August 4th, 2022 at 9:07:00 PM
permalink

I'm working on a strategy for ACE$ Bonus Poker, and I just noticed that the Wizard's page on ACE$ Bonus at https://wizardofodds.com/games/video-poker/tables/aces-bonus-poker/ may have two errors. I would appreciate it if other math-oriented people take a look and tell me who is right (or what I'm doing wrong).

First thing is, the Wizard says "use a 94 for four aces when there are no aces on the deal [...] These are weighted averages of 80 and 800, according to the probability of getting ACE$." But I get 92 instead of 94 when I try to do the math. This is my math:

If there no aces in the dealt hand and one hopes to get all of them on the redraw, there are two ways to get ACE$, one starting from the first position and another starting at the second position. But there are 120 different ways the four aces can land (the ace of diamond can land in any of the five places, the ace of spades in any of the remaining four, the ace of hearts in any of the remaining three, and the ace of clubs in any of the remaining two: 5*4*3*2 = 120). So we have a 2/120 chance of getting ACE$, and a 118/120 chance of getting aces in an undesirable position. 2/120*800+118/120*80 = 92.

The second problem happens when there is one dealt ace in a correct position. The Wizard says "With one ace, in a correct position, use 110". I arrive at the same number when I do my math: the remaining three aces can fall into the four open locations in 4*3*2=24 different ways, only one of which can result in ACE$ (because the first in-position ace only allows one of the two possible ACE$ happen). So we have 1/24*800+23/24*80=110. That's fine.

But let's look at the hand A♠Q♠T♦9♦8♦ and let's assume A♠ is in a correct position and Q♠ is out of the way, so it would not be blocking ACE$ from getting formed. If we consider holding A♠Q♠, the weighted averages change! Now, the remaining three aces can only fall in six different ways, one of which would form ACE$. So we'll have 1/6*800+5/6*80=200 instead of 110. So we can't assume a strategy with 110 for four aces would work when it may acually need to be 200 if two cards are held.

What I am doing wrong?

First thing is, the Wizard says "use a 94 for four aces when there are no aces on the deal [...] These are weighted averages of 80 and 800, according to the probability of getting ACE$." But I get 92 instead of 94 when I try to do the math. This is my math:

If there no aces in the dealt hand and one hopes to get all of them on the redraw, there are two ways to get ACE$, one starting from the first position and another starting at the second position. But there are 120 different ways the four aces can land (the ace of diamond can land in any of the five places, the ace of spades in any of the remaining four, the ace of hearts in any of the remaining three, and the ace of clubs in any of the remaining two: 5*4*3*2 = 120). So we have a 2/120 chance of getting ACE$, and a 118/120 chance of getting aces in an undesirable position. 2/120*800+118/120*80 = 92.

The second problem happens when there is one dealt ace in a correct position. The Wizard says "With one ace, in a correct position, use 110". I arrive at the same number when I do my math: the remaining three aces can fall into the four open locations in 4*3*2=24 different ways, only one of which can result in ACE$ (because the first in-position ace only allows one of the two possible ACE$ happen). So we have 1/24*800+23/24*80=110. That's fine.

But let's look at the hand A♠Q♠T♦9♦8♦ and let's assume A♠ is in a correct position and Q♠ is out of the way, so it would not be blocking ACE$ from getting formed. If we consider holding A♠Q♠, the weighted averages change! Now, the remaining three aces can only fall in six different ways, one of which would form ACE$. So we'll have 1/6*800+5/6*80=200 instead of 110. So we can't assume a strategy with 110 for four aces would work when it may acually need to be 200 if two cards are held.

What I am doing wrong?

August 4th, 2022 at 9:19:22 PM
permalink

There may be some other issues too. For example, if the dealt hand has a non-ace high card in the second, third, or fourth position and there is nothing better to hold in the hand, I believe we can't assume 92/94 for four aces anymore. If we hold that high card, ACE$ can never be formed, so we can't assume a weighted average and should only assign 80 to four aces in the strategy. Right?

August 5th, 2022 at 6:40:21 PM
permalink

Thank you for brining this to my attention.

I agree that 92 is the right weight pay for tossing everything, with all aces left in the deck.

As to the AQT98 question, let's rephrase it to a deal of AT98Q. I show the the player should use a win of 110 for four aces whether or not he holds the queen.

I am out of town and will correct my page at WoO when I get back.

Thanks again for the correction.

I agree that 92 is the right weight pay for tossing everything, with all aces left in the deck.

As to the AQT98 question, let's rephrase it to a deal of AT98Q. I show the the player should use a win of 110 for four aces whether or not he holds the queen.

I am out of town and will correct my page at WoO when I get back.

Thanks again for the correction.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

August 5th, 2022 at 9:23:30 PM
permalink

Thank you very much for the confirmation and the plan to fix the page.

Would you please tell me what's wrong with my result of 200 in case both the ace and the queen are held? I can't figure it out.

Quote:WizardAs to the AQT98 question, let's rephrase it to a deal of AT98Q. I show the the player should use a win of 110 for four aces whether or not he holds the queen.

Would you please tell me what's wrong with my result of 200 in case both the ace and the queen are held? I can't figure it out.

May 3rd, 2023 at 12:04:12 AM
permalink

Quote:Wizard

As to the AQT98 question, let's rephrase it to a deal of AT98Q. I show the the player should use a win of 110 for four aces whether or not he holds the queen.

link to original post

OK, I just finished coding my ACE$ Bonus hand analyzer. I didn't put any fancy logic in it, just running through all potential hands and calculating the best hold and its expected value. I fed it A♠T♦9♦8♦Q♠, and it told me the value of holding A♠Q♠ for 8/5 ACE$ Bonus is 0.583657.

Then I checked with the Wizard of Odds Video Poker Hand Analyzer set to Bonus Poker 8/5. First I gave four aces a win of 110, and it tells me holding A♠Q♠ has the value of 0.578107. I then gave four aces a win of 200, and tells me holding A♠Q♠ has the value of 0.583657.

This tells me 200 is the correct value in this scenario when considering holding A♠Q♠. (Note if that we want to consider just holding A♠, we need to use 110 instead to arrive at the correct expected value).

This hand is quite interesting. If Q♠ is in the way of forming the ACE$, basically in any position other than the last card, the correct hold is T♦9♦8♦ (in this case, since forming ACE$ is impossible, one should use a win of 80 when considering holding A♠Q♠). If it is in the last position, the correct hold is A♠Q♠.

Considering this, I think the recommendation on https://wizardofodds.com/games/video-poker/tables/aces-bonus-poker/ for calculating the strategy just based on the number of aces and if they are in position or not, is not optimal. ACE$ Bonus is more complicated than what we thought.

May 3rd, 2023 at 9:42:18 AM
permalink

I don't necessarily think he was going for optimal, I think he was going with something that balanced ease of remembering a strategy that can be executed while not giving up too much from the optimal return.

Last edited by: rsactuary on May 3, 2023

May 3rd, 2023 at 10:27:53 AM
permalink

Quote:rsactuaryI don't necessarily think he was going with optimal, I think he was going with something that balanced ease of remembering a strategy that can be executed while not giving up too much from the optimal return.

link to original post

You're absolutely right.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)