# Stats and game nerds, lend me your beanplates.

September 18, 2013 11:53 AM Subscribe

I have a list of 15 people. Each person has between 1 and 3 entries in a lottery, for a total of 35 entries. I need to select 9 people from the list of 15--nobody can win more than once.
What is the most transparent, most random, most low-tech way I can do this?

Assume that the drama and stakes level are such that accusations of tampering are a possibility, and that 3-5 names and/or entries may be added or withdrawn shortly before the lottery. Both these things seem to preclude premade names in a hat. I want anyone who wants to, to see and understand as much of the process as possible.

I have access to a standard assortment of RPG dice.

My current plan is to split the entries into 3 numbered groups of 12 (with one or more "free" spaces) and have different unaffiliated people throw a d12 until 3 unique people have been chosen from each group. Will this work? Will it matter whether someone with 3 entries has them all in one list or evenly split between lists? Will it matter if all of the smaller-entry people are in the same group or not? I have a vague suspicion that things get more complicated after a few people are selected (and their entries thereafter withdrawn), but I can't quite see it.

Assume that the drama and stakes level are such that accusations of tampering are a possibility, and that 3-5 names and/or entries may be added or withdrawn shortly before the lottery. Both these things seem to preclude premade names in a hat. I want anyone who wants to, to see and understand as much of the process as possible.

I have access to a standard assortment of RPG dice.

My current plan is to split the entries into 3 numbered groups of 12 (with one or more "free" spaces) and have different unaffiliated people throw a d12 until 3 unique people have been chosen from each group. Will this work? Will it matter whether someone with 3 entries has them all in one list or evenly split between lists? Will it matter if all of the smaller-entry people are in the same group or not? I have a vague suspicion that things get more complicated after a few people are selected (and their entries thereafter withdrawn), but I can't quite see it.

Is there any reason why you cannot draw slips of paper (one slip per entry) from a transparent bowl in front of everyone's face? If you happen to draw a person who has already won once, discard it and draw again.

I cannot imagine anything more transparent or low-tech.

posted by Tanizaki at 12:04 PM on September 18, 2013 [14 favorites]

I cannot imagine anything more transparent or low-tech.

posted by Tanizaki at 12:04 PM on September 18, 2013 [14 favorites]

The dice are a bit problematic, because of the 'free' spaces -- means someone in the free group has odds of winning of 1/11, while the others have 1/12.

You could make 35 scraps of paper with numbers 1 to 35 on them, and a master list of which name maps to which number (ideally a big printed sign that everyone can see, which is fixed before you do the drawing). Put them in a hat, and draw the numbers until you have 9 unique winners.

To go a bit farther, maybe get a set of poker chips, write the numbers on them in marker, and publicly put them in the hat before the drawing. That gets around the issue of uneven-sized scraps of paper.

Or you could avoid the drawing and roll 3d12 at once to get a number between 0 and 36. No grouping -- everyone in the same batch.

posted by PercussivePaul at 12:05 PM on September 18, 2013

You could make 35 scraps of paper with numbers 1 to 35 on them, and a master list of which name maps to which number (ideally a big printed sign that everyone can see, which is fixed before you do the drawing). Put them in a hat, and draw the numbers until you have 9 unique winners.

To go a bit farther, maybe get a set of poker chips, write the numbers on them in marker, and publicly put them in the hat before the drawing. That gets around the issue of uneven-sized scraps of paper.

Or you could avoid the drawing and roll 3d12 at once to get a number between 0 and 36. No grouping -- everyone in the same batch.

posted by PercussivePaul at 12:05 PM on September 18, 2013

Best answer: "roll 3d12 at once to get a number between 0 and 36."

But if you roll 3d12, you'll only get between 3 and 36. Furthermore, they will not have even distribution. You can read the wiki page on dice probability, but the short version is that the numbers in the middle will have a much higher probability of being chosen.

