When the Wright brothers

had to decide who would be the first to

fly their new airplane off a sand dune,

they flipped a coin. That was fair: we all know there’s an equal chance

of getting heads and tails. But what if they had

a more complicated contest? What if they flipped coins repeatedly, so that Orville would win as soon as

two heads showed up in a row on his coin, and Wilbur would win as soon as heads

was immediately followed by tails on his? Would each brother still have had an equal

chance to be the first in flight? At first, it may seem they’d still have

the same chance of winning. There are four combinations

for two consecutive flips. And if you do flip a coin just twice, there’s an equal chance

of each one — 25%. So your intuition might tell you

that in any string of coin flips, each combination would have

the same shot at appearing first. Unfortunately, you’d be wrong. Wilbur actually has

a big advantage in this contest. Imagine our sequence of coin flips

as a sort of board game, where every flip determines which

path we take. The goal is to get from start to finish. The heads/tails board looks like this. And this is the head/head board. There’s one critical difference. Heads/heads has a move that sends you

all the way back to the start that heads/tails doesn’t have. That’s why heads/heads

takes longer on average. So we can demonstrate that this is true

using probability and algebra to calculate the average number of flips

it would take to get each combination. Let’s start with the heads/tails board, and define x to be the average

number of flips to advance one step. Focus only on the arrows. It has two identical steps, each with a 50/50 chance of staying

in place or moving forward. Option 1: If we stay in place by getting

tails, we waste one flip. Since we’re back in the same place, on average we must flip x more times

to advance one step. Together with that first flip, this gives an average of x + 1

total flips to advance. Option 2: If we get heads

and move forward, then we have taken exactly one total flip

to advance one step. We can now combine option 1

and option 2 with their probabilities to get this expression. Solving that for x gives us an average

of two moves to advance one step. Since each step is identical, we can multiply by two and arrive

at four flips to advance two steps. For heads/heads,

the picture isn’t as simple. This time, let y be the average number

of flips to move from start to finish. There are two options for the first move,

each with 50/50 odds. Option 1 is the same as before, getting tails sends us back to the start, giving an average

of y+1 total flips to finish. In Option 2, there are two equally

likely cases for the next flip. With heads we’d be done after two flips. But tails would return us to the start. Since we’d return after two flips, we’d then need an average

of y+2 flips in total to finish. So our full expression will be this. And solving this equation

gives us six flips. So the math calculates that it takes an

average of six flips to get heads/heads, and an average of four

to get heads/tails. And, in fact, that’s what you’d see if you

tested it for yourself enough times. Of course, the Wright brothers didn’t

need to work all this out; they only flipped the coin once,

and Wilbur won. But it didn’t matter:

Wilbur’s flight failed, and Orville made

aviation history, instead. Tough luck, Wilbur.

## 100 Comments

If you enjoyed this video, check out the lessons in our “Can You Solve This Riddle?” playlist: http://bit.ly/1NJ3CwS

Is this weird that this question appeared in my test?…… no I’m serious it did show up on the test I took

Just. Why. Did. They. Put. It. In. The. Test?!?!

No

Now next time we do a coin toss to go home early at work, I'll insist on the two flip game. (And call heads – tails for myself!) #beatthesystem #math

i feel smart af

To all those who are saying it is misleading and made us assume that there was 1 coin, listen carefully here 0:24

" so that Orville would win as soon as two heads showed up in a row onhiscoin,and Wilbur would win as soon as heads was immediately followed by tails onhis?"1:06

B I G A D V A N T A G E

Okay here is what I don't get though.

The situation where You have heads and then another heads (which would not bring Wilbur back to the start and is the advantage stated here) would be the win for Orville, so how in the end it is still 50/50 that one wins, right?

I mean, the base that both need for winning is having heads first.

Afterwards it is again a 50/50 chance to either land on heads or tails.

Am I wrong in thinking this?

Why are you considering 2 different scenarios ….afterall they are flipping just one coin right!

That's wrong cuz if there are two heads it can't remain cuz the other one wins

There's even more of an advantage if it's tails-heads vs heads-heads. Tails-heads has a huge advantage because heads-heads can only win if the first 2 coins are heads.

I tried this out immediately after the video and i got 2 heads in a row. This being luck i tried again and yet again 2 heads in a row… hmmmmm…?

Terrible explanation. Misleading information.

But why don't you just do a normal coin flip

There is a flaw in this idea. When you first flip a heads you both are tied and you are both at the same part in the diagram. And say you flip a tails next. It wont matter whether or not you set the HH back because you will have already won, and if you flip a heads then the HH will have already won.

Okay, I understand this. Let's say you have HT. If you flip heads, then all you have to do is flip tails. No matter what you get, you'll never have to go back to start. It's either you flip tails and win, or flip heads again. If You're doing HH, you'll have to flip two heads consecutively. If you flip tails, you'll have to start over from 2 flips. In HT, you don't have to start over because you'll always be flipping heads, which counts as your 1st flip.

