The big open question is whether a set of 5 (permutation fair) 30-sided dice exists.
I’ve been working on that on and off since 2012. I picked it up again about a month ago and have made dramatic speed improvements to my search, but exhausting the whole space I’m searching will still take my computer an estimated 70 years.
But if you had those 3 dice but only 2 players you could not just have each player grab one of them and roll. If one of them happened to grab #1 they would only win 1/3 of the time instead of the desired 1/2.
With 2 players they would have to use just #2 and #3.
That's because the way I came up with those numbers is as follows.
1. Number the players 1, 2, and 3. We want #1 to win exactly 1/3 of the time. We could do that by given them a 3-sided die 1 1 H1, where all the numbers on the other dice are lower than H1 and higher than 1.
2. In the cases where #1 rolls 1, we want #2 to win half the time. Give them a 2-sided die 2 H2 where the remaining die has all numbers between 2 and H2.
3. Assuming the remain die is also 2-sided we will need a total of 6 different numbers. Using 1-6 our set of dice is (1 1 6), (2 5), (3 4).
4. Most people would probably prefer that they all have the same number of sides instead of 3, 2, 2. LCM of those is 6, so double the 3-sided and triple the two 2-sided: (1 1 1 1 6 6), (2 2 2 5 5 5), (3 3 3 4 4 4).
5. People might object to having the same number more than once on a die. We have 18 total sides so lets renumber from 1-18. Our 4 1s become 1-4, our 3 2s become 5-7, and so on, given the set of 3 6-sided dice at the start.
It seems pretty clear that this generalizes to more than 3 players, with the more players the more sides the dice will have. But all of those suffer from that annoyance of needed to exclude specific dice when you are trying to decide the starting order for less than the maximum number of players.
Do the dice in the article avoid that annoyance? I have no idea how I would go about making something like that.
Also note that my dice only determine who goes first. It would be really nice if they could be used to determine complete order. Mine fail for that because #1 is always either the highest or the lowest.
It would be possible to use #1s number on a losing roll to give their place: 1 2 means they go second and 3 4 they go third. You could even print something on the dice saying that, but I think most people would find it more elegant if it was a simple highest goes first, second highest second, and so on.
Do the dice in the article do that, or are they also just solving the who goes first problem?
Can somebody explain why not just make a die with 5! sides, and roll it once to decide the order? With each side having a unique order printed e.g. 12345 -> 12354 -> ...
Especially, that 120-sided dice are already invented and commercially available.
So it was more of a physical problem rather than a mathematical one [1]:
> Harshbarger says he and his colleagues always knew the dice were mathematically possible.
> The mystery was whether that mathematical solution could be translated into the physical geometry of a die — something that could actually be manufactured and rolled.
> “I knew there was a solution with something crazy like 1,440 sides for each die,” he said. “That's not makeable.”
I think it is not a matter of whether it is more of a "physical problem" or a "mathematical one". They knew a solution existed but it was physically unfeasible, which prompted them to mathematically optimize the solution by using fewer sides. They found a solution using only 120 sides, and a lower bound of 30 is known but brute-forcing it is still computationally expensive.
I think it's more about guaranteeing the whole sequence, rather than who goes first?
At least for two players, if you use a two sided die (a coin), have player one win ties on ones, player two win ties of twos - and otherwise highest wins - then that is trivially done?
I would have to do a little more math to see if it generalizes by induction... I'm not sure you would get a guaranteed sequence - but I think at least guaranteed fair winner works by just increasing the die (7, 9 and 11 would be tricky because if physics again... I suppose. Unless you just ignore highest tie for missing player (reroll on extremely rare 9 9s on a d10 for nine players)?
Ed: I suppose we break smaller ties, by letting closest and highest win (for ten players, 4, 6 and 7 roll 5 - 6 is closest and over/highest of the close players to 5, then come 7?)
Ed2: nevermind we end up biased towards "high" players that often win on "high" ties, like 5 or 6.
> I think it's more about guaranteeing the whole sequence, rather than who goes first?
What do you mean? The question is who goes first.
As a matter of practice, what happens in a board game is that everyone takes a position around the board before choosing who goes first. If turns proceed in a fixed sequence, that position will determine the sequence. If the order of turns is specified by the game (for example, many feature a turn order track), then that order will be used. You never need to decide on a sequence longer than one person.
