The Fair Game That’s Totally Unfair


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.

100 thoughts on “The Fair Game That’s Totally Unfair”

  1. I can't sell the meme cards but I hope you enjoy Woven Math! Wrap your body in sweet, sweet knowledge. https://represent.com/store/vsauce2

  2. But you didn't tell why rearranging the pattern helps. Did those sequences just happen to defeat each other in a specific way ?

  3. “Great choice there, Keanu”

    Thanks, Kevin😎

    I’m really enjoying this Keanu Reeves popularity. It’s nice hearing my name😅

  4. I just watched the recent LQG episode on PBS spacetime, and now I want to know: Can we simulate spacetime through non-transitive gamelogic and maybe the amount of variables which stringtheorie suggests as dimensions?

  5. Kevin: As Player 2 here, the method is very easy.
    Sprint guy: Isn’t it nice when you keep things simple?
    Me, not currently looking at the screen: Yes, it i- Hey you’re an ad!

  6. I guess you could make it so that each player has to write it down secretly at the same time… or do that, but then make it so you are deciding your opponents sequence, and vice versa

  7. Can someone please explain the top entry (BBB vs RBB) in the table of odds in favour of P2?
    By that I mean that the odds when P2 counter picks RBB vs P1's BBB are 7.5:1 when it should be 7:1, as the only way P1 will win is if 3 Black cards are dealt right at the beginning, as if a red card is drawn in any of the first 3 cards it would be a theoretical win for P2
    This means player 1 has a 1/8 chance of winning so the odds should be 7:1 not 7.5:1.

    Are these odds based on the fact that there are only 52 cards in the deck and there is a small chance that the game ends in a tie? And if so, would the odds be changed back to 7:1 if this game was played with rounds of coin tosses instead of cards?
    Or am I missing something here in my calculations?

  8. How to make the game fair: just have both players choose at the same time like Rock Paper Scissors. By the way this game is garbage.

  9. The solution for P1 is to just propose to write down your chosen pattern on a paper first before starting the game. That way P2 has no chance of knowing what pattern P1 chose and therefore would also have to guess.

  10. I would or really liked you to explain what was actually going on more than just saying the game is nontransitive. like why is it nontransitive and why is BBR a better choice than RBR to beat BRB/BRR

  11. I miss the old Vsauce2 video styles like "The Invention Of Blue" and "The Planet Behind Your Eyes" instead of these game/puzzle videos.

  12. 10:05 How can it be that BRR beats BRB with a 2:1 ratio, meanwhile RRB beats RBR with a 3:1 ratio? Shouldn't it be the same probability, since it's the same sequence just with flipped colours?

  13. I guess this is why we have a base 4 genome, and not base 2.
    And why the start codon is made of 3 different bases (AUG). Had it been AGA for example, 2 point mutations prior to AGA can move the codon back and start translation prematurely.

  14. The real trick is to know what player 1 did before it's your turn. The transitive / non-transitive thing does not matter at all. For rock-paper-scissors, if you know what the other player chooses beforehand, you can also always win.

  15. Is it weird everything that I watch i would have done like 3 days before at most
    (Unless we talking einstein mega super science)

  16. So… all you have to do to make the game fair is to choose sequence at the same time, or just not reveal it until both has chosen? Since it's entirely based on player 2 knowing both the trick and player 1s sequence, seams like a easy workaround to make it perfectly fair then.

    Still… having knowledge of how to turn a fair game so very tiled in ones favor might come in handy sometime….

  17. @Vsauce2 I'm not happy with this video for the reason recently mentioned by @Vsauce in the video named Laws And Causes. You did explain that the behaviour we see is called Non-Transitivity but you failed to explain the cause of that in this specific case. Looking at your chart near the end, I still do not know why the odds are they way you drew them.

  18. How can you know that a game is transitive or not ?
    And how do you calculate the odds of "duels" of triplets of cards ?

  19. I have a question: the advantage appears when the second player knows in advance the choice of the first player. What happens if player one and two have to blindly choose the sequence and then communicate it at the same time? Does the advantage still appear (as in tic tac toe, where I know for sure that I need to begin with a corner or the center)?

  20. Even if your opponent wasn't picking to counter pick you, wouldn't three of the same color always have a disadvantage, making this game unfair from the start?

    For why, if you take three of the same color, there are two points at which you can fail the streak, at zero streak, at one and at two. Basically you can completely reset to zero progress at any point in the streak. Unlike red black black, which can only go to zero progress from zero. Once a red card has been played, the furthest you can go back is to one red card. And even red black red (and black red black) should have an advantage over full one color, as the only possible full resets are at zero and two cards, as picking Red means you are always two cards away from winning, but picking two blacks in a row would reset you. Red red black and black black red can't completely reset after a streak of two, meaning winning is always one coin flip away after you get two correct. Basically every combination of cards has an advantage of some kind in terms of chain breaking points, except for pure one color which can break at any point.

    If someone knows why my thinking is wrong, feel free to tell me. I just can't currently see a flaw in that thinking.

  21. When Kevin has a 50/50 chance of properly pronouncing Keanu's name every time he says it. "Kee-Aw-New" vs "Kee-Aw-No"

  22. what are the odds that Kevin chose Keanu to be his celebrity friend just so he could make the reference at the end?

  23. I study computer sciences at an university and nobody explained transitivity as simply as you did, Kevin. Never thought of "liking food" and rock, paper, scissors to be such great examples for transitivity and non-transitivity!

  24. i can win rock paper scissors everytime by choosing rock

    rock beats scissors and scissors beats paper

    so rock can beat paper and scissors

  25. Wait according to my calculations, the RBB to BBB and BRR to RRR win ratio should be 8.5:1.
    I could be wrong so please correct me if I am.

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