The Game You Can Almost Always Win

Vsauce! Kevin here, with a homemade deck of 52 memecards to show you a game that should be perfectly fair… but actually allows you to win mostof the time.

How? BECAUSE.

There’s a hidden trick in a simple algorithmthat if you know it, makes you the overwhelming favorite even though it appears that bothplayers 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” becauseit’s binary: it has just two outcomes when you flip it, heads or tails, and each of thoseoutcomes are equally probable.

Although.

.

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it could land on its edge… in1993, 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 electrumcoins 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-craftedmeme legends.

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

Crying Carson.

It’s perfectly binary, and there’s noway 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 theoreticalcoin-flipping code and take advantage of a secret non-transitive property within thisgame? Yes.

Welcome to the Humble-Nishiyama RandomnessGame.

But before we get into that.

Look at my shirt! I’m really excited to announce the launchof my very own math designs.

This is Woven Math.

The launch of my very own store bridging recreationalmathematics and art.

This is the Pizza Theorem.

These are concepts that I’ve talked abouton Vsauce2 like this Pizza Theorem or also the Achilles and the Tortoise paradox.

And my goal here is to take cool math conceptsactually seriously and create soft, comfortable shirts I actually want to wear.

So there’s a link below to check them — thisis the first drop ever there will be more to come in the future — but I just reallylike the idea of blending clean sophisticated designs with awesome math.

And these shirts just look cool so that whenyou wear them people ask, “What is that shirt?” and then you get to explain awesome math conceptslike Achilles and the Tortoise or The Pizza Theorem.

So I think it’s great, I think that youwill too, check out the link below now let’s get back to our game.

Walter Penney debuted a simple coin-flippinggame in the October 1969 issue of The Journal of Recreational Mathematics, and then SteveHumble and Yutaka Nishiyama made it even simpler by using playing cards.

The first player chooses a sequence of threepossible outcomes from our deck of cards, like red, black, red.

And then the second player chooses their ownsequence of three outcomes.

Like black, black, red.

And then we just flip our red and black memecards and the winner is the one whose sequence comes up first.

So in this example, thanks to Guy Fieri, playerone 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 andblack cards after each draw, the probability won’t be exactly 50/50 on every draw — becauseeach time we remove one colored card, the odds of the opposite color coming next isslightly higher — but it’ll never be far from perfectly fair, and as we play it willcontinue to balance out.

So given that each turn of the card has aroughly equal chance of being red or black, and given that the likelihood of each sequenceof three is identical, the probability that both players have an equal chance of winningwith their red-and-black sequences has to be 50/50, too, right? Wrong — and to demonstrate why I’ve invitedmy 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 Keanucan choose: RRR, RRB, RBR, RBB, BRR, BRB, BBR, and BBB.

No matter what Keanu chooses, the probabilityof 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 goingto choose BBR.

Okay, now we'll just draw some cards and seewhich 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 nowthere's no way that I can lose.

Because of having these two black cards ina row, even if I pull five more black cards in a row, eventually I will get a red andI will win.

Ermergerd.

Ah hah.

There it is.

Minecraft Steve had sealed the victory forme.

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 pickedBRB, 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 sequencebecomes BBR.

I'll show you another example.

If Keanu had chosen red, red, red.

Then I would've just changed that middle Rto a B, put that at the beginning of my sequence, drop the last R, and my sequence would beBRR.

Just that little trick allows me to have anadvantage anywhere from about 2 to 1 up to 7.

5 to 1.

Which means in the worst possible scenariofor 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 optionsand their odds.

As Player 2, when we apply the algorithm we’rejumping into exactly the right place in a cycle of outcomes that Player 1 doesn’thave any control over.

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

How is this possible? How can I take something so seemingly fairto both players, so obviously 50/50, and turn it so strongly in my favor? The key is in recognizing that this game isnon-transitive.

So there ya go.

The end.

Wait… What is transitive? Think of it this way: you’ve got A, B, andC.

A beats B, and B beats C.

Therefore, A beats C.

Because if A beats B and B beatsC then obviously A can beat C.

That game sequence is transitive.

So like if you and your Keanu had transitivefood preferences, you'd rather have Pizza than Tacos, and you’d rather have Tacosthan Dog Food.

You’d also rather have Pizza than Dog Food.

Simple.

If you and Keanu somehow preferred Dog Foodto Pizza, then all of a sudden your food preferences become non-transitive.

In a non-transitive game, there is no bestchoice for the first player because there’s no super-powered A.

Instead, there’s a loopof winning choices… like rock, paper, scissors.

In rock paper scissors, rock — which we’llcall 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, andpaper B beats rock A, so B is better than A.

Scissors C loses to rock A and beats paperB — and we’ve got a loop of possible outcomes that goes on forever, with no one choice beingstronger than the other.

That’s non-transitive.

Since we’re in the flow chart mood here’sa flow chart that illustrates the player 2 winning moves in the Humble-Nishiyama Randomnessgame.

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

And so forth.

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

In the Humble-Nishiyama Randomness variationof 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 choicethat 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 guaranteesus 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 scissorsto multiply.

Benjamin Kirkup and Margaret Riley found thatbacteria compete with one another in a non-transitive way.

They found that in mice intestines, E.

colibacteria formed a competitive cycle in which three strains basically played a game of rockpaper scissors to survive and find an equilibrium.

Penney’s Game and its variations illustratehow even a scenario that seems perfectly straightforward, like unmistakably simple, should never betaken at face value.

There’s always room to develop, strategize, and improve our odds if we put in the effort and imagination required to understandingthe situation.

And that truly is…breathtaking.

And as always — thanks for watching.

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