ฝัง
- เผยแพร่เมื่อ 8 ต.ค. 2024
- Check out our Patreon page: / teded
View full lesson: ed.ted.com/les...
When you flip a coin to make a decision, there's an equal chance of getting heads and tails. What if you flipped two coins repeatedly, so that one option would win as soon as two heads showed up in a row on that coin, and one option would win as soon as heads was immediately followed by tails on the other? Would each option still have an equal chance? Po-Shen Loh describes the counterintuitive math behind this question.
Lesson by Po-Shen Loh, animation by Augenblick Studios.
Thank you so much to our patrons for your support! Without you this video would not be possible! Devin Harris, Tony Trapuzzano, Stephen Michael Alvarez, Tom Lee, Juliana, Jason Weinstein, Kris Siverhus, Alexander Walls, Annamaria Szilagyi, Morgan Williams, Abhijit Kiran Valluri, Mandeep Singh, Peter Owen, Sama aafghani, Vinicius Lhullier, Connor Wytko, سلطان الخليفي, Marylise CHAUFFETON, Marvin Vizuett, Jayant Sahewal, Joshua Plant, Quinn Shen, Caleb ross, Elizabeth Cruz, Elnathan Joshua Bangayan, Gaurav Rana, Mullaiarasu Sundaramurthy, Jose Henrique Leopoldo e Silva, Dan Paterniti, Jose Schroeder, Jerome Froelich, Tyler Yoshizumi, Martin Stephen, Justin Carpani, Faiza Imtiaz, Khalifa Alhulail, Tejas Dc, Govind Shukla, Benjamin & Shannon Pinder, Srikote Naewchampa, Ex Foedus, Sage Curie, Exal Enrique Cisneros Tuch, Ana Maria, Vignan Velivela, Ibel Wong, Ahmad Hyari, A Hundred Years, eden sher, Travis Wehrman, Minh Tran, Louisa Lee, Kiara Taylor, Hoang Viet, Nathan A. Wright, Jast3r , and Аркадий Скайуокер.
If you enjoyed this video, check out the lessons in our “Can You Solve This Riddle?” playlist: bit.ly/1NJ3CwS
TED-Ed Please do a History vs Frederick the Great or a video on Germany through history ! Or on Prussians !
❤
Did not enjoy this at all. Though the math is correct and would be true if you were trying to get your coin-sequence more often than your opponent, this does not count for this exact situation. To win, all you have to do is get your coin-sequence once. So the "average coin flips" don't matter. Because as soon as soon as you get heads, the next coin will be decisive. So both still have a 50-50 percent chance.
Were it the case that they did 100 coin flips and and out of this sequence they had to count how many times they got thier sequence, then yes: HT would be more common, because of you maths you described.
TED-Ed your videos become more boring each and every week i suggest you sell this TH-cam channel to someone who can make this channel more attractive like the way it was 2 years ago .
Cause the last time you posted a good video was in 2016 and since the beginning of 2017 this channel is going down and down and down every week
I'd get as many heads as possible
Amazing video. I was skeptical it could be explained in 4 minutes since Numberphile's video (consecutive coin flips) did not have a proof, and my own video was 10 minutes long. (Counter-Intuitive Probability. Coin Flips To HH Versus HT Are Not The Same!). But then I saw Po-Shen Loh's name and knew it would be rigorous and correct--great job!
PS: TED-Ed your frog riddle probability video is still wrong! Please see my video on the topic: "TED-Ed's Frog Riddle Is Wrong."
Hi everyone! Thanks for your comments about the video. This is a really interesting paradox because there is a critical nuance: there are two separate coins being flipped.
If there was only a single coin being flipped, then as many of the comments indicate below, the game would be fair. However, if each person flips a coin separately, then the game suddenly becomes unfair. The mathematical analysis in the video helps to dig into what is really going on, and the game board displayed in the middle of the video makes the intuition more clear. The many comments on this topic show that this paradox is really counterintuitive, because the apparently small change from one coin to two makes a critical difference. If you still are doubtful, try writing a computer simulation which checks what is the average number of coin flips before you get Heads-Heads immediately in a row, and what is the average number for Heads-Tails. The experiment will match the calculations!
Ultimately, that is the value of math: reality is complicated, and often defies human intuition. Math gives us a framework for clearly analyzing what is correct! Thank you all for watching and commenting.
Okay, I was wrong, I tought the video meant who wins more times instead of who gets it first.
Nope, its not about the different coins, both could have used te same random sequence of heads and tails and Wilbur would still have higher chance to win.
The confusion is with the argument that you wait until the first heads and then proceeds to determine in the next row who wins, being it H-T[Wilbur] or H-H[Orville]
If H-H-H were to be considered as Orville winning twice [the first 2Hs and the 2 last ones, then it would be a fair game.
Here is what I mean:
you wait until the first heads to show up, following:
HH 50% +1 for Orville
-HHH 25% Go back to where you started
-HHT 25% +1 for Wilbur and go back to where you started
HT 50% +1 for Wilbur
Go back to where you started
You can clearly see the advantage
Here is a 5000 digits binary random number generator:
artofmemory.com/3.1415926/t/binary.php
Now using ctrl+F and typing: "0 0" then ctrl+F : "0 1"
Reloading the page 3 times:
0 0: 820 , 834 , 840. Total of 2494 Comparing to the expected 2500
0 1: 1228 , 1247 , 1260 Total of 3735 Comparing to the expected 3750
The results from the second sequence were not too different from neither the first nor the second.
