Introduction to Advantage Play 15

More DIY (Continued)

In our previous Introduction to Advantage Play Article, we took a look at a game called Lucky Bells Keno, which is available on many Spielo brand machines in various land casinos and parlors.  Lucky Bells is a Video Keno game in which players receive win multipliers if they hit a certain number of bells on a winning combination, this is also a Progressive game, in many cases, if the player bets $0.50.

We analyzed the Seven-Spot version of this game in our last Article, so now it is time to analyze the Nine-Spot version of this game to determine what the break-even point of the Progressive is.

In the previous Article, we noted that the combination of 8/9 numbers and 9/9 numbers is significantly more likely to occur than 7/7 numbers.  However, that also means that the Progressive (at its base) is going to contribute more to the overall return of the game, thus, we must fully analyze the Nine-Spot to determine if it is more advantageous than the seven spot.

First, we must look at the Pays for the Nine-Spot and the multipliers:

Nine-Spot:

Match 4: 2

Match 5: 3

Match 6: 20

Match 7: 186

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

Just like we did previously, we are going to look at the overall probabilities for hitting the desired results and use those to determine the probabilities of hitting them both with and without multipliers to determine the overall return of the Base Game.

4/9: 0.114105182095472

5/9: 0.032601480598706

6/9: 0.005719557999773

7/9: 0.000591678413770

8/9: 0.000032592454996

9/9: 0.000000724276778

Once again, we will be using the same Scientific Calculator as before:

http://web2.0calc.com/

Probability of 4/9, overall: 0.114105182095472

nCr(6,2)*nCr(70,14)/nCr(76,16) = 0.2676629938833296

nCr(6,3)*nCr(70,13)/nCr(76,16) = 0.0876557172951255

nCr(6,4)*nCr(70,12)/nCr(76,16) = 0.0147352283384047

nCr(6,5)*nCr(70,11)/nCr(76,16) = 0.0011987982377007

nCr(6,6)*nCr(70,10)/nCr(76,16) = 0.00003662994615

0.114105182095472*0.2676629938833296= 0.0305417346572765321260164435712

0.114105182095472*0.0876557172951255= 0.010001971583669509531941221736

0.114105182095472*0.0147352283384047= 0.0016813659127720276030795735184

0.114105182095472*0.0011987982377007= 0.0001367890912085693003771612304

0.114105182095472*0.00003662994615= 0.0000041796666755930835188328

We will now round down to the sixth decimal place to determine the probability of hitting 4/9 without any multipliers:

.114105-.030542-.010002-.001681-.000137-.000004 = 0.071739

Now we can state what the return of 4/9 will be with or without multipliers:

4/9 (0,1 Bells): .071739*2= 0.143478

4/9 (2 Bells): .030542*4= 0.122168

4/9 (3 Bells): .010002*8= 0.080016

4/9 (4 Bells): .001681*16= 0.026896

4/9 (5 Bells): .000137 * 20 = 0.00274

4/9 (6 Bells): .000004 * 32 = 0.000128

We can now determine the overall return of hitting 4/9:

.143478+.122168+.080016+.026896+.00274+.000128 = 0.375426

Which gives us, so far:

0/9 = 0

1/9 = 0

2/9 = 0

3/9 = 0

4/9 = .375426

Now, we will analyze the overall return for 5/9, which has a probability of: 0.032601480598706

nCr(6,2)*nCr(69,13)/nCr(75,15) = 0.2542798441891631

nCr(6,3)*nCr(69,12)/nCr(75,15) = 0.0773248648996286

nCr(6,4)*nCr(69,11)/nCr(75,15) = 0.011998685932701

nCr(6,5)*nCr(69,10)/nCr(75,15) = 0.000894817255998

nCr(6,6)*nCr(69,9)/nCr(75,15) = 0.0000248560348888

0.032601480598706*0.2542798441891631= 0.0082898994069749854165050829486

0.032601480598706*0.0773248648996286= 0.0025209050828228043749250405916

0.032601480598706*0.011998685932701= 0.000391174926644918257503684906

0.032601480598706*0.00089481725599= 0.00002917236741054532526474894

0.032601480598706*0.0000248560348888= 0.0000008103435391879726481338928

With that done, we will now round to the sixth decimal place and determine the probability of hitting 5/9 without any multipliers:

.032601-.00829-.002521-.000391-.000029-.000001 = 0.021369

Now we can state what the return of 5/9 will be with or without multipliers:

5/9 (0,1 Bells): .021369 * 3 = 0.064107

5/9 (2 Bells): .00829 * 6 = 0.04974

5/9 (3 Bells): .002521 * 12 = 0.030252

5/9 (4 Bells): .000391 * 24 = 0.009384

5/9 (5 Bells): .000029 * 30 = 0.00087

5/9 (6 Bells): .000001 * 48 = .000048

We can now determine the overall return of 5/9:

.064107+.04974+.030252+.009384+.00087+.000048 = 0.154401

Which gives us, so far:

0/9 = 0

1/9 = 0

2/9 = 0

3/9 = 0

4/9 = .375426

5/9 = .154401

We will now analyze the return for 6/9, which has a probability of: 0.005719557999773

nCr(6,2)*nCr(68,12)/nCr(74,14) = 0.239538983656458

nCr(6,3)*nCr(68,11)/nCr(74,14) = 0.0672390129561987

nCr(6,4)*nCr(68,10)/nCr(74,14) = 0.0095641699463559

nCr(6,5)*nCr(68,9)/nCr(74,14)= 0.0006484183014479

nCr(6,6)*nCr(68,8)/nCr(74,14) = 0.0000162104575362

0.005719557999773*0.239538983656458= 0.001370057110229788256273984034

0.005719557999773*0.0672390129561987= 0.0003845774344504666682335428951

0.005719557999773*0.0095641699463559= 0.0000547028247278683921143772107

0.005719557999773*0.0006484183014479= 0.0000037086660832455570737713267

0.005719557999773*0.0000162104575362= 0.0000000927166520811532257392826

We will now round to the sixth decimal place, or the nearest one with a non-zero number, to determine the probability of hitting 6/9 without a multiplier:

0.00572-.00137-.000385-.000055-.000004-.0000001= 0.0039059

Now we can state what the return of 6/9 will be with or without multipliers:

6/9 (0,1 Bells): .0039059 * 20 = 0.078118

6/9 (2 Bells): .00137 * 40 = 0.0548

6/9 (3 Bells):  .000385 * 80 = 0.0308

6/9 (4 Bells): .000055 * 160 = 0.0088

6/9 (5 Bells):  .000004 * 200 = 0.0008

6/9 (6 Bells): .0000001 * 320 = 0.000032

We can now determine what 6/9 pays, in total:

.078118+.0548+.0308+.0088+.0008+.000032 = 0.17335

Which gives us, so far:

0/9 = 0

1/9 = 0

2/9 = 0

3/9 = 0

4/9 = .375426

5/9 = .154401

6/9 = .17335

We will now determine what 7/9 pays, it has a probability of: 0.000591678413770

nCr(6,2)*nCr(67,11)/nCr(73,13) = 0.2234355225703096

nCr(6,3)*nCr(67,10)/nCr(73,13) = 0.0574921812461615

nCr(6,4)*nCr(67,9)/nCr(73,13) = 0.0074343337818312

nCr(6,5)*nCr(67,8)/nCr(73,13)= 0.000453620366349

nCr(6,6)*nCr(67,7)/nCr(73,13) = 0.0000100804525855

0.000591678413770*0.2234355225703096 = 0.000132201975574271817425803192

0.000591678413770*0.0574921812461615 = 0.000034016882603906178221243855

0.000591678413770*0.0074343337818312 = 0.000004398734819470609661895624

0.000591678413770*0.000453620366349 = 0.00000026839737881514260622573

0.000591678413770*0.0000100804525855 = 0.000000005964386195872335302335

We will now round to the sixth decimal place, or the nearest one with a non-zero number, to determine the probability of hitting 7/9 without a multiplier:

.000592-.000132-.000034-.000004-.0000003-.00000001= 0.00042169

Now we can state what the return of 7/9 will be with or without multipliers, we must remember that the maximum pay for a non-jackpot is $1,000, which is 2000 credits, so that will apply:

7/9 (0,1 Bells): .000592 * 186 = 0.110112

7/9 (2 Bells): .000132 * 372 = 0.049104

7/9 (3 Bells):  .000034 * 744 = 0.025296

7/9 (4 Bells): .000004 * 1488 = 0.005952

7/9 (5 Bells):  .0000003 * 1860 = 0.000558

7/9 (6 Bells): .00000001 * 2000 = 0.00002

With that done, we can now determine the overall return of 7/9, with or without multipliers:

.110112+.049104+.025296+.005952+.000558+.00002= 0.191042

Which gives us, so far:

0/9 = 0

1/9 = 0

2/9 = 0

3/9 = 0

4/9 = .375426

5/9 = .154401

6/9 = .17335

7/9 = .191042

Finally, 8/9 and 9/9 are effectively the same thing, so we will combine those probabilities and multiply by the 2300 credit base win:

(0.000032592454996+ 0.000000724276778)*2300 = 0.0766284830802

Which gives us, so far:

0/9 = 0

1/9 = 0

2/9 = 0

3/9 = 0

4/9 = .375426

5/9 = .154401

6/9 = .17335

7/9 = .191042

8/9+9/9= .0766285

The total return of the base game is:

.375426+.154401+.17335+.191042+.0766285= 0.9708475

We have clearly determined that playing the Nine-Spot is the superior game.  Not only is there a greater probability of hitting the Progressive, but the base return is also greater.