That five dice double chain diagram looks really familiar... Hmm.. Yellow cuts blue. Blue covers red. Red crushes green. Green poisons white. White smashes yellow. Yellow decapitates green. Green eats blue. Blue disproves white. White vaporizes red. And as it always has, red crushes yellow.
@Horinius That's one of the early references that put the idea in the form of a game. The idea that such thing exists goes back a few years earlier. References: The Paradox of Nontransitive Dice, Richard P. Savage, Jr. The American Mathematical Monthly, Vol. 101, No. 5 (May, 1994), pp. 429-436 and; S. Trybula, On the paradox of n random variables, Zastos. Mat. 8 (1965), 143-154.
+deadlypendroppingby it probably didn't take more than a day to create the dice, film, edit, and post onto youtube. and five years later, people are still watching and commenting on this video.
+deadlypendroppingby does any of them actually do proper research? I thought that the jobs they had are more for community outreach and making maths more interesting to the general population.
@Horinius The set of three dice go way back to the original paper on the subject in the '70s. I spent some time designing the set of 5 to do what I want it to do. That is a tricky balancing act.
Michael S Yeah, translation isn't generally a one-to-one function. However, you do have "ao ringo" for green apples, "ao shingo" for green traffic lights (though some Japanese traffic lights are a bluer hue than the typical Western ones), "ao nori" for a kind of seaweed which is definitely green, etc.; but you're right, perspectives differ.
I love the red die from the Efron dice set. It's like it's been designed for fatalists. "Ah, you misthrown again! Could you be a bit more careful?" *rolling* "Three."
Wow adding an extra dice flips the cycle. And you turn that into a game. It's amazing how the numbers really fit together, after alot of designing for a 5dice game let alone a 3 dice game. Great work
I just googled it for you to save time: Spiegelhalter does not mean "mirror-holder". "Halter" comes from "Halde" which means hillside. There is a German name called "Winterhalter" meaning "the one living on a hillside, facing north and not getting lots of sun". The Spiegelhalter lived opposite to this on a sunny hillside. No wonder Mr. Spiegelhalter is very bright.
Now Spiegelhalter sounds like a funny profession. I'm a professional Spiegelhalter myself. (Spiegel -> german for mirror, Halter -> german for holder/thing that can hold stuff)
Wow! I have read of David Spiegelhalter - a distant cousin - in Spiegelhalter genealogies, but have never seen him. I am a member of his family's American branch.
Ricercar If you're being formal, of course you use someone's title and initials. But your buddy Dave--are you _really_ going to call him "Sir Dave"? With a straight face, I mean.
I would like to point out the reason this is true is because we start counting at 1 (an odd number), if the convention was to start counting at zero (an even number), then the evens would have the advantage by the same amount. If, alternatively, we decided to randomly alternate from starting at 1 or 0, then we would see a 50/50 split.
@pirsquared99 That's right. I had a few requests for this, especially after the Scam School episode - which is very good, people should check that out too. Yes, as you saw, the dotted line means it's about 50-50, the chances are slim in that case.
@kkriegg No it lowers! And with the three set of dice the probabilities reverse order so if A beats B with a greater than 50% probability, two As beat two Bs with a less than 50% probability. I wrote an article with more details linked in the description.
Game #3: In a 5 (points on circle) -ring, there are only 2 dyads (ways to pick 2 dice): 14 and 23. This limits the game to the basis of only 2 probability pentagrams. It's generalizable on all non-trivial rings and die counts. Polya did it, and Howard Hanson (Rochester Professor of Music).
Oh Grime! You were like this as a Kid weren't you? You were lucky enough to find that your obsession could become your occupation. This stuff does my head in, and I love it!!
I've got massive respect for Marcus Du Sautoy for promoting maths learning to the masses, but you're better when it comes to taking that to the next level of interest. You would be absolutely great as a RI Xmas Lecturer. Go and treat yourself to a biscuit or a bit of cake.
In the 2nd loop of the 5 dice problem, [so when you are using 2 dice.] The Red dice has approximately a 51.77% chance to beat the White dice. Its approximately 50% but the red dice actually WINS. The probability of the red dice winning is the probability that either the 9 face of ANY red dice is rolled or that the 0 face of ANY white dice is rolled. Any other combination not including one of those faces results in a white win so we dont have to calculate them. So the probability of the first red dice being a 9 is = 1/6 chance The probability that the 2nd red dice being a 9 and the first dice not being a 9 is = (5/6)x(1/6) The probability of the 3rd being 0 and first 2 not being 9 is = (5/6)x(5/6)x(1/6) The probability that the 4th being 0 and the first 2 not being 9 and 3rd not being 0 is = (5/6)x(5/6)x(5/6)x(1/6) You can stop calculating probabilities once you hit a 9 or a 0 because every branch on those paths lead to a Red victory so the entire brach is just a red win so calculating the further probability isnt necessary. [Thats why we stop at 1/6 for the first red dice but the 2nd we count the 1st red dice missing and then stop when the 2nd rolls a 9 etc for the other 2 dice] Thus, the probability that Red beats White is equal to the probability that either at LEAST one 9 OR 0 is rolled which is equal to (1/6)+[(5/6)x(1/6)]+[(5/6)x(5/6)x(1/6)]+[(5/6)x(5/6)x(5/6)x(1/6)] = 51.77% [about]
You can do a similar thing with Texas hold'em poker. You take three starting hands: AK offsuit (os), QJ suited (s) and 22. First your opponent chooses a hand, then you, and then you deal the full board and see who wins. No matter what your opponent chooses, you can always be ahead because in terms of the probability of a win when dealing the full board: 22 > AKos, AKos > QJs, QJs > 22. The margin is small though, so you'd have to play a lot of boards to see the difference!