But in all of the 3 games, Wilbur had an entirely different result
Thanks, Po-Shen! "Math is complicated and often defies human intuition" It seems like that's part of what makes it so fun. Thanks for making this exciting and brain-twisting lesson with us!
Please Read this -Po-Shen or TED_Ed-> Im very curious if i got this right Thanks in advance
I must intervene here and claim (if i understand this correctly), that the premiss of the original bet is the 1 thing in this conundrum between the brothers that is scewed; I would have assumed if i was the loosing brother that it was implied that if the winning brother (heads-tails) HAD to start over and could not simply continue his Head streak if he got Heads-Heads.
I might have missunderstood this, please i really want you to correct me if im wrong cause i love these little clever videos, Best regards, Adam
The video barely mentions that there's two separate coins, I spent the whole video scratching my head, thinking it was a shared coin. They kinda dropped the ball with this one.
When overthinking meets maths...
Haha
True that
maths is always overthinking, just more beautiful
🧠Smart
You somehow managed to complicate the explanation by trying to oversimplify it.
Yeah, I think Numberphile did a way better job at explaining this repeated coin flip problem
its quite simple...*easy explanation*: we firstly have to toss coins to the time it shows head...then
case1...if there is head we won but if not then we need to start again to get consecutive heads..
but in
case2 .. if we get tails we will win nd if not then it must be a head so we need not to start again..nd just have to get tails for once...
wow.thanks.now i understand it
Thanks
Thanks
I was very confused until I went back to the start of the video and realized they were flipping different coins.
Yeah that makes more sense. I missed that part and the whole thing was completely illogical.
It doesn't matter how many coins there are. Same probability with one coin. First flip, both people have 50% probability of getting heads. Second flip, whether the same coin or different coin, again 50% probability of getting the desired result. Where the twist comes is if both players on the second flip do not get the desired result. Then the advantage goes to the heads-tails player over the heads-heads player. Why? Because the desired results are P1 H-H and P2 H-T, right? First flip, both get heads, results logged. Okay, second flip, P1 gets tails, and P2 gets heads, the reverse of their desired result, again results are logged. P2 is still 1 flip away from victory. He got H-H but wanted H-T. You are still playing off a prior flip of H. You just need a T. 50% shot. P1 wanted got H-T, but wanted H-H. His prior flip was a T. He needs a minimum of two flips to get H-H, or 50% odds times 50% odds again = 25%. The number of coins is irrelevant.
@@100percentSNAFU What you are describing is them having two coins. For P1 to get tails and P2 to get heads on the same flip would require two coins obviously as one coin cant land on both heads and tails at the same time. If there is only one coin then as soon as it lands on heads it will mean that the next flip will end the game, as either P1 or P2 will win. Physically they can USE the same coin, but having "two coins" here means that P2 doesn't care what happens when P1 flips his coin, he only cares about his own flip. Same for P1 of course.
Yup, it's super ambiguous and doesn't seem to make sense until you realize they're both flipping their own coin. If it's one coin, it's obviously 50/50, but each flipping their own coin it seems inherently obvious that HT has a huge advantage. It's more about paying attention to semantics and that one detail, than the actual probability, which seems obvious once you figure out there are two coins.
If the guy on the left said HH and they are flipping the same coin then the guy on the right should do TH that way if T is flipped for the second of first flip the guy on the right instantly wins
You oversimplify so bad that it gets complicated.
Ree Godlevska wtf
I _still_ don't get it, I think I'm gonna *flip* out!
Sebastian Elytron Nice pun! But, if it helps, think about it over an extended period of flips as opposed to just 2. If you have 4 coins you have 16 combinations total. 7 of them feature two concurrent heads. 11 feature heads then tails. That means there are a few overlapping sequences when both happens. In fact, there are 4. In 3 of those heads-heads wins but, in one of them, heads-tails does. So this means that There are 6 cases where heads-heads wins, 8 where heads-tails wins, and 2 where neither occur.
Try it out for yourself: Tally the 16 possible combinations for 4 flips, see who'll win, and you'll get the same result :)
*laugh track* Everyone is comedian.
You flipping Sebastians are always a hazard!
Ok, so basically lets say mission one is the heads-then-tails, and mission two is heads-then-heads.
So, let's say you are MI2. Your objective is to flip the coin heads, then the next flip will be a heads. It has to be one right after the other. If you flip it heads first, good! Your first step is there, but the *next* flip right after it has to be heads too, to complete MI2. If you flip it again and it is tails, then you have to restart your mission. You have to get a heads again, and then a heads right after that. You have a less of a chance getting two in a row because you can easily have to restart MI2.
Ok, so let's say Harry has MI1, which is a heads, then a tails. This is easier, because all you have to do is get a tails right after a heads. He might have to flip twice to get a heads for the first step of MI1, but then he flips as much as he needs to get a tails. Because nothing can make him restart him mission.
