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Posted by colinprince 2 days ago

New type of dice guarantees no tie when deciding who goes first(www.cbc.ca)
84 points | 54 commentspage 2
toast0 4 hours ago|
If you want to buy these, they are commercially available https://mathartfun.com/dSpecial.html

(no affiliation)

I think there have been discussions about some of these sets here as well.

ninjalanternshk 2 hours ago|
What a cool company. They’ve got the coolest nerdiest things.

I’m loving the non-transitive dice. My son’s birthday is in (checks calendar) ten months, but he’s getting these dice now anyway. So cool.

leoqa 4 hours ago||
It didn’t mention the underlying theory? Is it just that these large dimensions reduce the collision probability?
trhaynes 4 hours ago||
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.
ixwt 4 hours ago||
If you had two dice, one with odds 1->11, and the other 2->12, and you made 1 beat 12, they would have even chances of winning then, right?
bombcar 4 hours ago||
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.
margalabargala 4 hours ago||
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?

arjvik 4 hours ago||
This is not exactly perfectly fair - appears that the first player wins 11/27 times and the other two win 8/27 each.
margalabargala 4 hours ago||
Ah. Now that I go through it it makes sense. You're right.
pfedak 4 hours ago||
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.

margalabargala 3 hours ago|||
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 :)

whycome 3 hours ago||
Didn’t they do the exact opposite of suggesting the problem was trivial?
cwillu 4 hours ago|||
Saying “I'm having trouble understanding” is not the same thing as “this is trivial”.
Dylan16807 4 hours ago|||
"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.
margalabargala 3 hours ago|||
That wasn't my intention.

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.

Petersipoi 3 hours ago||
Only the people who can't make eye contact with others interpreted your comment as hostile. Unfortunately, that describes a high number of people on this site.
webstrand 3 hours ago||||
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.
whycome 3 hours ago|||
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.
ChrisArchitect 2 days ago||
Normal url: https://www.cbc.ca/radio/asithappens/dice-mystery-board-game...
VyseofArcadia 2 hours ago||
Oh, Eric is a friend of mine. I've got a set of five non-uniform go first dice that he 3D printed for me.

Dude runs a mean D&D campaign too. And his Lego mural and sculpture portfolio are something to behold: http://www.ericharshbarger.org/lego/portfolio.html

sjrd 4 hours ago||
Better source: http://www.ericharshbarger.org/dice/go_first_dice.html

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?

vova_hn2 32 minutes ago|
> So well, is TFA just a big pile of slop?

It is a pile of slop and I am surprised that your comment is downvoted.

I don't know why would HN crown prefer this pile of journo fluff to the article by, you know, the guy who actually made the thing.

Fricken 2 hours ago||
You can just pull tokens out of a bag to generate a random sequence of virtually any length. Am I missing something? Why make it so complicated?
teo_zero 38 minutes ago|
How do you decide who is going to pull the tokens?
epispencer 2 hours ago|
If anyone wants a practical way to decide who goes first, in a single roll, with no possibility of a tie, this works:

For 2 players, the dealer rolls a standard 6-sided die, and the result determines who goes first:

  1, 2, 3 => Player A
  4, 5, 6 => Player B
For 3 players, the dealer rolls a standard 6-side die:

  1, 2 => Player A
  3, 4 => Player B
  5, 6 => Player C
For 4 players, the dealer rolls a standard d20 die:

  1, 2, 3, 4, 5 => Player A
  6, 7, 8, 9, 10 => Player B
  11, 12, 13, 14, 15 => Player C
  16, 17, 18, 19, 20 => Player D
For 5 players, the dealer rolls a standard d20 die:

  1, 2, 3, 4 => Player A
  5, 6, 7, 8 => Player B
  9, 10, 11, 12 => Player C
  13, 14, 15, 16 => Player D
  17, 18, 19, 20 => Player E
If the dealer has a tetrahedral (d4) die, they can use that instead for the 4-player problem.

This is really all you need to decide who goes first. It's perfectly fair. It uses standard dice that most people already have in their game drawer.

The article is actually about a different, much harder problem about the full playing order. You don't actually need that extra complexity to decide who goes first.