7-14-21

Looks like the simple strategy is:

1) find all the cards you could use to make a 7 hand (the goal is to qualify the 7 most times, I think)
2) Try to make 21 without busting the 7 hand
3a) Can you play 20-14, 20-13 or 20-12 without busting the 7 hand? Do it.
3b) Try to make a 14 without busting the 7 hand
4) Try to make a 20 without busting the 7 hand
5) Make the 14-hand as good as possible.

6) Then normally maximize the 21 hand, making 7 low, unless the 21 hand is reaaally bad (14-15), and you can play 6 or 7 in 7-hand.

I just worked the math using optimal strategy (brute force, no quantified strategy yet), and I calculate a base wager house edge of 0.540171515%. This looks dead on based on the numbers provided by the manufacturer.

I agree that "only" belongs before the percentage, or deleted.

Although it should be obvious, I think you should mention that the 7 hand does not have to be a lower total than the 14 or 21 hand, and that the 14 hand doesn't need to be lower than the 21 hand. I.E. Although you can make some stupid mistakes, you can't "Foul" the hand. Strategy may dictate doing something that seems counter-intuitive.

I'm unsure how you can do math for "optimal strategy" when you can't even decide on basic strategy.

Quote: Analysis - Bonus Bet section

Unlike most table games, in 7-14-21 the player's actions determine the outcome of the Bonus side bet. In some borderline cases, this will cause the player to deviate from the optimal strategy for the base game. The greater the ratio of the Bonus Bet to the 7, 14, and 21 bets, the more the player will try to win the Bonus Bet, at the expense of the other three.

The only case where this applies would possibly be "going for" two wins (and a loss) rather than two pushes and one win. The net on the basic bet is the same.

I think that once you settle on a strategy, then there won't be any exceptions for the bonus bet.

Quote: DJTeddyBear

I'm unsure how you can do math for "optimal strategy" when you can't even decide on basic strategy.

I let a progam decide the best way to sort the cards by running through each and every way to deal cards to the dealer. I have an excel file now which lists every possible dealt hand and how the hand should be sorted. So, I have a perfect strategy, but you just need to remember 4,915 different hands. The difficulty is putting into words how a player could make these decisions without my list.

Any guess on what the HE might become with the average Joe playing the game?

Someone could try simulating the various strategies that have shown up already:
1) copy the dealer
2) Wizard's from a couple pages back
3) mine from ^ there
and see what edges they play at.

Quote: CrystalMath

The difficulty is putting into words how a player could make these decisions without my list.

CrystalMath was kind enough to send me a spreadsheet of all 4,915 hands. Here is a hand that has caused me to rethink my basic strategy:

AA68TT

The strategy of making the 7 hand as good as possible would for setting this A6,A8,TT. However, the better play is AA,68,TT.

I think it is worth optimizing the 14 or 21 at the expense of the 7 if it bring the 14 or 21 to the highest value. Much like in pai gow (tiles) you strive to get the high hand to at least a high 9. If you can't, then you tend to follow a more balanced strategy most of the time.

One thing I can say is that if you should never bust more hands than you have to. There are ZERO exceptions to this rule.

All things considered, here is my latest strategy to run up the flagpole. The higher the rule, the higher the priority.

1. Never bust more hands than you have to.
2. Make 21 in the 21 hand, if you can.
3. Make 14 in the 14 hand, if you can.
4. Maximize the 7.
5. Maximize the 14.

Regarding another post, with 4,915 different hands it is not easy to test these strategies.

Miplet, CM, SH79 -- How is the strategy progress coming along?

Quote: odiousgambit

I wonder if you are taking into account the house way. The dealer has to start with making the best high hand, making his lower hands vulnerable. So it would be better to wreck the 21 if you are left with 7 and 14 . And playing 6 plus ace for the 7, even if this is leaving no ace for a 21.

As far as I can tell the Wizard is on the money with his strategy above.

edit: With the Wiz strategy, it seems if you are left with 2 hands that are lame, but not busted, it pays to strengthen one even to the point of busting the other.

I'm not even sure how common that situation is. From playing that online version for an hour or so, I will say it is fairly rare to even be able to place all 3 hands. So I figure the odds of placing a low hand, and then even having the choice for a high 21 and still making a mid is very low.


The strategy card says to maximize your low hand before considering the mid or high. Just making a low hand is huge, maximizing doesn't seem to add much. A 4 low and 14 mid is far better than 7 low and 12 mid.

Quote: Wizard

One thing I can say is that if you should never bust more hands than you have to. There are ZERO exceptions to this rule.

Consider this hand: AA779T
You could avoid busting with AA 77 9T, but the best strategy is 9A 77 TA, which is a bust on the 7 hand.

Quote: Wizard

Miplet, CM, SH79 -- How is the strategy progress coming along?

Not so well

Quote: Wizard

1. Never bust more hands than you have to.
2. Make 21 in the 21 hand, if you can.
3. Make 14 in the 14 hand, if you can.
4. Maximize the 7.
5. Maximize the 14.

I would switch 5 and 4 around. Still going off my math, but almost every improvement on the mid hand seemed better than going from 2 to 7 in the low hand.


My numbers would also add a step before #4.

3b) Make a 20 in the 21 hand if you can with a mid hand greater than 7.

mid 10 + high 20 seems better than mid 12 + high 18. An example would be A228TT Which I think should be played A2 28 TT rather than A2 2T 8T.

I needed to add the last part in for this hand 2222TT - 22 22 TT I think is worse than 22 2T 2T. I think this situation can only happen with zero aces, 2 very high cards(9+) and 4 very low cards(2 or 3).