D&D and WarHammer tabletop players just got their minds blown with this video. I guess. Also, I'm just wondering... Isn't there a simple calculation that determines the 'force' of the die? Perhaps by adding up the dots? As a layman, my instinct would be to add up the number of dots on the dice to determine how 'powerful' each one is.
+Generic Internetter never heard that term but you can figure out the average output value of the die by adding up the dots and dividing by the sides. which would be like the "force" of it but since the game isnt a test of the total output value, but rather how often you win then a 6 or a 2 has the same power when against a 1
At 0:32 the little text that pops up should be "Take that, Dog on a Skateboard!" When I read it, I thought you were suggesting I take a dog on a trip with me using a skateboard for locomotion.
@WhiteDragonTile Yeah, sorry, full credit to WhiteDragonTile who challenged me in the first place (this is an updated version of the dice, these are better).
@rg0057 Aren't you a grumpy-drawers. The constructions for 3, 4, and 5 dice sets are on the screen. If you want a general construction for n dice, there's quite a dull one that would work for up to 6 dice, but I don't think that would add anything to the sets I've already shown you. I do not know of any other general construction of non-transitive dice - which is why I don't mention any, they are not easy to construct, which is why finding the set of dice I describe is interesting.
In a 1v2 game it should be still more expected that you win one player but lose to the other, because the die advantage probabilities are so close to 50%. If the whole set winning probabilty doesn't exceed 70,7% for you, then it is likely that you will lose to at least one of the players since the probability of winning them both drops below 50%.
Not quite. We have an even amount of rounds so you forgot to calculate for draws. You can get to an advantage as low as 53,6% and still have a probability over 50% for not losing a game, while your probability of winning a set is just 46,8%. With an advantage of 7/12 (58,3%) like the one in the video for example, losing once is close to as likely as winning both (35,3% to 34,9%), win+draw is 25,3% and drawing both only 4,5%
It's possible I just changed the labelling. The colours have changed too, and some of the other details. Both the video and the article are correct. The article gets updated.
Another example of a non-transitive game is Rock, Paper, Scissors. Which is equivalent to the Red, Yellow, Green dice game (without any way to flip the order, unlike your cleaver one dice two dice trick). And Monkey, Pirate, Robot, Ninja, Zombie: a souped up version of Rock, Paper, Scissors. Which is equivalent to the Red, Yellow, Green, Blue, White dice game (again with out any way to flip the order).
@leungclj haha, have a look at the article in the description. I mention a game for three players that uses seven dice. But it's a trade, if you try to do it with fewer dice you'll have to add something in elsewhere, like I did with my extra rule.
I wonder if this video was the inspiration behind the latest Project Euler problem (#376)...I know that Dr. Grime and this video were the first things I thought of when I saw it!
It's simple: Scissors cuts Paper , Paper covers Rock , Rock crushes Lizard , Lizard poisons Spock , Spock smashes Scissors , Scissors decapitates Lizard , Lizard eats Paper , Paper disproves Spock , Spock vaporizes Rock , and, as it always has, Rock crushes Scissors.
I played 5 non-transitive dice game with my family, my mum and my brother Josh. Mum selects a 4, 4, 4, 4, 4, 9 dice, Josh selects a 0, 5, 5, 5, 5, 5 dice and I selected a 1, 1, 6, 6, 6, 6 dice. After 10 games, I realized that I won most amount of times than two members of my family. The result is, me 6, Josh 3 and mum 1.
My maths teacher did this with me the other week. First we did it with 3 dice, so it reminded me of Rock Paper Scissors, then they added more and more! We figured out the it works in alphabetical order (blue, magenta, olive, red, yellow) But also in number of letters (red, blue, olive, yellow, magenta) So blue beats magenta Magenta beats olive Olive beats red Red beats yellow But then yellow beats blue ALSO red beats blue Blue beats olive Olive beats yellow Yellow beats magenta Magenta then beats red! Really confused my year 8 brain as probabilities is numbers. Also if you double the, up, then something else happens - can’t quite remember this bit!
That's quite cool. However, I usually love games like this where I will always win but this one I think is too much trouble to remember and to get the dice. But as I say, well done, it's all very clever!!!
