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Problem From The Hardest Test - Use A Coin To Simulate Any Probability

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Description Problem From The Hardest Test - Use A Coin To Simulate Any Probability

Comments Problem From The Hardest Test - Use A Coin To Simulate Any Probability

Binary method 2 works for me It just seems a lot more intuitive than method 1 Comment from : Wyatt Stevens

Writing a problem in another teferrntisl always works nicely Fourrier representation of a curve or its polynomial approximation same for the digit representationbrThis works in base 6 with a usual cubic dicebror with a dodecahedron in base 12 which can be played with the throttle of the seconds in a chronometer/ time watch with 12 digits only, if nobody has too good a sense of timing or if the origin is dephased by a combination of 2 random base 12 numbers given by the players to "fight the timer" brThat would be interesting to see the probabilitirs of one dors have a good sense of time passing Comment from : Maelix Diogen

I got the first part but not the extension I guess I was trying to overthink it It seems obvious now that an biased coin can be made fair by simply taking turns! Comment from : Robert Beach

Interesting solutions! However, my immediate reaction to the question was to disregard heads or tails on the coin toss and rather look at its rotational position with regard to a reference line The deviation can be up to 2*pi radians Just partition the angle in the desired proportion to yield the right probability Comment from : Ronald Leemhuis

I think, I can see where, you’re going, this time Comment from : Christopher Reese

Plus that is just bad sort man ship Comment from : Christopher Reese

I can answer this question within 1 minutebrbrThe answer isbrbrbrbrbrbrbrNO Comment from : Davis Law

Correct me if I'm wrong here, but the question seems a little bit deceptive In both solutions, the probability of all the finite games do add up to 1, but neither of them actually guarantee the game ever terminates I assume the question was asked this way to make a solution possible Is this right? Comment from : RhodeXLX

"Can you figure it out?"brI can't even figure the question out! Comment from : JLvatron

how do yoou know the formula to find n at 3:25, how can you find it Comment from : Pho Lip

IMHO when you use the words "binary" and "finite" together, the idea of representing "any" number disappears Comment from : Bruno Grieco

But what about 1/ pi You never explain Comment from : pjmoran42

I can - probability! Comment from : Reijo P

The rules say that the game must end in a finite number of flips, but the solution includes the possibility "start over", which do not guarantee this Comment from : Jean-Francois Bouzereau

I wonder if this idea would help with a problem encountered in surveys Suppose you are polling random people to ask if they will vote Democrat or Republican Some fraction F of people will lie and give the wrong answer, but you don't know the value of F What is the most reliable estimate of the proportion of the population who will vote Democrat vs Republican? Comment from : Forest Pepper

So the expected value for the number of tosses is 2 for a fairy coin This is quick if you concern that a irrational number 0 Comment from : Bernd Mayer

Mid roll add for a short video Comment from : James Wylde

And why even introduce pi as a probabilityin real life you never or alnost never have irrational probabilitiescertainly not in coin flipsso it makes no sense to introduce it here Comment from : Leif

It diesnt make sense the x is still there within the average computation Comment from : Leif

If the game ends after finite with probability 1 that means it always ends with someone winninghow is that possible? What does it take to win? God this question is a hot mess Comment from : Leif

What does the last,sentence even mean? This question is,not clearly stated, Presh Howbdoes one win the game? By getting the right side on the last flip? How do you know what the last flip is? And what do you mean with probability 1? This game could easily be infinitethis question has serious problemsI can't be the only one who noticed this Comment from : Leif

Binary numbers means only using ones and zeroesso how can you have twos in the denominator? That's a mistakeplus what if the probability is out of an odd numbers of tossesso not a power of 2? Why did you just ignore this possibilityit should be n greater than x and equal to or less than y minus xthat's correct Comment from : Leif

I would imagine you would use base 3 instead if the probability had 3 outcome Comment from : Fayez Hawseea

Wonderful problem and solution But there is absolutely no way I would have figured it out So I'm just enjoying it Beautiful Comment from : ommadawnDK

