Once again, Tadashi-sensei explains a complex concept well before ever referring to it so the audience isn't disengaged by complicated sounding terms. He is a great educator.
One of the more profound quotes by an undergraduate professor that taught me a few years ago: "The problem with real life is that friendship is not transitive."
I once made an argument in one of my sketchbooks that the closer human relationships were to following equivalence relations, the stronger those relationships are. It's been many years and many ponderings, but I still think there may be some truth to it.
to be precise he has a set of five that have 2 different cycles. one of the two gets reversed if you play with two dice instead of one. this makes it fun because you can play with two other players and let them choose die so long as they let you decide how many dice will be used on the roll. you are guaranteed to have a stronger chance to win no matter what they choose. he made me wonder if it's possible to do it for more players (you plus 3 others). haven't found a solution but it's fun to try.
I love the way he uses language at times. "Careless enough to show it's 0" also gave me a new perspective on probability! I feel like I gain so much from his videos.
9,9,9,9,0,0 - same total value, but beats yours (6,6,6,6,6,6) This really could be an interesting game mechanic, if you distribute a fixed amount of points between a set of values, since it has nice rock-paper-scissor behavior. 10,10,4,4,4,4 - same value again, beats the above, but loses against yours. If we continue... 10,10,7,7,1,1 - beats all of the above, but... 8,8,8,8,2,2 - beats that, but not the others except for the 6's. Now.. given a fixed value, which distribution of points beats the most other configurations?
Dolkarr How does 10,10,4,4,4,4 beat 9,9,9,9,0,0? wouldnt it lose.... nvm it wins 55% So let's go with 36 as the number of given points to distribute. What combinations would be the best? 9,9,10,8,0,0 - beats 6,6,6,6,6,6, and ties 9,9,9,9,0,0, I'm too lazy to do the rest 11,11,11,2,1,0 - another combination that can beat the 10,10,4,4,4,4 *metagame intensifies*
Probably because 10 is not much compared to gazillion, which is supposed to be some very large number, so it doesn't really make any difference, thus putting it before the gazillion is meaningless and therefore funny, just like shooting a single bacteria from a giant cannon.
It doesn't have a fixed number of zeros. A gazillion is defined as "the smallest positive finite power of ten that is effectively infinite for practical purposes", i.e. slightly bigger than anything else you'll be dealing with, but not actually infinite. It's part of the family of numbers known as "Parker constants".
You could build strange magical weapons that use these weird dice instead of the ordinary roll. Maybe a shortsword from Mechanus always rolls the average of 3, for example. You could also have the dice be magical objects themselves in-fiction, and rolling certain faces on certain dies would have specific effects. Then you just give the actual dice set out as a prop ;)
@@safrprojects yeah that's what I wish had been mentioned. It's obvious that one of the two comparisons that weren't made in the video (333333 vs 555111) is equal, but the other one isn't as obvious. I had to do the math myself to see if it was equal.
"A is stronger than B and B is stronger than C but it's not necessarily means that A is stronger than C" I might use this example in a certain situations.
Love this video. I was just talking to someone the other day how logic and common sense are not the same thing, and sometimes logic completely contradicts common sense. This is a great example of that. If A is better than B, and B is better than C, then common sense would dictate A is better than C, but logically you cannot be certain about such a claim.
I like his final point. So many people think not paying attention to math growing up just means they will need a calculator. They don't realize how stuff like this seeps into so many things in life.
This has been known for a long time for pokémon enthusiasts. Charizard beats Venusaur. Venusaur beats Blastoise. But no-one would conclude that this means Charizard beats Blastoise!
i always knew that soccer (or any other sport) tournaments don't mean anything. depends on which team you play; if your strengths match the weaknesses of your opponents.
A video about non-transitive dice on Numberphile, without everyone's favorite star, Dr. James Grime?? Not even a mention, or a link? Cmon Brady... Tadashi is great, but Dr. Grime has a great set of non-transitive dice named after him.
1. There is a link to BUY Grime dice in the video description 2. I've been asking Dr Grime to a do a Grime Dice video with me for YEARS without success.
James Grime definitely made a video about non-transitive dice at some point. I was going to comment that he did it on Numberphile, but seeing this thread it makes more sense that he did it on his own channel. It also appears he did another video specifically about Grime dice which is actually shown on that Mathsgear website on the product page. I could understand why he wouldn't care to be redundant on the topic.
vaego haha before I researched those dice I had assumed it's some set of 10^25 dice with a dimension of 2^363728949-1 or something like that because that's often the scale of mathematical discoveries lol.
The 6,2 dice is the best though because it's the only one statistically stronger than 2 other dice... at least until people start using the all 3 dice to try and win the meta game.
To answer that last part: If all 4 dice were pitted together in one final battle to the death, The 6 and 5 die are the best with a 33% chance of winning. The 4 die has a 22% chance of winning, and the 3 die has an 11% chance of winning (don't pick that one!)
Thank you for saving my brain from a cognitive dissonance meltdown. I couldn't comprehend his analogy with drugs and implying that C>A>B>C. That's not possible with discrete data.
Here's a game for you: Two players discreetly select a die in order. The first player receives what they selected. If the second player requested the same, they pick again (knowing which die the first player chose). Since 6,2 is the best in a battle royale, and player-A knows this, but if player-B chooses the same, then for their second pick they're guaranteed to pick all-3's... so player-A could set an ambush by choosing all-3's. Unless the second player tries this as well and then learns to choose 4,0. So player-A might as well set a trap by starting on 4,0. But it will lose if player-B starts on 6,2. From a random Monty-Hall problem standpoint, it looks like player-B will win more often, but something tells me that in applied setting it would balance toward Rock-Paper-Scissors (50%).
