## Poll

4 votes (44.44%) | |||

3 votes (33.33%) | |||

2 votes (22.22%) | |||

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5 votes (55.55%) | |||

4 votes (44.44%) | |||

2 votes (22.22%) | |||

2 votes (22.22%) | |||

No votes (0%) | |||

2 votes (22.22%) |

**9 members have voted**

- Game is an optional feature on 3-play and 5-play video poker.
- The player must play 10 credits per line to invoke the feature. Wins will be based on 5 credits only.
- If the player is dealt 2 or 3 aces, then he will win additional hands above the 3 or 5 he paid for.
- The number of additional hands is not disclosed in the rules, except to state a maximum. For example, this maximum is 8 hands in 5-play Bonus Poker and 6 in Double Double Bonus.
- Regardless of how many hands the player has, they will be in the form of a vertical stack.
- Replacements cards are dealt to one hand at a time, starting at the bottom of the stack. If the player is dealt any additional aces on the draw, those aces will be sticky for all hands above.
- To invoke the feature, the player must hold the 2 or 3 dealt aces only. If the holds other cards, he won't get the bonus hands.

Example

In this hand, I was dealt two aces, which I held. Note you must click on hold, it's not automatic.

In the five hands I paid for, none of them improved.

I was awarded 7 out of a maximum of 8 bonus hands. The second bonus hand improved to a three of a kind. Note how the aces of hearts was replicated to all hands above it. Had one of those improved to four aces, that fourth ace would have replicated too.

Without knowing the probability distribution of extra hands, it will be impossible to analyze. However, I am expecting the returns from Action Gaming.

The question for the poll is would you play Stack 'em Up Poker?

https://wizardofvegas.com/forum/gambling/video-poker/34065-new-game-at-videopoker-com-stack-em-high/#post750498

The rules are kind of ambiguous but when playing DW versions it's the "2" and not aces that get stacked, just in case anyone tries to play it.

Quote:RealizeGamingI'm very curious about this game. Would this game take advantage of a different cabinet? We see slot games taking advantage of this all the time on the casino floors now, but I'm still waiting for video poker to catch up to it.

As I recall from G2E, it uses a second screen above the main one for the bonus hands. In other words, it would need a special cabinet. However, I may be wrong.

Contrary to what I wrote before, the number of extra plays is always the same for any given game, pay table, and number of original plays. This is good news is it makes the math quantifiable. I'm on it!

Please see my work in progress page on Stack 'em High poker for the number of extra plays.

Hand | 2 aces | 3 aces | 4 aces |
---|---|---|---|

1 | 0.875116 | 0.122109 | 0.002775 |

2 | 0.765827 | 0.223660 | 0.010513 |

3 | 0.670188 | 0.307450 | 0.022363 |

4 | 0.586492 | 0.375919 | 0.037590 |

5 | 0.513248 | 0.431190 | 0.055562 |

6 | 0.449151 | 0.475115 | 0.075734 |

7 | 0.393059 | 0.509303 | 0.097637 |

8 | 0.343972 | 0.535156 | 0.120872 |

9 | 0.301016 | 0.553890 | 0.145094 |

10 | 0.263423 | 0.566565 | 0.170012 |

11 | 0.230526 | 0.574098 | 0.195376 |

12 | 0.201737 | 0.577287 | 0.220976 |

Total | 5.593755 | 5.251742 | 1.154503 |

Can anyone confirm or deny?

I know there are 3 classes of two aces and 2 glasses of 3 aces on the deal, so this is just an example.

Quote:WizardCan anyone confirm or deny?

I get the same values. But just so we are all on the same page, here are where my numbers come from in case someone can spot a flaw in my logic.

First line:

The probability of no new aces is (45*44*43)/(47*46*45) = 0.8751156

The probability of getting an ace in position 1 is 2 * 45 * 44/(47*46*45). Multiply by 3 (the 3 possible ending positions) to get the overall probability of 1 ace = 0.12210916.

