7-14-21 Poker
Quote: Stroll Enterprises, LLCGetting Started: Each player places (3) equal bets as indicated on the table layout. All bets are made before the cards are dealt, including the optional bonus bet.
The Play:
You are only playing against the dealer, not the other players. Each player and the dealer are dealt 6 cards, face down. Aces may count as 1 or 11. Face cards count as 10.
Each player arranges their six cards into three pairs. You want to come as close to possible to 7, 14, and 21 without exceeding those values. You win the pair when it beats the dealer's pair. Ideally, you want your hand to "hit" 7, 14, and 21 (Perfect Triple) exactly. All winning pairs play 1 to 1 and a perfect triple pays 2 to 1.
Players may not show their hands to other players.
Bonus Bet:
You may make an OPTIONAL BONUS BET in addition to your standard 7-14-21 wager. This bet may be more or less than your standard 7-14-21 wager. All winning Bonus Bets are paid according to the posted pay table.
Bet the Bonus... win Big. Bet the BONUS, and if you hit the SUITED PERFECT TRIPLE (all cards, same suit) you win BIG, 1500 to 1
House rules for the dealer:
The dealer must first form the best 21 two card hand from the six cards dealt. Next the dealer must form the best 14 two card hand from the remaining four cards. The last two cards form the dealer's 7 hand.
Suggested Strategy for Player:
You should always try to arrange the cards in the best combination resulting in no busted hands. If this is not possible then you should attempt to make the best 7 hand first, then attempt to make the best 14 hand. The last two cards form your 21 hand.
*If there are two cards to form the best fourteen two card hand and one of these includes an ace, the dealer must choose the combination which does not contain an ace.
** Any Player's two card hand that exceeds 7 or 14 for these respective hands is a bust and automatically loses.
Pay Table:
Perfect Hand Award: 2 to 1
Win (per hand): 1 to 1
Optional Bonus Pay Table:
Suited Perfect Hand: 1,500 to 1
Perfect Hand Triple Win: (7, 14, 21): 60 to 1
Triple Win: 5 to 1
Win any Two Hands: 1 to 1.
Any thoughts on this game.
Where this game gets you is on the bust. You are very likely to bust on your 7 hand because you have to get two cards in your hand totalling 7 or less to make a hand there, and you are very tempted to use an ace to form a 21. Further, an Ace in your hand means that it's less likely for the dealer to make the 21 in that hand.
I can imagine that the only way to analyze this game is to simulate it.
I hope I'm not just stating the obvious when I observe that the house edge is primarily (only?) due to the fact that the player loses his bet any time the player and dealer both bust on the same hand.
The nerve of some people, to call a 21/BJ spinoff poker!
Quote: NareedPoker?
The nerve of some people, to call a 21/BJ spinoff poker!
Yeah, the game has nothing to do with poker, which is a strange way to name this game.
The game is fun! The 7 hand is really hard to make; most of the time I didn't have any combination of cards that would be under seven. I'd definitely be willing to give this game a try.
Also, is it really required that you be 21 to play the Shockwave game? I can't possibly see why...
Quote: ElectricDreamsYeah, the game has nothing to do with poker, which is a strange way to name this game.
Poker's still popular, which no doubt has helped games like TCP and PGP gain in popularity, not to mention the table versions of Hold 'em. But this one's a stretch.
Quote:The game is fun! The 7 hand is really hard to make; most of the time I didn't have any combination of cards that would be under seven. I'd definitely be willing to give this game a try.
The only ones I see are A-6, 2-5 and 3-4.
So your dream hand would be 3,4,6,8,10 (or J,K,Q) and an Ace, all the same suit. Which no doubt you'll see once in your lifetime, and the one time you don't make the sucker bet :P
The dealer is going to have 20 or better in the top hand almost every time.
Quote: dwheatleyThen the strategy is probably to make the best 14 hand, while qualifying the 7 hand if you can. Don't bother making a good 7 if you can make a better 14. The 21 is probably the hardest to win unless you have 21 (and a qualifying 7?), I'd put the junk in there.
The dealer is going to have 20 or better in the top hand almost every time.
I agree. It seems that taking an ace out of the 21 hand and swapping it with a 10 by moving it to the 7 hand so it qualify is almost always a good play. You usually lose 1/2 a bet in the top hand going from a push to lose or win to push, but make up for that by winning many more 7 hands.
Of course this is based off observational evidence not actual rigorous analysis.
http://www.ledgaming.com/LED_Home/html/index.shtml
The game is said to be patent pending, so there's probably an application out there with Lawrence DeMar or Scott Slomiany's name on it. They tend to put the math in their apps so you might be able to find it that way.
Quote: rdw4potusIt looks like the dealer always fills it's hand from 21 to 14 to 7. I wonder if that's why the suggested player strategy is to fill from 7 to 14 to 21. A qualifying 7 hand seems to be extremely strong. I agree that the best outcomes seem to come by filling 14, 7, and then 21; given that the 7 hand can qualify with an optimized 14 hand. If that condition can't be met, then 7, 14, 21 seems to be the best order.
I might be more ready to attribute the display sequence to programming rather than to the actual way the dealer sets his hand, but there was nothing I could find in that program that gave "the dealer's rules," like there are in PGP, Bac, etc.
In other words, the program decides where the cards will go, then (since it has to put something down first) just goes from 21 to 14 to 7 ... as opposed to "the dealer's rules" saying "first, these cards go to the 21, then these to the 14, then the rest."
Does that make sense?
