I was having dinner with a buddy, Al, who received a text from someone we both knew, Sandy. The text showed a picture of 8/5 Ace$ Bonus Poker along with a query whether I published a strategy for it?
Ace$ Bonus Poker is just like Bonus Poker, except the four aces each have a letter A, C, E, or S on them. If those four aces spell ACES in positions 1-4 or 2-5, then the hand pays 4,000 coins rather than 400.
I told Al to text back that the game paid 99.407%, which is about 0.23% superior to regular Bonus Poker. The game should be played exactly like Bonus Poker except for the hand aces full (e.g. AAAKK or AAA44, etc.). IF the three aces are in the appropriate positions to get the bonus (which usually isn’t the case, but can be), THEN you throw away the pair and go for the bonus, which you’re only going to hit 1/47 times (the same as drawing to 4-to-a-royal).
After a while, Sandy texted back. There are some REALLY close plays in Bonus Poker, such as A♠ 8♠ 4♥ 6♥ 7♥. In this hand holding the 467 is worth $2.4098 for the 5-coin dollar player while the A by itself was worth $2.4062. The difference is only 0.36¢, which is only about a third of a penny. Couldn’t the ace in appropriate position add enough to change the play?
I didn’t think so, but I said I’d go home and work it out and get back to Sandy. Since Al and Sandy are friends, I did this for free. I get all sorts of requests to figure out stuff for strangers that I turn down or say something like, I’ll do it for a fee of $xxx. Although I am a teacher and I do answer a lot of questions for strangers, I don’t see my role in life to be a research assistant for everybody who has a question. So, I pick my spots, so to speak. (And, of course, if the problem is interesting enough, I turn it into a column or two which pays me a bit.)
On the hand in question, holding one ace, you end up with four aces 44 times out of the 178,365 possible draws. That’s one time in 4,053.75. This calculation should be easy to perform for anybody with video poker software who can divide one number into another.
The next question is: How often does one “sequential” ace turn into four sequential aces? For simplicity, let’s assume the sequential ace is in the first position (which means the ace of clubs, which has a big yellow “A” on it). From here, we’re going to need to get the ace of diamonds (the one with a “C” on it) in the second position, which will happen 1/47 times. If that happens, we need the ace of hearts (the “E) in the third position which occurs 1/46 times, and if that happens, we need the ace of spades (the “$”) in the fourth position, which occurs 1/45 times.
Since they all need to happen together, we multiply them. That is, one sequential ace will become four sequential aces 1/47 * 1/46 * 1/45 = 1/97,290. Looking at it a little differently, since we get aces every 4,053.75 from that position, one in 24 of them are sequential and 23/24 of them aren’t.
The previous two paragraphs required some fairly simple probability calculations. Simple probability calculations are necessary in figuring out all sorts of video poker and other gambling problems. To succeed, you either need to be able to figure this out yourself or have someone who does and is willing to help you either free or for a fee.
So, from this position, once in 97,290 times, the four aces will be worth $4,000. Since the video poker software already assumes that these aces will receive $400, the bonus is worth $3,600. Once you get here, dividing $3,600 by 97,290 gives you $0.037, which is almost four cents. This is ten times as much as the difference between the two plays. So yes, there are some exceptions that my simplification misses.
I was actually surprised to find this out.
Still, to get the optimal value out of this, you need to look at all the hands that could be affected. Usually they are an ace with a 3-card straight flush with no high cards and one gap, such as 457 or 78T with external straight interference suited with the ace.
There are a number of qualifications in that sentence. For players who don’t seriously study penalty cards, this is pretty complicated. Many players aren’t going to be able to remember it precisely. If they have it written down somewhere it will take them some time to find it, look it up, and determine if the current hand qualifies.
So, I’m personally not going to take the time and energy to memorize this exception, at least at the current time. I rarely play this game. I’ve played 8/5 Bonus for several hundred hours in the past, but not in the past few years. I will have to relearn 8/5 Bonus before I play this game.
It is possible that Ace$ Bonus will become my “go to” game in a particular casino somewhere down the line, and in that case, I’ll relook at this article and reassess whether I’m going to add this refinement to my game. But in the meantime, in the few times I play this game, I’m content to use regular 8/5 Bonus strategy with the aces-full exception.
If you wish to make the opposite conclusion for yourself, be my guest.
Note: After I finished this article, I passed it by a player who graduated from Cal Tech to verify that my conclusions were correct. He responded:
I agree with your math, and want to add that there is another way of looking at it, via https://wizardofodds.com/games/video-poker/tables/aces-bonus-poker/
“To create a strategy for this game, you can use my video poker strategy maker, but use a 94 for four aces when there are no aces on the deal. With one ace, in a correct position, use 110. With two aces, in a correction position, use 200. With three aces, in a correction position, use 440. These are weighted averages of 80 and 800, according to the probability of getting ACE$. If you have aces that are out of position then use 80.”
Once more, the wizardofodds.com site provides useful information. The way to get the most value out of this site on this particular problem is to enter in the appropriate number for aces and look at the exceptions to the basic strategy. The ones that say the right play is the ace and not some other hold are the ones you want to look at.
I appreciate learning about the Wizard’s elegant solution to this problem and am glad to pass it on to you.