This reminds me of the False Positive Paradox, for some reason; although, I don't think it's actually related. But it gives me the same sort of pleasure when the counter-intuitive understanding finally clicks in my mind.
I object to the analogy with medicine. The quantity, percentage, etc. of people cured can be expressed by a number, so of course a>b>c implies a>c. The situation is quite different than that of these interesting dice.
Some of the concepts in this shed some light on the paper i did as an undergraduate. I now appreciate both this video and my own work as well. Now if I only had the time, I would have used some of the ideas here to simplify my calculations massively.
I would love a follow-up to this comparing the dice in Super Mario Party. Each character has a unique dice to give them all a different feel. It would be interesting if you could determine who's the strongest!
I noticed something when I compared the dice to their non-adjacent die. 333333 vs 555111 is even odds, half the time you win and half the time you lose. But with 662222 vs 444400, 662222 has a slight edge over 444400.
It can't be even. Both dice could be mapped onto 3 sided dice without changing the probabilities. When you roll 2 d3, there are 9 possible outcomes which is an odd number. If there are no draws then one must win more than the other. 662222 beats 444400 5/9ths of the time.
662222 has a 1/3 probability to roll a 6, which always wins. It has a 2/3 probability to roll a 2, in which case it wins 1/3 times. 1/3 + (2/3)*(1/3) = 0.555... 662222 indeed has a slight edge.
You might like to map it out. Just did it for interest because i like to see well balanced games too. The dices with the 5 and 6 win over 2 other dices. But the dices with the 3 and 4 only win over one other dice. That is still useful for games. The 3 and 4 can be healer and tank like in mmos. The 6 and 5 range and melee attack.
My math teacher in highschool was educated in philosophy, but he said that the two subjects are so closely related, that one compliments the other in a surprising amount of ways.
That Guy Nah, friendo, he had a point. A lot of real logical problems can be quantized with mathematics. It doesn't even have to be particularly complicated. Besides, it's not as if he wasn't educated in math as well. He had a masters in philosophy, and a bachelor in mathematics, so he had his credentials in order.
To sum up what he said at the end of the vid: A could equal 2/6. B could be greater with 3/6. If B is better than A only by 1/6 and B is greater than C, then C could be less than or equal to A.
When I saw this video, my first thought was to design a rock-paper-scissors type game around it. I immediately came up with a game in which two people randomly draw a die from the set of four presented in this video, and then roll their selected dice to determine the winner. Unfortunately, I found that in a series of random matchups, the [6/2] die is overall the strongest. [4/0] beats [3], odds 2:1 [5/1] beats [4/0], odds 2:1 [6/2] beats [5/1], odds 2:1 [3] beats [6/2], odds 2:1 [3] and [5/1] are matched 1:1. [6/2] beats [4/0], odds 5:4. Based on the above, we can see that [3] and [5/1] have average performance; they are weak to one die, strong against another, and evenly matched by another. But [6/2] performs slightly better than all the others; it is weak to one die, but strong against the other two dice. Conversely, [4/0] performs slightly worse than all the others; it is strong to one die, but weak against the other two dice. But luckily, there's a way to fix this! You can even out this power balance if you make a small alteration: change the [4,4,4,4,0,0] die to a [4,4,4,4,0,X] die. X, in most cases, is equivalent to a 0, and loses to any other roll... *except* a 6, which it trumps. In a matchup between this die and [6/2], the chance of rolling 6 vs. X is 1-in-18; this alters the odds exactly enough that each die is equally likely to win against the other. So, by changing one of the 0's to an X, and giving the X the properties stated above, you have a game in which every die is strong against one, weak against another, and equalled by another. A perfect balance. ...Okay, so it would be a pretty boring game on its own. But you could use it as the basis for a tabletop RPG's combat system!
@@avocette Ah, that's where you've confused yourself and confused me. The 6/2 dice is not 666622, but rather 662222 Against 44440X, it'll work like this: 4 dice rolls a 0 with 1/6 chance, guaranteed loss. Rolls an X with 1/6 chance, wins with 1/3 chance if 6 dice rolls a 6. Rolls a 4 with 2/3 chance, wins with 2/3 chance if 6 dice rolls a 2. So let's see if the math checks out this time: 2/3 * 2/3 is 4/9. That's just under a half. This means that the X roll has to be at least 1/18, which will put it at 1/2 chance to win or lose. 1/6 * 1/3 is 1/18. Oh, that's why. The original comment states that the 4 dice is strong against 1 and weak to 2. This change does, indeed, make it strong against 1 and weak against 1, with it now being equal to the 6 dice.
I love watching Tadashi videos. He is full of fascinating insights and thoughts. Discovered him before I saw him on Numberphile though, doing a series of lectures in Cape Town at the African Institute for Mathematical Sciences, lecturing on topology.
"662222" is the strongest of the four dice. It beats "333333" one third of the time (its weakest game). It beats "555111" two thirds of the time (its strongest game). It beats "444400" 55.56% of the time. That makes "444400" the weakest of the four dice, since it loses more often to its similar strength opponent.
6 is the strongest because in the entire, uniterated meta-game it beats 4 and 5 where 3 ties with 5. So, while the loop is a non-transitive tournament, there are still two more edges to the full graph. It is in these two edges we can make a ranking. Additionally, if we zoom out and consider pick rates for the prior information, even when considering this non-transitive loop, we can rank the four dice as equally strong, according to the definition given.
@@reesverleur9546 Sure. 6 beats 5 as show in the video. 6 beats 4 because: The probability of A rolling 0 and B rolling 6 = 1/3*1/3 = 1/9 P of A=4 and B=6 = 2/3*1/3 = 2/9 P of A=0 and B=2 = 1/3*2/3 = 2/9 P of A=4 and B=2 = 2/3*2/3 = 4/9 As you can see, only one of these events ends up with A winning, event #4 which has a probability of 4/9, less than 0.5. Thus, A is weaker than B. 3 is equally strong as 5 because 3 is a constant die and the 1's and 5's are equally probable on the 5 die. Hence, a both dice are equally likely to win when head-to-head.
