.

Let me see if my method is right:
If the deal is, for example, As Ks Qs Jh 10d, then there are three possible ways for the draw to be a Royal Flush:
(1) The first two cards are Js 10s (in either order); this has a probability of 1 / 1081
(2) The first four cards are Ah Kh Qh 10h; this has a probability of 1 / 178,365
(3) The first four cards are Ad Kd Qd Jd; this also has a probability of 1 / 178,365
The three possibilities are independent of each other, since the first card determines the suit of the eventual Royal Flush, so the probability of getting a Royal Flush from that hand is 1 / 1081 + 2 / 178,365.

Let's consider the "4" situation, where you say the probability is 1/47.

Isn't it possible that the next 5 cards in the draw are a royal to one of the other three suits? So, shouldn't the probability of 4 to a royal on the deal, where the 5th card is a 9 or less, 1/47 + 3*(1/1533939) ? Of course, many other lines will have this same issue of possible full royals on draw.

I'm not enough of a math guy to do the calculations, but an easy way to think of this is to just make the best 5 card hand out of the first 10 cards dealt. And then discount cases where you have multiple ways of getting the same best hand.

Quote: DJTeddyBear

I'm not enough of a math guy to do the calculations, but an easy way to think of this is to just make the best 5 card hand out of the first 10 cards dealt. And then discount cases where you have multiple ways of getting the same best hand.

I would not go down that path. The peeker can not pick and choose which cards he wants from the predestined five cards on the draw. If this situation is:

Deal: Th Jh Qh Kh 2s
Draw 3c Ah 4c 5c 6c

Then there is zero chance at a royal. The peeker cannot say "I'll take the Ah." The 3c is first in line for the draw and nothing can change that.

Does a game exist that shows you the next five cards or is this a Altuve vs Judge for MVP type of discussion?

Quote: DJTeddyBear

I'm not enough of a math guy to do the calculations, but an easy way to think of this is to just make the best 5 card hand out of the first 10 cards dealt. And then discount cases where you have multiple ways of getting the same best hand.

That will not work - here's why:

Suppose you are dealt As Ks Qs Js 3c
Also suppose the "next five cards," in order, are 2d 10s 8c Jc 3h.
The "best five cards" make a Royal Flush, but you can't get the 10 of spades unless you discard at least one of the cards in your hand that make that Royal Flush.

Quote: Wizard

Let's consider the "4" situation, where you say the probability is 1/47.

Isn't it possible that the next 5 cards in the draw are a royal to one of the other three suits? So, shouldn't the probability of 4 to a royal on the deal, where the 5th card is a 9 or less, 1/47 + 3*(1/1533939) ? Of course, many other lines will have this same issue of possible full royals on draw.

You are right, and when you take those into account, I get 1 / 20,292.64

Quote: billryan

Does a game exist that shows you the next five cards or is this a Altuve vs Judge for MVP type of discussion?

As far as I know, it is just a math problem. I can't see how knowing the answer would help in any video poker game, except one that had a bug that somehow revealed the draw cards.

BTW, it is my understanding the draw cards are not predestined but the rest of the deck is shuffled over and over while the game waits for you to make a decision.

BTW, I couple hours of programming could determine the answer to question number 1. However, it would help to have some kind of reason for the question before I fussed with it.

Sigh, thanks for pointing out the error in my logic. That just confirms that I should avoid these complex math scenarios.