Who makes these questions?? Comment from : Ranit Chatterjee

Your 1/3 example failedbrYou could hypothetically continue to play forever by continually getting the "start over"brYou can't have any result which is "start over"brbrLikewise your method of flipping until you get a heads fails as you can just keep getting tails forever Sure it is extremely unlikely, but for any given finite number there will be a probability that the game still needs to continue, rather than the 0 your problem demandsbrbrLikewise your method of flipping until it differs fails as you can keep having it match, meaning you need to go on foreverbrbrSo you completely failed to actually solve the problem and instead solved a MUCH MUCH MUCH easier problembrbrLots of problems are easy to solve, if you throw out the hard part Comment from : Jeffrey Black

it's not hard the rule is 2 flip a coin 2 times and p is 3/4 Comment from : Neymar

The biased coin must have a probablility of heads greater than zero and less than 1; otherwise, you could flip forever and never see HT or TH (For example a two-headed coin is VERY biased, and obviously could not be used here) Comment from : Ken Haley

The method used for the irrational probablility doesn't make any sense, if you divide by powers of 2 you DON'T get the numerator, that is if you divide by powers of 10 Comment from : Rasheed Alawashiz

Just a question, but just going to one of the early examples what is the finite bound on the number of coin flips required to finish the game with a probability of winning of 1/3? If it's finite there must be a finite bound on the number of flips but I can't come up with a finite number in which the game is guaranteed to end Comment from : Matthew Karr

question is not clear Comment from : Srikanth Tupurani

That is an absurd, weasely interpretation of finitely many tosses It relies on technical definitions of finiteness and probability and is utterly divorced from English Comment from : Paul Provencher

I think Presh needs to get laid more :) Comment from : pauldzim

I thought about method 2 and the extension Comment from : John Adams

"hey this is pressure locker" Comment from : Long Plays

wow! good one Comment from : Akil Rangwalla

Oh, here we go; I'm definitely going to play a D&D game with binary expansions written out for the different probabilities Comment from : Jonathan Fowler

3:16 the subtitles is wrong It's 1/pi not 1/2 Comment from : Girl More

The solutions presented do not satisfy the conditions given: If the probability is indeed an irrational number, there is no end to the binary sequence representing it Thus, the game cannot end in a finite number of coin flips Therefore, the prerequisite is not met for irrational numbers, eg 1/pi Comment from : Uwe Meyer-Gruhl

I've watched a bunch of these They can be amusing, interesting, challengingbut this onebrbrThis solution is awesome Comment from : Mephistahpheles

This was amazing, there's no way in hell I'd have thought of that Comment from : Gretgor

back in time I was asked to solve a riddle which on the surface looks similar to this one I never solved it and now, afterr seeing this video, I'm starting to wonder if that version wasn't stated correctly to me or if it really has a solutionbrbryou were given 2 coins and you had to simulate any probability with a illimitate but finite number of flips (so the start again procedure doesn't work nor the idea of flipping a coin until you get a head) before you start flipping the coins if needed you can choose to tweak both the coins and set them with any p,q probability Comment from : nemesisk79

This solution doesn't end the game in finite steps with probability 1brWe prove these things in Real Analysis as below:brYou write any finite natural number N (denoting number of steps) however large and I'll write one sequence of tosses with steps (N+1) which doesn't end the game It implies that there doesn't exist a natural number N however large for which game ends with probability 1 in N stepsbrQED Comment from : Manish Tiwari

Hello peeps telephone truly sounds lnke$sweevbox Comment from : Noe Guerin

Great Now I can throw away all of my n-sided RPG dices :P All I need is a coin Comment from : Sci Twi

public Comment from : trater oy

i have no idea whats going on Comment from : Anton Nguyen

I don't even understand the problem this time Comment from : Fransamsterdam

I do not think you can finish the game in finite numbe of steps when you introduce "start over" steps I think it get even more unlikely with irrational numbers such as 1/pi Comment from : NetAndyCz

How do you put 1/5 as a decimal in binary? I still don't get it Comment from : rosgori

very neat trick for normalizing the biased coin! makes me think of a Hadamard gate (like the quantum gate) Comment from : empmachine

being a programmer, I immediately thought of method 2: basically generating a random number but I really like method 1 better you only have to look at one of the digits in number p that's so cool Comment from : mellamobob

Let's give proper credit to Jon Von Neumann for inventing the solution to the last part of this video, the part about getting a fair coin flip from a biased coin Comment from : Tehom

Your two last videos are much better than videos like "many adults are stumped by this riddle" Please keep going this way! Comment from : Дмитрий Зырянов

great coin toss random number generator! you enriched my life! Comment from : p0gr

