## Poll

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**15 members have voted**

Knockout 52 is a unique card game that will premier at the Strat casino in Las Vegas on May 17, 2024. There is no other game it can be easily compared to. Briefly, there is a layout on the table with the 13 ranks in a deck. The dealer will deal one card at a time to each position on the layout. Play stops when a card from the deck matches the rank on the next spot on the layout. If the player goes 13, 26, or 39 cards without a match, then he starts over at the first position. The object is to correctly predict how long it will take for a match to be formed or no match at all.

This game won the best new table game award at the 2024 Global Table Games and Game Protection Conference. I made a video of the game at the show, which should be up before the big launch.

For all the rules and analysis, please see my new page at WoO on Knockout 52. I welcome all comments and questions.

The question for the poll is would you play Knockout 52?

Quote:heatmapwhen are we going to see YOUR game? and is this a very low house edge or am i just not good at reading your tables?

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Which is my game? Lowest house edge of 3.54% doesn't seem so low to me.

https://youtu.be/BrrQjMYewdg?si=9xJ5F5-XONR4RJlP

Similar concept and from the same inventor, but Knockout 52 is far simpler.

Easier to explain, easier to play and easier to deal.

When I first saw Match 52 I thought there was no way that would see the light of day in a casino.

When I saw Knockout 52 in March at the gaming conference, I knew it was a winner!

——

Hey Wiz! Where’s the poll choice reminder of the next eclipse? 🤪

If you can’t wait until then, and you’re in the area, it’s launching on 5/3 at Firekeepers in Battle Creek MI.Quote:WizardOn May 17, Knockout 52 will begin a field trial at the Strat.

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Quote:WizardQuote:heatmapwhen are we going to see YOUR game? and is this a very low house edge or am i just not good at reading your tables?

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Which is my game? Lowest house edge of 3.54% doesn't seem so low to me.

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a game you invent

I ran the simulation and it shows that your result is correct. I must be getting worse at math...

Quote:CodingforestWhy is the probability for a round 1 knockout not just 1-(12/13)^13

I ran the simulation and it shows that your result is correct. I must be getting worse at math...

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Let's look at the case of no match is the first two cards.

The exact probability is 1-(11/13)*(47/51) - (1/13)*(48/51) = 0.147813

Your approximation would give 1-(12/13)^2 = 0.147929

My simple explanation is if the first card was not an ace, then the remaining cards are ace rich for the second card. That makes it less likely a 2 is the second card.

Much for the same reason the first two cards dealt from a deck are more likely to be opposite in color than the same.

I haven't pursued this game because I figured no casino would want to have a game that takes so long to conclude. They always want quick deal games.

How long on average does this game take for a single deal and do you see this as being prohibitive and if no, why not?

Quote:darkozI thought of a game which in deals was similar in that there was a possibility the entire deck could be dealt before a game concluded.

I haven't pursued this game because I figured no casino would want to have a game that takes so long to conclude. They always want quick deal games.

How long on average does this game take for a single deal and do you see this as being prohibitive and if no, why not?

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I think the average round deals 13.5 cards, it plays quick since there are no decisions beyond the initial bet.

Quote:CodingforestWhy is the probability for a round 1 knockout not just 1-(12/13)^13

I ran the simulation and it shows that your result is correct. I must be getting worse at math...

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It would be 1 - (12/13)^13 if they shuffled the cards back into the deck immediately after they are drawn.

Let's work on the probability that it is not a first-round knockout. 13 things have to happen:

1. The first card is not an Ace; the probability is 12/13, or 48/52.

2. The second card is not a 2; there are 51 cards left in the deck, of which 47 are not 2s, so the probability is 47/51.

3. The third card is not a 3; there are 50 cards left in the deck, of which 46 are not 3s, so the probability is 46/50.

4. The fourth card is not a 4; there are 49 cards left in the deck, of which 45 are not 4s, so the probability is 45/49

...

13. The 13th card is not a King; there are 40 cards left in the deck, of which 36 are not Kings, so the probability is 36/40.

The probability of not getting a first-round knockout is (48 x 47 x 46 x ... x 36) / (52 x 51 x 50 x ... x 40) = (39 x 38 x 37 x 36) / (52 x 51 x 50 x 49) = 6327 / 20,825.

The probability of a first-round knockout = 1 - 6327 / 20,825 = 14,498 / 20,825, or about 0.69618.

Note that 1 - (12/13)^13 = about 0.64674.