I don't think he meant just dice, but rather people following this channel tends to follow James Grime and Matt Parker and they also did similar videos. It can sometimes be hard to remember which channel you saw it on, because of the similarity, hence the confusion and the comment.
There are several videos on UA-cam with people who often appear on numberphile showing off nontransitive dice, but none on the actual numberphile channel. I was confused at first too.
The same flag went up for me even though I've been living in Japan for over 6 years. Tadashi is Japanese and he almost definitely developed that habit before making his way west. In Japan, pop culture from the west has informed Japanese (and other countries, too, I imagine) that the middle finger has a negative connotation, so everybody here knows that it's not good to toss your lone middle finger up at someone, but in using it for anything else, it goes totally unnoticed because it doesn't have the same meaning in Japanese culture. Actually, being the longest finger, it's logically the most convenient and useful to use for a number of tasks, including pointing, referencing, and glasses up-pushing. :D It took a while to get used to, but now I've gotten quite used to it. In this case, it was difficult to miss, though. hahaha
+ItchyKneeSon I, a North American, also use my middle finger to push up my glasses, but I have _all_ my fingers extended when I do so. XD +Artm Borisovskiy I didn't notice it either. XD
This can be used in say....paradox interactive games, where simulations show that one type of army composition kills more men than the other, but doesn't necessarily win as often...A niche case, but it can happen. Usually more losses means a much lower chance of winning, but when looking at simulation results, one should try to find sweetspots to avoid a too qualitative army and instead go for More army elsewhere. I hope i'm making some sense. There's also dice involved, but finding 2 identical armies with 2 identical dice modifiers is rare, and winning/losing tends to snowball based on too many variables.
Something interesting I noticed: Each winning die wins 2/3 of the time, in all four cases. Similarly, opposite dice (333333 and 555111, 444400 and 662222) are exactly evenly matched, such that each one would win against the other 50% of the time.
That's an interesting observation. However, actually, the 662222 die is slightly stronger than the 444400 die. At first glance, it may seem like 50-50, but the 662222 die wins 20/36 possible outcomes, whereas the 444400 die wins only 16/36 times. So one could argue that 662222 is the strongest of them all, as it is the only one that wins over two different dice.
+Evan Aubry that assumes that the dice are blind-picked. In the real world equivalent, such as videogame balancing, things like draft pick make things interesting.
You could actually play a form of Rock Paper Scissors with this: Every player has a full set of dies. All choose one of their respective dies in secret and then "throws" them. Now what you actually care about is what die was chosen by whom so actually throwing them would be a bad idea. What it really means is the person with one actually on top first (at least one gap in the cycle) wins the whole thing or you do winner out of three or something.
Clinton votes > Bernie votes doesn't mean Trump votes > Bernie votes, as they aren't the same thing; primary votes and general election votes were cast in different contexts. Is that what you mean?
I got wondering what the relation was between two dice on opposite sides of the circle, and figured out that they always have a 1/2 chance, so if A>B>C>D>A, then A=C and B=D. It could have been helpful to have said something about this in the video, but it doesn't really matter because like this you can have some exercise yourself.
It can't be even. Both dice could be mapped onto 3 sided dice without changing the probabilities. When you roll 2 d3, there are 9 possible outcomes which is an odd number. If there are no draws then one must win more than the other. 662222 beats 444400 5/9ths of the time.
I agree with Straw, the 1st and the 3rd die are equal. No matter how many you throw, both types will win as many times as the others. And sums also have the same average value
And, perhaps more importantly, this also works in politics. When you ask people "Do you prefer A or B", "Do you prefer B or C" , "Do you prefer C or A", whether ABC are politicians or cheese, you may end up with a cycle. Which leads to the question, "how fair is system that forces it down to a single choice ?" And this is as true when deciding for a president as when deciding in group whether you go to a Greek, Chinese or Indian restaurant.
I'm familiar with nontransitive dice but can someone explain the medicine example? If A cures more than B and B cures more than C, then why wouldn't A cure more than C? It's not like nontransitive dice where the comparison between two can change over time. Assuming the studies were done properly the results should more or less stay the same, so it's not really like rolling B multiple times. You're doing (effectively) one trial for each and seeing which number is the highest.
If you're confident (at some level) that A cures more people than B and that B cures more people than C, I'd say you can conclude you're confident (at the same level) that A cures more people than C, right? I don't think the medicine example works.
It could have to do with confidence intervals, but you also have to take demographics into account. Was A and B compared on women while B and C was compared on men? Or young vs old? Or different social standings? Or any other of probably hundreds of factors that could make the two results incomparable.
That makes sense to me but I assumed the studies in question would be equivalent. If that's not the case it's hard to say the results are really comparable at all.
If we're thinking of a drug as basically being a coin flip (it has a certain probability of curing any sick person it's given to), then "A cures more people than B" basically means "A has a higher success probability than B", so you're just ranking numbers, so yes, you can't get non-transitivity. Either he chose a bad example, or he was imagining some more complicated model that would make it work. I think one place where you can get nontransitivity is in voting. If when asked to choose between candidates A and B, most people prefer B, and when asked to choose between B and C, they mostly prefer C, that doesn't mean most people will prefer C to A. But that's kind of obvious.
whoooaaa! you gotta guess what die the other person will throw and then throw yours so it's two layers of chance. you'd have to know how every other dice pair's win/lose relationship too though to know how that game would work and I'm too lazy to figure it out right now
Now discuss this in various ranked voting algorithms. I'd like to see algorithms that resolve cycles in ranked voting. Is Tarjan's algorithm a candidate?
