Talossa National Lottery

[Caleb Frenibuerg]

Right! Here we go. I've tried my best to sort this out but this is the best outcome I can figure out.

Sales of all tickets will go towards:
- 10% will be donated to a charity of the Burgermeisters choice at the time of each draw.
- 15% will be be forwarded to the Burgermeister to assist in the running of Talossa.
And the rest will be equally divided amongst all tickets that have 4 numbers matched.

Citizens will be able to purchase 1 ticket.

Each ticket will have a base price of $10USD. You can add extra lines of 4 numbers up to a maximum of 4 times, so a max of 5 lines per ticket. Any extra lines will cost an additional $2.50USD.

If you purchase a ticket, you must select  4 numbers. The numbers will range from 1-59.

If you match 1 number, you win nothing.
If you match 2 numbers, you receive a 25% discount on your next ticket.
If you match 3 numbers, you receive 1 free ticket into the next draw, although a donation towards a ticket would be much appreciated.
If you match all 4 numbers, as above, all funds that remain once the deductions have been removed will be divided equally amongst all tickets that have 4 numbers matched.

Payments will be taken by PayPal, or if you can sort another form of payment out, let me know when you message to purchase a ticket.

[Caleb Frenibuerg]

Current staff members are as follows:

Caleb Frenibuerg, Head of Talossa National Lottery

Mic'haglh Autófil, Assistant to the Head

Breneir Tzarcomprada, Head of Marketing

[Baron Alexandreu Davinescu]

Wow, look at this tycoon get to work!  Two staffers already, the most for a private enterprise in years!  Love it!

[Sir Ian Plätschisch]

It's great to see new citizens taking initiative!

With that made clear, I have some concern about how the mathematics of this lottery are going to shake out.

The number of unique combinations of four numbers between 1-59 (assuming the order doesn't matter, which my understanding of the rules) is the sum of:

Cases in which all four numbers are distinct (ex: 1 2 3 4): 59 choose 4 = 455,126
Cases in which one number appears twice (ex. 1 2 3 3): (59 choose 3) * 3 = 97,527
Cases in which one number appears thrice (ex. 1 2 2 2): (59 choose 2) * 2 = 3,422
Cases in which one number repeats four times (ex. 1 1 1 1): 59 (pretty obvious)
Cases in which two numbers appear twice (ex. 1 1 2 2): 59 choose 2 = 1,711

For a total of 557,845 possibilities. Since each line can only contain one of these possibilities, each line has only a 1/557,845 = .000001793 chance of winning on a particular draw.

The pricing of the tickets clearly encourages people to buy five lines on each ticket, so assuming each ticket has five unique lines, the chance of a ticket winning on a particular draw is 5/557,845 = .000008963

Similarly, suppose twenty tickets (with five unique lines each) are sold, which is pretty optimistic. The chance that anyone wins on a particular drawing is 100/557,845 = .0001793

Suppose that all of the tickets remain in effect until one of them wins. The number of drawings required before any of the tickets wins is a geometric random variable with distribution:

Pr(D = d) = .0001793 * (1 - .0001793)^(d - 1)

For example, the probability that the first win occurs on the 10th drawing is:

Pr(D = 10) = .0001793 * (1 - .0001793)^9 = .0001790

The mean of this distribution is 1/.0001793 = 5,577

If a drawing were held every week, we would not expect anyone to win for over 100 years!

If anything, the Treasury would surely benefit from being able to loan out the proceeds at interest while it waited for anyone to win.

I am happy to join this project as Chief Actuary to help come up with a more workable design.

[Caleb Frenibuerg]

It's great to see new citizens taking initiative!

With that made clear, I have some concern about how the mathematics of this lottery are going to shake out.