The green dice vs the red dice is the most interesting when you throw two of each. As the green dice has 5 fours and the red dice has 5 threes so it seems intuitive that green will usually win. However because red will win if green rolls any ones or if red rolls any sixes the actual odds of red winning are 1 - (5/6)^4 = about 52%.
This is similar to (well, actually the same as) using 9 cards numbered 1 through 9. The 1-6-8 cards have Red backs; the 3-5-7 cards have White backs; and the 2-4-9 cards have Blue backs. Then, playing one card of one color against one card of a different color, one number will be higher than the other and in 55% of the plays Red beats White White beats Blue Blue beats Red
@@thomaseckles4686 none. All cards placed face down. Each player picks one card from the pile. To have an advantage, player 1 lets player 2 pick first. If player 2 picks a white card, player 1 picks red. If player 2 picks a blue card, player 1 picks white. If player 2 picks a red card, player 1 picks a blue card. In each of these scenarios player 1 will win 5 times out if 9 plays, on average.
I bought the dice and the chains and colors illustrated in the instruction are different then this video. They are: red 4-9 blue 2-7 olive 5-0 yellow 3-8 magenta 1-6 not considering that the colors do not much, not only because are different, in this video the winning chain is different from the purchase you show: 4-9 3-8 2-7 1-6 5-0 which is right? Or I have to calculate myself which is the correct chain?
Allow me to pick a nit here: the singular form of the word "dice" is "die." Granted, we don't usually use the singular, because dice generally come in pairs, like scissors. But it is worth remembering that there is a grammatical form just _waiting_ for the odd occasion when it's appropriate to use it.
B. Xoit fair point made. you are right "dice" is acceptable as the singular form of "dice". that being said i still support us using "die" it makes our language more precise to be able to determine whether the speaker is referring to a single unit or multiple. in the same way we dont say "shoes" means the same as "shoe". the language can change and i think a language in which we can tell whether it is singular or plural is better than a language where we cannot.
@GetMeThere1 1. I'm really bad at justifying things using 'real life', but you can think of nature as a non-transtie chain. It is something we want to avoid in elections. There's probably something really convincing I can't think of. 2. Patterns don't hold for triples etc. It's quite a balancing act, and these were carefully designed so a pattern held when you doubled the dice. But it's delicate so there's no particular pattern to what beats what after that.
Is it signficant that, in the case of the original three dice, that the sides of all three dice add up to 21? This doesn't appear to be the case for the other games presented.
This is like picking a starter in Pokemon games, the opponent always picks the one with a type-advantage over yours. It goes: Fire > Grass, Grass > Water, Water > Fire. I guess It's also like rock-paper-scissors now that I think of it.
the remarkable thing is that they will all tie, because they all have the same average roll (3.5). This is also the same average roll as a standard d6, so they will tie that as well After doing the math, any of the non-transitive dice will have a 5/12 (41.6%) chance of winning against a standard d6, a 5/12 (41.6%) chance of losing, and a 1/6 (16.6%) chance of tieing (sp?). This is the same for all three dice. For the two-die game, the key to finding the probabilities is to find the probabilities of each possible sum. On a d6, these odds are: (2: 1/36) (3: 2/36) (4: 3/36) (5: 4/36) (6: 5/36) (7: 6/36) (8: 5/36) (9: 4/36) (10: 3/36) (11: 2/36) (12: 1/36) For the red die they are: (6: 25/36) (9: 10/36) (12: 1/36) For yellow: (4: 9/36) (7: 18/36) (10: 9/36) For green: (2: 1/36) (5: 10/36) (8: 25/36) Going through, you will find that all dice have an average roll of 7, meaning that THEY will tie as well. If you roll all five dice at once (plus a d6), the odds become much more different. Each die has a different average roll. Red: 4.83 Yellow: 4.66 Green: 4.5 Blue: 4.33 White: 4.16 D6: 3.5 There is no contest. Using a standard D6 is futile, BUT if you change the numbers on that to make it 2-7 instead of 1-6, it will tie with the green die.
+Some Random Fellow "the remarkable thing is that they will all tie, because they all have the same average roll (3.5)." Wasn't it established already that this shouldn't matter? I think it's also wrong, let me calculate: When it comes to the red die, it has a 1/6 chance of winning by definitions, when as when it rolls a 6, nothing gets higher. in the other 5/6 of cases the chance of winning is actually very low as the other 2 have to roll their low number, which in the case of the green die is already 1/6, and in the case of the yellow die 1/2. But in total, the chance of the red die winning is 1/6+5/6*1/6*1/2=1/6+5/72=12/72+5/72=17/72. And this is already lower than 1/3 (24/72). When it comes to the yellow die, the chance of winning becomes 5/6*1/2=5/12 because in the case where red doesn't roll a 6 it always wins half the time when it rolls a 5, and it always loses half the time when it rolls a 2 (as it's always lower than whatever the red die will roll), nonetheless its chance of winning is a very big chance (30/72). When it comes to the green die, there's the same case of course that it'll always lose to red's 6, and it'll always lose in the case it rolls a 1 for which the chance is 1/6 and it will always lose when yellow rolls a 5 (1/2 chance), in the end its chance of winning will be 5/6*5/6*1/2=25/72, which is just over a third. Concluding: Red: 17/72 Yellow: 30/72 Green: 25/72 So Yellow is the best one in this case, actually by a rather large margin.
