
Nontransitive dice - adenadel
https://en.wikipedia.org/wiki/Nontransitive_dice
======
latkin
You can purchase a set of non-transitive dice here:
[https://mathsgear.co.uk/products/non-transitive-grime-
dice](https://mathsgear.co.uk/products/non-transitive-grime-dice) (I am not
affiliated in any way with this store)

These particular ones are fun because the winning order reverses if you double
the count of dice. i.e. A beats B beats C beats A, but 2 As lose to 2 Bs lose
to 2 Cs lose to 2 As.

I wrote up a blog a while back that explores the various possibilities and
chains using Mathematica: [http://latkin.org/blog/2015/01/16/non-transitive-
grime-dice-...](http://latkin.org/blog/2015/01/16/non-transitive-grime-dice-
via-mathematica/)

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creatine_lizard
This reminded me of Penney's Game
([https://en.wikipedia.org/wiki/Penney's_game](https://en.wikipedia.org/wiki/Penney's_game)).

~~~
fenomas
Freaky! I find that result even more surprising than the intransitive dice.
Thanks for posting.

~~~
allenz
You might find the result less surprising after you solve a riddle by Martin
Gardner:

> A young man lives in Manhattan near a subway express station. He has two
> girlfriends, one in Brooklyn, one in the Bronx. To visit the girl in
> Brooklyn, he takes a train on the downtown side of the platform; to visit
> the girl in the Bronx, he takes a train on the uptown side of the same
> platform. Since he likes both girls equally well, he simply takes the first
> train that comes along. In this way, he lets chance determine whether he
> rides to the Bronx or to Brooklyn. The young man reaches the subway platform
> at a random moment each Saturday afternoon. Brooklyn and Bronx trains arrive
> at the station equally often—every 10 minutes. Yet for some obscure reason
> he finds himself spending most of his time with the girl in Brooklyn: in
> fact on the average he goes there 9 times out of 10. Can you think of a good
> reason why the odds so heavily favor Brooklyn?

The idea shows up again in the Elevator paradox, which has a delightful
article on Wikipedia:
[https://en.wikipedia.org/wiki/Elevator_paradox](https://en.wikipedia.org/wiki/Elevator_paradox)

~~~
tgb
I see an immediate solution to that riddle and it matches the idea of the
Wikipedia page you link. But I don't see any connection to Penney's game. Can
you explain?

~~~
frumiousirc
The operative words in the rules of both riddles are "appears first".

~~~
allenz
Yup, Penney's game is surprising because at first glance the probabilities of
two sequences of the same length are unrelated. But if more than half of the
sequences overlap, then one sequence will tend to arrive before the other. As
the proportion of overlap tends to 100%, Player B has a 2:1 advantage over
Player A: [https://i.imgur.com/eKujwrK.png](https://i.imgur.com/eKujwrK.png)

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amalcon
I find nontransitive dice to be a clear demonstration of the effects of
premature rounding. The nontransitivity is only possible because, after each
iteration, the result is rounded to a victory for one die. If the totals were
summed over time, they could clearly be ranked by expected value.

You can see this result in other places, also. It's especially visible in
sports, for example, or in the stock market.

~~~
CydeWeys
Negative. Non-transitive dice work even when the expected value of each die is
the same. The first example in the Wikipedia article exhibits this:

    
    
       Die A has sides 2, 2, 4, 4, 9, 9.
       Die B has sides 1, 1, 6, 6, 8, 8.
       Die C has sides 3, 3, 5, 5, 7, 7.
    

The expected value of each die is 5, yet A beats B, B beats C, and C beats A.
Other examples in the article have sets of transitive dice where the expected
value is not the same; transitivity still holds.

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amalcon
I'm confused, because this reads to me like exactly what I was trying to say,
but you seem to be disagreeing with me. Could you clarify?

