Posted by Mission146
Jul 11, 2015

## More DIY

For this edition of Introduction to Advantage Play, we're actually going to get to look at a play that I was asked about by Forum Member TeddyS, so I want to give a shot out to him for asking me about this play and giving me permission to convert his question into an Article.  I'm titling this Article, "More Do-It-Yourself," because, rather than just do the Math on this Keno Progressive and give an answer, this actually presents me an opportunity to teach people how they may analyze Progressives on unique Video Keno games.

The game in question is Lucky Bells Keno and it is a game by which a player selects anywhere from 2-10 spots with Base Pays according to the number of Spots selected vs. those hit.  In addition to the Base Pay, after the player has selected his/her Spots, the game will randomly select six other numbers upon which a bell will be placed.  If the numbers hit on the Bells AND the player has a winning combination of his/her picks, then that win will be multiplied:

The Progressive applies for any bet of \$0.50-\$2.00 when the player plays any number of spots ranging from 7-10 Spots.  The first thing I want to look at is a strong probability of the Progressive hitting, so I have narrowed that down to playing for 7/7 spots and either 8/9 or 9/9 spots.  8/8 as well as 9/10+10/10 are too remote for the Progressive to have any real value on those numbers of Picks.

With that said, let's take a look at the Base Pays for the Progressive (\$1,150 at a \$0.50 bet) as well as the Base Pays for the rest of the Picks for the Seven-Spot and Nine-Spot.  Since hitting 8/9 OR 9/9 is more likely than hitting 7/7, the return, with multipliers, is going to have to be considerably better for us to want to go after the Seven Spot game:

(All Pays indexed to \$0.50 = 1 Unit for simplicity)

Seven-Spot

Match 3: 1

Match 4: 3

Match 5: 16

Match 6: 92

Match 7: 2,300 (\$1,150 Base Progressive)

Nine-Spot:

Match 4: 4

Match 5: 6

Match 6: 40

Match 7: 372

Match 8: 2,300 (\$1150 Base Progressive)

Match 9: 2,300 (\$1150 Base Progressive)

Multipliers:

Two Bells = 2x

Three Bells = 4x

Four Bells = 8x

Five Bells = 10x

Six Bells = 16x

The Rules of the game are such that any Non-Jackpot pay may not exceed \$1,000.  That is important to note because, as a result, hitting the 7/9 with either Four, Five or Six Bells would all pay \$1,000 as it would exceed 2,000 units.

The WizardofOdds Keno Calculator:

http://wizardofodds.com/games/keno/calculator/

Cannot be used to analyze this game, however, it can give us a good start.  Mainly, we will be able to use it to determine the probabilities of each number of hits for however many spots we will end up playing:

Seven Spot:

3/7: 0.174993241448938

4/7: 0.052190966747929

5/7: 0.008638504841036

6/7: 0.000732076681444

7/7: 0.000024402556048

The first thing that we are going to have to do is determine the probabilities of hitting each particular number of Picks with x amount of Lucky Bells, however, there is an exception for either 0-1 Bells because they would be irrelevant to the pay.  After we do that, we have to subtract the probability of getting the Hits + Bell Hits from the overall probability of matching each number of Picks to determine the probability of hitting the number of picks and not improving.

The way that we are going to do this is by using an online scientific calculator that I like with the nCr (Combinatorial) function.  Other than that, it’s just basic addition, subtraction and multiplication!

http://web2.0calc.com/

We will start out with 3/7 Hits on the Seven-Spot game.  The result of having three matches is that there are only seventeen remaining numbers that could possibly be Bells, because they have already been matches.  There are six total Bells which leaves 71 numbers that are neither Bells nor matches out of 77 total unused numbers:

Three Hits with Two Bells:

nCr(6,2)*nCr(71,15)/nCr(77,17) = 0.279713622

That might look confusing, but you see that if you add the numbers in the respective spots in the first two sets of parenthesis, the result is the same as in the last set of parenthesis.  In other words, Six Bells + 71 Non-Bells = 77 Remaining Numbers AND 2 Bells Hit + 15 Hits that were not Bells = the remaining 17 numbers to be selected.

