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. 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.