I think you didn't get the joke... Die is the singular form of "Dice" which Tadashi is using. No one wishes death upon the phrase "好きです". This person seems to like puns and dice.
"We want to avoid arithmetic."
-Tadashi Tokieda
Wee I hope so, I can't even count to ten gazillion!
@@dielaughing73 but it's NUMBER phile
@@MarcTelang he's from arithmephile
"If the other one were careless enough to show its zero." I love this guy's choice of phrasing!
Yeah, that would be pretty careless.
5:04 There it is. The legendary anime, glass-setting middle finger. It appears once every few months to even years.
I applaud. He is great.
yeah =))
Once again, Tadashi-sensei explains a complex concept well before ever referring to it so the audience isn't disengaged by complicated sounding terms. He is a great educator.
It really isn't that complicated
I'm a simple man.
When I see a Tadashi video I watch it, my mind is blown, and then I hit "Like" in a math-induced stupor.
+
I'm a simple man.
When I see descriptions of what other simple men do, I like, and comment - telling my story as a simple man.
@@Klayhamn its just that simple
false.
One of the more profound quotes by an undergraduate professor that taught me a few years ago:
"The problem with real life is that friendship is not transitive."
I once made an argument in one of my sketchbooks that the closer human relationships were to following equivalence relations, the stronger those relationships are. It's been many years and many ponderings, but I still think there may be some truth to it.
5:04 did you just give us the finger?
No, it was meant specifically for you ;)
XPossseidon
I know XD
Secreto Terra
I'd feel honored to get the finger from Tadashi.
As a glasses wearer myself...
That sometimes happens and I instantly panic if anyone took it personal.
Sapphire Crook
I'd be surprised if anyone took that as an insult.
Surprisingly, I first learned about this phenomenon when compairing stats for guns in Borderlands 2.
I just can't get enough of his lessons, I need more Brady! Please!
we're working on it
Derek Leung Thanks a lot! I'm gonna watch it later today =D
Professor Tadashi has a great general maths lecture with toys that's definitely worth checking out. I had great fun with Tadashis toy lecture.
Neat! Would love to see a follow-up with James Grime, who has his own set of these with additional interesting properties.
Well it’s wierd to see you here
@@mzadro7 nah man
Music theory is just group theory mod12
to be precise he has a set of five that have 2 different cycles. one of the two gets reversed if you play with two dice instead of one.
this makes it fun because you can play with two other players and let them choose die so long as they let you decide how many dice will be used on the roll. you are guaranteed to have a stronger chance to win no matter what they choose.
he made me wonder if it's possible to do it for more players (you plus 3 others). haven't found a solution but it's fun to try.
Yet again Tadashi blows my mind
I love the way he uses language at times. "Careless enough to show it's 0" also gave me a new perspective on probability! I feel like I gain so much from his videos.
This needs to be a game where you design die and try to beat your opponents' die.
More tadashi!
666666
9,9,9,9,0,0 - same total value, but beats yours (6,6,6,6,6,6)
This really could be an interesting game mechanic, if you distribute a fixed amount of points between a set of values, since it has nice rock-paper-scissor behavior.
10,10,4,4,4,4 - same value again, beats the above, but loses against yours.
If we continue...
10,10,7,7,1,1 - beats all of the above, but...
8,8,8,8,2,2 - beats that, but not the others except for the 6's.
Now.. given a fixed value, which distribution of points beats the most other configurations?
+
Dolkarr How does 10,10,4,4,4,4 beat
9,9,9,9,0,0? wouldnt it lose.... nvm it wins 55%
So let's go with 36 as the number of given points to distribute. What combinations would be the best?
9,9,10,8,0,0 - beats 6,6,6,6,6,6, and ties 9,9,9,9,0,0, I'm too lazy to do the rest
11,11,11,2,1,0 - another combination that can beat the 10,10,4,4,4,4
*metagame intensifies*
If you allow negative numbers, I propose 1 gazillion 5 times and -5 gazillion once. Wins 5 out of 6 against everyone here.
I don't know why I found that "ten gazillion" thing so funny... :)
Why not m8? :)
Because of Godzilla? Tadashi.... Godzilla.... Tadashi... Godzilla ;) you see the connection?
I want one of those 10 gazillion dice.
Litigious Society Imagine playing Snakes and Ladders with that :P
Probably because 10 is not much compared to gazillion, which is supposed to be some very large number, so it doesn't really make any difference, thus putting it before the gazillion is meaningless and therefore funny, just like shooting a single bacteria from a giant cannon.
Tadashi needs to host a podcast, I could listen to him all day.
He has a great voice and an excellent command of English... a natural.
+
I prefer that he gets some rest and lives a long life to be around and do his work
I knew it! I knew gazillion was a number! A mathematician just said so!
Who want's to be a Gazillionaire? I know I do.
finally
+
what is that number? how many zeros? ya i thought so
It doesn't have a fixed number of zeros. A gazillion is defined as "the smallest positive finite power of ten that is effectively infinite for practical purposes", i.e. slightly bigger than anything else you'll be dealing with, but not actually infinite. It's part of the family of numbers known as "Parker constants".
Spent most of the video thinking "how could these dice be incorporated into D&D..."
Easy sand off everything between 5 and 18 from the d20 and let people choose wether to use d20 or d6 for rolls
You could build strange magical weapons that use these weird dice instead of the ordinary roll. Maybe a shortsword from Mechanus always rolls the average of 3, for example.
You could also have the dice be magical objects themselves in-fiction, and rolling certain faces on certain dies would have specific effects. Then you just give the actual dice set out as a prop ;)
Amazing, you know the teacher is good when the solution seems obvious once explained :O
5:1 and 3:3 is equal.