Say Harry flips once. He gets a heads. He flips again and gets another heads. So, since he got two heads in a row, he hasn't completed his mission. *BUT*, as you can see, he doesn't have to restart his mission, because the first step is landing a heads, which he has repeated twice. So, if his second flip is a heads right after his first heads, he is just repeating the first step. Which counts as a first step. So he can never be forced to restart his mission, as the worst that could happen to him is repeat the first step, not mess up (like you can in MI2) and have to flip a couple times in order to restart. If Harry 'messes up' MI1, he has already restarted his mission and doesn't have to flip again to restart.
Hope this helps!
Sebastian Elytron COIN you not make another bad pun?
Every single TED-Ed video is like a perfectly scripted essay, hitting all the main points with supporting details and engaging writing style. It is amazing that so many people put so much work into these videos just to educate us about topics we might never have otherwise even considered.
Number-crunching - Here are some *cumulative* probabilities for winning within a given number of flips:
*Orville (Heads-Heads)*
*If his first flip was H* -
2 flips: 50.00% | Ways to win in exactly 2 flips (starting with H): {HH}
3 flips: 50.00% | No way to win in exactly 3 flips (starting with H)
4 flips: 62.50% | Ways to win in exactly 4 flips (starting with H): {HTHH}
5 flips: 68.75% | Ways to win in exactly 5 flips (starting with H): {HTTHH}
... (remember these probabilities are cumulative)
*If his first flip was T* -
2 flips: 0%
3 flips: 25.0% | Ways to win in exactly 3 flips (starting with T): {THH}
4 flips: 37.5% | Ways to win in exactly 4 flips (starting with T): {TTHH}
5 flips: 50.0% | Ways to win in exactly 5 flips (starting with T): {TTTHH, THTHH}
...
*Wilbur (Heads-Tails)*
*If his first flip was H* -
2 flips: 50.00% | Ways to win in exactly 2 flips (starting with H): {HT}
3 flips: 75.00% | Ways to win in exactly 3 flips (starting with H): {HHT}
4 flips: 87.50% | Ways to win in exactly 4 flips (starting with H): {HHHT}
5 flips: 93.75% | Ways to win in exactly 5 flips (starting with H): {HHHHT}
...
*If his first flip was T* -
2 flips: 0%
3 flips: 25.00% | Ways to win in exactly 3 flips (starting with T): {THT}
4 flips: 50.00% | Ways to win in exactly 4 flips (starting with T): {THHT, TTHT}
5 flips: 68.75% | Ways to win in exactly 5 flips (starting with T): {THHHT, TTHHT, TTTHT}
...
They both have an equal chance of winning in exactly two flips, but Wilbur has a greater chance of winning in exactly 3 flips, 4 flips, etc., so on average (and especially if both of them were unlucky to begin with), Wilbur will win in fewer flips.
one problem. Wilbur doesn't win HHT, HHHT, etc. Even though these contain HT, if HH occurs before HT then Orville wins the round.
so your Wilbur (Heads-Tails) section is only true on two flips.
Wilbur does win HHT as they are using separate coins
Once its clear there are separate coins, this can be seen in a much easier way. In essence, both are waiting for the first head on their coin. If you are looking for HT, you can keep getting heads until you get the tail you want, and then you are done. However, if you are looking for HH, if you don't get the second head you want, you need to start over. The state diagrams at 1:23 show this as well, for HT, once "get out" of the start state, you are never forced back in. However, for HH, once you "get out" of the start state, you have a 50% chance of being forced back in. Of course, this reasoning doesn't get you the exact number of expected flips, but it may be an easier way to see why its not equal.
Can't believe how many people didn't pick up on the two coins but I think there is an interesting scenario where you can play the game with ONE coin and it will trick you.
I know it's a bit late to add this to the discussion but I think it works and wait to stand corrected.
Imagine 4 players (still works with 2 but 4 is more interesting) flip the same coin until one of the players hits their own chosen 3-coin combo.
Player A: HHH, Player B: HTH, Player C: TTT, player D: THH
On the face of it, it seems fair; each combo has a 1 in 8 chance. If the first flip is H, A and B are off the marks, if it's T, C and D. Then the second flip decides who advances and one more flip gives them a 50/50 chance of finishing. Note there is no possibility of a coin flip with 100% chance of a winner as there would be with TTH and TTT.
In reality, players A and C have a huge disadvantage. Every time they hit the wrong coin they go back to the start and have a 1 in 8 chance of getting the combo right in the next three flips (not counting for someone else winning before they do)
If the first two flips are HH, player A will now have a 50/50 chance of winning but player B, will be right behind him, if the next coin is T, player B now only needs a H to win in 4 flips.
However, if my maths (as I'm British) is correct, player D has the best chance overall since after an initial T they can never be knocked back to the start. Obviously, the fact that some winning combos overlap complicates the matter but I'm pretty sure this trick works.
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.
I think the conundrum is somewhat misleading. If the question was "who has the most points after a large number of flips" with each brother earning a point when ever their corresponding "HH" or "HT" combinations occurr, than yes the brother with "HT" would have the statistical advantage and earn more points, but as it stands in order for the second brother with "HH" combination to go back to start, and thus loose his progress towards the win, a "HT" combination would have to occur which would stop the game immediately as the "HT" brother would have won.
