Mathematicians roll dice and get rock-paper-scissors
quantamagazine.org
quantamagazine.org
Except that this is a definition of "stronger" quite different from the one in common use:
> [Die] A beats [die] B on 58% of rolls.
> B beats C on 58% of rolls.
> C beats A on 69% of rolls.
By any conventional metric, one of these dice is the strongest, and it's C. If you released something like this into the wild, you'd expect to come back later and find that C had done the best and A had done the worst.
import random
dice = [
[3,3,3,3,3,6],
[2,2,2,5,5,5],
[1,4,4,4,4,4]
]
wins = [0,0,0]
for t in range(10000):
die1 = random.randrange(3)
face1 = random.randrange(6)
die2 = random.randrange(3)
face2 = random.randrange(6)
result1 = dice[die1][face1]
result2 = dice[die2][face2]
if result1 > result2:
wins[die1] += 1
elif result2 > result1:
wins[die2] += 1
print(wins)
edit: The reason C and B look equally good in the code above is that B gets many fewer draws than the other dice do (because it's split 3/3 whereas they're both split 5/1). Those draws that don't happen are a win for B and a loss for [the other] B, but if you only count the win, it looks like B is benefiting. B wins about the same amount as C does, but it loses a lot more. A loses about the same amount as B does, but it wins a lot less.
Intransitive dice are still surprising and interesting to me, I Wonder if there are 3 such dice which are equally strong when played at the same time ?
Another fun dice concept are go first dice: https://en.m.wikipedia.org/wiki/Go_First_Dice
Contrast with a scenario where two dice are our normal 1-6 flavor and one die is all 6’s. In that case the all-6 die is stronger because it doesn’t matter which other dies it goes up against.
But there is a die you can pick that is uniformly better than any alternative. That die is C. If your opponent is playing correctly, C will give you the optimal outcome. And if your opponent isn't playing correctly, C will also give you the optimal outcome. This is not true of the other two dice.
> If Gates had chosen first, then whichever die he chose, Buffett would have been able to find another die that could beat it (that is, one with more than a 50% chance of winning).
It seems like you're thinking about this outside of the relevant context, which is a math problem, loosely inspired by real life games.
What's optimal about that is that it's the best outcome you can theoretically achieve. That's what "optimal" means.
The counterintuitive part is that the first player cannot win more often than they lose.
The point they are making is that the three dice don't satisfy the transitive property.
A > B and B > C, but A is not > C.
On a related note, the mechanic where your opponent gets foreknowledge of your strategy and they're free to make whatever changes they want, but you're not allowed to change your strategy and you don't get any knowledge of theirs, isn't really relevant to much.
Even with that mechanic, C is very clearly the strongest die. It's just that someone who chooses C first will lose as badly as someone who chooses B first. (And much less badly than someone who picks A first.) But someone who chooses C second will win by much larger margins than anyone else can.
Never? I suggest doing a survey of some people.
Example 1) You're in a prison block with 100 inmates and 20 gaurds who are all average-sized and skill humans except for 1 inmate who is Mike Tyson (in his prime, the heavy weight boxing champion of the world). The prison warden says "We're going to have a boxing match. The prisoners may select a fighter to represent them from the inmates, and I will select a fighter from the guards. If the prisoner fighter wins, everyone goes free. However, if the guard wins, we add 2 years to everyone's sentences."
Here is just one case where it's very important to know that Mike Tyson is statistically the "strongest" (fighter in this case) over any other individual human candidate (gaurd or prisoner) in the set of candidates.
Example 2: A small zoo has only 3 types of animals: an Elephant, a monkey, and a wolf. Each animal is pregnant and will have a baby this month. The zoo attendants have a $1000 pool going and you have to bet on which baby will weigh the most when born. In this case, you clearly wager on the elephant as they are statistically "the strongest" over all other individual candidates with respect to the body mass of their offspring.
Example 3: Back to your tennis example (I play and follow tennis, not that that matters much). Hyperbollically, if you have 5 players and each plays each other 100 times, and player #3 beats all other players 90/100 times, whereas all other matches result in a 50/50 split. In this case, yes, player #3 is the strongest player and is favored to win over any of the other 4 players in the set in a head-to-head match.
Yes, very often we rate and characterize the "strongest" as being the candidate that is statistically more likely to beat any other individual candidates in a set.
There are some games that have a single property that is clearly sortable, like weight. In this case clearly the there is one winner. (Actually, here you have a distribution of weight, like the average and the dispersion. So it may be more complicated.)
There are more complicated games. When I was young we had cards with the properties of cars, like weight, length, price, and other stuff. The game was to pick the card at the top and pick a property, and compare with the same property of the top card of the other player. Some cards were stronger than other. I don't remember the strongest card, but I guess there was one. Anyway, that card didn't win against all cards, it was just good against most card, but you could be unlucky if it was at the top of your deck and the other player selected one of the bad properties of the best card.
So it approximates wins = 10000 * 2/9 * [58%+31%, 42%+58%, 42%+69%] = 20000/9 * [89%, 100%, 111%].
(Note these percentages are rounded, e.g 21/36 to 58%).
This is just averaging the win and loss percentages of each coin against the 2 others, not "a conventional metric" like expected value of a roll which are all the same as a regular die (3.5).
Note that your math is incorrect; B gets more wins from being paired against itself than either A or C do.
edit: here are results from sets of 1000 single-elimination 256-man tournaments where everyone brings a random die (in this case, draw/draw is equivalent to win/loss):
[140, 424, 436]
[144, 394, 462]
[160, 417, 423]
[123, 419, 458]
[174, 411, 415]
[150, 397, 453]
Obviously, in reality, people wouldn't bring random dice to this event. More people would bring C dice, because C is stronger than the other two.
All you need to do is look at each pair of dice and compare. Especially for some sets of dice the comparison is trivial. Like look at the ones in the article with five 3s and five 4s, you don't even need to think to figure out which one wins more.
See for example: http://www.borrett.id.au/computing/petals-bg.htm or the umpteen versions of http://math.bme.hu/~petz/vnsumming.html
I cannot wait until my child is old enough to be fooled by them.