Yet if you compare the other opposites, 662222 is stronger than 444400
@@safrprojects yeah that's what I wish had been mentioned. It's obvious that one of the two comparisons that weren't made in the video (333333 vs 555111) is equal, but the other one isn't as obvious. I had to do the math myself to see if it was equal.
This all needs to be implemented in a board game of some kind.
@@Aggressiphyst Throw away the board.... how about if you somehow made hand symbols to represent the dice. And maybe just use three of them.
@@safrprojects Indeed. 662222 beats out 444400 20/36 times, or 5/9 times, for anyone who was interested and didn't want to do the math.
"A is stronger than B and B is stronger than C but it's not necessarily means that A is stronger than C"
I might use this example in a certain situations.
@@eltodesukane No it isn't.
i honestly love the tadashi videos
so do we
probably my favorite guy for this channel
what's the probability?
When Tadashi says "nifty"
This is something I study a lot in game design. It's important to look at breakpoints like this when you look at balancing statistics.
Got the answer, most powerful dice is the one with 10 gazillion on all faces. Ez
sorry, i think you missed the one with 10 gazillion 1 on it.
fishecod
Must be quite a huge dice ._.
Still would rather have a 10 gazillion-sided die with 10 gazillion on all faces
Should you try to throw it, you would be about to die.
or maybe oo on all faces!
I always knew "10 gazillion" was a legit number! :-)
yeah that part made me smile ... which hurts because I just got my wisdom teeth removed ._.
Of course it is ! Economists use it all the time, as in: "The US debt is 10 gazillion dollars" :)
Oh yeah Tadashi be my waifu
Hej Simon
Anka Hej Ankan
ZimoNitrome, du er en af de fugtigste svenskere jeg kender på den anden side at Øresund! Keep it up!
Tadashi could literally make a video of him just telling the entire world how terrible of a person I am and I would still love it. Dude's awesome.
Love this video. I was just talking to someone the other day how logic and common sense are not the same thing, and sometimes logic completely contradicts common sense. This is a great example of that.
If A is better than B, and B is better than C, then common sense would dictate A is better than C, but logically you cannot be certain about such a claim.
@@eltodesukane No it isn't.
It's always nice to start the day with Tadashi-san.
No, he's more of a teacher role. So sensei would fit better.
The 5,1 dice and the all 3's dice relationship could be a clue as well. One dice can't not be declared stronger than the other.
6:02 Two comparisons were missed out: The left die (662222) beats the right die (444400); The upper die (555111) and the lower die (333333) are equal.
Gonna like before watching because Tadashi is always the dankest.
Yeh, his videos are awesome. Love this guy.
Now I can tell my math teacher that one of the most liked Numberphile mathematicians said "A Gazillion"
I found that ironic when he used 10 gazillion.
After watching the video, I can say that your confidence is well placed.
Does anyone else have non-transitive preferences between mathematicians?
I like his final point. So many people think not paying attention to math growing up just means they will need a calculator. They don't realize how stuff like this seeps into so many things in life.
I'd love a d20 with 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,10 gazillion
The ultimate critical hit
Tadashi's lessons never ceased to amaze me! I learn every single time I watch his video!
This has been known for a long time for pokémon enthusiasts.
Charizard beats Venusaur.
Venusaur beats Blastoise.
But no-one would conclude that this means Charizard beats Blastoise!
Avoiding arithmetic? Not using a 10 syllable word and expecting us to know it? Fun easy presentable demonstration? Cartoons? BEST TEACHER EVER
wow! that is so counterintuitive!
I love these videos with Tadashi! So simple and elegant.
Numberphile, I really love your vids. Mr. Tokieda explained that extremely well. Thank you!
At first a I thought "o, another non transitive dice video..."But Tadashi makes everything worth watching
Wow... The necessity for cross validation of statistical models just clicked for me in a new way. Mind blown!
This would make a really fun game of rock paper scissors because even if you pick the stronger dice you still have to make the roll.
Rock paper scissors is exactly what I was thinking too
i always knew that soccer (or any other sport) tournaments don't mean anything. depends on which team you play; if your strengths match the weaknesses of your opponents.
I have no idea what those dice are made of, but the sound they make when they hit that table is very pleasing.
Very counterintuitive and interesting!
This guy is the most interesting mathematician to listen to u
On numberphile..
A video about non-transitive dice on Numberphile, without everyone's favorite star, Dr. James Grime?? Not even a mention, or a link? Cmon Brady... Tadashi is great, but Dr. Grime has a great set of non-transitive dice named after him.
1. There is a link to BUY Grime dice in the video description
2. I've been asking Dr Grime to a do a Grime Dice video with me for YEARS without success.
I think he made a video in his own channel a while ago
James Grime definitely made a video about non-transitive dice at some point. I was going to comment that he did it on Numberphile, but seeing this thread it makes more sense that he did it on his own channel. It also appears he did another video specifically about Grime dice which is actually shown on that Mathsgear website on the product page. I could understand why he wouldn't care to be redundant on the topic.
vaego haha before I researched those dice I had assumed it's some set of 10^25 dice with a dimension of 2^363728949-1 or something like that because that's often the scale of mathematical discoveries lol.
Tadeshi Blows My Mind Every Time. He does it so graciously too.
Tadashi is a boss.
Dice A, B, C, and D
B has win rate of 66.7% against A
C has 66.65% win rate against B
D has 66.65% against C
and A has 66.7% against D
The 6,2 dice is the best though because it's the only one statistically stronger than 2 other dice... at least until people start using the all 3 dice to try and win the meta game.
And then there would be a small number of people running the niche 4,2 strat to counter the 3s
just doing the maths for those who want it,
the dice are:
3 3 3 3 3 3 (3), 4 4 4 4 0 0 (4), 5 5 5 1 1 1 (5), 6 6 2 2 2 2 (6)
3 beats 6, ties 5, loses 4
4 beats 3, loses 5, 6(5/9th of the time)
5 beats 4, ties 3, loses 6
6 beats 5, 4, loses 3
To answer that last part: If all 4 dice were pitted together in one final battle to the death, The 6 and 5 die are the best with a 33% chance of winning. The 4 die has a 22% chance of winning, and the 3 die has an 11% chance of winning (don't pick that one!)
