The last game played in the session is a regular bingo with a series of allowable patterns that segues into a coverall game. We currently limit the maximum coverall jackpot at 55 numbers or less for $500. After the jackpot is is hit, the following week it drops to 52 numbers or less for $350 and increases weekly (53 or less for $400, 54 or less for $450) until we get back to the $500 level. The problem with doing that is that the big spenders won't come back until the jackpot gets back to the $500 level.

I want to make the proposal to the club trustees that we change the weekly increases in a fashion that will 1.) keep the regulars from missing weeks, and 2). attract new players. I want to propose that after the $500 is paid out, the following week's jackpot will be 51 numbers or less for $900, then increment as 52 numbers or less for $800, 53 numbers or less for $700, 54 numbers or less for $600 and capping at 55 numbers or less for $500.

I don't know how to crunch the numbers to be able to tell the trustees what the odds are for the house to lose as we start to pick up new attendees. Would anyone be willing to help me with this?

I need to know odds of the payouts I specified earlier with 25, 30, 35, 40, 50, 60 attendees.

Quote:grolmundHi all! I belong to a fraternal organization that has a Bingo session once a week. It is a small gathering, open to the public, not well publicized and due to attrition by death and illness, is shrinking in size to the point that it is difficult to find club members willing to be involved. We currently have approximately 25 regulars and we want to increase attendance to make this lucrative again.

The last game played in the session is a regular bingo with a series of allowable patterns that segues into a coverall game. We currently limit the maximum coverall jackpot at 55 numbers or less for $500. After the jackpot is is hit, the following week it drops to 52 numbers or less for $350 and increases weekly (53 or less for $400, 54 or less for $450) until we get back to the $500 level. The problem with doing that is that the big spenders won't come back until the jackpot gets back to the $500 level.

I want to make the proposal to the club trustees that we change the weekly increases in a fashion that will 1.) keep the regulars from missing weeks, and 2). attract new players. I want to propose that after the $500 is paid out, the following week's jackpot will be 51 numbers or less for $900, then increment as 52 numbers or less for $800, 53 numbers or less for $700, 54 numbers or less for $600 and capping at 55 numbers or less for $500.

I don't know how to crunch the numbers to be able to tell the trustees what the odds are for the house to lose as we start to pick up new attendees. Would anyone be willing to help me with this?

I need to know odds of the payouts I specified earlier with 25, 30, 35, 40, 50, 60 attendees.

Well, in Bingo it's not so much how many attendees as how many cards are being played. Do you limit the cards any one person can have in that final game?

I do think your idea of more money for less numbers is a good one, but why not lock that in? If it's good for one week, it should be good for all weeks.

Look at the table for coverall from WizardofOdds.com while waiting for the math guys to answer. I'm not one of them, but I may be able to figure out some of what you want.

Balls drawn (or less) | Probability | Happens 1 card in | 200 cards in play 1 game in |
---|---|---|---|

50 | 0.00000471508113370243 | 212085.4279373309 | 1060.427139686655 |

51 | 0.00000890626436366014 | 112280.520673405 | 561.402603367025 |

52 | 0.00001654020524679740 | 60458.74190101201 | 302.2937095050601 |

53 | 0.00003022865096828490 | 33081.19839856506 | 165.4059919928253 |

54 | 0.00005441157174291290 | 18378.44355471108 | 91.8922177735554 |

55 | 0.00009653665954387760 | 10358.7590944794 | 51.79379547239701 |

56 | 0.00016893915420178600 | 5919.290911129335 | 29.59645455564667 |

57 | 0.00029180399362126600 | 3426.957895916022 | 17.13478947958011 |

58 | 0.00049778328323627800 | 2008.906352778216 | 10.04453176389108 |

59 | 0.00083912039174115500 | 1191.724107580188 | 5.958620537900939 |

60 | 0.00139853398623526000 | 715.0344645480956 | 3.575172322740478 |

This info is from the link above in the 1st 2 columns; the last 2 are derived from the probability numbers. Do NOT take these as correct until one of the math guys confirms I did this correctly, please.