But if they win as soon as their combination appears ( not if it appears more often ) then they should have the same chance of winning because they both start with the same face, if heads is thrown, then the next coin flip would immediately decide the winner and make the scenario a 50 50. overall it was a good video about probability and patterns but the premise used to explain it doesn't work

Moral of the story, tails never fails

I’m a geography major, so this kinda broke my brain.

The situation you proposed doesn't work like this, because you only have to win once. Because after a heads is flipped, it's a 50/50 chance to win, and if a tails is flipped, it's a 0/0 chance to win on the next turn.

It will be the same because when there will be tail it doesn't mean anything for both and when there will be head than each has 50% chance.

The video is wrong.

But to get HH you don't start over after getting HT because the winner will have already been decided…

If the first flip is T. Nobody is ahead and the flips start over.Once the coin flips to H for the first time, both players are one step forward.

The step after the first H will declare a winner. At that point, it becomes a 50/50 chance for H or T after the first H.

If they only flipped one coin with the same win condition, they would have the same chances

This scenario was extremely poorly clarified. It went from one coin to two coins, and two separate scenarios, with one coin only being able to reach HT, and one only being able to reach HH. How it was presented made it seem as if either both were flipping the same coin (as people generally do in any coin toss), or that either brother reaching either scenario would result in a win for one of them. Judging by this comments section they really need to point out first that each brother has his own coin, second that Orville reaching HT would mean nothing, and third Wilbur reaching HH would mean nothing.

If they were flipping a shared coin, rather than flipping their own coins independently, the probability would go back to 50/50

Understand the math but I don't think you asked the right question, it's not asking who will get more combinations it asked who will get it first

If it were, "if I get two different coins after flipping twice, I win. If you get heads twice, you win".

Then one person would win with HT or TH, and the other with only HH.

The only way to make that fair would be letting the person needing HH to also win with TT.

But the video doesn't say this…

I actually paused it and thought of the concept you mentioned good videos anyway

And this is assuming each flip can only be part of one pair. If flips 2 and 3 could pair up, that's more math then.

h u h

This is a simplification of Penney's Game.

Here's a Wikipedia article, to explain it better than I can:

https://en.wikipedia.org/wiki/Penney%27s_game

😐😐😐😐😐😐😐😐🤔🤔🤔🤔🤔🤔🤔❓❔❔❓❓❔❔❓

1:05 holy damn that head zoom scared me

It didn't clearly explain that each brother flips seperate coins. If it was the same coin then it would be 50/50 because when a head is flipped then the next flip is a 50/50 win for each person.

If it is going to be just math problem instead of a riddle, the question could have been more clear. For the longest time I thought the first winning combination would count BUT in fact they each throw their own coin, making it irrellevant what the other person flips. Of course takes less flips if a failed second throw, doesnt take you back to throw one.

i dot get yor idle but i get my idle

25% Red guy (If is get first Tails His dot have chance) 0% Win

25% ////////// (Restart: 2X Tails) 25% Restart

50% Blue Guy (IF i gat Tails I will have cance to win or to restart the game) 50% Win

Dot get it ?Tails Tails = Restart

Tails Heads = Blue guy

Heads Tails = Blue Guy

Heads Heads = Red guy

I don't think that's actually true, you calculated how much on average their combination would show up if you flipped the coin many times, but I thought the game was whoever gets their combination first wins, not whoever gets it more. So yeah, if you flipped the coin for a while Head-Tails would probably show up more than Head-Heads, but if the game stops as soon as one of them wins it is actually the same odds

if they used the same coin it would still be 50/50 : ^)

This is wrong because there is a 1% difference depending on which one is heads up. Sorry to change your math my friend

But the whole video is entirely wrong. Both participants require heads to start with so only the second flip matters, which is a 50/50. It really doesn't matter the frequency over a series because the first occurrence ends the game and they are equally likely to be first.

No offense but your definition is bad.

Let Bela and Andrei be two different people like the wright brothers.

Solution:

Let state 1 be one correct coin flip, and state 0 be no correct coin flips, and state 2 be all correct coin flips.

The reason Bela's expected value is more is because if we flip into state $1$ for Andrei, he has a 1/2 chance of going to state 2 or going to state 1. Where as Bela in state 1 has a 1/2 chance of going to state 2 or going to state 0. Therefore the expected value for Bela's should be higher.

This would only work if they were flipping different coins causing this to be pointless. Nevethteless, its interesting.

Wish this was explained since I'm good at maths yet I'd have to watch this over if I want to understand it.

I've just tried this, 6 flips, 4 heads heads, only 2 heads tails, not an average case!