Thank you for saving my brain from a cognitive dissonance meltdown. I couldn't comprehend his analogy with drugs and implying that C>A>B>C. That's not possible with discrete data.
Here's a game for you:
Two players discreetly select a die in order. The first player receives what they selected. If the second player requested the same, they pick again (knowing which die the first player chose).
Since 6,2 is the best in a battle royale, and player-A knows this, but if player-B chooses the same, then for their second pick they're guaranteed to pick all-3's... so player-A could set an ambush by choosing all-3's. Unless the second player tries this as well and then learns to choose 4,0. So player-A might as well set a trap by starting on 4,0. But it will lose if player-B starts on 6,2.
From a random Monty-Hall problem standpoint, it looks like player-B will win more often, but something tells me that in applied setting it would balance toward Rock-Paper-Scissors (50%).
This reminds me of the False Positive Paradox, for some reason; although, I don't think it's actually related. But it gives me the same sort of pleasure when the counter-intuitive understanding finally clicks in my mind.
James grimes did the same thing on his own channel, and now it's the third video of exactly the same dice.
Yes he did
This one's more powerful though.
James Yang Each one beats the other, non-transitively?
I object to the analogy with medicine. The quantity, percentage, etc. of people cured can be expressed by a number, so of course a>b>c implies a>c. The situation is quite different than that of these interesting dice.
"...This is something that should be know by all CITIZEN..." lol
League of Legends?
Laugh Out Loud
Lots of love
and then he flips the bird ahah
logarithmic or linear
Some of the concepts in this shed some light on the paper i did as an undergraduate. I now appreciate both this video and my own work as well. Now if I only had the time, I would have used some of the ideas here to simplify my calculations massively.
I would love a follow-up to this comparing the dice in Super Mario Party. Each character has a unique dice to give them all a different feel. It would be interesting if you could determine who's the strongest!
The baritone of this man's voice is worth listening to without focusing on the actual video
I noticed something when I compared the dice to their non-adjacent die. 333333 vs 555111 is even odds, half the time you win and half the time you lose. But with 662222 vs 444400, 662222 has a slight edge over 444400.
Are you sure? I think it's even odds as well
It can't be even. Both dice could be mapped onto 3 sided dice without changing the probabilities. When you roll 2 d3, there are 9 possible outcomes which is an odd number. If there are no draws then one must win more than the other. 662222 beats 444400 5/9ths of the time.
662222 has a 1/3 probability to roll a 6, which always wins. It has a 2/3 probability to roll a 2, in which case it wins 1/3 times. 1/3 + (2/3)*(1/3) = 0.555...
662222 indeed has a slight edge.
555111 cannot be mapped onto a 3 sided die. 333333 and 555111 can be mapped to a two sided die/coin as 33 and 51.
love that '10 Gazillion' example to explain averages
Professor Tadashi got me thinking again - this time about how to improve the design of a game I'm working on :)
You might like to map it out. Just did it for interest because i like to see well balanced games too. The dices with the 5 and 6 win over 2 other dices. But the dices with the 3 and 4 only win over one other dice. That is still useful for games. The 3 and 4 can be healer and tank like in mmos. The 6 and 5 range and melee attack.
This is what makes rock-paper-scissors the game that it is
Tadashi has taught me more than all of my math teachers ever have.
And you're English teachers...
The irony...
My math teacher in highschool was educated in philosophy, but he said that the two subjects are so closely related, that one compliments the other in a surprising amount of ways.
That Guy Nah, friendo, he had a point. A lot of real logical problems can be quantized with mathematics. It doesn't even have to be particularly complicated.
Besides, it's not as if he wasn't educated in math as well. He had a masters in philosophy, and a bachelor in mathematics, so he had his credentials in order.
IchBinEin Fair enough, but I still hold my point that philosophy is nothing like maths.
To sum up what he said at the end of the vid: A could equal 2/6. B could be greater with 3/6. If B is better than A only by 1/6 and B is greater than C, then C could be less than or equal to A.
When I saw this video, my first thought was to design a rock-paper-scissors type game around it. I immediately came up with a game in which two people randomly draw a die from the set of four presented in this video, and then roll their selected dice to determine the winner.
Unfortunately, I found that in a series of random matchups, the [6/2] die is overall the strongest.
[4/0] beats [3], odds 2:1
[5/1] beats [4/0], odds 2:1
[6/2] beats [5/1], odds 2:1
[3] beats [6/2], odds 2:1
[3] and [5/1] are matched 1:1.
[6/2] beats [4/0], odds 5:4.
Based on the above, we can see that [3] and [5/1] have average performance; they are weak to one die, strong against another, and evenly matched by another.
But [6/2] performs slightly better than all the others; it is weak to one die, but strong against the other two dice.
Conversely, [4/0] performs slightly worse than all the others; it is strong to one die, but weak against the other two dice.
But luckily, there's a way to fix this! You can even out this power balance if you make a small alteration: change the [4,4,4,4,0,0] die to a [4,4,4,4,0,X] die.
X, in most cases, is equivalent to a 0, and loses to any other roll... *except* a 6, which it trumps. In a matchup between this die and [6/2], the chance of rolling 6 vs. X is 1-in-18; this alters the odds exactly enough that each die is equally likely to win against the other.
So, by changing one of the 0's to an X, and giving the X the properties stated above, you have a game in which every die is strong against one, weak against another, and equalled by another. A perfect balance.
