Vsauce!

Kevin here, with a homemade deck of 52 meme cards to show you a game that should be perfectly

fair… but actually allows you to win most of the time.

How?

BECAUSE.

There’s a hidden trick in a simple algorithm that if you know it, makes you the overwhelming

favorite even though it appears that both players have a perfectly equal 50/50 chance

to win.

Well, that’s not very fair.

What is fair?

We can consider a coin to be “fair” because it’s binary: it has just two outcomes when

you flip it, heads or tails, and each of those outcomes are equally probable.

Although... it could land on its edge… in 1993, Daniel Murray and Scott Teale posited

that an American nickel, which has a flat, smooth outside ridge, could theoretically

land on its edge about 1 in every 6,000 tosses.

But for the most part, since the first electrum coins were tossed in the Kingdom of Lydia

in 7th century BC, they’ve been pretty fair.

As are playing cards, like this deck of hand-crafted meme legends.

When you pull a card, you get a red or a black card.

Crying Carson.

It’s perfectly binary, and there’s no way for a card to like, land on its edge.

It’s either red or it's black.

It's 50% like a snap from Thanos.

Given that, is it possible to crack the theoretical coin-flipping code and take advantage of a

secret non-transitive property within this game?

Yes.

Welcome to the Humble-Nishiyama Randomness Game.

But before we get into that.

Look at my shirt!

I’m really excited to announce the launch of my very own math designs.

This is Woven Math.

The launch of my very own store bridging recreational mathematics and art.

This is the Pizza Theorem.

These are concepts that I’ve talked about on Vsauce2 like this Pizza Theorem or also

the Achilles and the Tortoise paradox.

And my goal here is to take cool math concepts actually seriously and create soft, comfortable

shirts I actually want to wear.

So there’s a link below to check them -- this is the first drop ever there will be more

to come in the future -- but I just really like the idea of blending clean sophisticated

designs with awesome math.

And these shirts just look cool so that when you wear them people ask, "What is that shirt?"

and then you get to explain awesome math concepts like Achilles and the Tortoise or The Pizza

Theorem.

So I think it’s great, I think that you will too, check out the link below now let’s

get back to our game.

Walter Penney debuted a simple coin-flipping game in the October 1969 issue of The Journal

of Recreational Mathematics, and then Steve Humble and Yutaka Nishiyama made it even simpler

by using playing cards.

The first player chooses a sequence of three possible outcomes from our deck of cards,

like red, black, red.

And then the second player chooses their own sequence of three outcomes.

Like black, black, red.

And then we just flip our red and black meme cards and the winner is the one whose sequence

comes up first.

So in this example, thanks to Guy Fieri, player one would've won this game because player

one chose red, black, red.

Here's a little bit of a nitpick.

Because we’re not replacing the red and black cards after each draw, the probability

won’t be exactly 50/50 on every draw -- because each time we remove one colored card, the

odds of the opposite color coming next is slightly higher -- but it’ll never be far

from perfectly fair, and as we play it will continue to balance out.

So given that each turn of the card has a roughly equal chance of being red or black,

and given that the likelihood of each sequence of three is identical, the probability that

both players have an equal chance of winning with their red-and-black sequences has to

be 50/50, too, right?

Wrong -- and to demonstrate why I’ve invited my best friend in the whole wide world.

Where are you best friend?

Keanu Reeves.

Ah, alright.

Keanu, you're a little tall.

Hold on.

How's this?

Okay, Keanu will be player 1.

There are only 8 possible sequences that Keanu can choose: RRR, RRB, RBR, RBB, BRR, BRB,

BBR, and BBB.

No matter what Keanu chooses, the probability of that sequence hitting is equal to all the

other options.

For player 1, there really is no bad choice, one choice is as good as the next.

So let’s say Keanu chooses BRB.

Great choice there, Keanu.

Now that I know your sequence, I’m going to choose BBR.

Okay, now we'll just draw some cards and see which sequence appears first.

Red, Tommy Wiseau.

Some red Flex Seal action.

So far nobody has an advantage.

