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February 3rd, 2022 at 7:15:10 AM
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Prairie Thunder Keno is a keno variant I noticed at DraftKings. You may play the demo for free here.

I explain the rules in my new Prairie Thunder Keno page at Wizard of Odds. The help files claim a return of 93.45%. However, you can see I get 92.33% to 92.97%. I'd sure appreciate a second opinion from any other the math geniuses of the forum.

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan

February 3rd, 2022 at 7:37:41 AM
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Doing a combinatorial analysis by hand and stretching the Free Games to a point beyond all reason would take forever for a game that likely can't be played at an advantage in the first place.

I think it suffices for me to just conclude that the help file stated RTP is wrong on the grounds that I would be shocked if they could get the three-spot to the ten-spot to the same percentage return to the 1/100th of one percent.

The Help File also states:

Anyway, the last time I did one of these sorts of things, it was for the purpose of a Keno Game that had a progressive element. I actually wrote about that analysis on LCB, though I was asked not to specifically name the game by the original person who asked me for help. I can link to that if anyone wants.

That being said, it took me the better part of a full day to complete that analysis because base games and free games with multiple elements take forever to do the only way that I know how to do them. There are a great many conditional probabilities involved, as you know, because each number that hits one of the relevant Free Games or multiplier symbols cannot hit one of the player's selected numbers.

With this, I would have to do an analysis based on every possible combination of multipliers that could be hit, so I imagine that this analysis would take even longer than nearly a full day!

It keeps redirecting me to the DraftKings PA site, so I can't see the other Keno games because it seems PA hasn't any. If you can find other ones, does it say the same thing for the RTP on those ones? It could be they are just taking the mean average of all of their Keno Games and putting it on the help file.

I think it suffices for me to just conclude that the help file stated RTP is wrong on the grounds that I would be shocked if they could get the three-spot to the ten-spot to the same percentage return to the 1/100th of one percent.

The Help File also states:

Quote:When multiplier spots are hit (during Free Games) with a win, there is a random weighted drawing for each hit symbol's multiplier value (either 2x, 3x or 4x). The weightings are determined by the number of the player's picked spots.

Anyway, the last time I did one of these sorts of things, it was for the purpose of a Keno Game that had a progressive element. I actually wrote about that analysis on LCB, though I was asked not to specifically name the game by the original person who asked me for help. I can link to that if anyone wants.

That being said, it took me the better part of a full day to complete that analysis because base games and free games with multiple elements take forever to do the only way that I know how to do them. There are a great many conditional probabilities involved, as you know, because each number that hits one of the relevant Free Games or multiplier symbols cannot hit one of the player's selected numbers.

With this, I would have to do an analysis based on every possible combination of multipliers that could be hit, so I imagine that this analysis would take even longer than nearly a full day!

It keeps redirecting me to the DraftKings PA site, so I can't see the other Keno games because it seems PA hasn't any. If you can find other ones, does it say the same thing for the RTP on those ones? It could be they are just taking the mean average of all of their Keno Games and putting it on the help file.

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219

February 3rd, 2022 at 8:10:00 AM
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Quote:Mission146Doing a combinatorial analysis by hand and stretching the Free Games to a point beyond all reason would take forever

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That element of it was actually pretty easy to analyze.

Here are the number of combinations of catching 0 to 5 horseshoes.

Horseshoes | Combinations | Probability |
---|---|---|

0 | 5,461,512 | 0.227184 |

1 | 9,752,700 | 0.405686 |

2 | 6,501,800 | 0.270457 |

3 | 2,017,800 | 0.083935 |

4 | 290,700 | 0.012092 |

5 | 15,504 | 0.000645 |

Total | 24,040,016 | 1.000000 |

The general formula for the number of combinations of catching h horseshoes is combin(20,h)*combin(60,5-h), where combin(x,y)=x!/(y!*(x-y)!).

The ratio of total free games initiated by one free spin, including that free spin itself, is 1/(1-4*p(3)-10*p(4)-15*p(5)), where p(x) = probability of catching h horseshoes. I can prove this upon request, but it's a common formula seen in infinite sum problems.

In this case, that ratio is 1.873843347.

So, for catching 4 free horseshoes in an initial spin, the player will play out, on average 10*1.873843347 = 18.73843347 free spins.

Including 5-horseshoe bonus, I get an average number of free spins per bonus of 19.21282419 and a bonus probability of 0.012737263.

“Extraordinary claims require extraordinary evidence.” -- Carl Sagan

February 3rd, 2022 at 8:16:32 AM
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Well...that's certainly a quicker way of doing that than the way I did it...thank you!

I basically analyzed the Expected Free Games Added on a per draw basis, which gives you the Expected Free Games Added for the initial set, which I added to the Free Games and did the Expected Free Games Added for the Expected Free Games already added from the original set, etc. etc...

Basically, in my other analysis of something of this nature, I think I stopped doing my repeat method of solving for that when I got to the point that the Expected Free Games Added (from the previous) was .0001 Free Games, or something. I figured that got me close enough.

I basically analyzed the Expected Free Games Added on a per draw basis, which gives you the Expected Free Games Added for the initial set, which I added to the Free Games and did the Expected Free Games Added for the Expected Free Games already added from the original set, etc. etc...

Basically, in my other analysis of something of this nature, I think I stopped doing my repeat method of solving for that when I got to the point that the Expected Free Games Added (from the previous) was .0001 Free Games, or something. I figured that got me close enough.

https://wizardofvegas.com/forum/off-topic/gripes/11182-pet-peeves/120/#post815219