The number of unique combinations of four numbers between 1-59 (assuming the order doesn't matter, which my understanding of the rules) is the sum of:

Cases in which all four numbers are distinct (ex: 1 2 3 4): 59 choose 4 = 455,126
Cases in which one number appears twice (ex. 1 2 3 3): (59 choose 3) * 3 = 97,527
Cases in which one number appears thrice (ex. 1 2 2 2): (59 choose 2) * 2 = 3,422
Cases in which one number repeats four times (ex. 1 1 1 1): 59 (pretty obvious)
Cases in which two numbers appear twice (ex. 1 1 2 2): 59 choose 2 = 1,711

For a total of 557,845 possibilities. Since each line can only contain one of these possibilities, each line has only a 1/557,845 = .000001793 chance of winning on a particular draw.

The pricing of the tickets clearly encourages people to buy five lines on each ticket, so assuming each ticket has five unique lines, the chance of a ticket winning on a particular draw is 5/557,845 = .000008963

Similarly, suppose twenty tickets (with five unique lines each) are sold, which is pretty optimistic. The chance that anyone wins on a particular drawing is 100/557,845 = .0001793

Suppose that all of the tickets remain in effect until one of them wins. The number of drawings required before any of the tickets wins is a geometric random variable with distribution:

Pr(D = d) = .0001793 * (1 - .0001793)^(d - 1)

For example, the probability that the first win occurs on the 10th drawing is:

Pr(D = 10) = .0001793 * (1 - .0001793)^9 = .0001790

The mean of this distribution is 1/.0001793 = 5,577

If a drawing were held every week, we would not expect anyone to win for over 100 years!

If anything, the Treasury would surely benefit from being able to loan out the proceeds at interest while it waited for anyone to win.

I am happy to join this project as Chief Actuary to help come up with a more workable design.

I would thoroughly appreciate any help.
I was a bit unsure myself to be fair... Thank you for pointing these things out.

[Mic’haglh Autófil, SMC EiP]

It's great to see new citizens taking initiative!

With that made clear, I have some concern about how the mathematics of this lottery are going to shake out.

The number of unique combinations of four numbers between 1-59 (assuming the order doesn't matter, which my understanding of the rules) is the sum of:

Cases in which all four numbers are distinct (ex: 1 2 3 4): 59 choose 4 = 455,126
Cases in which one number appears twice (ex. 1 2 3 3): (59 choose 3) * 3 = 97,527
Cases in which one number appears thrice (ex. 1 2 2 2): (59 choose 2) * 2 = 3,422
Cases in which one number repeats four times (ex. 1 1 1 1): 59 (pretty obvious)
Cases in which two numbers appear twice (ex. 1 1 2 2): 59 choose 2 = 1,711

For a total of 557,845 possibilities. Since each line can only contain one of these possibilities, each line has only a 1/557,845 = .000001793 chance of winning on a particular draw.

The pricing of the tickets clearly encourages people to buy five lines on each ticket, so assuming each ticket has five unique lines, the chance of a ticket winning on a particular draw is 5/557,845 = .000008963

Similarly, suppose twenty tickets (with five unique lines each) are sold, which is pretty optimistic. The chance that anyone wins on a particular drawing is 100/557,845 = .0001793

Suppose that all of the tickets remain in effect until one of them wins. The number of drawings required before any of the tickets wins is a geometric random variable with distribution:

Pr(D = d) = .0001793 * (1 - .0001793)^(d - 1)

For example, the probability that the first win occurs on the 10th drawing is:

Pr(D = 10) = .0001793 * (1 - .0001793)^9 = .0001790

The mean of this distribution is 1/.0001793 = 5,577

If a drawing were held every week, we would not expect anyone to win for over 100 years!

If anything, the Treasury would surely benefit from being able to loan out the proceeds at interest while it waited for anyone to win.

I am happy to join this project as Chief Actuary to help come up with a more workable design.

Assuming my math is right, if we used three numbers, that time period drops to about 8.25 years. If we only used two, it would drop to about 18 weeks. No?