In the game with rounding, A beats B, B beats C, C beats A. In the game
without rounding, where totals are summed, they are evenly matched and it's
down to chance. That's exactly the effect I was referring to.

~~~
CydeWeys
There are many sets of non-transitive dice where the average is not the same
though (I should have used one of those examples). Here's one:

    
    
       A: 4, 4, 4, 4, 0, 0 (avg: 8/3)
       B: 3, 3, 3, 3, 3, 3 (avg: 9/3)
       C: 6, 6, 2, 2, 2, 2 (avg: 10/3)
       D: 5, 5, 5, 1, 1, 1 (avg: 9/3)
    

There's no reason that the average values need to be the same to have non-
transitive dice. To modify the original three to have the same winning
properties, but radically different averages, just imagine doing this:

    
    
       Die A has sides 2, 2, 4, 4, 99, 99.
       Die B has sides 1, 1, 6, 6, 8, 8.
       Die C has sides 3, 3, 5, 5, 7, 7.
    

The expected value of an individual die roll doesn't play into it at all.

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amalcon
It doesn't play into this game _because of the rounding_. Again, you seem to
be attempting to disagree with me, by echoing the exact same things I just
said.

~~~
tripzilch
That's because the second person that disagrees with the same thing is
expected to be more right on average.

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dws
The non-transitivity of Rock-Scissors-Paper is easy to understand, partly
because it's so simple, but mostly because you're likely never played outside
the usual rules, even if adding Lizard-Spock.

Non-transitive dice screw with the 'nature' of dice that most of us expect. To
get to the mathematical intuition, one may have to get past a deeply-ingrained
feeling that something about these dice just isn't right. That's a big part of
the fun.

~~~
kybernetikos
I think the numeric and gambling aspect helps too. If you're used to
probability, you probably start thinking about expected value automatically in
these situations, but it isn't the case that the die with the highest expected
value wins most of the time against a die with lower expected value. (e.g.
2,2,2,2,2,2s vs 1,1,1,1,1,100).

~~~
hammock
Mean EV doesn't work because the dice have different variances? Trying to wrap
my head around the rationale here

~~~
hawkice
Variance isn't the best lens here, better to look at the distributions. It's
also a discrete problem, so EV isn't as precise as looking at all the possible
combinations.

~~~
hammock
Variance describes distribution...

Discreteness is another layer. What are the exact defining characteristics of
this problem though? You could take two dice 1-2-3-4-5-6 and 2-3-4-5-6-7 and
the typical mean EV calculation would work fine...

1-2-3-4-5-7 and 2-3-4-5-6-8 have different variances but are probably not
intransitive (haven't checked the math) since they are translations of each
other

I'm just thinking out loud here and trying to narrow it down...maybe someone
reading wants to help me :)

~~~
hawkice
Variance does describe the distribution, but not well enough, in this case. It
reduces the vector of possibilities to a scalar, and hides the most important
aspect, the non-transitivity. Numbers have transitive comparisons, so there's
essentially no way a scalar value _could_ capture that in any straightforward
way. Basically, abandon all single-value summaries. They'll do you no good in
understanding this phenomenon. But the detailed analysis isn't too
complicated, so it's okay.

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adenadel
Brought to my attention from Timothy Gowers' blog

[https://gowers.wordpress.com/2017/04/28/a-potential-new-
poly...](https://gowers.wordpress.com/2017/04/28/a-potential-new-polymath-
project-intransitive-dice/)

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diminoten
Does anyone have any board game recommendations that take advantage of this
dice configuration?

~~~
qznc
Good question. It seems this non-transitive games are suited to trick people,
but can you make skill-based game out of the mechanics?

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Frondo
This video is where I first learned of nontransitive dice:

[http://www.youtube.com/watch?v=u4XNL-
uo520](http://www.youtube.com/watch?v=u4XNL-uo520)

Grime dice! And James Grime seems like such a fun, happy, charismatic guy, I'd
love to have a beer with him and play dice with him, even knowing the outcome.
:)

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iamwil
Or if you have a 3D printer, you can print out your own.

[http://www.thingiverse.com/thing:17782](http://www.thingiverse.com/thing:17782)

I made it a while back. There's 3 sets. Efron's dice, Miwin's dice, and
Grime's dice.

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purplejacket
OMG, I just finished working through an example of this with my student, like
literally minutes ago, and look what Hacker News brings me ...

~~~
bbcbasic
Well what were the odds of that, eh?

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spapas82
It was a little difficult to believe that such a property (dice being
transient) was possible. But a little simulation in python 2 (using the
Wikipedia example) makes everything crystal:

[https://gist.github.com/spapas/9b29e1dec752b85fc1d6ad660af7f...](https://gist.github.com/spapas/9b29e1dec752b85fc1d6ad660af7fef3)

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agumonkey
Reminds me of political system bugs where a candidate might win only by
transitivity.

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diegoperini
Quantum Rock/Paper/Scissors :)