Eugene Bright Assuming there is only one coin. The example hinted at each brother having a coin
I feel that in order to preserve the statistical advantage, and thus the premise of the video, the Wilbur brother would have to win with a "TH" instead of "HT" combination. This way "HT" reset condition for Orville would not trigger the automatic win for Wilbur and "TTTH" game would correctly provide Wilbur with an advantage without prematurely ending the game.
@Raymond You are so right! Having each brother use a separate coin as a subtle, and important, difference. With each brother winning based on their own coins, the premise of the video stands true.
That was my thought first too. But in the beginning, it is said, that each brother flips his own coins (0:25). So when one gets reseted the other one doesn't win.
Precisely!
The interesting thing is that if they throw only one coin, then the 2 brothers would have equal chances. That's because we wait for the first head, and the coin after it determines who wins
Ok guys for those of you who were initially confused like me, there is a crucial bit of info in the video that wasn't made clear: Both brothers are flipping their own coins until one person gets their desired sequence. This whole time I thought they were just flipping ONE coin. The script doesn't make it clear but the visuals do. Hope this helps
Im watching this instead of studying for my trig exam.
Edit (2 hours later): Still havent bothered. Heading to class, exam in 40 (good thing im just auditing this course lol)
Edit: I answered 4 of 15 problems lol
Edit: it’s been 2 years or so but I’ve only moved 2 semesters forward in my maths. Calc 3 next term
like, im procrastinating so hard that I didnt go to bed last night and i still havent started... lol
Now you can perhaps calculate your probability of passing the test.
Uriah Siner trig exam ?
Haha i have an easy year 11 algebra exam
Why would you pay money for a class if you won't get credit for it?
Its a mystery... 🤔
1:05 holy damn that head zoom scared me
The people who are saying this is wrong probably missed that there were *two* coins involved: one for Wilbur and one for Orville.
It is true that the game is fair if there were only one coin involved. But this is not the case if Wilbur and Orville both have separate coins.
The expected number of coin flips required to get two heads in a row is higher than the expected number of flips required to get a head followed by a tail. So, in this sense, two heads in a row is more rare.
Wouldn't the change just be 50/50? Correct me if I'm wrong, but suppose your first throw is heads... Then the next coinflip would determine who of the Wright Brothers would win, since if the second throw were to be heads, Orville would win and if tails, Wilbur would win. Both events have an equal probability of happening.
The other option is that you would throw tails first. In that case, neither of them would gain anything, since neither of their sequences (heads, heads and heads, tails) start with tails. They would keep flipping again and again until a heads showed up, in which case you'd just be back at the first situation.
You're missing the part that there are two coins. That means there were *two* first flips: one for Wilbur, one for Orville, *two* second flips, *two* third flips and so on. If Orville gets a head followed by a tail from his coin, Wilbur doesn't get anything. In order for Wilbur to win, he needs to get a HT sequence from *his* coin.
They are actually competing to see who can get their sequence in a fewer number of coin flips.
Two flips. Only one coin.
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 *own* coin.
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 *his* own 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
I totally didn't understand that they were flipping coins on their own.
If it was just one coin (which I thought the whole video through) this wouldn't matter, since after the coin showed heads once it's a 50 50 chance.
You explained this very poorly.
Tycho Wozniaki no lol
Zach Lee Yes lol
You have to pay attention to the imagry and word choice at 0:21 he says "on HIS coin" twice and shows them flipping seperate coins leading to seperate sequences.
People keep saying things like this but I don't understand what difference it makes whether they used one coin over and over, or a different coin on each flip.... Oh wait, I think I understand. You thought Orville flipped a coin and then gave it to Wilbur and then his flip could decide what happened to both. Or the equivalent of that. I thought their flips were separate the whole time, so they both had to flip at least twice before anything could be decided.
there was only one coin
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
like someone on the comment section said, just watch the number file's video on Consecutive Coin Flips. Basically, when you have a string of H and T, like
THTHHHTTT
you have two HT but only one HH.
Notice that we ignore one of the HHs in the "HHH" because we start over after we find a sequence of either HT or HH. As a result, you get more HTs than HHs in a string of Hs and Ts or less waiting time for HTs.
"it's an ingenious solution to a problem that never should have existed in the first place" - James May
What a tangled conundrum we weave.
This effect is due to the implied rule that once you flip a HH the second H is discarded for the HH camp while retained for the HT camp. If you allow HHH to count as 2 moves for the HH camp, they will have equal probability.
The way I understand it, they flip two coins, as people in the comment section note, meaning the two events (HH) occurs and (HT) occurs are independent: given that you win when you get (HH), you can flip (HT) without losing.
That said, let’s look at all the outcomes of flipping a coin three times:
HHH, HHT, HTH, HTT,THH,THT,TTH,TTH. In total HH occur 3 times, HT 4 times. Notice that HHT is not excluded when you count the times HT occurs, because again, HH and HT are independent. You flip the coin only hoping to get HT, not fearing about getting HH. It’s easy to generalize this example to demonstrate what the video is trying to say.