...Okay, so it would be a pretty boring game on its own. But you could use it as the basis for a tabletop RPG's combat system!
late comment, but can you explain more in detail how (44440X) beats (666622)?
@@avocette Ah, that's where you've confused yourself and confused me. The 6/2 dice is not 666622, but rather 662222
Against 44440X, it'll work like this:
4 dice rolls a 0 with 1/6 chance, guaranteed loss.
Rolls an X with 1/6 chance, wins with 1/3 chance if 6 dice rolls a 6.
Rolls a 4 with 2/3 chance, wins with 2/3 chance if 6 dice rolls a 2.
So let's see if the math checks out this time:
2/3 * 2/3 is 4/9. That's just under a half. This means that the X roll has to be at least 1/18, which will put it at 1/2 chance to win or lose.
1/6 * 1/3 is 1/18.
Oh, that's why. The original comment states that the 4 dice is strong against 1 and weak to 2. This change does, indeed, make it strong against 1 and weak against 1, with it now being equal to the 6 dice.
I read your whole explanation.
I love watching Tadashi videos. He is full of fascinating insights and thoughts. Discovered him before I saw him on Numberphile though, doing a series of lectures in Cape Town at the African Institute for Mathematical Sciences, lecturing on topology.
This guy's ability to explain things in a language that's foreign to him is extremely impressive.
"662222" is the strongest of the four dice.
It beats "333333" one third of the time (its weakest game).
It beats "555111" two thirds of the time (its strongest game).
It beats "444400" 55.56% of the time.
That makes "444400" the weakest of the four dice, since it loses more often to its similar strength opponent.
This man could teach rocket science to preschoolers.
Any time my friends get into powerscaling debates with 3 or more characters I’ll just show them this video
5:03 Tadashi, how dare you sir!
his gift to you
6 is the strongest because in the entire, uniterated meta-game it beats 4 and 5 where 3 ties with 5. So, while the loop is a non-transitive tournament, there are still two more edges to the full graph. It is in these two edges we can make a ranking. Additionally, if we zoom out and consider pick rates for the prior information, even when considering this non-transitive loop, we can rank the four dice as equally strong, according to the definition given.
Would you mind giving your calculation? I had a feeling this was true from some cursory calculations but I wasn't able to totally convince myself.
@@reesverleur9546 Sure. 6 beats 5 as show in the video. 6 beats 4 because:
The probability of A rolling 0 and B rolling 6 = 1/3*1/3 = 1/9
P of A=4 and B=6 = 2/3*1/3 = 2/9
P of A=0 and B=2 = 1/3*2/3 = 2/9
P of A=4 and B=2 = 2/3*2/3 = 4/9
As you can see, only one of these events ends up with A winning, event #4 which has a probability of 4/9, less than 0.5. Thus, A is weaker than B.
3 is equally strong as 5 because 3 is a constant die and the 1's and 5's are equally probable on the 5 die. Hence, a both dice are equally likely to win when head-to-head.
10 gazillion is my favourite number
The day comes closer where this channel is taken over by our king Tadashi
I love Tadashi. But this is like the third video on nontransitive dice.
it's our first!?
I think he just meant dice, but dont listen to him, more dice videos please.
I don't think he meant just dice, but rather people following this channel tends to follow James Grime and Matt Parker and they also did similar videos. It can sometimes be hard to remember which channel you saw it on, because of the similarity, hence the confusion and the comment.
There are several videos on UA-cam with people who often appear on numberphile showing off nontransitive dice, but none on the actual numberphile channel. I was confused at first too.
I wish I could listen to this man lecture all day. Just perfect explanations.
excellent. very interesting to learn about the cicle.
5:04 WTF? :-)
The same flag went up for me even though I've been living in Japan for over 6 years. Tadashi is Japanese and he almost definitely developed that habit before making his way west. In Japan, pop culture from the west has informed Japanese (and other countries, too, I imagine) that the middle finger has a negative connotation, so everybody here knows that it's not good to toss your lone middle finger up at someone, but in using it for anything else, it goes totally unnoticed because it doesn't have the same meaning in Japanese culture. Actually, being the longest finger, it's logically the most convenient and useful to use for a number of tasks, including pointing, referencing, and glasses up-pushing. :D It took a while to get used to, but now I've gotten quite used to it. In this case, it was difficult to miss, though. hahaha
ItchyKneeSon
Didn't notice the gesture until read comments about it. We don't use it much in Russia either.
+ItchyKneeSon I, a North American, also use my middle finger to push up my glasses, but I have _all_ my fingers extended when I do so. XD
+Artm Borisovskiy I didn't notice it either. XD
NoriMori That's called 'dazzle fingers'.
+ItchyKneeSon By who? Why?
I'm loving all the dice related videos!
These set up some fun experiments for students to explore.
Cheers
this needs to be a game mechanic
Rock, paper & scissors?
similar but better.
A lot of strategy games use a similar mechanic but deeper.
Rock, paper, scissors?
Sort of.
It's called rock-paper-scissors ;)
This can be used in say....paradox interactive games, where simulations show that one type of army composition kills more men than the other, but doesn't necessarily win as often...A niche case, but it can happen. Usually more losses means a much lower chance of winning, but when looking at simulation results, one should try to find sweetspots to avoid a too qualitative army and instead go for More army elsewhere. I hope i'm making some sense. There's also dice involved, but finding 2 identical armies with 2 identical dice modifiers is rare, and winning/losing tends to snowball based on too many variables.
These dice are akin to the Fire, Water, Electric, and Grass types in Pokémon.
333333 Fire
004444 Water
111555 Electric
222266 Grass
Grass doesn't win or lose to electric tho
Think defensively. Grass resists electric but not the other way around.
it's been that way since gen 1
Rock, bug, psychic, fighting?