So don't do this.

posted by ethidda at 12:09 PM on September 18, 2013

But if you roll 3d12, you'll only get between 3 and 36. Furthermore, they will not have even distribution. You can read the wiki page on dice probability, but the short version is that the numbers in the middle will have a much higher probability of being chosen.

So don't do this.

posted by ethidda at 12:09 PM on September 18, 2013

Oh, morning lack of coffee fail, hehe. Do not roll 3d6 :)

posted by PercussivePaul at 12:13 PM on September 18, 2013 [1 favorite]

posted by PercussivePaul at 12:13 PM on September 18, 2013 [1 favorite]

Best answer: Deck of cards. Give everyone markers and one, two or three cards, they write their names on them and hand them back. You shuffle the 35 cards and turn them over one-by-one until you have nine unique names. Same basic idea as PP's poker chips or Tanizaki's scraps of paper.

posted by RobotHero at 12:16 PM on September 18, 2013 [2 favorites]

posted by RobotHero at 12:16 PM on September 18, 2013 [2 favorites]

Best answer: Something like Tanizaki's answer is the one I'd use.

If you need to be very transparent, print the names in large, clear text onto index cards (three cards per person) and lay out the correct number of cards clearly in rows on a table. Allow your audience to inspect the cards. You can easily add or remove cards (again transparently) before the lottery begins.

Then jumble all of the cards in a dark bag, or a box with a hand-sized hole in the top. Assign an impartial person to draw cards one at a time. Repeat names get discarded.

That would be both low-tech and transparent at every step.

posted by pipeski at 12:18 PM on September 18, 2013 [1 favorite]

If you need to be very transparent, print the names in large, clear text onto index cards (three cards per person) and lay out the correct number of cards clearly in rows on a table. Allow your audience to inspect the cards. You can easily add or remove cards (again transparently) before the lottery begins.

Then jumble all of the cards in a dark bag, or a box with a hand-sized hole in the top. Assign an impartial person to draw cards one at a time. Repeat names get discarded.

That would be both low-tech and transparent at every step.

posted by pipeski at 12:18 PM on September 18, 2013 [1 favorite]

The most transparent, low-tech way is names in a hat, exactly what you have excluded, although I'm not clear on why you think it is not an option. It is good enough for multi-million dollar state lotteries (numbered balls in a big bingo hopper is essentially the same technique). So here's a suggestion:

Choose a time and place to do the drawing. Anyone who wants to show up and observe the proceedings is welcome to do so. You should absolutely have at least 2 other people there to witness the whole thing, people who all the entrants would agree are reasonably disinterested in the outcome. Also, get your webcam or cellphone set up to record the whole thing in one uninterrupted shot, ideally broadcast live online and recorded so it is available for anyone to inspect after the fact. While on camera, and in front of your witnesses, write down everyone's name on a slip of paper. For someone who has multiple entries, they get multiple slips (i.e. if Alice has 3 entries, you create 3 slips with Alice's name on them). After you have all the slips written out, show them neatly arrayed on the table for a few seconds so everyone agrees there are the correct number of slips and names. Put the slips in a hat or bowl or whatever. Mix vigorously. Have one of your witnesses reach in and draw a slip out. You have your first winner. Keep drawing slips to determine your further winners. If you draw a slip with a name that was already drawn, just throw it aside. Keep drawing until you have your 9 unique names/winners.

Important points to pay attention to if anyone is worried about tampering:

- Any of the entrants can attend in person and watch.

- At least 2 impartial witnesses are in attendance.

- Everything is captured on video, live-broadcast, and recorded for later review, in one continuous shot, ideally from a stable camera that is not moved or touched during the drawing.

- The impartial witnesses do the actual drawing.