But even if that wasn't the case, the article couldn't be more explicit:
> Eric Harshbarger was asked by a board game designer if he could come up with dice that would determine who goes first — without the possibility of a tie.
> The idea was simple: settle the first turn quickly and get on with the game.
This from the article appears somewhat questionable:
>> “It was a question that did not have an obvious answer and that's something that a mathematician will often jump at.”
The problem they're bragging about solving is using dice to quickly and unambiguously select one of five options with equal probability.
The obvious answer should be that you roll a single 10- or 20-sided die, divide by 2 or 4, round up, and there you have it.
I think I'm having trouble understanding why this is so complicated/requires so many sides.
If I imagine a 3-sided die, for simplicity, you should be able to have this result if the sets are [1,5,9],[2,6,7],[3,4,8]. And so on for larger numbers of players. Why doesn't this work?
You might want to check that your proposed solution works at all before suggesting the problem is trivial. What's the probability that the first die goes first with your numbers?
You also can't generally "and so on" constrained combinatorial arrangements like this.
I certainly would have, had I suggested the problem trivial! I'll keep that in mind for if I suggest things are trivial rather than suggesting I lack understanding.
> You also can't generally "and so on" constrained combinatorial arrangements like this.
I know you can't generally but in the specific case I proposed you can (3 sets of 3, 4 sets of 4, 5 sets of 5, each dice taking one ordinal of each set).
You might want to check that your proposed generalization criticism applies to the generalization at all before suggesting the arrangement doesn't work :)
"I'm having trouble understanding why this is so complicated" could be interpreted in a non-dismissive way but it defaults to dismissive. And when it comes with a supposed solution attached, and that solution is really simple, that reinforces it sounding dismissive.
I by no means was suggesting that the trivial solution I, a non-mathematician, thought up in 20 seconds was somehow out of reach to a math professor who spent years on the problem. I knew I was wrong.
I didn't see why until I actually went through the solutions by hand.
Why does that default to dismissive for you? It absolutely did not come off that way for me and I’m wondering if it’s because of experience or culture or something else.
I didn't read it as dismissive. I think the "Why doesn't this work?" at the end is key, implying they know they're probably wrong, but don't know why. It's pretty common to form and present hypothesis like this, hoping that anyone who knows the actual theory can easily provide a counterexample showing why it's wrong. It's not at all intended to be dismissive.
My naive first idea was that (for two people) one set would have even numbers and the other odds. But then the even number person is more likely to win. So it's something around which numbers are on which dice.
I was wondering that too - it seems you start with a linked list of sorts, and then evenly distribute the links to the dice. But I’ve obviously not thought it out.
* https://en.wikipedia.org/wiki/Go_First_Dice
As well as the pages of the project:
* http://gofirstdice.ericharshbarger.org/
A physical example of dice (USD 35):
* https://www.mathartfun.com/thedicelab.com/GFD5.html
* https://www.youtube.com/shorts/yMtTqiAhol8
* UK store: https://mathsgear.co.uk/collections/dice/products/go-first-d...
In addition to the above 5-player go first, they also have 4- and 3-player go first:
* https://www.mathartfun.com/dSpecial.html
I’ve been working on that on and off since 2012. I picked it up again about a month ago and have made dramatic speed improvements to my search, but exhausting the whole space I’m searching will still take my computer an estimated 70 years.
With 2 players they would have to use just #2 and #3.
That's because the way I came up with those numbers is as follows.
1. Number the players 1, 2, and 3. We want #1 to win exactly 1/3 of the time. We could do that by given them a 3-sided die 1 1 H1, where all the numbers on the other dice are lower than H1 and higher than 1.
2. In the cases where #1 rolls 1, we want #2 to win half the time. Give them a 2-sided die 2 H2 where the remaining die has all numbers between 2 and H2.
3. Assuming the remain die is also 2-sided we will need a total of 6 different numbers. Using 1-6 our set of dice is (1 1 6), (2 5), (3 4).
4. Most people would probably prefer that they all have the same number of sides instead of 3, 2, 2. LCM of those is 6, so double the 3-sided and triple the two 2-sided: (1 1 1 1 6 6), (2 2 2 5 5 5), (3 3 3 4 4 4).