I like a die with four 5s and two 1s, a die with four 3s and two 4s, and a die with two 6s and four 2s. The odds work out differently, but the same rules apply with both sets of one and two dice. I wonder how many different sets of these dice can be made.
"But you can always make nontransitive dice, for any given number of dice."- Dr. S.B. How do we know that to be true? (i hear a chorus of "proof! proof! proof!")
@Gnomeslayer778 I've tried to fix that problem with the Maths Gear version. Using the colours, one chain is ordered alphabetically, the other chain is ordered by word length.
that is awesome. I dont know how to design these games, but surely if you can do a 3 player system, you its probably possible do a 4 player system or an n player system?
This is an interesting case of natural properties defying immediate intuition. It's tough to argue that the counterintuitive non-transitivity is false while immediate intuition is true. But lately I've been arguing that Cantor's non-intuitive work may actually not be as correct as it is accepted to be. I made a video titled "Does Counting Exist?" to illustrate this point. It might be a good idea for legitimate mathematicians to correct any errors in the video.
Stumbling upon this channel over a decade after this video was posted and I am so happy I did, what wonderful content!
That five dice double chain diagram looks really familiar... Hmm.. Yellow cuts blue. Blue covers red. Red crushes green. Green poisons white. White smashes yellow. Yellow decapitates green. Green eats blue. Blue disproves white. White vaporizes red. And as it always has, red crushes yellow.
Spock
I love you
Can you say that again?
🖖
It's exactly the same as RPS-5?
02:33 So, it's like Pokemon starters... The second one always gets the better choice.
Girish Manjunath
People usually think of Rock Paper Scissors but that's also true
In Rock Paper Scissors you pick at the same time.
In Pokémon your rival picks afterwards just to screw you over.
@@MrMctastics also in RPS you always win, whereas in pokemon you area only more likely to win. so this game is more like pokemon.
Did he say the Cambridge Professor of Ornithology? (9:13)
That's hysterical.
He really quacks me up!
I'm interested in birds too:)
very nice introduction to expected values and abnormalities with it.
@Horinius That's one of the early references that put the idea in the form of a game. The idea that such thing exists goes back a few years earlier.
References: The Paradox of Nontransitive Dice, Richard P. Savage, Jr. The American Mathematical Monthly, Vol. 101, No. 5 (May, 1994), pp. 429-436 and;
S. Trybula, On the paradox of n random variables, Zastos. Mat. 8 (1965), 143-154.
Spiegelhalter looks like he thinks James should do less UA-cam and more research, LMAO
deadlypendroppingby No way! He loves it.
singingbanana hehe, okay!
+deadlypendroppingby it probably didn't take more than a day to create the dice, film, edit, and post onto youtube.
and five years later, people are still watching and commenting on this video.
Generic Internetter Yup, I think it's cool, too, to have a little fun at work, too. Besides it advertises math.
+deadlypendroppingby does any of them actually do proper research? I thought that the jobs they had are more for community outreach and making maths more interesting to the general population.
@Horinius The set of three dice go way back to the original paper on the subject in the '70s. I spent some time designing the set of 5 to do what I want it to do. That is a tricky balancing act.
12 years later and this video is still brilliant.
i have never been happy to get a lecture before and you sir are a genuine master of making the most advance stuff seem interesting
This is absolutely mind blowing...I've never seen anything like this..Thanks for sharing this Jim ~ !
10 years
@@1.4142 sup
@@1.4142 and 6 months
Spielgelhalter gives a funny look at 1:24. I wonder why. Could it be that Grime is calling an orange die red?
+jimpozcaner
I was thinking that too
He's secretly Japanese.
The true Satoh ... except that it's green the Japanese call blue ...
Or blue that they call green, from another perspective.
Perhaps you could translate ao as "cyan"
Michael S Yeah, translation isn't generally a one-to-one function. However, you do have "ao ringo" for green apples, "ao shingo" for green traffic lights (though some Japanese traffic lights are a bluer hue than the typical Western ones), "ao nori" for a kind of seaweed which is definitely green, etc.; but you're right, perspectives differ.
I love the red die from the Efron dice set. It's like it's been designed for fatalists.
"Ah, you misthrown again! Could you be a bit more careful?"
*rolling*
"Three."
Wow adding an extra dice flips the cycle. And you turn that into a game. It's amazing how the numbers really fit together, after alot of designing for a 5dice game let alone a 3 dice game.
Great work
Right. All the figures are in the article in the description.
I just googled it for you to save time: Spiegelhalter does not mean "mirror-holder". "Halter" comes from "Halde" which means hillside. There is a German name called "Winterhalter" meaning "the one living on a hillside, facing north and not getting lots of sun". The Spiegelhalter lived opposite to this on a sunny hillside.
