How Probability Helps Us Understand an Unbeatable Game

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Let's say I tell you I've got two dice, and when I roll one of them, we'll call it die A, it beats the other one, we'll call that die B, 58% of the time. First of all, you'd know something fishy is going on, right? It should not be the case that between two fair dice, one of them wins so often.
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But, hear me out, I'm going to let you choose whichever of these dice to roll and try to beat me. So if you think one is better than the other, you're free to pick that die first. There is a third die, actually, we'll call that die C, but die B beats that one 58% of the time as well, so surely you wouldn't want to pick die C.
Well, surprising as it may be, even though die A beats die B most of the time, and die B beats die C most of the time, die C actually beats die A at an even higher clip than either of the first two. Given the dice I'm using in this simulation, die C will beat die A about 69% of the time.
This is the beauty of what are called intransitive dice. These dice are not unfair in the sense that they're weighted (in fact, if they were weighted, it's less likely that this property ever emerges). Instead, the dice all have the same expected value, but very different modal probabilities. It's that difference that I explore here.
If you'd like to play around with the simulation itself, I modeled it out in @Desmos , and you can find that graph here: www.desmos.com/calculator/bs1....
If you'd like to see the sample spaces, those were also modeled in Desmos:
+ Fair Dice Sample Space: www.desmos.com/calculator/mlo...
+ Intransitive Dice Sample Space: www.desmos.com/calculator/czl...
If you'd like the "net" of a die, you can see that here: www.desmos.com/calculator/upt....
Finally, if you'd like to read more about intransitive dice, this Quanta article inspired me to make the original simulation: www.quantamagazine.org/mathem....
#probability #probabilitytheory #intransitivedice #thegameyoucannotwin
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[@slapthesloper]

There is a game for the Nintendo switch called Mario party 10 where you roll dice to move around a board but what’s super cool is each character has its own unique die along with a regular die that you can choose to roll if you want.
They all have different means and variances with values from 0 to 10 on each side and some of the 0s have additional minor negative effects to you in the game (losing coins)
As a math major and a good video game player I quickly realized I should always use the character with sides 0/lose coins,0/lose coins,0,8,9,10 (bowser) because it has the highest mean and moving around the board quickly is ideal while also me being good at the games offset the frequent loss of coins.
However there are still niche cases where you may want a character with the dice 3,3,3,3,4,4 if there is some specific space you want to have a higher chance of landing on.
I think this imbalanced die concept is an easy way to add a ton of complexity into games and wish it was more common

[@polymathematic]

very cool! i had no idea.

[@KnThSelf2ThSelfBTrue]

i feel like you can extrapolate from this lesson to understand how to design a mathematically diverse meta for a competitive game.

[@parmesanzero7678]

That was actually my first thought!

[@SSNewberry]

Already been done.

[@devsutong]

rock, paper and scissor in the realm of mathematics

[@carstenlechte]

So, the really clever person asks "what about C vs A are you not telling me?"

[@Wulf169]

One could make die 1,1,1,6,6,6 or 1,1,3,4,6,6 from 1,2,3,4,5,6 standardized dice each pair summing to 7 and be swapped in and out for each other. but you can also go 1,2,3,5,5,5 instead 2,2,2,5,5,5 you put down. You could readily build a table showing all possibilities, a set of 6 dice that evaluate to 21 given the numbers 1-6 can be used. 2,2,2,4,4,7 if expand the rule using numbers 1-7 but still only 6 dice that must equal 21.
Also, if one changes the rules slightly so there were multiple winners. Then A>B>C and A>C>B all equates to A winning 1 bet every time with those outcomes and b and c would hold in different intervals 1 bet. You end up effectively with a win, draw, lose instead of a win, lose, lose. See we can bring that draw back if we want.

[@buzzynut]

Gerrymandering: Dice Edition!

[@camronstringer9745]

do you recommend any books to use to study for the putnam?

[@polymathematic]

i've never studied for the putnam (for myself), so i'm not sure! i'll have to look into it.

[@jofx4051]

I only can recommend website which probably most people into math know... Art of Solving Problem

[@SC-dm1ct]

A lot of cheap d6s are already weighted, simply because they weren't adjusted. What I mean is the side with more indents has less weight on that side than the sides with fewer indents. There's less mass there.

[@jimburnsactual]

Correction: That A beats C most of the time

[@bozhidarmihaylov]

Neutrino Dice 😊

[@MF-kr4hf]

11:00 I don't get why he can just assume C is a 1 and C is a 4 for his calculations!?

[@polymathematic]

When I show the net for the die, die C only has faces that are 1 or 4. So we calculate the probabilities for each of those two faces possibilities only.

[@MictheEagle]

I think we better use the term "rolls higher than" (as in, "If A rolls higher than B and B rolls higher than C) rather than the word "beats" (as in, "If A beats B and B beats C") if it is to be clear what values we are comparing.
C=1,4,4,4,4,4 A=6,3,3,3,3,3
When rolled, there are 36 possible outcomes (6x6). C's side is greater than A's side 25 times (5x5) while A's side is greater than C's only 11 times (1x6 + 5x1).

[@vpambs1pt]

here you meant the the converse right? 1:10, A beats C by transivity.

[@yourfutureself4327]

💚

[@iLuminoM]

I chose b

[@marceames4670]

I have bridge I’d like to sell you

[@devsutong]

then let me choose a 😂

[@CheckmateSurvivor]

Just posted the most difficult puzzle in the world. Please give it a try.

[@brinleyhamer729]

101 69

[@Crazyapple16]

Bro didnt vsauce 2 kevin make the same exact video years ago

[@polymathematic]

being that i'm in this video, it seems pretty unlikely!

[@SSNewberry]

There are a lot of instances where A

[@polymathematic]

sure! in this particular case, there's B>C>A. but there's no C>B>A (again, with these particular dice).

[@SSNewberry]

@@polymathematic I took some time from studying for LSAT tests to read the rest of the entries. I knew what you were doing here but you did a fine job explaining it.