Imagine doing math for a living

In Soviet Russia, coin flip

you!Oooohh I never knew Wilbur was first to try but Orville was first to fly

I figured out which had more of a change of happening in less than five seconds.

By throwing the same coin, the outcome would be 50-50 after the first heads. But by throwing different coins, a "heads tails" combo has a higher chance of occurring before a "heads heads" combo. This is happens because of one thing, a failure on the second flip (after the first heads) is easier to come back from if your win condition is "heads tails".

The following is a summary of what the video is sayingTo move on to the second phase of flipping, both players need to flip heads first, making them equal. They then each have the same chance of flipping the side that they need to win. Sounds even, right? Well if your win condition is "heads heads", then you are out of luck. If you flip tails as your second flip, you need to slip at least TWICE more to get heads followed by another heads. But, if your condition is "heads tails", a failed flip is another heads, putting you right back where you were before this flip, only requiring at least ONE flip to win, as a tails would lead to having a heads tails combo.

I hope this comment helps explain the video if it was not clear. Try reading this while looking at the diagrams for maximum knowledge absorption.

in this senario its still 50/50 isnt it? because both wait on heads and the flip after first head is 50/50

If there's only one coin and they're waiting to see which combination comes up first, they each have a 50/50 chance–so it's counter-intuitive that this changes when there are two separate coins.

~~hellu~~While the average might be unfair, it wouldn't matter if the brothers actually played the game:

If they start out with some number of tails they can ignore those, since the winning sequence starts with heads for either brother.

Eventually the first heads will come.

Now there are 2 possible follow ups:

Heads->bro1 wins

Tails->bro2 wins

So the probability for either brother remains 50-50

Please correct me if I'm mistaken or confirm if I'm into something.

Cheers

On the testing coins picture, the coins flipped for heads – tails flips head – heads first. The heads – tails is only statistically faster than heads – heads if they're either using different coins, which seems to be the case (which confuses us since it's not directly stated, only shown in the animation), or you're going for multiple wins.

The reason for this is because the second paths cannot be taken. After heads is flipped, either Orville or Wilber wins next flip, and this is true because the first flip is linked. Another way this problem could have been presented is if Wilber won on tails – heads instead of heads – tails; however, this would have made the solution more intuitive as after flipping tails once it's impossible to flip heads twice before flipping heads once, while after flipping heads, you can still flip tails without flipping heads again, this gives Orville a 20% chance to win

Still don't get it. In this example, the triggering factor is the first heads, because no matter how often tails shows in a row, neither of them can win if it isn't heads first.

After that, the next flip decides the game with 50/50, or am I missing something?

This video wrong, because it applies the math as if both brothers flip the coins for themselves, while the situation stated they fliped coins together that counted for both.

The way the situation was stated, it's fair again because after the first heads (which both need), it's either heads again or tails. Either way 1 person won, so it doesn't matter wheather the other path in that diagramm goes to the start or stays where it is, as the game is over by that point.

I think this is wrong. Asking the questions which will happen first and which will happen more often are two different questions although they seem the same. The game could be stated that they will flip a coin until they get heads and then the next coin flip will decide the winner because both of the sequences start with a head. This makes it obvious that it is still a 50 50 chance.

Weird, expected time to win does not tell you who is more likely to win. Take for instance a game where player 1 always finishes in 2 seconds, and player 2 has 80% chance to finish in 1 second and 20% chance to finish in 11 seconds. Then player 1's average time is 2 seconds, player 2's average time is 3 seconds, yet player 2 would win most of the times.

In the case of this video, if they share the coin, their chance of winning will still be 50-50, despite the calculated average times to win. They should have done Head-Head versus Tails-Head, cause then the second actually does have a bigger chance to win (and the calculation in the video for expected time remains the same).

I thing you were wrong

Actually they still have 50% chance of winning

They both need to have the head first to continue. So the probability for the next head or tail still 50 %

I only got 2 flipsssss to disseapearing my coin, lol

Heads is heavier because it has more part going out so it is easier for it to go down

The way this problem is worded is misleading. They should emphasize that the condition has changed from flipping one coin to each of the two brothers flipping a coin of their own(I feel there was an intent to mislead by de-emphasize this change). And in this case, it should also be made clear that the brother who flips his winning pattern in a fewer number of flips wins. Not that most of us would be confused into thinking that the faster "flipper" would have a better chance to win; rather, it would have clarified the condition for most of us that they are flipping separately and that when one brother flips the winning pattern of the other would not conclude the game. I feel this is as much a logic puzzle in rhetoric as it is in probability. Tricky.

If they flipped the same coin it's still a 50/50 chance, since the game would be over the first flip after a heads was seen.

Was this video serious?

Oh my goodness.

Did you run this by a statistician?

Maybe run it by one now. See what they say.