You should run this procedure by all of your entrants to ensure they all agree this is fair, but this is about as transparent and low-tech as you can get.

posted by stoffer at 12:20 PM on September 18, 2013

Choose a time and place to do the drawing. Anyone who wants to show up and observe the proceedings is welcome to do so. You should absolutely have at least 2 other people there to witness the whole thing, people who all the entrants would agree are reasonably disinterested in the outcome. Also, get your webcam or cellphone set up to record the whole thing in one uninterrupted shot, ideally broadcast live online and recorded so it is available for anyone to inspect after the fact. While on camera, and in front of your witnesses, write down everyone's name on a slip of paper. For someone who has multiple entries, they get multiple slips (i.e. if Alice has 3 entries, you create 3 slips with Alice's name on them). After you have all the slips written out, show them neatly arrayed on the table for a few seconds so everyone agrees there are the correct number of slips and names. Put the slips in a hat or bowl or whatever. Mix vigorously. Have one of your witnesses reach in and draw a slip out. You have your first winner. Keep drawing slips to determine your further winners. If you draw a slip with a name that was already drawn, just throw it aside. Keep drawing until you have your 9 unique names/winners.

Important points to pay attention to if anyone is worried about tampering:

- Any of the entrants can attend in person and watch.

- At least 2 impartial witnesses are in attendance.

- Everything is captured on video, live-broadcast, and recorded for later review, in one continuous shot, ideally from a stable camera that is not moved or touched during the drawing.

- The impartial witnesses do the actual drawing.

You should run this procedure by all of your entrants to ensure they all agree this is fair, but this is about as transparent and low-tech as you can get.

posted by stoffer at 12:20 PM on September 18, 2013

Best answer: In case you want to stick with the dice route. You can roll a random number from 1-36 using two d6. Here's how:

1. Roll the first die. Subtract 1 and multiply by 6. So, 1=0. 6=30

2. Roll the second die. Add the value to that of the first die.

posted by vacapinta at 12:21 PM on September 18, 2013 [1 favorite]

1. Roll the first die. Subtract 1 and multiply by 6. So, 1=0. 6=30

2. Roll the second die. Add the value to that of the first die.

posted by vacapinta at 12:21 PM on September 18, 2013 [1 favorite]

One thing you might consider - if your drawing starts with the cheap stuff and ramps up to the big prize, if one of the multi-entry people is drawn for a first prize and then drawn again for a larger prize, I would immediately offer that person the choice of which prize they want to keep and then draw again to fulfill the uniqueness for whichever one they didn't choose. I would even do it again if their name comes up again for the next big prize and the next and the next. Of course, you do this in full view of everyone and the cameras.

From a fairness and transparentness perspective, I don't really know if that's a good idea, but from the perspective of the person who bought a lot of tickets to have a good chance of winning a good prize, I feel cheated if I win a coffee mug and then I'm not eligible to win the TV even though I currently own the majority of the tickets in the hopper.

posted by CathyG at 12:22 PM on September 18, 2013

From a fairness and transparentness perspective, I don't really know if that's a good idea, but from the perspective of the person who bought a lot of tickets to have a good chance of winning a good prize, I feel cheated if I win a coffee mug and then I'm not eligible to win the TV even though I currently own the majority of the tickets in the hopper.

posted by CathyG at 12:22 PM on September 18, 2013

Best answer: If you have the regular set of rpg dice, you can combine pretty much anything with a d% to get a range of numbers. If you have 35 entries, give every entry a number from 10-49. Roll a d4 and a d10; the 4 is your tens place, and the d% is your ones. Reroll anything that isn't associated with an entry, or any entry from someone who's already won. If you get >40 entries, swap out the d4 for a d6.

posted by specialagentwebb at 12:45 PM on September 18, 2013 [1 favorite]

posted by specialagentwebb at 12:45 PM on September 18, 2013 [1 favorite]

Best answer:

Let's call them oners, twoers, and threers, based on the number of entries they have. Their entries are all split between three groups, A, B, and C.

If A, B, or C don't all have the same number of oners, twoers and threers, there will be a slight imbalance of odds between groups.

Imagine a scenario with 31 oners, one twoer and one threer.

Suppose you put all three threeer entries in group A, the twoers in group B, and group C is all oners.

Your group C will pick three entries, and each oner in group C will have 3/12 odds of being picked.