No wonder Mr. Spiegelhalter is very bright.
I finally saw that 50-50 part of the two-dice version for the five-die set lolz
I cant belive you have been doing this for more than 10 years now
These dice have intrigued me for a few years now. I need to either make or order a set. These would be fun as just to have around.
Now Spiegelhalter sounds like a funny profession. I'm a professional Spiegelhalter myself. (Spiegel -> german for mirror, Halter -> german for holder/thing that can hold stuff)
7:53 , anyone else thinking rock paper scissors lizard spock
*looks at comments* yes, clearly :3
Yep
@Supreme4321 That's right, red is just an easier word to say when speaking quickly.
Wow! I have read of David Spiegelhalter - a distant cousin - in Spiegelhalter genealogies, but have never seen him. I am a member of his family's American branch.
Ron Spiegelhalter I'll pass this on. He's Sir Professor David Spiegelhalter OBE FRS these days. And lovely.
singingbanana Shouldn't that be Professor Sir David Spiegelhalter OBE FRS? I thought 'Sir' always went with the given name.
Ricercar Correct.
Ricercar If you're being formal, of course you use someone's title and initials. But your buddy Dave--are you _really_ going to call him "Sir Dave"? With a straight face, I mean.
+Ron Spiegelhalter He kind of looks like Bill Murray.
Beating your friends is not okay. #noviolence
#FRIENDSLIVESMATTER
#YesAllFriends #SoNoOneToledYouLifeWasGonnaBeThisWasClapClapClapClapClap
I would like to point out the reason this is true is because we start counting at 1 (an odd number), if the convention was to start counting at zero (an even number), then the evens would have the advantage by the same amount. If, alternatively, we decided to randomly alternate from starting at 1 or 0, then we would see a 50/50 split.
@ericsurf6 Glad you like it Eric. This should be right up your street. Try it out, they're easy to make or buy.
@pirsquared99 That's right. I had a few requests for this, especially after the Scam School episode - which is very good, people should check that out too. Yes, as you saw, the dotted line means it's about 50-50, the chances are slim in that case.
@FHomeBrew Dice may also be used as the singular. Especially in casual conversation. So ner! :)
@kkriegg No it lowers! And with the three set of dice the probabilities reverse order so if A beats B with a greater than 50% probability, two As beat two Bs with a less than 50% probability. I wrote an article with more details linked in the description.
Game #3: In a 5 (points on circle) -ring, there are only 2 dyads (ways to pick 2 dice): 14 and 23. This limits the game to the basis of only 2 probability pentagrams. It's generalizable on all non-trivial rings and die counts. Polya did it, and Howard Hanson (Rochester Professor of Music).
I like the description that says "one die" which is very meaningful in the context of playing dice games with friends :))
Oh Grime! You were like this as a Kid weren't you? You were lucky enough to find that your obsession could become your occupation. This stuff does my head in, and I love it!!
7:08 "Richard Of York Gave Battle In ...Wayne"
I've got massive respect for Marcus Du Sautoy for promoting maths learning to the masses, but you're better when it comes to taking that to the next level of interest. You would be absolutely great as a RI Xmas Lecturer. Go and treat yourself to a biscuit or a bit of cake.
In the 2nd loop of the 5 dice problem, [so when you are using 2 dice.] The Red dice has approximately a 51.77% chance to beat the White dice. Its approximately 50% but the red dice actually WINS.
The probability of the red dice winning is the probability that either the 9 face of ANY red dice is rolled or that the 0 face of ANY white dice is rolled. Any other combination not including one of those faces results in a white win so we dont have to calculate them.
So the probability of the first red dice being a 9 is = 1/6 chance
The probability that the 2nd red dice being a 9 and the first dice not being a 9 is = (5/6)x(1/6)
The probability of the 3rd being 0 and first 2 not being 9 is = (5/6)x(5/6)x(1/6)
The probability that the 4th being 0 and the first 2 not being 9 and 3rd not being 0 is = (5/6)x(5/6)x(5/6)x(1/6)
You can stop calculating probabilities once you hit a 9 or a 0 because every branch on those paths lead to a Red victory so the entire brach is just a red win so calculating the further probability isnt necessary. [Thats why we stop at 1/6 for the first red dice but the 2nd we count the 1st red dice missing and then stop when the 2nd rolls a 9 etc for the other 2 dice]
Thus, the probability that Red beats White is equal to the probability that either at LEAST one 9 OR 0 is rolled which is equal to
(1/6)+[(5/6)x(1/6)]+[(5/6)x(5/6)x(1/6)]+[(5/6)x(5/6)x(5/6)x(1/6)] = 51.77% [about]
There's no general construction that I know of. But you can always make nontransitive dice, for any given number of dice.