[Sir Ian Plätschisch]

Assuming my math is right, if we used three numbers, that time period drops to about 8.25 years. If we only used two, it would drop to about 18 weeks. No?

I got about seven years for three numbers, but about the same for two numbers.

Remember that both depend on 100 unique entries eligible on each draw.

[Caleb Frenibuerg]

Perhaps they just pick 1 number? I'm trying to work through this.

[Mic’haglh Autófil, SMC EiP]

Perhaps they just pick 1 number? I'm trying to work through this.

I like the idea of picking multiple numbers, since that also allows you to sort of "tier" the prizes. Also, depending on how many choices you let people make (pick one number, but still have multiple entries per ticket?), you don't need many tickets sold before you start approaching a point where it becomes certain that you have a winner week after week. Just my 2 cents, of course, but I like the idea of having it build up a bit on occasion.

Alternatively, to alter the odds some, you could also avoid having repeat numbers in the draw. (Think of how actual lotteries work -- only one ball with a given number in the drum, usually.)

There's also the possibility of using a "bonus" ball. Say you pick two numbers from one through 59, but then also a third number from, say, 1-8. Getting all three is necessary for the grand prize, but not as hard as getting all three when all three numbers could be from 1-59.

[Baron Alexandreu Davinescu]

I'm not a math guy, but I guess I don't understand why one number doesn't win. That's what I would have assumed - you buy one or more tickets for a dollar each, and if you buy the winning number, you get whatever the pot is for that drawing. Or split it if someone else picks the same number. Pretty low stakes, very simple. Although maybe a dollar wouldn't be enough of a pot if only a handful of people enter, so it might make sense to start off with pricier tickets.

But you guys are engineers and math people, so there are probably good reasons that I don't understand.

[Sir Ian Plätschisch]

I'm not a math guy, but I guess I don't understand why one number doesn't win. That's what I would have assumed - you buy one or more tickets for a dollar each, and if you buy the winning number, you get whatever the pot is for that drawing. Or split it if someone else picks the same number. Pretty low stakes, very simple. Although maybe a dollar wouldn't be enough of a pot if only a handful of people enter, so it might make sense to start off with pricier tickets.

But you guys are engineers and math people, so there are probably good reasons that I don't understand.

I endorse this design. The only mark against it is that it's a bit boring. In that case you could reduce the number of options from 59 while keeping multiple "balls"

[Mic’haglh Autófil, SMC EiP]

I'm not a math guy, but I guess I don't understand why one number doesn't win. That's what I would have assumed - you buy one or more tickets for a dollar each, and if you buy the winning number, you get whatever the pot is for that drawing. Or split it if someone else picks the same number. Pretty low stakes, very simple. Although maybe a dollar wouldn't be enough of a pot if only a handful of people enter, so it might make sense to start off with pricier tickets.

But you guys are engineers and math people, so there are probably good reasons that I don't understand.

I endorse this design. The only mark against it is that it's a bit boring. In that case you could reduce the number of options from 59 while keeping multiple "balls"

Pretty much the other option, if 59 isn't as important. The key is to balance between winners coming up too frequently and winners coming up not frequently enough.

[Sir Ian Plätschisch]

I'm not a math guy, but I guess I don't understand why one number doesn't win. That's what I would have assumed - you buy one or more tickets for a dollar each, and if you buy the winning number, you get whatever the pot is for that drawing. Or split it if someone else picks the same number. Pretty low stakes, very simple. Although maybe a dollar wouldn't be enough of a pot if only a handful of people enter, so it might make sense to start off with pricier tickets.

But you guys are engineers and math people, so there are probably good reasons that I don't understand.

I endorse this design. The only mark against it is that it's a bit boring. In that case you could reduce the number of options from 59 while keeping multiple "balls"

Pretty much the other option, if 59 isn't as important. The key is to balance between winners coming up too frequently and winners coming up not frequently enough.

Right...maybe we figure out how frequently we want a winner first, then design the game accordingly