Ra Better explanation than the video :)
I think you meant TTT for that last one. Also, an HH pattern *can* be found twice in the first scenario HHH, though I know that doesn't really count 🙃
There's only 3 HT there..
The HHT is counted as HH as HH appeared first
@Marcus Yang read the rest of the comment; it counts because the two flip goals are independent, which means that the HHT flip looking for HT counts because if you're looking for HT, you can ignore any and all consecutive Hs that come before HT
I’m surprised it didn’t ask to pause the video to figure it out first! I always pause the Ted Ed riddle videos as if I’m ever going to figure out the answer 😂
because its not riddle, it's a math problem
Hey I saw your comment on a sci show video. And I looked at you channel. Pretty cool, albeit, I had a pretty awesome Chemistry teacher.
Your*
Daniel Varzari hey! We all watch the same videos then 😂 thank you! I had a decent chemistry teacher once I got to uni, I wasn’t as lucky as you for secondary school!
Science with Katie You have a good taste in videos, learning about poop has its uses.
To everybody still not getting it, it's not about the last flip itself, there is a 50-50 chance of them winning after getting the first heads, but the question is what happens if they lose, in HT the next flip can be the winning one, but on HH you getting a tails as the second one means you need to start over again. Which means at least two more flips to win, and it's not about who has the most chances of winning, but about who wins with the least flips possible.
After looking at the first flow chart I understood it in one way you guys can try
See, if either get a tails at first, they have to restart so it's fair till there. Once they get a heads each, if hh guy gets tails he has to go back to the start for two consecutive heads. But if ht guy gets a heads, he just has to keep Flipping till it's a tail. See if he gets h then h then tails he is still the winner. After a single heads he can just keep going till he gets a tails. Other guys chain is broken by a tails.
Hope someone understands this
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
Only had to pause for two minutes for this one. Pretty proud
The equations solving animation is really satisfying...
Hey whatever you explained was the expected time. But it doesn't mean the experiment with greater expected time would occur later in a sequence of random experiments . The probability of either of the sequence occuring first here are equally likely in this case. Please have a look.
This is a great explanation but I find it obscures the intuitive way of thinking about it. When I saw this problem I immediately thought the heads-tails was more likely. This is because the problem represents finding a pattern in a series not just two coin flips. For example if you could only flip two coins per try then the other player tries then these two combinations would be equal probability. Since its a series its intuitive to consider one coin before the next two. Starting with heads requires one more step, getting heads while starting with tails requires two more steps heads-heads. Meanwhile heads-tails only requires one step from the initial coin regardless of heads or tails. Notice my thinking is based on discrete things. "only requires one" etc. Which is easier to think about than "there is a 25% probability after the second step but can repeats x number of times"
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 on _*_his_*_ coin,_
_and Wilbur would win as soon as heads was immediately followed by tails on _*_his?_*_ "_
It doesn't change anything if they started with tails they restart and if they start with heads they now can get either head or tails
Guys, this is wrong... H-H and H-T in a best of 1 is actually a 50-50
Because if it flips T noting happens because both start with H
Then after the first T flips (or on the 1st flip) if gets H both are with 1/2 of their goal
If both are with H done, then on the next flip, if flips H one wins, if T other wins so its exactly a 50-50
So this would only apply if instead of H-T was T-H because H-H would only win if the 1st two flips were H-H
OR
If this was a best of more than 1 since H-H needs less flips to get a winstreak than H-T
Edit: *Ignore this, I thought there was only 1 coin involved*
Just made the same kind of mistake in my comment xD (deleted that one)
there is
Is this your mistake tho?
IMO video is really bad.
Not clearly enough presented in the video for sure - also, IMO the video still uses invalid reasoning. One cannot simply compare two average (first) occurrences of anything and draw any conclusions about which comes first with higher probability. I now made a whole comment about that myself: th-cam.com/video/IAiNqQi30-Y/w-d-xo.html&lc=UgwieJIagfAzd5N86Mx4AaABAg
This makes no sense, are you assuming they are throwing different coins?
Fuvity yes he is.
Even using the same coin, Wilbur still has a highe odd of winning.
Henrique Sato No he doesnt. In the case of heads - tails, the x = 4 formula doesn't apply because if on the second flip u don't get tails, then u don't get to just flip again since Orville would have won already because that's two heads in a row.
Unless I'm missing something, I agree with FistDaMonkey: The game ends before these rules make a difference. If they were playing to get the winning combination multiple times, sure. But if they're playing to just get the combination once, then the game always ends after the first heads - nothing else matters.
The Story of Life what are you talking about? This is maths, math is exact.
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
This is the first ted-ed puzzle that I figured out!
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.
Well this is just not true. If the first flip is tails, then it's a pointless flip because both sequences are 2 flips away from happening; and once you land on heads once, you're gonna be one flip away from finding a "winner", since your current sequence is H- and one person needs H-H and another needs H-T. So its bascially just flipping a coin except you need to land on Heads before you can find a winner
What you said would be completely true if there were only one coin involved. But there are actually two coins: one for Wilbur and one for Orville. They are competing to see who can get their sequence in a fewer number of coin flips.
What you said only hold true if they were flipping one coin to decide. Each brother is flipping a separate coin and that means each person gets a different result (one person might get heads while the other gets tails).