You can do a similar thing with Texas hold'em poker. You take three starting hands: AK offsuit (os), QJ suited (s) and 22. First your opponent chooses a hand, then you, and then you deal the full board and see who wins. No matter what your opponent chooses, you can always be ahead because in terms of the probability of a win when dealing the full board: 22 > AKos, AKos > QJs, QJs > 22. The margin is small though, so you'd have to play a lot of boards to see the difference!
This is fascinating! Also, I miss David Spiegelhalter! I hope he posts some more in the future.
02:37 "As you have seen", *three crosses pop up* "There was some witchcraft involved."
D&D and WarHammer tabletop players just got their minds blown with this video. I guess.
Also, I'm just wondering... Isn't there a simple calculation that determines the 'force' of the die? Perhaps by adding up the dots? As a layman, my instinct would be to add up the number of dots on the dice to determine how 'powerful' each one is.
+Generic Internetter never heard that term but you can figure out the average output value of the die by adding up the dots and dividing by the sides. which would be like the "force" of it
but since the game isnt a test of the total output value, but rather how often you win then a 6 or a 2 has the same power when against a 1
@squaltch Yeah, "Richard of York gave battle in vain" is more poetic! :)
I saw a lecture by David Spiegelhalter today :)
At 0:32 the little text that pops up should be "Take that, Dog on a Skateboard!"
When I read it, I thought you were suggesting I take a dog on a trip with me using a skateboard for locomotion.
OOOOOOH. l wondered wether he had a small stroke
I shouldn't play this with my sisters... it would piss them off pretty bad...
But very cool! Thanks for the video, Dr. James!
I saw a diagram of the Chinese elements for the first time and this video INSTANTLY popped to mind. The resemblance is uncanny.
@WhiteDragonTile Yeah, sorry, full credit to WhiteDragonTile who challenged me in the first place (this is an updated version of the dice, these are better).
@rg0057 Aren't you a grumpy-drawers. The constructions for 3, 4, and 5 dice sets are on the screen. If you want a general construction for n dice, there's quite a dull one that would work for up to 6 dice, but I don't think that would add anything to the sets I've already shown you.
I do not know of any other general construction of non-transitive dice - which is why I don't mention any, they are not easy to construct, which is why finding the set of dice I describe is interesting.
Nice work James~ I'd really like to see a follow-up to this video reviewing Miwin's Dice and some games that could be played with those.
@singingbanana haha. Yes, seeing that diagram made the basics of probability finally click!
In a 1v2 game it should be still more expected that you win one player but lose to the other, because the die advantage probabilities are so close to 50%. If the whole set winning probabilty doesn't exceed 70,7% for you, then it is likely that you will lose to at least one of the players since the probability of winning them both drops below 50%.
Not quite. We have an even amount of rounds so you forgot to calculate for draws. You can get to an advantage as low as 53,6% and still have a probability over 50% for not losing a game, while your probability of winning a set is just 46,8%.
With an advantage of 7/12 (58,3%) like the one in the video for example, losing once is close to as likely as winning both (35,3% to 34,9%), win+draw is 25,3% and drawing both only 4,5%
Ah, yeah, totally excluded that, dunno why.
It's possible I just changed the labelling. The colours have changed too, and some of the other details. Both the video and the article are correct. The article gets updated.
Another example of a non-transitive game is Rock, Paper, Scissors. Which is equivalent to the Red, Yellow, Green dice game (without any way to flip the order, unlike your cleaver one dice two dice trick). And Monkey, Pirate, Robot, Ninja, Zombie: a souped up version of Rock, Paper, Scissors. Which is equivalent to the Red, Yellow, Green, Blue, White dice game (again with out any way to flip the order).
... or Rock, Paper, Scissors, Lizard, Spock!
@@mmaarriiaannnnee33Either way, you still can't flip the one loop.
What started out as a board game with James quickly escalated to a scene from Scanners.
@leungclj haha, have a look at the article in the description. I mention a game for three players that uses seven dice. But it's a trade, if you try to do it with fewer dice you'll have to add something in elsewhere, like I did with my extra rule.
this is interesting game theory, where A beats B which beats C, and C beats A, a bit like rock paper scissors in dice form! :)
I wonder if this video was the inspiration behind the latest Project Euler problem (#376)...I know that Dr. Grime and this video were the first things I thought of when I saw it!
if you don't have dice you can use a deck of cards and choose between AK, J10, 22
@IB112 A website called Grand Illusions sell them.
And I thought Rock-Paper-Scissors-Lizard-Spock was confusing. :)
this is essentially the same thing
It's simple:
Scissors cuts Paper
,
Paper covers Rock
,
Rock crushes Lizard
,
Lizard poisons Spock
,
Spock smashes Scissors
,
Scissors decapitates Lizard
,
Lizard eats Paper
,
Paper disproves Spock
,
Spock vaporizes Rock
,
and, as it always has, Rock crushes Scissors.
Loved the video. Just one small thing: the singular of 'dice' is 'die', so you pick one die, not 'one dice'.
I played 5 non-transitive dice game with my family, my mum and my brother Josh.