In group A, you have to re-choose a result that has more than one of the same threer. I'm not figuring out the exact numbers right now, but that means all the oners in group A have slightly higher odds of being picked than the oners in group C.

For group B, the odds of any specific oner being picked will be somewhere between group A and group C.

Now, if there were just oners and threers, you could get all three groups evenly balanced by spreading the threers across the groups, but you've also got twoers, and balancing them will only work if the number of twoers is a multiple of 3.

Plus you've got the problem of there's 35 entries instead of 36, there's going to be a slight advantage to being in a smaller group instead of a larger group.

Also if you spread them across three groups, and you do get the same person come up in more than one group, then you have to pick which group has to re-roll, whichever group you re-roll the oners in that group get an extra shot at the prize, so you have to pick between those groups randomly, I guess.

I think even if you find a way to do the grouping and recombining such that it's fair, and it's transparent in the sense that everyone can see how it works, it might be too complex for everyone to

CathyG has a point that if the prizes are all unique then it will matter which order you give them out. If you want that evenly weighted, you could then do a random draw from those 9 names to assign which specific prize they get.

posted by RobotHero at 1:17 PM on September 18, 2013

*My current plan is to split the entries into 3 numbered groups of 12 (with one or more "free" spaces) and have different unaffiliated people throw a d12 until 3 unique people have been chosen from each group. Will this work? Will it matter whether someone with 3 entries has them all in one list or evenly split between lists? Will it matter if all of the smaller-entry people are in the same group or not?*Let's call them oners, twoers, and threers, based on the number of entries they have. Their entries are all split between three groups, A, B, and C.

If A, B, or C don't all have the same number of oners, twoers and threers, there will be a slight imbalance of odds between groups.

Imagine a scenario with 31 oners, one twoer and one threer.

Suppose you put all three threeer entries in group A, the twoers in group B, and group C is all oners.

Your group C will pick three entries, and each oner in group C will have 3/12 odds of being picked.

In group A, you have to re-choose a result that has more than one of the same threer. I'm not figuring out the exact numbers right now, but that means all the oners in group A have slightly higher odds of being picked than the oners in group C.

For group B, the odds of any specific oner being picked will be somewhere between group A and group C.

Now, if there were just oners and threers, you could get all three groups evenly balanced by spreading the threers across the groups, but you've also got twoers, and balancing them will only work if the number of twoers is a multiple of 3.

Plus you've got the problem of there's 35 entries instead of 36, there's going to be a slight advantage to being in a smaller group instead of a larger group.

Also if you spread them across three groups, and you do get the same person come up in more than one group, then you have to pick which group has to re-roll, whichever group you re-roll the oners in that group get an extra shot at the prize, so you have to pick between those groups randomly, I guess.

I think even if you find a way to do the grouping and recombining such that it's fair, and it's transparent in the sense that everyone can see how it works, it might be too complex for everyone to

**understand**that it's fair.CathyG has a point that if the prizes are all unique then it will matter which order you give them out. If you want that evenly weighted, you could then do a random draw from those 9 names to assign which specific prize they get.

posted by RobotHero at 1:17 PM on September 18, 2013

Here's my easy RPG dice solution.

It requires one 1d20 and any other die, (I use 6 sided in this example). It also requires a very low stakes treat (a mini pack of M&Ms, a K-cup of coffee, Cookies or whatever).

When rolling both dice, if the 6 sided dice is 4-6 add 20 to the 20 sided die, otherwise don't add anything.

This gives you a random number between 1-40.

Number the entries 1-35.

Now roll the dice.

If the number is between 1-35, that person wins something, yay! for them.

If the number is between 36-40 everyone gets a low stake treat.

Roll again

Same rules as above, except if a person wins twice they get a low stakes treat the 2nd+ time they win (or perhaps something a bit nicer, like a $1 Lottery ticket or whatever).

Repeat until you have 9 winners.

Plan on getting 3-5 rolls between 36-40

Plan on getting 2-4 multiple winners

Chances are pretty good that everyone will get a treat, so that might help defuse some of the drama.