Yeah actually this is wrong.. this only applies to for example a best of 3, to get a win streak, H-H only needs 3 coinflips his way (H-H-H) but if H-T wants a 2 win streak aswell it would need 4 coinflips (H-T-H-T) so in theory you can prove like this that in a best of x beeing x>1, H-H would have a higher chance of winning..
In this case I think its 50-50 exactly as you said
You are right, but in the video they talk about throwing two coins independently. So if say Orville would throw head, then tails with his coin, it would not matter. It's just about who would throw their own sequence first. At least I think that's what they mean, they're not very clear about it.
And also, there was an exception wich maybe they swaped by mistake, if instead of H-T was T-H then it would be correct because If H-H doesnt win with the first 2 coinflips beeing H-H then its garantee that when it does happen T-H would've already won
this video was very helpful to me and my schoolwork status.I watched this on my macbook and its the only video that hasn't been excruciatingly glitchy.
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
This doesn’t work in a contest between two people. Yes, separately the HT combination has a higher probability of succeeding than HH, but if they’re flipping one coin, as soon as the first H is flipped, the next flip determines the winner, giving them a 50/50 shot.
The little animations on the coins were so cool, and the entire thing is mindblowing
The chances are still equal.
All tails before the first heads are irrelevant as both are sent back to the start.
The first flip after the first heads will decide the winner.
This flip has a 50% chance of tails and 50% chance of heads.
There is a diifarence between expected value and probability. The first brother will indeed needs more filps on avrege in order to win, but he will
win at the same precentage of the trials.
Key distinction: they are each flipping their own set of coins.
The example at the beginning of the video was misleading: they have two coins, and they are each racing to see who flips their sequence first. They're actually not using the same coin. So for one brother if he keeps getting heads it's only a matter of time before he gets tails and wins, while the other brother if he gets heads, if the next one isn't heads he might get a string of tails before he gets another heads and has an opportunity of winning again. Hope that makes things clearer.
Thank you. That actually really cleared it out a lot.
2 coins with 2 possibilities is 2^2. AKA 4
TT
TH
HH
HT
Therefore, if they flip their coin twice, they will have a 25% (1 out of 4) of getting a win
For people that got confused: watch Numberphile Consecutive Coin Flips video.
Gleb YES! Just string of H and T shows the difference much better.
Finally I (think I) understand what it is. The probability of HH and HT doesn't change. Each of them is 0.25. The main point is, which outcome (HH or HT) will appear first. Because HH averagely needs 6 flips, it is slower than HT (which needs 4 flips).
As the hi hi user said, as soon as you get heads, the next coin will be decisive. So both still have a 50-50 percent chance. Instead of Heads-Tails and Heads-Heads, there had to be Heads-Heads and Tails-Heads, in this case it was better to choose Tails-Heads
I think a more susceptible explanation to this conundrum is looking into the occurrence of these patterns over a period of time the probability would decrease. The case scenario presented doesn't really mean much because if you do the math both have the same probability of first occurance but it is correct that in the case of multiple occurances heads tails would win
Good thing it wasn't a coin flip with Harvey Dent.
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
That was some serious math which went from over of my head.
East or west ted-ed is the best
For Ted videos, we may need two new buttons on TH-cam;" Understood","want a simpler video on same topic" 😂
Another enlightening Video! THANKS!
The end was a bit dark, but it was a nice idea for me to implement this in a short python script with flip counter documentation to prove, that with n >= 100 runs there will be 1.5x more flips with the heads/heads setup than with the heads/tails setup. Consequently, this also should work with the constellation tails/tails having 1.5x more flips to get from end to start compared to tails/heads. Let's try and see.
quick maffs🤯
HAHAHAHAHA nice, but not quick at all
.... if continuously flipping coins, it’s still 50:50 chance because assuming H for coin 1 (T means back to start for both), the next flip is either H or T. So someone wins. Theres no second loop for either option because there’s always a winner and thus end game.
With two coins and a race this doesn't apply.
Since both people are racing to get their combination first instead of just flipping and then counting the number of times they got it, they're actually equally likely to win.
We begin by flipping either heads or tails as our first coin. If we flip tails, nothing happens and we flip again. As soon as a heads is flipped, the next coin must be either heads or tails. If it's heads, HH wins, if it's tails, HT wins. Thus the winning player is determined just by that last coin, giving us a 50/50 win distribution.
The reason your math doesn't apply is because we're asking who will win first, so as soon as HT flips heads at his second turn, he lost because HH won, so it doesn't matter where that flip took him on the chart
Ok I probably don't got this right but anyway here is how I think about this:
Both of them need to have an head first so until there is no head both of them can't get their pattern. After the first head is there it is a 50/50 chanse for both of them because it ends 100% after the flip after the first head. So pls correct me when I'm wrong but for me it seems fine
the problem is that they flip the coins repeatedly.
After the first head, both have a fair 50/50 chance to win.
If you get tails next, Wilbur gets 1+ point and both sequences reset.
However, if you get heads again, Orville would win, and only his sequence would reset, so right after that, there is a 50% winning chance for Wilbur [tails], and a 50% chance to reset both sequences [heads again], wich means Wilbert has an advantage.