5. People might object to having the same number more than once on a die. We have 18 total sides so lets renumber from 1-18. Our 4 1s become 1-4, our 3 2s become 5-7, and so on, given the set of 3 6-sided dice at the start.
It seems pretty clear that this generalizes to more than 3 players, with the more players the more sides the dice will have. But all of those suffer from that annoyance of needed to exclude specific dice when you are trying to decide the starting order for less than the maximum number of players.
Do the dice in the article avoid that annoyance? I have no idea how I would go about making something like that.
Also note that my dice only determine who goes first. It would be really nice if they could be used to determine complete order. Mine fail for that because #1 is always either the highest or the lowest.
It would be possible to use #1s number on a losing roll to give their place: 1 2 means they go second and 3 4 they go third. You could even print something on the dice saying that, but I think most people would find it more elegant if it was a simple highest goes first, second highest second, and so on.
Do the dice in the article do that, or are they also just solving the who goes first problem?
Especially, that 120-sided dice are already invented and commercially available.
> Harshbarger says he and his colleagues always knew the dice were mathematically possible.
> The mystery was whether that mathematical solution could be translated into the physical geometry of a die — something that could actually be manufactured and rolled.
> “I knew there was a solution with something crazy like 1,440 sides for each die,” he said. “That's not makeable.”
[1] https://www.cbc.ca/radio/asithappens/dice-mystery-board-game...
At least for two players, if you use a two sided die (a coin), have player one win ties on ones, player two win ties of twos - and otherwise highest wins - then that is trivially done?
I would have to do a little more math to see if it generalizes by induction... I'm not sure you would get a guaranteed sequence - but I think at least guaranteed fair winner works by just increasing the die (7, 9 and 11 would be tricky because if physics again... I suppose. Unless you just ignore highest tie for missing player (reroll on extremely rare 9 9s on a d10 for nine players)?
Ed: I suppose we break smaller ties, by letting closest and highest win (for ten players, 4, 6 and 7 roll 5 - 6 is closest and over/highest of the close players to 5, then come 7?)
Ed2: nevermind we end up biased towards "high" players that often win on "high" ties, like 5 or 6.
What do you mean? The question is who goes first.
As a matter of practice, what happens in a board game is that everyone takes a position around the board before choosing who goes first. If turns proceed in a fixed sequence, that position will determine the sequence. If the order of turns is specified by the game (for example, many feature a turn order track), then that order will be used. You never need to decide on a sequence longer than one person.
But even if that wasn't the case, the article couldn't be more explicit:
> Eric Harshbarger was asked by a board game designer if he could come up with dice that would determine who goes first — without the possibility of a tie.
> The idea was simple: settle the first turn quickly and get on with the game.
This from the article appears somewhat questionable:
>> “It was a question that did not have an obvious answer and that's something that a mathematician will often jump at.”
The problem they're bragging about solving is using dice to quickly and unambiguously select one of five options with equal probability.
The obvious answer should be that you roll a single 10- or 20-sided die, divide by 2 or 4, round up, and there you have it.
(no affiliation)
I think there have been discussions about some of these sets here as well.
If I imagine a 3-sided die, for simplicity, you should be able to have this result if the sets are [1,5,9],[2,6,7],[3,4,8]. And so on for larger numbers of players. Why doesn't this work?
You also can't generally "and so on" constrained combinatorial arrangements like this.
> You also can't generally "and so on" constrained combinatorial arrangements like this.
I know you can't generally but in the specific case I proposed you can (3 sets of 3, 4 sets of 4, 5 sets of 5, each dice taking one ordinal of each set).
You might want to check that your proposed generalization criticism applies to the generalization at all before suggesting the arrangement doesn't work :)
I by no means was suggesting that the trivial solution I, a non-mathematician, thought up in 20 seconds was somehow out of reach to a math professor who spent years on the problem. I knew I was wrong.
I didn't see why until I actually went through the solutions by hand.
TFA claims it's "new" in 2026, but the current state of the art seems to still be that of 2022.
I bought actual dice like these in 2024 from https://mathsgear.co.uk/collections/dice/products/go-first-d...
So well, is TFA just a big pile of slop?