March 2nd, 2019 at 7:04:10 AM
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Somebody asked me about the variance of Cleopatra Keno. I had to dust off my college statistics books to help me with that one. As a reminder, it plays like conventional spot keno, except if the last ball drawn contributes to a win, the player gets 12 free games with a 2x multiplier. Free games do not earn more free games. Let's use the 3-10-56-180-1000 pick-8 pay table as an example.

To begin, recall var(x + y) = var(x) + var(y) + 2*cov(x,y)

In this case of this game, var(entire game) = var(base game) + var(bonus) + 2*cov(base game,bonus)

Let's start with the variance of the base game.

As a reminder, the variance equals exp(x^2) - (e(x))^2.

So, var(base game) = 19.530214 - 0.593301^2 = 19.178208.

Next, let's do the variance of the bonus, given that the player won the bonus in the first place. Recall var(ax) = a^2 * var(x). In the bonus there are 12 doubled free games. So the variance of a bonus would be...

12 * 2^2 * 19.178208 = 920.554000.

However, the player doesn't always win the bonus. This was a tricky step. It would have been nice to just multiply that by the probability of winning the bonus of 0.021644, but you can't.

Let's do the easy part first. The average bonus win is 12*2*0.593301 = 14.239212. The expected win from the bonus on any given spin is prob(bonus)*(average bonus) = 0.021644 * 14.23921236 = 0.308198. Not that we directly care, but the overall return of the game is exp(base game) + exp(bonus) = 0.593301 + 0.308198 = 0.901498.

To find the exp(x^2) of the bonus, recall variance = exp(x^2) - (e(x))^2.

To rearrange:

exp(x^2) = var(x) + (e(x))^2.

In the case of the bonus, given a bonus win in the first place:

exp(x^2) = 920.554000 + 14.239212^2 = 1123.309169.

Now we're ready to find the overall variance from the bonus (including when the player doesn't win it):

We already know exp(bonus win) = 0.308198.

exp((bonus win)^2) = 0.021644 * 1123.309169 = 24.313239.

Thus the variance of the bonus on each spin is 24.313239 - 0.308198^2 = 24.218253.

Next, let's do the covariance. Why is there any covariance, you might ask? It's because the player has to hit a winning ball on the last draw to trigger the bonus. Given the fact that the last ball won makes it more likely the player won money in the base game on that spin. In other words, winning the bonus is correlated to winning something on the base spin.

We'll need to know the expected win in the base game, given that the bonus was won. Here is that table:

From our college statistics class, we know cov(x,y) = exp(xy) - exp(x)*exp(y).

Exp(xy) = 0.021644 * 6.757734 * 14.239212 = 2.082719.

So, the overall covariance is 2.082719 - 0.593301 * 0.308198 = 1.899865.

Thus, our bottom line is:

var(base game) + var(bonus) + 2*cov(base game,bonus) = 19.178208 + 24.218253 - 2*1.899865 = 47.19619.

Here is a summary of the key numbers.

I hope the one who asked me (I'm not sure if I can state his name) is happy, this took hours, including running a simulation to confirm the answer.

To begin, recall var(x + y) = var(x) + var(y) + 2*cov(x,y)

In this case of this game, var(entire game) = var(base game) + var(bonus) + 2*cov(base game,bonus)

Let's start with the variance of the base game.

Catch | Pays | Winning combinations | Probability | Return | exp win^2 |
---|---|---|---|---|---|

0 | 0 | 2,558,620,845 | 0.088266 | 0.000000 | 0.000000 |

1 | 0 | 7,724,138,400 | 0.266464 | 0.000000 | 0.000000 |

2 | 0 | 9,512,133,400 | 0.328146 | 0.000000 | 0.000000 |

3 | 0 | 6,226,123,680 | 0.214786 | 0.000000 | 0.000000 |

4 | 3 | 2,362,591,575 | 0.081504 | 0.244511 | 0.733533 |

5 | 10 | 530,546,880 | 0.018303 | 0.183026 | 1.830259 |

6 | 56 | 68,605,200 | 0.002367 | 0.132536 | 7.422014 |

7 | 180 | 4,651,200 | 0.000160 | 0.028882 | 5.198747 |

8 | 1000 | 125,970 | 0.000004 | 0.004346 | 4.345661 |

Total | 28,987,537,150 | 1.000000 | 0.593301 | 19.530214 |

As a reminder, the variance equals exp(x^2) - (e(x))^2.

So, var(base game) = 19.530214 - 0.593301^2 = 19.178208.

Next, let's do the variance of the bonus, given that the player won the bonus in the first place. Recall var(ax) = a^2 * var(x). In the bonus there are 12 doubled free games. So the variance of a bonus would be...

12 * 2^2 * 19.178208 = 920.554000.

However, the player doesn't always win the bonus. This was a tricky step. It would have been nice to just multiply that by the probability of winning the bonus of 0.021644, but you can't.

Let's do the easy part first. The average bonus win is 12*2*0.593301 = 14.239212. The expected win from the bonus on any given spin is prob(bonus)*(average bonus) = 0.021644 * 14.23921236 = 0.308198. Not that we directly care, but the overall return of the game is exp(base game) + exp(bonus) = 0.593301 + 0.308198 = 0.901498.

To find the exp(x^2) of the bonus, recall variance = exp(x^2) - (e(x))^2.

