Posted by Mission146
Jul 12, 2015

More DIY (Concluded)

In the two previous Introduction to Advantage Play Articles, we analyzed a game called Lucky Bells Keno to determine when the Progressive is at a positive point.  That game is unique in that the multipliers that can be awarded in the game for winning combinations involve Bells that are intentionally not placed on numbers the player has selected.

For this article, we are going to analyze a Progressive game in which there are Bonus Games that are triggered when the final number hits on a winning combination, but that final number is, in fact, one of the numbers the player has selected.  For those of you who found the Math on the last two Articles difficult, you can breathe a sigh of relief, as the Math is easier on this one.

While it is true that I have not seen a Cleopatra Keno game with a Progressive before, we are going to use a theoretical Cleopatra Keno Progressive in this article because it will provide a simple way to understand how to calculate something of this nature.

In Cleopatra Keno, the player selects 2-10 numbers, and if the player gets a winning combination with the last ball hitting, the player is awarded 12 Free Games in which all wins are doubled.

Because this is a theoretical Progressive for Cleopatra Keno, we are going to go ahead and base it on a Six-Spot Keno game.  Furthermore, we are also going to say that the Progressive is awarded in the event that the player hits 6/6, however, the Progressive amount is not doubled on the Free Games. 

The first thing that we need to determine is the likelihood of any win, however, the Wizard of Odds Keno Calculator does that for us:


3/6: 0.129819547541067

4/6: 0.028537917778424

5/6: 0.003095638538677

6/6: 0.000128984939112

We will also use a base paytable found on the Wizard of Odds Cleopatra Keno page, the NYNY paytable:


3/6: 2

4/6: 5

5/6: 34

6/6: 375

The first thing that we will need to determine is the pays of the Bonus Games, fortunately, that's really easy because we are going to use the Base Paytable, with the associated Pays/Probabilities multiplied by 12, because that's how many Free Games you get.  Remember, all wins except 6/6 will be doubled on this theoretical Cleopatra Keno game:

(0.129819547541067*4)+(0.028537917778424*10)+(0.003095638538677*68)+(0.000128984939112*375) * 12 = 12.762361688946528

That means that the average Free Games will be worth 1.59559 units in this game. 

Even though the Bonus probabilities are given on The Wizard of Odds page, the point of this exercise is to learn how to figure them out ourselves.  The probability of hitting the last number will always be 6/80, so we simply need to know the probabilities of hitting 2-5 more numbers:

nCr(5,2)*nCr(74,17)/nCr(79,19) * 6/80 = 0.01947293213116

nCr(5,3)*nCr(74,16)/nCr(79,19) * 6/80 = 0.0057075835556848

nCr(5,4)*nCr(74,15)/nCr(79,19) * 6/80 = 0.0007739096346691

nCr(5,5)*nCr(74,14)/nCr(79,19) * 6/80 = 0.0000386954817335

We can now determine the probability of hitting the selected wins, but not hitting the Bonus Games:

3/6: 0.129819547541067-0.01947293213116= 0.110346615409907

4/6: 0.028537917778424- 0.0057075835556848= 0.0228303342227392

5/6:  0.003095638538677-0.0007739096346691= 0.0023217289040079

6/6: 0.000128984939112- 0.0000386954817335= 0.0000902894573785

The overall return of the game, not including Bonus Games:

(0.129819547541067*2)+(0.028537917778424*5)+(0.003095638538677*34)+(0.000128984939112*375)= 0.555949746456272

We now need to know the overall probability of hitting Bonus Games:



We can now multiply the Unit return of the Bonus Games by the probability of the Bonus Games to get the overall return of the Bonus Games:



(Please keep in mind that the Bonus Return will differ from Wizard of Odds because we are not multiplying the Six Spot win by two)

The total return, is:



Okay, so now we need to isolate the Six Spot win to determine how much the Progressive increase would need to be to have a positive play.  The first step in doing this is to determine how likely it is to hit the Six-Spot during Free Games, but not Free Games that resulted from us already hitting 7/7 because we would have already won the Progressive:



Now, if the Six Spot has a probability of 0.000128984939112, then the probability of it not hitting is one minus that and it would need to fail to hit twelve consecutive times in Bonus Games:

(1-0.000128984939112^12= 0.9984532783082449698138485002375831205435568156612024096

Therefore, the probability of it hitting during Bonus Games is:

1-0.9984532783082449698138485002375831205435568156612024096= 0.0015467216917550301861514997624168794564431843387975904

When multiplied by the probability of actually getting Bonus Games, other than a 6/6 Bonus Games:


*0.0259544253215139= 0.0000401442726418215726699627389884613960500197345534249754032546

Then, we combine that result with the probability of hitting it during the Base Game:



We need to make up roughly 0.1123166444 units in EV, so:

0.1123166444/0.00016912921175382157266996 = x

x = 664.08778965684584202287132555804

That means the result would need to pay about 664 more units, or 1,039 units total, to be break even.

We can prove that by taking the total Base Return:


And adding to that the overall probability of hitting a Seven Spot multiplied by 664:

0.8876833555717956203932673550272+(664*0.00016912921175382157266996) = 0.9999851521763331446461207950272

Hopefully, after these last three Articles, anyone will be able to look at a Video Keno Progressive, the base paytables and Rules of any Bonus Games and be able to use that information to determine at what point the game is positive.


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