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.

Quote:DaveAringI originally posted this question under the MATH category, but it was suggested that I put it here so that it would be more likely that the Wizard would see it and respond to it. Consequently, here is my repost...

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.

I worked only on your simplified question. If the player always tries for a royal (even by holding a single ten), I get these numbers:

n-to-the-royal | P(deal) | P(draw royal) | P(deal)*P(draw royal) | Relative Prob. |
---|---|---|---|---|

5 | 0.0000015 | 1.0000000 | 0.00000154 | 0.0355 |

4 | 0.0003617 | 0.0212766 | 0.00000770 | 0.1776 |

3 | 0.0166374 | 0.0009251 | 0.00001539 | 0.3552 |

2 | 0.2394035 | 0.0000617 | 0.00001476 | 0.3408 |

1 | 0.6661126 | 0.0000056 | 0.00000373 | 0.0862 |

0 | 0.0774833 | 0.0000026 | 0.00000020 | 0.0047 |

1.00000000 | 0.00004333 | 1.0000 |

So for the royal-at-any-cost strategy, a royal's most likely starting hand would be three-to-a-royal with a probability of 35.52%.