DaveAring
DaveAring
Joined: Aug 23, 2020
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August 23rd, 2020 at 4:24:58 PM permalink
I will try to make this question clear. I find it a very difficult question to ask. It is a video poker question that has absolutely no relevance, but I WOULD like to know the answer (if possible). Here is the question...

I just hit a royal. Which number (0-5) has the highest probability for the number of cards I held while going for the royal?

The answer will be between 0 and 5. For example, if I was dealt and held four cards to the royal, it would be a 47:1 shot that I got a royal. If I held 0 cards to the royal it would be (approximately) 650,000:1. Obviously, the odds for 1, 2, 3, and 4 held cards would be in between those odds. However, that is NOT the question. Simple question, but please consider the following when calculating the answer.

1. You may ignore WHICH game I was playing. It (essentially) should be ignored because in some games the strategy MIGHT dictate that you do not hold a single high card (or even two high cards) to the royal.

2. Since there is no strategy (that *I* know of) where you would hold a single 10, for the purpose of this question, you may consider dealt hands with a single 10 to be the same as a dealt hand with NO high cards even though a 10 is a card that would be part of a completed royal. To be sure, if dealt a hand with a single 10, you might hold 3344 and not go for the royal.

3. *I* believe that the TWO major factors that come into play are:
a. How often a player would be dealt a hand where they would hold 0, 1, 2, 3, 4, or 5 cards to the royal.
b. How often each of those different hands turned into a royal.

Referring to #3 above, if you held four cards to a royal you would have a chance of 1 in 47 to draw the card you need. However, you are not dealt four to the royal as often as (say) two cards to the royal. Therefore, you would have more chances going for the royal while holding two cards than one, but the odds of actually getting the cards you need are higher when holding two cards.

As I said above, I am ignoring strategy for a specific game. However, that presents a conundrum. If you are dealt a hand... AAAKQ where the AKQ are the same suit, you have three cards to the royal. In most every game, you would hold the three aces instead of the three to the royal. Therefore, when calculating the number of hands with three to the royal, would/should that hand count as holding three to the royal? So, even though I said NOT to consider specific game strategies, it IS going to be a factor in determining the answer. The question here is, HOW do we take that fact into consideration OR do we just ignore the fact?

Lastly, if it is too difficult to answer my specific question, I would like to simplify the original question and state that regardless of the game being played, let's assume that NO MATTER WHAT HAND IS DEALT, you go for the royal on EVERY hand. i.e. if dealt AAAAK, you would hold the king and the ace whose suit matches the king.

Anyone care to chime in with their thoughts?

Thanks.
ksdjdj
ksdjdj
Joined: Oct 20, 2013
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August 24th, 2020 at 3:24:04 AM permalink
Hi and welcome,

I probably can't help you for most of your post, but below is a link to a thread that may help answer the last part of your post.

"Royal at all costs strategy" Thread
heatmap
heatmap
Joined: Feb 12, 2018
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August 24th, 2020 at 4:36:46 AM permalink
Quote: ksdjdj

Hi and welcome,

I probably can't help you for most of your post, but below is a link to a thread that may help answer the last part of your post.

"Royal at all costs strategy" Thread



My brain was very confused because I honestly thought it was a question about that and that other thread is very concentrated on that one subject I thought there is no way itís the same thing
rsactuary
rsactuary
Joined: Sep 6, 2014
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August 24th, 2020 at 6:34:15 AM permalink
Ask and answered by the Wizard in one of his columns many years ago. I'll try and find it, but the answer, if I understand your question correctly, is that you are most likely to get a royal from being dealt 3 and drawing the remaining 2.

ETA: I can't find it, but hopefully someone else can.
Joeman
Joeman
Joined: Feb 21, 2014
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August 24th, 2020 at 6:36:24 AM permalink
Quote: DaveAring

I will try to make this question clear. I find it a very difficult question to ask. It is a video poker question that has absolutely no relevance, but I WOULD like to know the answer (if possible). Here is the question...

I just hit a royal. Which number (0-5) has the highest probability for the number of cards I held while going for the royal?

The answer will be between 0 and 5. For example, if I was dealt and held four cards to the royal, it would be a 47:1 shot that I got a royal. If I held 0 cards to the royal it would be (approximately) 650,000:1. Obviously, the odds for 1, 2, 3, and 4 held cards would be in between those odds. However, that is NOT the question. Simple question, but please consider the following when calculating the answer.