Mum selects a 4, 4, 4, 4, 4, 9 dice, Josh selects a 0, 5, 5, 5, 5, 5 dice and I selected a 1, 1, 6, 6, 6, 6 dice. After 10 games, I realized that I won most amount of times than two members of my family. The result is, me 6, Josh 3 and mum 1.
@singingbanana yes I've noticed, you fixed them so that the last two dice aren't equally matched. :)
The lottery is still played in the UK, even in Cambridge, so I'm wondering how good he is at helping the public understanding of risk ;)
I remembered seeing the 4 dice version some time ago, searched for silver and found this gold.
Two professors one doc
DO NOT WANT.
David's acting skills are #WIN
Fascinating.
My maths teacher did this with me the other week.
First we did it with 3 dice, so it reminded me of Rock Paper Scissors, then they added more and more!
We figured out the it works in alphabetical order (blue, magenta, olive, red, yellow)
But also in number of letters (red, blue, olive, yellow, magenta)
So blue beats magenta
Magenta beats olive
Olive beats red
Red beats yellow
But then yellow beats blue
ALSO
red beats blue
Blue beats olive
Olive beats yellow
Yellow beats magenta
Magenta then beats red!
Really confused my year 8 brain as probabilities is numbers.
Also if you double the, up, then something else happens - can’t quite remember this bit!
note* i was using the colors and dice listed in the link provided
This is incredible. Do you ever tour the Americas? I'd definitely come see your tour in say Oregon/Washington area.
That's quite cool. However, I usually love games like this where I will always win but this one I think is too much trouble to remember and to get the dice. But as I say, well done, it's all very clever!!!
@Tossphate Nom nom.
The green dice vs the red dice is the most interesting when you throw two of each. As the green dice has 5 fours and the red dice has 5 threes so it seems intuitive that green will usually win. However because red will win if green rolls any ones or if red rolls any sixes the actual odds of red winning are 1 - (5/6)^4 = about 52%.
So cool that you have Bill Murray in your video
This is similar to (well, actually the same as) using 9 cards numbered 1 through 9.
The 1-6-8 cards have Red backs; the 3-5-7 cards have White backs; and the 2-4-9 cards have Blue backs.
Then, playing one card of one color against one card of a different color,
one number will be higher than the other and in 55% of the plays
Red beats White
White beats Blue
Blue beats Red
How many of each number are there in the respective players hand
@@thomaseckles4686 none. All cards placed face down. Each player picks one card from the pile. To have an advantage, player 1 lets player 2 pick first. If player 2 picks a white card, player 1 picks red. If player 2 picks a blue card, player 1 picks white. If player 2 picks a red card, player 1 picks a blue card. In each of these scenarios player 1 will win 5 times out if 9 plays, on average.
I bought the dice and the chains and colors illustrated in the instruction are different then this video.
They are:
red 4-9
blue 2-7
olive 5-0
yellow 3-8
magenta 1-6
not considering that the colors do not much, not only because are different, in this video the winning chain is different from the purchase you show:
4-9
3-8
2-7
1-6
5-0
which is right? Or I have to calculate myself which is the correct chain?
@dammilo yeah, and it's a version with a twist, so you can always win, if u can choose the rules ( 1 or 2 dices)
Math Rules =)
I root for the Cambridge Professor of Ornithology.
hoot
Allow me to pick a nit here: the singular form of the word "dice" is "die." Granted, we don't usually use the singular, because dice generally come in pairs, like scissors. But it is worth remembering that there is a grammatical form just _waiting_ for the odd occasion when it's appropriate to use it.
+B. Xoit it is "die" 0.o
i looked it up to double check, "dice" is the plural form of "die" for the definition of "die" we are talking about
+B. Xoit interesting etymology. but it in no way affects the reality of the meaning of the words. we are not speaking middle english
+BigBen Hebdomadarius
Well, Oxford Dictionaries trumps Guy on Internet.
Dice is the common singular form of (plural) dice.
B. Xoit fair point made. you are right "dice" is acceptable as the singular form of "dice".
that being said i still support us using "die" it makes our language more precise to be able to determine whether the speaker is referring to a single unit or multiple. in the same way we dont say "shoes" means the same as "shoe". the language can change and i think a language in which we can tell whether it is singular or plural is better than a language where we cannot.
Yes, but there's deer and deer, so... it's not like English is consistent on disambiguation.
0:45 I don't have two friends :')
@thenosa87 He was trying to teach people how probability works. It isn't a competition.
@StuziCamis True.
@GetMeThere1 1. I'm really bad at justifying things using 'real life', but you can think of nature as a non-transtie chain. It is something we want to avoid in elections. There's probably something really convincing I can't think of.
2. Patterns don't hold for triples etc. It's quite a balancing act, and these were carefully designed so a pattern held when you doubled the dice. But it's delicate so there's no particular pattern to what beats what after that.
He explained that you have to be able to choose the rules after your opponents have picked their dice, because that avoids this problem.