Now, to get the overall probability of having Three Hits with Two Bells we must multiply the probability of any Three Hits by the probability of getting Three Bells:

0.17499324* 0.27971362 = 0.04894799

Okay, so now that you likely have the concept down, we’re going to go ahead and quickly compute the probabilities of 3/7 Matches with 3-6 Bells:

nCr(6,3)*nCr(71,14)/nCr(77,17) = 0.09814513

nCr(6,4)*nCr(71,13)/nCr(77,17) = 0.01776765

nCr(6,5)*nCr(71,12)/nCr(77,17) = 0.00156596

nCr(6,6)*nCr(71,11)/nCr(77,17) = 0.00005219

0.17499324 * 0.09814513 = 0.01717473

0.17499324 * 0.01776765 = 0.00310922

0.17499324 * 0.00156596 = 0.00027403

0.17499324 * 0.00005219 = 0.00000913

Finally, we have to subtract the probabilities of getting the multipliers from the overall probability of getting 3/7 without any winning multipliers, to get the probability of Matching 3/7 and only getting the Base Pay.  For simplicity, we’re going to now begin rounding off after the sixth decimal place.

0.174993-0.048948-0.017175-0.003109-0.000274-0.000009 = 0.105478

Now, we can state what the returns are for hitting three out of seven with x multipliers:

3/7 (0,1 Bells) = 0.105478 * 1 = 0.105478

3/7 (2 Bells) = .048948 * 2 = 0.097896

3/7 (3 Bells) = .017175 * 4 = 0.0687

3/7 (4 Bells) = .003109 * 8 = 0.024872

3/7 (5 Bells) = .000274 * 10 = .00274

3/7 (6 Bells) = .000009 * 16 = .000144

Therefore, the overall return for all 3/7’s hit is approximately:

.105478+.097896+.0687+.024872+.00274+.000144 = 0.29983 UNITS

Thus, we can start a new chart with the Total number of units each hit is worth just to keep track:

0/7 = 0

1/7 = 0

2/7 = 0

3/7 = .29983

The way that I am doing this should be fairly clear by now, so I am going to continue without much in the way of further notes.

Probability of 4/7, overall: 0.052190966747929

nCr(6,2)*nCr(70,14)/nCr(76,16) = 0.26766299

nCr(6,3)*nCr(70,13)/nCr(76,16) = 0.08765572

nCr(6,4)*nCr(70,12)/nCr(76,16) = 0.01473523

nCr(6,5)*nCr(70,11)/nCr(76,16) = 0.00119880

nCr(6,6)*nCr(70,10)/nCr(76,16) = 0.00003663

0.05219097*0.26766299=0.01396959

0.05219097*0.08765572=0.00457484

0.05219097*0.01473523=0.00076905

0.05219097*0.00119880=0.00006257

0.05219097*0.00003663=0.00000191

With that done, we’ll round off to the sixth decimal place again and determine the probability of hitting 4/7 without a multiplier:

0.052191-.01397-.004574-.000769-.000063-.000002 = 0.032813

Now we can state what the return of 4/7 will be with or without multipliers:

4/7 (0,1 Bells): 0.032813*3 = 0.098439

4/7 (2 Bells): 0.01397 * 6 = .08382

4/7 (3 Bells): .004574 * 12 = 0.054888

4/7 (4 Bells): .000769 * 24 = 0.018456

4/7 (5 Bells): .000063 * 30 = 0.00189

4/7 (6 Bells): .000004 * 48 = 0.000192

The sum of these numbers is:

.098439+.08382+.054888+.018456+.00189+.000192= 0.257685 UNITS

Which gives us, so far:

0/7 = 0

1/7 = 0

2/7 = 0

3/7 = .29983

4/7 = .257685

Now, we shall move on and determine it for 5/7 Matches:

Probability of 5/7, Overall: 0.008638504841036

nCr(6,2)*nCr(69,13)/nCr(75,15) = 0.25427984

nCr(6,3)*nCr(69,12)/nCr(75,15) = 0.07732486

nCr(6,4)*nCr(69,11)/nCr(75,15) = 0.01199869

nCr(6,5)*nCr(69,10)/nCr(75,15) = 0.00089482

nCr(6,6)*nCr(69,9)/nCr(75,15) = 0.000024856

Now, we will determine the probability of getting the 5/7 with the accompanying Bells:

0.00863850*0.25427984= 0.00219660

0.00863850*0.07732486= 0.00066797

0.00863850*0.01199869= 0.00010365

0.00863850*0.00089482= 0.00000773

0.00863850*0.00002486= 0.00000021

With that done, we’ll round off to the sixth decimal place again and determine the probability of hitting 5/7 without a multiplier:

.008639-.0022-.000668-.000104-.000008-.0000002=0.0056588

Now we can state what the return of 5/7 will be with or without multipliers:

5/7 (0,1 Bells) : .0056588*16 = 0.0905408

5/7 (2x): .0022 * 32 = 0.0704

5/7 (4x) .000668 * 64 = 0.042752

5/7 (8x) .000104 * 128 = 0.013312

5/7 (10x) .000008 * 160 = .00128

5/7 (16x) .0000002 * 256 = .0000512

The sum of these numbers is:

.0905408+.0704+.042752+.013312+.00128+.0000512= 0.218336 UNITS

Which gives us, so far:

0/7 = 0

1/7 = 0

2/7 = 0

3/7 = .29983

4/7 = .257685

5/7 = 0.218336

Now, we shall move on and determine it for 6/7 Matches:

The overall probability of 6/7 Matches is: 0.000732076681444

nCr(6,2)*nCr(68,12)/nCr(74,14) = 0.23953898

nCr(6,3)*nCr(68,11)/nCr(74,14) = 0.06723901

nCr(6,4)*nCr(68,10)/nCr(74,14) = 0.00956417

nCr(6,5)*nCr(68,9)/nCr(74,14)= 0.000648418

nCr(6,6)*nCr(68,8)/nCr(74,14) = 0.000016210

Now, we will determine the probability of getting 6/7 with the accompanying Bells:

0.00073208*0.23953898= 0.00017536

0.00073208*0.06723901= 0.00004922

0.00073208*0.00956417= 0.00000700

0.00073208*0.00064842= 0.00000047

0.00073208*0.00001621= 0.00000001

With that done, we’ll round off to the sixth decimal place, or nearest one that actually has a non-zero, and determine the probability of hitting 6/7 without a multiplier:

0.000732-.000175-.000049-.000007-.0000005-.00000001= 0.00050049

Now we can state what the return of 6/7 will be with or without multipliers:

6/7 (0,1 Bells) : .00050049*92 = 0.04604508

6/7 (2x): .000175 * 184 = 0.0322

6/7 (4x) .000049 * 368 = 0.018032

6/7 (8x) .000007 * 736 = 0.005152

6/7 (10x) .0000005 * 920 = .00046

6/7 (16x) .00000001 * 1472 = 0.00001472

The sum of these numbers is:

.04604508+.0322+.018032+.005152+.00046+.00001472 = 0.1019038 UNITS

Which gives ont-variant: normal; font-weight: normal; text-decoration: none; vertical-align: baseline; background-color: transparent;">us, so far:

0/7 = 0

1/7 = 0

2/7 = 0

3/7 = 0.29983

4/7 = 0.257685

5/7 = 0.218336

6/7 = 0.1019038

Fortunately, since the Bells do not matter on the Seven Spot, we can simply multiply the probability by the base Progressive amount, in units, and get:

0.000024402556048*2300 = 0.0561258789104

Thus, our Return Table is:

0/7 = 0

1/7 = 0

2/7 = 0

3/7 = 0.29983

4/7 = 0.257685

5/7 = 0.218336

6/7 = 0.1019038

7/7 = 0.0561258789104

Our overall Base Return, not including Progressive increase, per unit bet is:

0.056126+0.1019038+0.218336+0.257685+0.29983 = 0.9338808

Which is a base return of 93.38808%, approximately, for a base House Edge of 6.61191%

In order to overcome this House Edge, we would need to return 0.0661191 more units on the Progressive hit, and we can easily determine what the increase to the Progressive must be to make this happen:

0.0661191/0.000024402556048 = x

x = 2709.515342161012296264

Therefore, the increase to the Progressive must be approximately 2710 units, or \$1,355 for a total value of 5,010 units, or \$2,505 for the Progressive to be positive in this scenario.  Here is the proof:

0.1019038+0.218336+0.257685+0.29983 + (5010 * 0.000024402556048) = 1.00001160580048

There will be a follow-up to this Article in which we will analyze the Nine-Spot for this game and determine whether or not the Progressive is playable at a lower point than with the Seven-Spot.  Furthermore, we will look at a theoretical game of Cleopatra Keno so that we can learn how to analyze a Keno Bonus game in which the special feature involves hitting numbers that you have actually chosen.