But both have the same chance to get that advantage.
You can have both to use the same sequence of heads and tails, and Wilbur would still have the advantage.
TED this was huge underestimation
just because one event happens more often doesn't mean it happens first as the two events are not statistically independent
It was not clearly explained, but the set up of the problem is that they're doing their coin flips separately. This makes them independent.
Yeah but, the probability of head-tail will be higher.
Oh, thank you. that explains it. I forgot that they flipping their coins separately. If it was done on the same sequence it would 50 / 50.
The video doesn't explain it in a clear cut way though because who starts first matter. If it was the HT guy first he would win about 67% of the time, meanwhile if it was the HH guy first it would be 53%. Either cases the HT guy is more lucky.
sghaier mohamed how do you calculate their chances of winning?
Easy but amusing application of Markov chains, you could actually calculate the probability of winning for each player after the same number of iterations, using Markov chain.
on the second step of both flowcharts, the unwanted coin will lead to the other flowchart's goal, not back to the start.
I get the point. When one bets on getting two heads, and the other one bets on getting head-tails, the other one has a better chance of winning because when you get heads on your first, then all you have to do is get tails without starting over
i think i got it .
1) In the first flip both have 50% of possibilities to obtain Head
So let's assume that both of them get head .
Now both players are just one step away from victory . There is however a difference
2) Second Flip : Player 1 gets Head and Player 2 gets Tail so none of them won .
Player 1 however is still just one step away from winning since he got Head and je just needs to get Tail to win .
Player 2 however having obtained tail at the second flip , now is two steps away from winning .
Let's analyze any scenario for each Player :
Player 1 :
First Flip has two scenarios , he either gets Head or Tail
1a) H
1b) T ( Let's ignore this possibility for both players so that things are easier)
2a) H T . He won after two flips
2b) H H He didn't win but still he just needs to next coin to be Tail to win
3a) H H
3b) H T
So after two flips , in the worst case scenario for Player 1 he has 50 % possibility to obtain Tail in the next flip and thus winning the game .
Player 2 :
Same as above , he either gets Head or Tail
1a) H
1b) T
2aa) H T . He didn't win and now he needs the next flip to be Head and then again Head . So he needs two more flips in his favor to win , so he has 25% of possibilities .
2ab) H H . He won after two flips
So after two flips , in the worst case scenario for Player 2 he has 50% of possibility to obtain Head in the next flip and thus Continuing the game , not Winning it .
The difference is all here : Player 1 in the worst case scenario has 50% of possibility to WIN the game
Player 2 in the worst case scenario has 50% of possibility to CONTINUE the game
You don’t make it clear that two coins are being used. While HH is expected to need more flips that HT, on one coin there will be a 50/50 chance. This is because there is a causal relationship with HH winning and HT occurring soon after because the next T flip will cause a HT to occur. If HT comes first, HH can not possibly occur on the next flip. This is the reason the expected number of flips are so different.
Fantastic video! I actually understood this one off the bat for once!
Edit: Edited to clarify, the for once reflects how the puzzles usually break my brain, not how clearly you explain the logic. Y'all always do a fantastic job.
If you're averaging how often that happens over many flips, this makes sense. But if the first one wins, it's still 50-50. Any number of flips may occur until heads comes up. All is even. Note that we have our heads, the game is on. Heads wins for one and tails wins for the other. 50-50.
Heads is heavier because it has more part going out so it is easier for it to go down
MIND BLOWN!!!
Happy Chinese New Year from Malaysia!!
I love the animations!
It took me so long but now I GET IT. The one guy has a bigger chance of winning because when he gets the tails, then he can get whether heads or tails. With heads he wins, with tails he's still one step from the winning (maybe the next one will be heads).
Okay.. this video is NOT as wrong as some believe, the answer is NOT that there is a 50-50 winning chance for either. But I’m NOT happy with your proof either, TED-Ed. Take this game for example:
A flips a coin repeatedly. He scores as soon as his 1st two coins are HH, HT or TH or as soon as his 999,994th coin is either H or T (so anything).
B flips a (different) coin repeatedly, independent from A. B scores as soon as he gets any three coins.
Both A and B are flipping coin after coin, both at the same time. Whoever scores first wins.
So, A will score after 2 flips in 75% and after 999,994 flips in the other 25% of cases, on average (3* 2 + 1* 999,994)/4 ) = 250,000 flips.
And B will always score after 3 coin flips.
Now we have our averages of 250,000 flips for A and 3 flips for B. Nonetheless A obviously wins in 75% of the games, because in 75% of cases he scores first (after 2 flips, whereas B needs 3) and in 25% of cases he scores last (after 999,994 flips while B still needs only 3).
What I want to say - just looking at the average flips PROVES NOTHING! You’d need to tackle the problem differently to make the reasoning work.
EDIT: As for a solution that WOULD be correct: Use the same approach as in the video, just combine the two "boards" into one in the following way.
You need to run both games in parallel, so have 4 nodes called (Start,Start) (Start,H) (H,Start) and (H,H) and have 4 outbound arrows on each of those for the 4 possible throws of: (1) Both get H, (2) both get T, (3) one get H the other gets T, (4) one gets T the other gets H. Then add 3 more nodes: "HT wins", "HH wins", and "draw" and point the 16 arrows each on the correct next one of the 7 nodes. Then do the whole calculation for that "board" and you can quantify if and how much better HT is than HH.