Something interesting I noticed: Each winning die wins 2/3 of the time, in all four cases. Similarly, opposite dice (333333 and 555111, 444400 and 662222) are exactly evenly matched, such that each one would win against the other 50% of the time.
That's an interesting observation. However, actually, the 662222 die is slightly stronger than the 444400 die. At first glance, it may seem like 50-50, but the 662222 die wins 20/36 possible outcomes, whereas the 444400 die wins only 16/36 times. So one could argue that 662222 is the strongest of them all, as it is the only one that wins over two different dice.
+Evan Aubry that assumes that the dice are blind-picked. In the real world equivalent, such as videogame balancing, things like draft pick make things interesting.
So this is similar to stone/paper/scissor.
similar but not intuitive it deserves to be pointed out !
Woah, Tadashi talking about actual math! This is rare on Numberphile...
Didn't we already see a video on non-transitive dice featuring Dr. James Grime (aka singingbanana)?
I guess it was never on Numberphile. It was on Dr. Grime's personal channel (singingbanana) and the Maths Gear channel (mathsgear).
You could actually play a form of Rock Paper Scissors with this:
Every player has a full set of dies. All choose one of their respective dies in secret and then "throws" them.
Now what you actually care about is what die was chosen by whom so actually throwing them would be a bad idea.
What it really means is the person with one actually on top first (at least one gap in the cycle) wins the whole thing or you do winner out of three or something.
Seeing this peculiarity being demonstrated in the US Presidential election this year.
+
Clinton votes > Bernie votes doesn't mean Trump votes > Bernie votes, as they aren't the same thing; primary votes and general election votes were cast in different contexts. Is that what you mean?
No. Clinton beats Sanders in primary. Trump (potentially) beats Clinton. Sanders would have clobbered Trump.
I got wondering what the relation was between two dice on opposite sides of the circle, and figured out that they always have a 1/2 chance, so if A>B>C>D>A, then A=C and B=D.
It could have been helpful to have said something about this in the video, but it doesn't really matter because like this you can have some exercise yourself.
He just flipped me off 5:02
Isn't the 1st and 3rd die the same (50/50) because it's 555111 and 333333 therefore each die wins half the time and loses the other half :/
It can't be even. Both dice could be mapped onto 3 sided dice without changing the probabilities. When you roll 2 d3, there are 9 possible outcomes which is an odd number. If there are no draws then one must win more than the other. 662222 beats 444400 5/9ths of the time.
I agree
I agree with Straw, the 1st and the 3rd die are equal. No matter how many you throw, both types will win as many times as the others. And sums also have the same average value
Straw Hat Roger, nice catch! Yes, the dice 555111 vs 333333 have a probability of 1:1 for predicting outcome of each roll.
Tim, no, 555111 cannot be mapped to 3 sided dice, what three numbers, 551, 511, 555, 111??? it could be reduced to a 2 sided/coin tho 51
And, perhaps more importantly, this also works in politics.
When you ask people "Do you prefer A or B", "Do you prefer B or C" , "Do you prefer C or A", whether ABC are politicians or cheese, you may end up with a cycle. Which leads to the question, "how fair is system that forces it down to a single choice ?"
And this is as true when deciding for a president as when deciding in group whether you go to a Greek, Chinese or Indian restaurant.
I'm familiar with nontransitive dice but can someone explain the medicine example? If A cures more than B and B cures more than C, then why wouldn't A cure more than C? It's not like nontransitive dice where the comparison between two can change over time. Assuming the studies were done properly the results should more or less stay the same, so it's not really like rolling B multiple times. You're doing (effectively) one trial for each and seeing which number is the highest.
If you're confident (at some level) that A cures more people than B and that B cures more people than C, I'd say you can conclude you're confident (at the same level) that A cures more people than C, right?
I don't think the medicine example works.
It could have to do with confidence intervals, but you also have to take demographics into account. Was A and B compared on women while B and C was compared on men? Or young vs old? Or different social standings? Or any other of probably hundreds of factors that could make the two results incomparable.
That makes sense to me but I assumed the studies in question would be equivalent. If that's not the case it's hard to say the results are really comparable at all.
If we're thinking of a drug as basically being a coin flip (it has a certain probability of curing any sick person it's given to), then "A cures more people than B" basically means "A has a higher success probability than B", so you're just ranking numbers, so yes, you can't get non-transitivity. Either he chose a bad example, or he was imagining some more complicated model that would make it work.
I think one place where you can get nontransitivity is in voting. If when asked to choose between candidates A and B, most people prefer B, and when asked to choose between B and C, they mostly prefer C, that doesn't mean most people will prefer C to A. But that's kind of obvious.
***** maybe you can point us to a specific paragraph or section in the 16 page document.
Grime dice explained by Tadashi? What a world to live in.
this is like rock paper scissors
whoooaaa! you gotta guess what die the other person will throw and then throw yours so it's two layers of chance.
you'd have to know how every other dice pair's win/lose relationship too though to know how that game would work and I'm too lazy to figure it out right now
Now discuss this in various ranked voting algorithms. I'd like to see algorithms that resolve cycles in ranked voting. Is Tarjan's algorithm a candidate?
Diesuki desu.
Please, do not wish death upon Suki Desu. :D
dai*suki desu / だいすきです。/ 大好きです。
I think you didn't get the joke...
Die is the singular form of "Dice" which Tadashi is using.
No one wishes death upon the phrase "好きです".
This person seems to like puns and dice.
I've seen these dice done by someone else but I can't remember who.
That was Numberphile regular James Grime (Singingbanana) :)
You can buy his dice online, and they're really cool :)
link in the video description
Another numberphile host, James Grime.
watch?v=u4XNL-uo520
Yup, in maths gear shop. (Grime, Mould and Parker, actually)
Oskarpuzzle has a version of these that he made into a sort of game
I'd like to buy a die with a 3 on every side