The numbers in the last column will change as you add cards to the game. The others should be static no matter how many cards are in play. It appears that if you sell 700 cards, that is your break-even point for a winner in 60 or less balls nearly every night. I could be wrong. But if you were to base your pays on the table above (and 200 cards in play), you wouldn't see very many nights with winners at 55 or less balls; if you played once a week with 200 cards, the jackpot would only hit once a year on average.

If you could be more specific about how many cards are in play per person with those attendee numbers, I could give you a table more closely built to your question, but I also want forum confirmation I'm giving you accurate numbers before I try to compute that.

Quote:beachbumbabsSo, using that table, let's say you have 25 players, each daubing 8 cards, so 200 cards are in play. These would be averages over a long time; could never happen, could happen back-to-back, or anything in between.

This info is from the link above in the 1st 2 columns; the last 2 are derived from the probability numbers. Do NOT take these as correct until one of the math guys confirms I did this correctly, please.

I heard that...

Your numbers look good to me, Babs.

Quote:beachbumbabsSo, using that table, let's say you have 25 players, each daubing 8 cards, so 200 cards are in play. These would be averages over a long time; could never happen, could happen back-to-back, or anything in between.

Balls drawn (or less) Probability Happens 1 card in 200 cards in play 1 game in 50 0.00000471508113370243 212085.4279373309 1060.427139686655 51 0.00000890626436366014 112280.520673405 561.402603367025 52 0.00001654020524679740 60458.74190101201 302.2937095050601 53 0.00003022865096828490 33081.19839856506 165.4059919928253 54 0.00005441157174291290 18378.44355471108 91.8922177735554 55 0.00009653665954387760 10358.7590944794 51.79379547239701 56 0.00016893915420178600 5919.290911129335 29.59645455564667 57 0.00029180399362126600 3426.957895916022 17.13478947958011 58 0.00049778328323627800 2008.906352778216 10.04453176389108 59 0.00083912039174115500 1191.724107580188 5.958620537900939 60 0.00139853398623526000 715.0344645480956 3.575172322740478

This info is from the link above in the 1st 2 columns; the last 2 are derived from the probability numbers. Do NOT take these as correct until one of the math guys confirms I did this correctly, please.

The numbers in the last column will change as you add cards to the game. The others should be static no matter how many cards are in play. It appears that if you sell 700 cards, that is your break-even point for a winner in 60 or less balls nearly every night. I could be wrong. But if you were to base your pays on the table above (and 200 cards in play), you wouldn't see very many nights with winners at 55 or less balls; if you played once a week with 200 cards, the jackpot would only hit once a year on average.

If you could be more specific about how many cards are in play per person with those attendee numbers, I could give you a table more closely built to your question, but I also want forum confirmation I'm giving you accurate numbers before I try to compute that.

I believe the probabilities for the last column should be…

1 – [(1 – P)

^{200}]

where P is the probability from your second column

In words, it’s the probability that a single card does not win raised to how many cards are in play. This is the probability that none of the cards in play wins. So subtract from 1 to get the probability that at least one of the cards wins.

To get in the game session, attendees must buy one game packet. If an attendee wants to buy two packets, so much the better but, there are two or three attendees who will buy a second packet. We allow 16 seconds between balls so they have to hustle to cover their cards. We also allow and actively try to sell additional strips of three for all of the games. During the final game, the coverall game, at least half of the players buy an extra strip or two. So, that said, I can say for now I have 23 players with 9 cards (192 cards), 2 players with 18 cards (36 cards), and about half of the players will buy an extra strip of three cards so let's say 13 players with an additional 3 cards (39 cards) for total of 267 cards in play.

Now I'm "getting" that I have to think about how many cards can be in play rather than how many attendees. Brainf@rt on my side.