Black, where are you fingers?

Uh oh.

Sad, sad Keanu.

You should be sad once you realize that now there's no way that I can lose.

Because of having these two black cards in a row, even if I pull five more black cards

in a row, eventually I will get a red and I will win.

Ermergerd.

Ah hah.

There it is.

Minecraft Steve had sealed the victory for me.

Sorry, my most excellent dude.

But I win.

Because regardless of what player one chooses, what matters is the sequence that player 2

picks.

As player 2, the method here is very easy -- I just put the opposite of the middle color

at the front of the line, so when Keanu picked BRB, I changed his middle R to a B, and then

put that in the front of my sequence.

So I just dropped the last letter and my sequence becomes BBR.

I'll show you another example.

If Keanu had chosen red, red, red.

Then I would've just changed that middle R to a B, put that at the beginning of my sequence,

drop the last R, and my sequence would be BRR.

Just that little trick allows me to have an advantage anywhere from about 2 to 1 up to

7.5 to 1.

Which means in the worst possible scenario for me, I win 2 out of 3 times.

And in the best, it’s nearly 8 out of 9.

Look, I’ll write out all the choice options and their odds.

As Player 2, when we apply the algorithm we’re jumping into exactly the right place in a

cycle of outcomes that Player 1 doesn’t have any control over.

The best Player 1 can do is choose an option that’s the least bad.

How is this possible?

How can I take something so seemingly fair to both players, so obviously 50/50, and turn

it so strongly in my favor?

The key is in recognizing that this game is non-transitive.

So there ya go.The end.

Wait…

What is transitive?

Think of it this way: you’ve got A, B, and C. A beats B, and B beats C. Therefore, A

beats C. Because if A beats B and B beats C then obviously A can beat C. That game sequence

is transitive.

So like if you and your Keanu had transitive food preferences, you'd rather have Pizza

than Tacos, and you’d rather have Tacos than Dog Food.

You’d also rather have Pizza than Dog Food.

Simple.

If you and Keanu somehow preferred Dog Food to Pizza, then all of a sudden your food preferences

become non-transitive.

In a non-transitive game, there is no best choice for the first player because there’s

no super-powered A. Instead, there’s a loop of winning choices… like rock, paper, scissors.

In rock paper scissors, rock -- which we’ll call A -- loses to paper, which we'll call

B. B is better than A. But A beats scissors, which is C. So A is better than C.

But B loses to C, so C is better than B, and paper B beats rock A, so B is better than

A.

Scissors C loses to rock A and beats paper B -- and we’ve got a loop of possible outcomes

that goes on forever, with no one choice being stronger than the other.

That’s non-transitive.

Since we’re in the flow chart mood here’s a flow chart that illustrates the player 2

winning moves in the Humble-Nishiyama Randomness game.

So if you follow the arrows you can see that like RBB beats BBB and like BBR beats BRB.

And so forth.

With the odds added, you can clearly see how some sequence scenarios go from bad to worse.

In the Humble-Nishiyama Randomness variation of Penney’s Game, we know what sequence

of card colors player one has chosen first, so we can jump in the most advantageous part

of the non-transitive loop and make a choice that gives us a significant advantage.

By recognizing that the game is non-transitive, we take seemingly-obvious fairness and find

a paradoxical loophole that nearly guarantees us success.

To everyone who doesn’t recognize the intransitivity, it just kinda looks like we’re extremely

lucky.

And why does all of this matter?

Because bacteria play rock paper scissors to multiply.

Benjamin Kirkup and Margaret Riley found that bacteria compete with one another in a non-transitive

way.

They found that in mice intestines, E. coli bacteria formed a competitive cycle in which

three strains basically played a game of rock paper scissors to survive and find an equilibrium.

Penney’s Game and its variations illustrate how even a scenario that seems perfectly straightforward,

like unmistakably simple, should never be taken at face value.

There’s always room to develop, strategize, and improve our odds if we put in the effort

and imagination required to understanding the situation.

And that truly is…breathtaking.

And as always -- thanks for watching.