1. You may ignore WHICH game I was playing. It (essentially) should be ignored because in some games the strategy MIGHT dictate that you do not hold a single high card (or even two high cards) to the royal.

2. Since there is no strategy (that *I* know of) where you would hold a single 10, for the purpose of this question, you may consider dealt hands with a single 10 to be the same as a dealt hand with NO high cards even though a 10 is a card that would be part of a completed royal. To be sure, if dealt a hand with a single 10, you might hold 3344 and not go for the royal.

3. *I* believe that the TWO major factors that come into play are:
a. How often a player would be dealt a hand where they would hold 0, 1, 2, 3, 4, or 5 cards to the royal.
b. How often each of those different hands turned into a royal.

Referring to #3 above, if you held four cards to a royal you would have a chance of 1 in 47 to draw the card you need. However, you are not dealt four to the royal as often as (say) two cards to the royal. Therefore, you would have more chances going for the royal while holding two cards than one, but the odds of actually getting the cards you need are higher when holding two cards.

As I said above, I am ignoring strategy for a specific game. However, that presents a conundrum. If you are dealt a hand... AAAKQ where the AKQ are the same suit, you have three cards to the royal. In most every game, you would hold the three aces instead of the three to the royal. Therefore, when calculating the number of hands with three to the royal, would/should that hand count as holding three to the royal? So, even though I said NOT to consider specific game strategies, it IS going to be a factor in determining the answer. The question here is, HOW do we take that fact into consideration OR do we just ignore the fact?

Lastly, if it is too difficult to answer my specific question, I would like to simplify the original question and state that regardless of the game being played, let's assume that NO MATTER WHAT HAND IS DEALT, you go for the royal on EVERY hand. i.e. if dealt AAAAK, you would hold the king and the ace whose suit matches the king.

Anyone care to chime in with their thoughts?

Thanks.

Welcome, Dave and congrats on the royal hit! I can't help you with the math here, but I can offer some thoughts.

First of all, let me say that I have been contemplating a similar question for some time. I have been reluctant to ask it here because, as you noted, its a purely academic question that doesn't have much bearing on the actual playing of VP. I know there are those on this board that could answer the question, and they might see it a as a mathematical challenge, or may have the same curiosity we do. So, there's a good chance someone will come along who will answer the question. Personally, I think this would have made for a fun Ask The Wizard question.

I'm not sure if your #1 and #2 were constraints you wanted to add to the problem, or if they were meant to simplify the math. I would think it may be easier to constrain the problem if you were to specify a game (e.g Jacks or Better) and a strategy. I think it would be the least "messy" to use either a basic strategy for a specific game, or to use a "royal-at-all-costs-strategy." Personally, I would like to see the answer for Double Double Bonus using basic DDB strategy.

At any rate, I do like the question, and hopefully someone will provide an answer.
"Dealer has 'rock'... Pay 'paper!'"
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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Thanks for this post from:
Joeman
August 24th, 2020 at 6:54:18 AM permalink
Assuming (a) you always "play for the Royal," and (b) if the highest card in the dealt hand is a 10, then you discard the 10s, I get this:
HeldHandsRoyals
544
490819.32
341,25638.16
2592,44036.54
11,587,3608.9
0376,9920.25

The three columns are (a) the number of cards in the Royal, (b) the number of dealt hands out of the 2,598,960 possible hands, and (c) the expected number that become Royal Flushes.
As you can see, being dealt 3 is the most likely result, followed closely by 2.

However, this does not take into account playing best strategy - for example, you do not break up a King-high straight flush, nor do you break up any "paying hand" with 3 to a Royal (for example, with Ace of spades, Queen of spades, Jack of spades, Jack of hearts, 3 of clubs, you hold the pair of Jacks).
rsactuary
rsactuary
Joined: Sep 6, 2014
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August 24th, 2020 at 8:31:03 AM permalink
Again, this was answered in an "Ask the Wizard" column many years ago.
Joeman
Joeman
Joined: Feb 21, 2014
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August 24th, 2020 at 8:48:18 AM permalink
Quote: rsactuary

Again, this was answered in an "Ask the Wizard" column many years ago.

Lo and behold, it looks like this has been discussed here as well, including percentages for JoB, DB, DDB, and NUSD, courtesy of CrystalMath!

I think this along with TDG's analysis of the "royal-at-any-cost" strategy above answers OP's question.
"Dealer has 'rock'... Pay 'paper!'"

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