The probability of getting 2 aces is 1 minus the sum of the no and one ace = 1 - 0.8751156 - 0.12210916 = 0.00277524

Second line:

If we have just 2 aces, the three probabilities are the same as above. Call these A.

If we have 3 aces, the probability of ending with 2 aces is zero. The probability of ending with 3 is 45*44/(46*45) = 0.956521739. The probability of ending with 4 is just the remainder = 0.04347826. Call these B.

If we have 4 aces, the probability of ending with 2 or 3 is zero and the probability of 4 is 1. Call these C.

Now we have to take into account the likelihood of A, B and C. That is, we apply the starting probabilities. Thus we get 0.8751156A + 0.12210916B + 0.00277524C which gives

0.765827 0.22366 0.010513

Row 3 would be similar but starting with these values for the likelihood of the current number of aces. And so on.

Quote:GaryJKoehlerI get the same values.

I didn't see any flaws and think we did it the same way. The fact that we agree says a lot. I have a feeling any error I made is probably in setting up the various situations the player can get two aces. What the player throws away affects the subsequent hands.

Let me go a bit further.

If the player gets AAxyz (two aces and three singletons), here is the value of the subsequent 12-hand bonus.

Hand | Pays | Expected | Win |
---|---|---|---|

High pair | 1 | 4.556604125 | 4.556604125 |

Two pair | 2 | 1.021776788 | 2.043553576 |

Three of a kind | 3 | 4.917540672 | 14.75262202 |

Full house | 8 | 0.34957575 | 2.796605997 |

Four of a kind | 80 | 1.154502665 | 92.36021322 |

Total | 12 | 116.5095989 |

EV of bonus with AAxxy = 116.5562273

EV of bonus with AAxxx = 116.6479072

Blended average of AA bonuses = 116.5185234

In conventional 8-5 Bonus, I show, on average, all hands with two aces on the deal have an EV of 1.856810.

I know I'm skipping a lot of steps, but here is the final table showing the value of the feature:

Deal | Probability | Feature win | VP win | Add win | Value |
---|---|---|---|---|---|

two aces | 0.039930 | 116.518523 | 9.284049 | 107.234474 | 4.281853 |

Three aces | 0.001736 | 291.276848 | 33.361938 | 257.914910 | 0.447761 |

Total | 0.041666 | 4.729614 |

The bottom right cell shows the feature, in 8-5 Bonus, is worth 4.73 credits. However, the player must pay 5 credits for it. That gives the additional bet a return of 94.6%. I doubt it is that low. It should be greater than the 99.17% return of 8-5 Bonus.

Quote:Wizard

If the player gets AAxyz (two aces and three singletons), here is the value of the subsequent 12-hand bonus.

Hand Pays Expected Win High pair 1 4.556604125 4.556604125 Two pair 2 1.021776788 2.043553576 Three of a kind 3 4.917540672 14.75262202 Full house 8 0.34957575 2.796605997 Four of a kind 80 1.154502665 92.36021322 Total 12 116.5095989

EV of bonus with AAxxy = 116.5562273

EV of bonus with AAxxx = 116.6479072

Blended average of AA bonuses = 116.5185234

I can confirm all the numbers in the section above. Bottom line, I also calculate that the blended average of AA bonuses = 116.5185234.

Quote:WizardThe bottom right cell shows the feature, in 8-5 Bonus, is worth 4.73 credits. However, the player must pay 5 credits for it. That gives the additional bet a return of 94.6%. I doubt it is that low. It should be greater than the 99.17% return of 8-5 Bonus.

I'm unclear as to why you think that this feature should improve the return to player. Think of the additional 5 credits that you pay for this feature as akin to a side-bet in a carnival game. A return of 94.6% then seems to be reasonable.

Quote:gordonm888I'm unclear as to why you think that this feature should improve the return to player. Think of the additional 5 credits that you pay for this feature as akin to a side-bet in a carnival game. A return of 94.6% then seems to be reasonable.

Thank you for the confirmation.

The reason I am skeptical of the bottom line is Action Gaming video poker variants almost without fail make the return higher if the player bets more to pay for the feature.