Quote:DJTeddyBearIf you can’t wait until then, and you’re in the area, it’s launching on 5/3 at Firekeepers in Battle Creek MI.Quote:WizardOn May 17, Knockout 52 will begin a field trial at the Strat.

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I just added a table for eight decks to my page at WoO.

Quote:zbrownson

I think the average round deals 13.5 cards, it plays quick since there are no decisions beyond the initial bet.

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I show the average cards dealt is 12.906 with one deck and 12.825 with eight decks.

Something about they needed some mechanical features otherwise they were being rejected.

Is that no longer the case?

]]Quote:darkozSo what happened with the issue years ago that patents were being denied for new strictly card based games?

Something about they needed some mechanical features otherwise they were being rejected.

Is that no longer the case?

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You can still create a game without a patent. Some patents have gone through for card games in the last few years, however they are usually written to be played in electronic format then add an embodiment where they could be played on a physical table. I can't speak to what percentage might get approved this way, but I know some have, at the end of the day it still doesn't answer the question of if it is enforceable, but people will take the risk.

I assume that is calculated off the top of the shoe. If you start a new game when the shoe contains nothing but aces, the probability for matching in round #1 goes to 100%. I am guessing any shoe that has depleted some ranks unevenly favors the player.Quote:WizardQuote:zbrownson

I think the average round deals 13.5 cards, it plays quick since there are no decisions beyond the initial bet.

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I show the average cards dealt is 12.906 with one deck and 12.825 with eight decks.

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She said, 60. Frankly, when she mentioned it at the show, I thought there was no way that could be true. But at ~13 cards to get resolution, maybe 60 hands per hour is possible.

Oh. Who is Angela? She’s in this photo. Figure it out. 🤪

Quote:MentalI assume that is calculated off the top of the shoe. If you start a new game when the shoe contains nothing but aces, the probability for matching in round #1 goes to 100%. I am guessing any shoe that has depleted some ranks unevenly favors the player.Quote:WizardQuote:zbrownson

I think the average round deals 13.5 cards, it plays quick since there are no decisions beyond the initial bet.

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I show the average cards dealt is 12.906 with one deck and 12.825 with eight decks.

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I wrote a sim for the game. I deal consecutive games through the shoe with no burn card between games. The first table is a sanity check. There should be almost no difference between dealing Game #0 from the top of the shoe or dealing Game #30. I don't think the shoe composition is biased much depending on whether it is a new shoe or whether many games have been dealt from it.

Game | Rnd 1 | Rnd 2 | Rnd 3 | Rnd 4 | No Match | Sum Prob |
---|---|---|---|---|---|---|

0 | 0.64623028 | 0.22848800 | 0.08097850 | 0.02864830 | 0.01565490 | 1.00000000 |

10 | 0.64639378 | 0.22841150 | 0.08090430 | 0.02865410 | 0.01563630 | 1.00000000 |

20 | 0.64634281 | 0.22848921 | 0.08085300 | 0.02863160 | 0.01568340 | 1.00000000 |

Then I define a concept of a ragged shoe. I count the ranks remaining in a shoe. The shoe is ragged if the max rank count is at least twice the min rank count. For example, assume there are 130 cards remaining in the shoe. There might be 15 sixes and only 7 kings. This qualifies as a ragged shoe. (Obviously, the new shoe is never ragged.) The biggest difference that I see is that 'No Match' is more probable for a ragged shoe. This isn't what I expected.

Game | Rnd 1 | Rnd 2 | Rnd 3 | Rnd 4 | No Match | Sum Prob |
---|---|---|---|---|---|---|

10 | 0.61687106 | 0.21595275 | 0.10837787 | 0.03768794 | 0.02111039 | 1.00000000 |

20 | 0.63427502 | 0.22440833 | 0.08931462 | 0.03201239 | 0.01998965 | 1.00000000 |

30 | 0.63892955 | 0.22897851 | 0.07681894 | 0.03143449 | 0.02383850 | 1.00000000 |

I push this concept further and use (max >= (3*min)) as the cutoff. Now I am seeing bigger differences. Maybe there are not enough super ragged shoes after only 10 games and my sample size is too small. It still seems that ragged shoes favor 'No Match'.