Regardless of how you go about it ... Good Luck!

posted by forforf at 4:07 PM on September 18, 2013

It requires one 1d20 and any other die, (I use 6 sided in this example). It also requires a very low stakes treat (a mini pack of M&Ms, a K-cup of coffee, Cookies or whatever).

When rolling both dice, if the 6 sided dice is 4-6 add 20 to the 20 sided die, otherwise don't add anything.

This gives you a random number between 1-40.

Number the entries 1-35.

Now roll the dice.

If the number is between 1-35, that person wins something, yay! for them.

If the number is between 36-40 everyone gets a low stake treat.

Roll again

Same rules as above, except if a person wins twice they get a low stakes treat the 2nd+ time they win (or perhaps something a bit nicer, like a $1 Lottery ticket or whatever).

Repeat until you have 9 winners.

Plan on getting 3-5 rolls between 36-40

Plan on getting 2-4 multiple winners

Chances are pretty good that everyone will get a treat, so that might help defuse some of the drama.

Regardless of how you go about it ... Good Luck!

posted by forforf at 4:07 PM on September 18, 2013

I would definitely just use a pack of cards and write everyone's names on cards the required number of times. If you wanted a more technical solution, you can download R for free, and then write the following code

set.seed(101)

#The number in the brackets needs to be picked randomly. If you get everyone to pick a number from 1-100 and then add them all together, this will defeat accusations of bias reasonably effectively

x=c("bob","sue","james",....,"mike")

#Your list of friends

weights=c(3,1,2,....,1)

#The number of chances each of your friends have

select=sample(x,9,replace=FALSE,prob=weights)

select

#This produces a list of names, chosen randomly and uniquely with weighting equal to their number of chances

But this answer is probably over technical- the card solution will work just as well and provide more drama.

posted by Cannon Fodder at 12:41 AM on September 19, 2013

set.seed(101)

#The number in the brackets needs to be picked randomly. If you get everyone to pick a number from 1-100 and then add them all together, this will defeat accusations of bias reasonably effectively

x=c("bob","sue","james",....,"mike")

#Your list of friends

weights=c(3,1,2,....,1)

#The number of chances each of your friends have

select=sample(x,9,replace=FALSE,prob=weights)

select

#This produces a list of names, chosen randomly and uniquely with weighting equal to their number of chances

But this answer is probably over technical- the card solution will work just as well and provide more drama.

posted by Cannon Fodder at 12:41 AM on September 19, 2013

Response by poster: Cards and/or poker chips are a great idea that I was totally missing. Since I have playing cards I'll probably use them. Barring a card shark, that's the sort of thing that everyone will agree is appropriately random. I'm making a note of some of the dice techniques too as a backup.

The "lottery" is for a chance to be a vendor at my craft group's small annual fair, so there aren't any runner-up or grand prize types of prizes.

Thanks!

posted by tchemgrrl at 10:36 AM on September 19, 2013

The "lottery" is for a chance to be a vendor at my craft group's small annual fair, so there aren't any runner-up or grand prize types of prizes.

Thanks!

posted by tchemgrrl at 10:36 AM on September 19, 2013

This thread is closed to new comments.

Put every entry in a row in Excel so that column A reads like:

Jane

Jane

Tom

Tom

Tom

Sally

Peter

Peter

etc.

Then next to your top entry, put =rand()

Fill down

Copy that column and paste as values

Sort by that column

Then you can remove duplicates to leave each name once in order of their first appearance, or you can find the top 9 names by just like... reading down the list (e.g., if you get Jane and then Jane again, just skip her the second time, and keep going until you have 9 people).

You can do this on a projected screen or something so everyone watches. That allows you to do it all last minute and everyone to watch that everything is on the up and up. And it allows Excel to be your random number generator.

You can do this in Google Docs too if Excel isn't handy... and you can share the doc so people who maybe aren't present (if they're not) can watch it happen also.

posted by brainmouse at 12:02 PM on September 18, 2013