To rearrange:

exp(x^2) = var(x) + (e(x))^2.

In the case of the bonus, given a bonus win in the first place:

exp(x^2) = 920.554000 + 14.239212^2 = 1123.309169.

Now we're ready to find the overall variance from the bonus (including when the player doesn't win it):

We already know exp(bonus win) = 0.308198.

exp((bonus win)^2) = 0.021644 * 1123.309169 = 24.313239.

Thus the variance of the bonus on each spin is 24.313239 - 0.308198^2 = 24.218253.

Next, let's do the covariance. Why is there any covariance, you might ask? It's because the player has to hit a winning ball on the last draw to trigger the bonus. Given the fact that the last ball won makes it more likely the player won money in the base game on that spin. In other words, winning the bonus is correlated to winning something on the base spin.

We'll need to know the expected win in the base game, given that the bonus was won. Here is that table:

Catch | Pays | Winning combinations | Probability | Return |
---|---|---|---|---|

0 | 0 | - | 0.000000 | 0.000000 |

1 | 0 | - | 0.000000 | 0.000000 |

2 | 0 | - | 0.000000 | 0.000000 |

3 | 0 | - | 0.000000 | 0.000000 |

4 | 3 | 472,518,315 | 0.753119 | 2.259358 |

5 | 10 | 132,636,720 | 0.211402 | 2.114019 |

6 | 56 | 20,581,560 | 0.032804 | 1.837010 |

7 | 180 | 1,627,920 | 0.002595 | 0.467036 |

8 | 1000 | 50,388 | 0.000080 | 0.080310 |

Total | 627,414,903 | 1.000000 | 6.757734 |

From our college statistics class, we know cov(x,y) = exp(xy) - exp(x)*exp(y).

Exp(xy) = 0.021644 * 6.757734 * 14.239212 = 2.082719.

So, the overall covariance is 2.082719 - 0.593301 * 0.308198 = 1.899865.

Thus, our bottom line is:

var(base game) + var(bonus) + 2*cov(base game,bonus) = 19.178208 + 24.218253 - 2*1.899865 = 47.19619.

Here is a summary of the key numbers.

Exp base win | 0.593301 |

Exp bonus | 0.308198 |

Return of game | 0.901498 |

Average base win given bonus won | 6.757734 |

Average bonus | 14.239212 |

Prob bonus | 0.021644 |

Variance base win | 19.178208 |

Variance bonus, given bonus | 920.554000 |

win^2 in bonus | 1123.309169 |

Variance bonus | 24.218253 |

covariance | 1.899865 |

total variance | 47.196191 |

I hope the one who asked me (I'm not sure if I can state his name) is happy, this took hours, including running a simulation to confirm the answer.

Last edited by: Wizard on Mar 2, 2019

It's not whether you win or lose; it's whether or not you had a good bet.

March 2nd, 2019 at 10:13:29 AM
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As usual, a post that took me hours to write gets no replies.

It's not whether you win or lose; it's whether or not you had a good bet.

March 2nd, 2019 at 10:21:53 AM
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Wiz this is probably great information but what I take away from it is I should randomly pick 8 numbers hit the start button and watch my credits slowly disappear.

March 2nd, 2019 at 10:30:34 AM
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I saw a Cleopatra Keno yesterday, with a Blackjack game available. Blackjack pays 2 for 1, same as any other win, so no.

March 2nd, 2019 at 10:36:51 AM
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I used to have a Keno strategy when picking 2 numbers and winning would pay 15 to 1. I'd better stick to 17 to 1 payouts on Roulette instead.

Lots of people playing 8 to 10 spot penny or nickel Keno to win the big prizes though.

Lots of people playing 8 to 10 spot penny or nickel Keno to win the big prizes though.

March 2nd, 2019 at 10:38:46 AM
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Quote:WizardAs usual, a post that took me hours to write gets no replies.

Come on Mike, it's only been 3 hours since you posted it. :P

And well done! I haven't done much work with covariance. I took college statistics, but I definitely didn't learn covariance well and definitely hadn't seen it applied to a gaming example with careful detail. This is a fairly difficult calculation for most adults to do, and I think you guided the reader very well. So if even the person who asked didn't learn anything, at least I did!

Also writing any post on here that uses a table is a royal PITA and will take forever to write unless you have a nice script to add rows and columns.

March 2nd, 2019 at 11:55:23 AM
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Thanks guys for the replies, although I'm sure they were sympathetic. To be honest, it was good for me to review those formulas and I'll get mileage from it as a future "ask the wizard" question.

It's not whether you win or lose; it's whether or not you had a good bet.

March 2nd, 2019 at 12:29:04 PM
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Quote:WizardAs usual, a post that took me hours to write gets no replies.

Sorry. I hang out at a website that taught me Keno was for suckers so I tend not to read articles about the game.

Have you tried 22 tonight? I said 22.

March 2nd, 2019 at 9:58:43 PM
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I find Cleopatra Keno to be the funnest keno game in a casino

4 card Cleo is even more funner

4 card Cleo is even more funner

"Man Babes" #AxelFabulous

March 3rd, 2019 at 12:15:26 AM
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I've been using http://miplet.net/table/ for a long time. Just copy and paste from a spreadsheet.Quote:tringlomaneAlso writing any post on here that uses a table is a royal PITA and will take forever to write unless you have a nice script to add rows and columns.

“Man Babes” #AxelFabulous