Game | Rnd 1 | Rnd 2 | Rnd 3 | Rnd 4 | No Match | Sum Prob |
---|---|---|---|---|---|---|

10 | 0.60032892 | 0.20814575 | 0.12181880 | 0.04429191 | 0.02541463 | 1.00000000 |

20 | 0.63117778 | 0.22217239 | 0.08940069 | 0.03357313 | 0.02367602 | 1.00000000 |

30 | 0.63924652 | 0.22851303 | 0.07460260 | 0.03235944 | 0.02527841 | 1.00000000 |

I repeated the calculations for a smooth shoe. (max <= (1.5*min))

Game | Rnd 1 | Rnd 2 | Rnd 3 | Rnd 4 | No Match | Sum Prob |
---|---|---|---|---|---|---|

0 | 0.64637232 | 0.22856776 | 0.08085106 | 0.02856392 | 0.01564496 | 1.00000000 |

10 | 0.63850844 | 0.22876911 | 0.08577770 | 0.03034509 | 0.01659963 | 1.00000000 |

20 | 0.63874590 | 0.23098660 | 0.08429455 | 0.02970863 | 0.01626430 | 1.00000000 |

30 | 0.63557017 | 0.23737301 | 0.08079363 | 0.02995183 | 0.01631135 | 1.00000000 |

Note that I am using number of games dealt as a proxy for penetration. Maybe I should be using number of decks dealt instead.

I did this all in less than 2 hours, so there could be coding errors. Still, I am only changing the definition of ragged/smooth and the probabilities seem to be affected.

Question? Comments? Can anyone corroborate my results?

EDIT: I am disavowing much of what I wrote above. Because I was using number of games as a proxy for penetration, I often ran out of cards in the simulations above. There are usually enough cards to do thirty games from one 8-deck shoe. But sometimes, all 416 cards will be used up during game #30. This corrupted my results. I switched to using the number of decks to measure penetration in my new version, so I will never run out of cards in the shoe. I will write a new post later.

The last row is included for a reference point. It is calculated for zero penetration and using every shoe.

Pen | Rnd 1 | Rnd 2 | Rnd 3 | Rnd 4 | No Match | Ragged% |
---|---|---|---|---|---|---|

2.5 | 0.64109 | 0.22847 | 0.08450 | 0.02754 | 0.01840 | 0.008% |

3.5 | 0.64597 | 0.22939 | 0.08065 | 0.02834 | 0.01565 | 0.422% |

4.5 | 0.64628 | 0.22858 | 0.08095 | 0.02867 | 0.01550 | 6.361% |

5.5 | 0.64667 | 0.22835 | 0.08083 | 0.02852 | 0.01559 | 36.609% |

6.5 | 0.64646 | 0.22845 | 0.08084 | 0.02859 | 0.01564 | 75.545% |

0.0 | 0.64637 | 0.22852 | 0.08089 | 0.02855 | 0.01566 | 0.000% |

Quote:MentalHere are my results from my updated study. The first column is penetration measured in numbers of decks. The last column is the percent of shoes that are very ragged at the start of the first game after the penetration mark is achieved. Very ragged means (max >= (min*3) where max is the most frequent rank left in the shoe and min is the least frequent rank. The first two rows have a very small percentage of ragged shoes, so the results are not very precise (and probably not meaningful).

The last row is included for a reference point. It is calculated for zero penetration and using every shoe.You need a lot of penetration to get to high percentages of ragged shoes. Playing from a raggedy unbalanced shoe does not seem to change the probabilities for any of the bets to any significant degree. This means the game is probably not countable. The penetration can never exceed 7 of 8 decks because you need 52 cards to play a game.

Pen Rnd 1 Rnd 2 Rnd 3 Rnd 4 No Match Ragged% 2.5 0.64109 0.22847 0.08450 0.02754 0.01840 0.008% 3.5 0.64597 0.22939 0.08065 0.02834 0.01565 0.422% 4.5 0.64628 0.22858 0.08095 0.02867 0.01550 6.361% 5.5 0.64667 0.22835 0.08083 0.02852 0.01559 36.609% 6.5 0.64646 0.22845 0.08084 0.02859 0.01564 75.545% 0.0 0.64637 0.22852 0.08089 0.02855 0.01566 0.000%

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It seems like a naturally balancing game, because you have to play through all 13 ranks each round it seems to eliminate that key card effect.