EDIT2: This is how this would look: imgur.com/a/hJcdx
Someone would still have to do the calculations for winning probabilities of either and for the draw.
You're wrong! In your example not each step's move has the same chance . In TED-ED's example each move has the same chance(50%) so they can look at the average.
You really should have clarified that the brothers had separate coins to flip independently from eachother with unambiguous language.
Dumont made history.... These guys made history in the catapult industry.
Otavio Biamino it was more a first step rather than achive ment. More people wanted to fly and not throw planes of cliffs
if you found this explanation interesting, you should look up the KMP substring search computer science algorithm. It uses the same 'mechanism'
Before I watch this, I think the HT player has a higher win rate than the HH player. My quick reasoning is this:
consider a string of three coins: (c1, c2, c3)
Cases:
(1) HHH -- HH
(2) HHT -- HH
(3) HTH -- HT
(4) HTT -- HT
(5) THH -- HH
(6) THT -- HT
(7) TTH -- 0
(8) TTT -- 0
The HH player "wins" in 3 cases, the HT player "wins" in 4 cases.
Case (2) is the one where they both win, although HH wins first.
But in a longer string, the ratio of "wins" for the HT player feels roughtly like 4/8, compared to HH's 3/8.
I'll watch it now and see how good my intuion is.
So my logic of useing the string of just 3 coins wasn't too bad. Obviously my ratio of 4/3 is not the correct 6/4 , but I wasn't expecting to get the right number with just 3 coins, only the right direction, and I correctly predicted the average winner using this approximation with no algebra :)
With HT once you get a head you never have to start over from the beginning ever again. You keep flipping till you get a single tail which is all you need from that point onwards.This is why on average it would be quicker than HH which sets you back further in the chain comparatively - forcing you to start over from the beginning multiple times on average.
Who didn't understand this? 😆😄👆👆👆👆
Ruby Duby Hey thanks a lot for that explanation! You explained it better than the video itself as i was wondering why we need to start again but that is because for HH the sequence is broken by a tail coming in between and for HT it is not when a Head comes in between! I hope it is correct? :-P
yes and you re-explained it in fewer words. nice
Ruby Duby hehe was just trying to reconfirm :-D
It matters if you're counting how many times a head-head or head-tail happen after like 100(or other arbitrary number) flips.
But if you're flipping until getting a head-head or head-tail, the probability of winning is still 50%
Also, did you count the fact that if Head-Head happened, it would not go back to start, rather it will go to the second spot?
You know what? I don't flipping care.
halfapineapple
You get an award.
halfapineapple
That joke wasn’t even *worth* it
Thank you, thank you. Xo
oh i get it
chuckle i'm in danger
videos like these make me wanna be a mathematician
The biggest riddle is to understand that it's about a race between two different coins. The only hint is visual, and I always expect the narration to be self sufficient.
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.
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%.
me: why is a simple kid's game became sooo algebraic. (its supposed to be just-for-fun)
my stomach and headfeels weird
this is a great video BTW. keep on the cool and good works of hidden knowledge behind simple things. I thank You for this
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.
Yeah... it isn't quite clear, but they introduce 2 coins that are flipped in the same interval (th-cam.com/video/IAiNqQi30-Y/w-d-xo.html)
if they would flip just 1 coin and and one was winning when 2 heads show up, the other when tail after head shows up, it would still be 50-50%.
With 2 coins there are 2 Mealy machines / state diagrams.
With 1 coin it would be 1 with 2 end states, that split at the first throw. (50-50%)
Exactly, I was confused at first because I didn't realize there were two coins.
What would they do if both happened at the same time though?
The point of the video was to show how probability works and to find averages within them. The contest isn't important.
When i went to
your website it just says 404 not found
The link got messed up by the following parenthesis. You have to remove the trailing closing paren, then it works ;-)
EDIT: And.. the link only links back to exactly this video anyways xD
To summarize: A failed second flip on the heads-tails game is a successful first flip, while a failed second flip on the heads-heads game is also a failed first 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.
Am I missing something here? The game always ends after the first heads. Whichever coin lands after that heads is the only thing that determines the outcome, is it not?
You can get tails, tails, tails, heads, and then there's a 50/50 chance of either of them winning with the next coin.
If they were planning to keep playing, yeah one would have an advantage in the longer term - but if they're only playing to the first heads, heads or head,tails, then the flow charts shown don't matter.
People are remarkably inattentive.
The video EXPLICITLY shows multiple times that there are two coins:
1) At 0:21, there is a visual of two hands flipping TWO coins.
2) The simple heads-or-tails scenario is described at 0:13 as "flip A coin," SINGULAR! The new scenario is described at 0:22 as "flipped -coinS-," PLURAL!
3) 0:28 and 0:33 call each brother's own coin "HIS coin." Not THEIR coin... HIS coin.
Wow... even heads and tails has a scientific explanation that I can’t understand
RIP Wilbur
Thats...actually insane. Thanks for the lesson!