To simplify this even further, any time you need the same result twice in a row it will be less probable. This is magnified with a sample size of two (heads and tails). The first flip is always a 50% shot for both players. So is the second flip. But when you fail on the second flip is where it changes. P1 needs H-H, P2 needs H-T. Let's say both get H on flip 1. Then say both fail flip 2…P1 gets T, P2 gets H. P2 is still only one flip away from H-T, because their last flip was an H. P1 must get H-H. Their prior flip was a T, so they need a minimum of two flips to get H-H, which the probability is now at 25% to P2's 50%.

Ehh? Until a heads is flipped, neither of them can win. When the heads is flipped, they each have a 50 percent chance of winning on the next flip….

I'd just flip my coin real fast.

My intuition was: "heads heads is more rare!"

Actually, if the contest involved a single coin and whoever flipped heads first won, then whoever flips the coin first has a greater chance of winning than the person that flips it second

this is basic automata theory. why people in the comment can't understand this is beyond me.

What did the quote in the beginning of the video mean??

amazing. I see it now but did not at all – at first. geat video

This video is completely wrong. So disappointed

Lol this is soo confusing

Deterministic Pushdown Automata? Is that you?!

If they would be using one coin, their chances would be equal but they are doing it separately. He should have said that in the beginning. It took me 10 min to figure out.

WhAt

MONEY>3$$$$$$$$ 1d$

Why is this so hard for people to understand? The probability that you will get 2 of the same outcomes in the row is smaller than 2 different outcomes. The probability gets smaller the more times in a row you are trying to get the same outcome. It's very simple. The video made the mistake of using a coin flip as an example, as you conflate the 50/50 with the two flips. A better example would be trying to flip heads 5 times in a row or something like that.

Nice video but the problem should be stated more precisely: Po-Sheh clarified each player tosses their own coin. But even like so there are two ways the game could be ruled: if player one gets head-head (potential win) but player 2 gets head-tail at the same time this is a draw, so what you do? continue tossing using the previous tosses, or start over?

I calculate the probabilities and it goes like this:

prob. palyer head-head wins =39/121=32.2…% (without having a draw before)

prob player head-tail wins=65/121 (without having a draw before).

rob to get a draw = 17/121

So if you start over when a draw:

prob. HH player wins= 39/104

prob HT player wins = 65/104

but if you continue tossing after first, second, or any number of draws (tougher calculation this one):

prob HH player wins = 47/114

prob HT player wins = 67/114

if you found this explanation interesting, you should look up the KMP substring search computer science algorithm. It uses the same 'mechanism'

I'll let siri to do the work!😏

That’s what bullies get

Who wants me to explain

I ❤️ deep stuff

Remember, HH means if you get tails at anypoint you must restart, for HT if you get H and then H again you only need tails to win, that's because HHT still counts for HT but HTH doesn't count

If they're only going until the first instance of HH or HT, then the resulting string will be T…TH (with any non-negative number of tails) followed by either a head or a tail. For the first occurrence both will be determined by this single last coin flip making the probability still 50%. For a total count in a fixed number of flips it would be different or even if they were looking for TH instead of HT.

can i do math ia on this?

Tldr

When trying for HT, if you fail on the second throw and get HH, you're already halfway to completing the next HT

If you're trying for HH and get HT, you haven't gotten started on the next HH at all

What about 3 flips

HH – HT – TH – TT

So getting (HH = 1/4 = 25%)

And getting (HT = 1/4 = 25%)

To me..

I still don't get it !

Can anyone explain to me please ?

Make it short and clear kindly

Just flip the damn coin

originally I thought each brother would flip 2 coins, (or the same coin twice and keep track of results), and they either got the desired result or they didn't. Then they would start fresh with another 2 flips. In that case, the probabilities would be equal.

But each brother flipping UNTIL they got their desired result, really does favor HT.

1:51. Did…did he run out of breath?

If they both had 1 attempt then it is equal. This is only the case if the use the 2nd flip from the first turn as a part of the second turn

The people's comments are the ones confusing…

The people's comments are the ones confusing…

what if it lands on side

Clarifications to everyone who suggests that both players should have an equal chance of being the first to flip a sequence of HH compared to HT.

Each player is flipping their own coin and the winner is the person who achieves their goal (HH or HT) in the fewest number of flips of their

owncoin.The point is that if your goal is to achieve a sequence of HH with your own coin for example, you DO NOT automatically lose if you happen to flip a sequence of HT (Opponent's goal). The same rule applies that you don't win automatically if your opponent flips a sequence of HH with

hisown coins.The common misconception (Probably because the rules weren't explained clearly enough in the video) is that the game is only played with 1 coin which applies to both players. In that case both players are equally likely to win as after a Heads is flipped, there is a 50% of the next coin being flipped as Heads or Tails, and thus achieving the respective player's goal