Anyway, there's nothing but some play supporting these rules-of-thumb, so they're pretty weak, but here's what I gleaned:
* a qualifying 7 is strong to the point where anything non-Ace might win ... 4 or 5 is nearly as likely a winner as 6 and 7
* use your 21 as your throwaway whenever possible. it seems that 20 loses just as often as 15.
* any two perfect sets are very strong, but prioritize the 14 then the 21 but don't lose a qualifying 7 to do that. A qualifying 7 plus one pefect > 7-bust plus perfect 14 and 21
* two aces can make a very strong hand because of the flexibility
Hope that's helpful ... interesting game! Especially the time I got dealt a "monster" - something like AAKKKK9 - and had one push and two losses!
Quote:House rules for the dealer:
The dealer must first form the best 21 two card hand from the six cards dealt. Next the dealer must form the best 14 two card hand from the remaining four cards. The last two cards form the dealer's 7 hand.
That is the "house way", so to speak. I found that the best strategy was to make 21 or 14 if possible, then, make any seven hand (since the dealer will usually bust), then make a 20, 19 or 13 if possible, as the 14 seems really easy to get for the dealer.
It's difficult to analyze because there 20,358,320 possible 6 card hands.
Quote: boymimbo
It's difficult to analyze because there 20,358,320 possible 6 card hands.
There is only a million (10^6) meaningfully different ones though. That's not too many.
But for each of the million player's hands there is a million dealer's hands to compare against. making it the total of 1 billion comparisons.
I wrote a quick brute-force solution yesterday, but killed it after ~20 minutes of running. Waiting for a few spare minutes now to think about finding some kind of a short cut, that would speed it up.
Ahh, perhaps another example of the milliard/billion/trillion disagreement commented on in the "Good losing streak story" thread :-)Quote: SOOPOOA million hands compared to a separate million hands yields a trillion results, not a billion. ...
Quote: SOOPOOA million hands compared to a separate million hands yields a trillion results, not a billion. Trillion is a number I only understand when discussing government deficits.
Indeed. However, I just realized, that there are less, than a million hands, because different permutations of the same set of cards make the same hand. I am having a stupid moment, and can't decide what the actual number is though. Is it 10^6/6! ?
That doesn't sound right, because it's not a whole number though ...
Quote: DocAhh, perhaps another example of the milliard/billion/trillion disagreement commented on in the "Good losing streak story" thread :-)
No disagreement here. SOOPOO is right, I just misspoke.
Quote: DocAhh, perhaps another example of the milliard/billion/trillion disagreement commented on in the "Good losing streak story" thread :-)
Who was the senator who said, "A billion here, a billion there, and pretty soon, you're talking real money"?
Of course, anyone who watched Carl Sagan on the Discovery Channel would know what "billyuns and billyuns" were.
It's only in our new improved modern economy that the goverment has started to talk about blowing trillions of dollars, not billions. I suspect that "three trillion" goes down easier than "three thousand billion", or even "three million million"---to most people, a "trillion" is a complete abstraction, because there's nothing in their lives to relate that number to.
Quote: weaselmanI just realized, that there are less, than a million hands, because different permutations of the same set of cards make the same hand. I am having a stupid moment, and can't decide what the actual number is though. Is it 10^6/6! ?
That doesn't sound right, because it's not a whole number though ...
Figured it out.
F(n,k) = SUM(j:1 to k; F(n-1,j))
F(1,k)=k
F(n,1)=1
The actual number of the truly unique combinations is suprisingly low - only 5005.
Quote: weaselmanFigured it out.
F(n,k) = SUM(j:1 to k; F(n-1,j))
F(1,k)=k
F(n,1)=1
The actual number of the truly unique combinations is suprisingly low - only 5005.
This does not seem reasonable to me. Does this imply that the likelihood of getting AAAA22 (in any order) is one in 5005? I'm not believing that.
There are 52 x 51 x 50 x 49 x 48 x 47 / (6 x 5 x 4 x 3 x 2 x 1) total combinations of unique cards. That number is 20,358,320. But because we ignore the suit of the cards, the question becomes more complicated.
My brute force method gives me 4,915 unique value combinations. However, the odds of getting each combination is not 1/4,915. It is much more difficult to get an A A A A 2 2 then it is to get a 10 - 10 - 10 - 10 - 10 - 10.
The number is less than 5,005, because you can't get 5 or 6 Aces through Nines. This takes away 90 combinations.
So now, I have to figure out how many combinations make up each combination, and what the probability of each dealer hand will be.
-Tim
i haven't analyzed the game at all, just passing the message along.Quote:Actually, the house edge on the base game when played properly is about 2% using the current paytable.
The key is to use a strategy that is roughly opposite of the dealer. First make sure you don't bust your 7 hand (the dealer busts the 7 hand slightly more than 70%). Then, play the very best 14 hand you can without busting the 7 hand (this often means playing something like a 2, 3 or 4 in the 7 hand to get the 14 hand near the top. If it means not busting the 7 hand, play your ace down their notwithstanding how much you want to use it to make a 21.
The bonus bet does have a higher house edge but the base game is a 98% game.
Quote: PapaChubbyThis does not seem reasonable to me. Does this imply that the likelihood of getting AAAA22 (in any order) is one in 5005? I'm not believing that.
No. There are 5005 different combinations, but they are not equally likely. Generally, the more unique cards are in the combination, the more is the likelihood of getting one of its variations (also, 10 is a special case).
The 5005 is the number of combinations to consider for different splitting strategy. The likelihood of each needs to be calculated separately (I am still thinking about a good analytical way to do it, but haven't made much progress yet).
BTW, I did think of a non-recursive way to count the combinations meanwhile :) Picking k cards out of N with repetitions in any order is COMBIN(N+k-1,k)