This is very interesting But I wonder which of the two methods is faster? That is, which method on average will yield a result in the fewest number of coin flips? Somehow I suspect the second method is faster, as this is basically what computers do, and computers are typically made to be as efficient as possible Comment from : Jacob Dawson

The answer to this is obviously no, it can't be done There are uncountably infinite different values of p between zero and one, and many of them have values represented by infinite and unrepeating sequences of digits Comment from : David Lloyd-Jones

I have a feeling that the binary method is the same as just using the inverse function theorem by generating a uniform number between zero and one by using coin-flips This was my thought at the start of the problem, anyway Comment from : James Pedid

Dudereally shouldn't use the finite amount of flips requirement That makes it impossible to use start over, as you have chance of an infinite loopbrbrBefore any smartasses come along with probability zero of that happening, Doesn't work That's the result of limit function, which uses an x tending to infinity Problem is, infinity isn't allowed here due to the "finite" requirement Therefore using this limit can not be used to make any claim on the (in)possibility of this question Comment from : Gloweye

Both solutions don't satisfy the "equal exactly p" in case of irrational numbers (because rely on rational approximations) Comment from : Arthur Dgn

Method 2 is simply the Base2 version of a standard D&D tactic:brUsing multiple d10s to represent individual digits of a number, roll under the probabilitybrbrMost commonly limited to whole percentages, it can easily be extended to any level of precision by rolling further d10s for additional digits as neededbrbrMy proposed solution was attempting to convert coin flips into d10s, but your solution of expressing the probabilities in Base2 seems far easier :) Comment from : MumboJ

I didn't do any math or formula or w/e but as a software developer I already knew the answer was binary Comment from : Robot Selene

Your first solution for the x/y probability does not end in a finite number of coin flips There is a propability of it never ending since the repeat result exist, so it is not an allowed solution by your rules of " The game has to end in a finite number of coin flips with probability 1" Comment from : Chocolate Vault

I throw the coins 2 times: friends wins: 1 heads 2 heads , I win: 1 heads 2 tails ^ 1 tails 2 tails ^ 1 tails 2 heads brthus I have a chance of 3/4 to win and he has a chance of 1/4 valid? Comment from : veryagilehands

great! Comment from : qwerty uiop

Awesome video, and you can even use the content in your life :D One of your best videos, keep on the good work! Even though I have no clue how you come up with this (and many people of the comments do as well), I understood how it works :3 What is your job by the way? :p Comment from : Languste

This is astounding! My gosh! ^^ Comment from : Sleep Mast R

this one is good Comment from : jerckï72

Now these are the types of problems your channel should be focusing on! I still solved it within a minute, but this was so much more fun than the facebook crap that your channel was spouting before Comment from : TheOriginalFayari

with the given rules in answer for the biased coin would that even work if the bias was 100 always one side? cause you'd always end up reflipping Comment from : HadienReiRick

Very nice problem and i liked the binary solution Keep up with those videos and dont upload easy problems Good job! Comment from : Maximos Stratis

I think this video and its comments are a perfect representation of how nobody really understands probability (not saying I do) Comment from : Carl Lewis

If YouTube had a playback speed greater than 2x, I would use it for these videos But great problem! Comment from : John Chessant

Seems a very complicated way to explain binary(flip) > binary(p) for a loss, binary(flip) < binary(p) for a win, keep increasing accuracy whilst they are equal Comment from : Christopher Burke

are you a maths teacher? Comment from : King Fayo

lol i cant even determine how to get to 1/3 not even Talking about irrational p Comment from : Bloomex

I hated probability calculations Comment from : Adventures with Frodo

a tough problem with interesting solutions bravo! Comment from : Trung Nguyễn Quang

how is it not infinite? you could just keep getting start over ad infinatum Comment from : Nathan Deyak

interesting content Comment from : realcygnus

THE KEY IS IN THE SET UP ALWAYS SET THE COIN ON THE DESIRED OUTCOME Comment from : Shawn Sumpter

Is this solution guaranteed to have a finite number of tosses? Seems like it does not Am I missing something? Comment from : jjmalm

2:20 I don't get it, what if you get TT infinitely many times in a row? IT IS POSSIBLE Comment from : Derek Liu

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