Yes, the game inventors were probably aware of this naturally balancing effect. The 8-deck shoe cannot become seriously unbalanced by random depletion of ranks. However, I can force the shoe to be extremely unbalanced, and then all the probabilities go out of whack.Quote:zbrownsonQuote:MentalYou need a lot of penetration to get to high percentages of ragged shoes. Playing from a raggedy unbalanced shoe does not seem to change the probabilities for any of the bets to any significant degree. This means the game is probably not countable. The penetration can never exceed 7 of 8 decks because you need 52 cards to play a game.

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It seems like a naturally balancing game, because you have to play through all 13 ranks each round it seems to eliminate that key card effect.

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Imagine I have a stack of cards with 32 aces on top, 32 deuces just below, 32 threes below that, ..., and 32 kings on the bottom. Now, I pick a number, Num, of cards off the top of the stack and shuffle just those cards and put them in the shoe. For example, if Num=128, then the shoe only contains ranks ace through four.

Num | Rnd 1 | Rnd 2 | Rnd 3 | Rnd 4 | No Match |
---|---|---|---|---|---|

32 | 1.00000 | 0.00000 | 0.00000 | 0.00000 | 0.00000 |

64 | 0.74605 | 0.18942 | 0.04813 | 0.01220 | 0.00420 |

96 | 0.70137 | 0.20939 | 0.06263 | 0.01866 | 0.00795 |

128 | 0.68205 | 0.21679 | 0.06893 | 0.02197 | 0.01026 |

160 | 0.67102 | 0.22078 | 0.07260 | 0.02389 | 0.01171 |

192 | 0.66380 | 0.22320 | 0.07502 | 0.02521 | 0.01277 |

224 | 0.65914 | 0.22476 | 0.07661 | 0.02603 | 0.01346 |

256 | 0.65527 | 0.22585 | 0.07795 | 0.02682 | 0.01412 |

288 | 0.65311 | 0.22651 | 0.07852 | 0.02730 | 0.01455 |

320 | 0.65072 | 0.22718 | 0.07943 | 0.02778 | 0.01489 |

352 | 0.64910 | 0.22770 | 0.08005 | 0.02799 | 0.01517 |

384 | 0.64752 | 0.22820 | 0.08051 | 0.02835 | 0.01543 |

416 | 0.64622 | 0.22858 | 0.08096 | 0.02856 | 0.01568 |

For Num=32, the game always ends after one card is dealt, so the game always ends in Round 1. If Num=64, then the game has an approximately 75% chance of ending in round 1. There is roughly a 25% chance each that the two cards off the top of the shoe are AA, A2, 22, or 2A. The game only goes to round 2 in the latter case.

The game is naturally balancing, but extreme random fluctuations will change the probabilities slighty. These extreme situations are so unlikely that they would have no significant effect on the game. This is true even with penetration of 7 out of 8 decks.

Quote:GenoDRPhI was playing around with the online game on the developer's website. A couple fo times I won with an ace dealt on the first round. I then thought that maybe there is a possibility of some side bets on betting which position wins the hand. For example, if I bet on round 1 to win, with an ace round 1 wins with a dealt ace, then I would win the round bet as well as the side bet on which card. Maybe even a range of winners, like one bet for cards A-7, and another bet of 8-K. I leave the mathematics to those smarter on such topics than I am. With a table layout of 6 spaces, the side bet layout could get unwieldy, I suppose.

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So if you lookup a game called "Match 52" it was this game with a million side bets (for every single position), had to be played on a craps or roulette table and was a mess. So this was the much better version of it. I could still see some form of a progressive added based on a single position, or streak of cards, or something.

I did only analyze round 1 betting to hit the 3% house edge. After some thinking... there may be something to do with no match.

The best best is on Round 1, so I bet the minimum $10 on that every time. Over the hour, I lost $25. However, they gave me some swag, mainly a t-shirt and baseball hat.

I got the impression from the dealer that the game was not getting much play.

In other news, the Strat now charges $20 for parking regardless of the time. Signage seemed to indicate parking was free for hotel guests. They seemed to not have the procedure down yet and I snuck out without paying. $20 to park at the Strat? I tend to think I will be returning for a while.

Quote:WizardIn other news, the Strat now charges $20 for parking regardless of the time. Signage seemed to indicate parking was free for hotel guests. They seemed to not have the procedure down yet and I snuck out without paying. $20 to park at the Strat? I tend to think I will NOT be returning for a while.

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So on losing rounds, did they continue to deal out cards?Quote:WizardThe entire hour I was the only one at the table.

The best best is on Round 1, so I bet the minimum $10 on that every time.

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