[267x30]/25=321 cards for 30 players

[267x35]/25=374 cards for 35 players

[267x40]/25=427 cards for 40 players

[267x50]/25=534 cards for 50 players

[267x60]/25=641 cards for 60 players

1 – [(1-p)

^{n}]

n is the number of cards in play

p is the probability in the second column of Babs’s table

For example, if you have 641 cards in play, the probability that at least one person will win in 55 or less balls is:

= 1 – [(1-0.0000965)

^{641}]

= about 6%

We don't allow sleepers when we play. Wouldn't I want to use the probability numbers from the tables column called "bingo on last ball"?

Quote:PeeMcGeeSimply use the function in my last post:

1 – [(1-p)^{n}]

n is the number of cards in play

p is the probability in the second column of Babs’s table

For example, if you have 641 cards in play, the probability that at least one person will win in 55 or less balls is:

= 1 – [(1-0.0000965)^{641}]

= about 6%

BeachBumBabs and PeeMcGee,

We don't allow sleepers when we play. Shouldn't I use the probability numbers from the tables column called "bingo on last ball"?

Quote:grolmundBeachBumBabs and PeeMcGee,

We don't allow sleepers when we play. Wouldn't I want to use the probability numbers from the tables column called "bingo on last ball"?

I’m not well versed with bingo terminology, but you would use that column if your players can only win if they coverall on that exact number of balls.

In your original post, you said that jackpot are awarded on coveralls of X balls or less. Therefore, you would use the “Bingo on or Before Last Ball”.

Quote:PeeMcGeeI’m not well versed with bingo terminology, but you would use that column if your players can only win if they coverall on that exact number of balls.

In your original post, you said that jackpot are awarded on coveralls of X balls or less. Therefore, you would use the “Bingo on or Before Last Ball”.

A "sleeper" is one who misses their opportunity to call Bingo on the last number that was called. Bingo machines that control the light boards always flash the the last number called and in the verify software that produces the verified bingo onscreen, if the last number called is not in the bingo pattern displayed, and flashing, it is a sleeper. And a sleeper is not awarded the win. But I understand now why I need to use the second column

I'm curious why anyone would deliberately hold off on calling a bingo if "sleepers" were allowed. What are you protecting against in prohibiting this practice?

I'm in PA. Pa's anti-smoking law, twisted as it is, allows private clubs to decide to be smoking, or not. Ours has decided to be smoking. The anti-smoking law also dictates that if you are a smoking establishment, you cannot publicly advertise Bingo UNLESS, you have a non-smoking room (meeting room or rental hall) where you can have the Bingo activities. At this time, we cannot make that accommodation. So, it has to be word of mouth until we figure out a way to work around the law.

As to not changing the game, when the pot is low, and with our current attendance of 25 people, we lose 12% of our attendees until the pot gets back up. That is what will drive the change.

I suspect it's different in the US than UK, but here because of the legal definition of bingo being skill, the player has to make the call (without prompting by the caller) and if they miss they cannot win - and you cannot win the full house prize if the two-line prize has not been claimed. Thus players do miss - I've done it and I've also been the recipient winning when someone else should have.Quote:beachbumbabs...why anyone would deliberately hold off on calling a bingo...

As to the numbers, it is of course possible that two-people could win on the same number (and presumably the prize would be shared) but 1-(1-p)^n or similar is a reasonable estimate.

Your problem of trying to offer big prizes is one that small bingo clubs in the UK have. One method is to have linked games, another is there is a national game, and the last is some kind of progressive. I've seen the full-house in N numbers a few times, but also the build-up mechanism more often (usually on a quiet night as a feature) - and yes the club does fill up a bit more when the times are right (but that's good marketing as say the players usually only come in on Sundays so they try another night and buy other books, cups of teas, play the fruit machines etc. and might even come back.)

The other method is to use a Monopoly like board - the manager was a chippy so very good at building things - whenever you called a full house you moved along the track (e.g. B=1, I=2...O=5). The last square was £500, but there were other prizes along the route. (Obviously you can devise a similar method.)