Probably my favorite response to any comment ever on UA-cam
David Spiegelhalter looks like Bill Murray's doppelganger in some shots.
Is it significant that, in the case of the original 3 transitive dice, that all 3 dice have sides that add up to 21?
Is it signficant that, in the case of the original three dice, that the sides of all three dice add up to 21? This doesn't appear to be the case for the other games presented.
This is like picking a starter in Pokemon games, the opponent always picks the one with a type-advantage over yours. It goes: Fire > Grass, Grass > Water, Water > Fire.
I guess It's also like rock-paper-scissors now that I think of it.
This was brilliant.
Oskar Van Davender did this ages ago and with far more dice xD
What happens when you throw all three die? Which will win in the long run?
the remarkable thing is that they will all tie, because they all have the same average roll (3.5). This is also the same average roll as a standard d6, so they will tie that as well
After doing the math, any of the non-transitive dice will have a 5/12 (41.6%) chance of winning against a standard d6, a 5/12 (41.6%) chance of losing, and a 1/6 (16.6%) chance of tieing (sp?). This is the same for all three dice.
For the two-die game, the key to finding the probabilities is to find the probabilities of each possible sum. On a d6, these odds are: (2: 1/36) (3: 2/36) (4: 3/36) (5: 4/36) (6: 5/36) (7: 6/36) (8: 5/36) (9: 4/36) (10: 3/36) (11: 2/36) (12: 1/36)
For the red die they are: (6: 25/36) (9: 10/36) (12: 1/36)
For yellow: (4: 9/36) (7: 18/36) (10: 9/36)
For green: (2: 1/36) (5: 10/36) (8: 25/36)
Going through, you will find that all dice have an average roll of 7, meaning that THEY will tie as well.
If you roll all five dice at once (plus a d6), the odds become much more different. Each die has a different average roll.
Red: 4.83
Yellow: 4.66
Green: 4.5
Blue: 4.33
White: 4.16
D6: 3.5
There is no contest. Using a standard D6 is futile, BUT if you change the numbers on that to make it 2-7 instead of 1-6, it will tie with the green die.
Some Random Fellow
Wow. Thanks for the detailed and very long explanation. Really appreciate it.
+Some Random Fellow
"the remarkable thing is that they will all tie, because they all have the same average roll (3.5)." Wasn't it established already that this shouldn't matter?
I think it's also wrong, let me calculate:
When it comes to the red die, it has a 1/6 chance of winning by definitions, when as when it rolls a 6, nothing gets higher. in the other 5/6 of cases the chance of winning is actually very low as the other 2 have to roll their low number, which in the case of the green die is already 1/6, and in the case of the yellow die 1/2.
But in total, the chance of the red die winning is 1/6+5/6*1/6*1/2=1/6+5/72=12/72+5/72=17/72. And this is already lower than 1/3 (24/72).
When it comes to the yellow die, the chance of winning becomes 5/6*1/2=5/12 because in the case where red doesn't roll a 6 it always wins half the time when it rolls a 5, and it always loses half the time when it rolls a 2 (as it's always lower than whatever the red die will roll), nonetheless its chance of winning is a very big chance (30/72).
When it comes to the green die, there's the same case of course that it'll always lose to red's 6, and it'll always lose in the case it rolls a 1 for which the chance is 1/6 and it will always lose when yellow rolls a 5 (1/2 chance), in the end its chance of winning will be 5/6*5/6*1/2=25/72, which is just over a third.
Concluding:
Red: 17/72
Yellow: 30/72
Green: 25/72
So Yellow is the best one in this case, actually by a rather large margin.
I get the same result as PauLtus B.
This dice game reminds me of the manga Gamling Maverick Kaiji which I feel is pretty clever.
I like a die with four 5s and two 1s, a die with four 3s and two 4s, and a die with two 6s and four 2s. The odds work out differently, but the same rules apply with both sets of one and two dice. I wonder how many different sets of these dice can be made.
Very interesting. Thank you. It has got me thinking!
"But you can always make nontransitive dice, for any given number of dice."- Dr. S.B. How do we know that to be true? (i hear a chorus of "proof! proof! proof!")
Counter-example - for two dice you cannot make non-transitive dice (single die game).
I think it was meant that you can make nontransitive dice for any given number of dice thrown, rather than any given number of players
@Gnomeslayer778 I've tried to fix that problem with the Maths Gear version. Using the colours, one chain is ordered alphabetically, the other chain is ordered by word length.
If you pick yellow, you can't beat both white and blue because they only interact on the second loop and yellow is in the middle of the path.
that is awesome.
I dont know how to design these games, but surely if you can do a 3 player system, you its probably possible do a 4 player system or an n player system?
This is an interesting case of natural properties defying immediate intuition. It's tough to argue that the counterintuitive non-transitivity is false while immediate intuition is true. But lately I've been arguing that Cantor's non-intuitive work may actually not be as correct as it is accepted to be. I made a video titled "Does Counting Exist?" to illustrate this point. It might be a good idea for legitimate mathematicians to correct any errors in the video.