3 Discoveries in Mathematics That Will Change How You See The World 

Yours too? That's 2 now let's see if we can get a third person whose dad didnt say that and see who can count.

All 3 of my dads would say that

There are 10 types of people in the world. Those who understand binary and those who don’t.

There's 1 types of people, those who know binary and those who don't.

That's deeper than it seems

The questions posed in this in regards colorblindness I totally understand my best friend has proposed the prospective situation. What his of green perception to my green there's no way to confirm or deny that.

The trouble with explaining colorblindness to people is that Certain Shade blend together there are certain shades of green very light look yellow to me are things like moss which are red sort of green throwing some Ferns and I can't find my golf discs

That happened to a co-worker of mine, too.

There are women who have 4 different cones..

@@khironkinney1667 I have know idea what you wrote. It is extremely hard to understand. I would recommend either editing it with correct grammar and punctuation. Otherwise if English isn’t your first language then just write in the language your best in.

Wait a minute. Combined they have a 2/3 chance. Break them up and they don't each retain a 2/3 chance of winning! The pairing is arbitrary. Option 1. Say you have two picks. The first pick has a 1/3 chance of being right. If you were wrong, your second pick would have a 50% chance of being right. Regardless of which one you pick. In total you had a 2/3 chance of winning. The doors could have been 2 red and 1 blue door or 2 blue and 1 red door. It doesn't matter how they are paired. Option 2. You can pick one door or pick a pair of doors. Naturally you would pick a pair of doors (say #2 & #3) because that would also give you a 2/3 chance of winning. Then the host says surprise, door three just disappears, it's gone. So does the 1/3 chance that that door being the winner. I don't see why the 1/3 chance that door three had would transfer to the remaining door in your pair, any more than it would transfer to door number 1. There are no transfer of odds.

The fact that the host won't show you if you win ruins that

@@djtjm7000 It actually does matter because the door that is shown the contestant is *always* a loser door. If it were random, you might be right, but it's not. And that's why this math trick is deceptive. People act like the last two doors are a 50/50 chance because they act like the elimination of the first door is random *but it's not*. If you picked a winning door, the eliminated door will be a loser because both other doors are losers. But if you picked a losing door, the remaining door that doesn't get opened *is always the winning door*. Given there is a 2/3 chance you picked a losing door to begin with, there is a 2/3 chance the remaining door is the winning door, because the door that gets opened is not random.

@@djtjm7000 I tend to agree with you. When one door is removed, the odds remaining turn into 50/50 that I picked correctly the first time versus the remaining door. I would argue that the 1/3 that got removed gets split into the two remaining doors, since we don't know which of them is the winning door. It is kind of like flipping a penny. You flip it 9 times and it comes heads (unlikely but possible, presuming 50/50 change heads/tails). The next flip still has a 50/50 chance of coming heads.

You explained it better than Simon.

And it's irrelevant to the question

@@fudhater8592 It wasn't a question, it was a statement.... in a video about statements. Thank you for trying to sound smart at my expense. Better luck next time.

@@TROOPERfarcry It's irrelevant to the question of "What are the odds if you stay or switch". Try to keep up

@@fudhater8592 You're like a smartass, _hold-the-smart._ If you eat enough fiber to dislodge your head from your rectum, you'll realize that in order for the Monty Hall problem to work, the host has to know where the prize is. IE, it's not random. Do try to keep up, dear.

Is math a plural word?

Ah! Now it makes sense! Thanks:)

Exactly! Really well explained

great explanation. I was just thinking that since he is not going to show you the winner, and he is not going to show the one you picked, so seems that there was a good chance that he was forced to show the one he did. so I was on the right track, but your explanation clearly explained the why while I was only circling around the edges of the reason! nicely done.

Wonderful explanation, it makes complete sense now. Thanks!

This. This explains it. Thank you. Now I understand why the probabilities change the way they do…too bad my stats teacher couldn’t have used this in class…

Not sure where it falls, but my default is to stand still and let the other person move around me. Saves a bit of time, but I lose out on the two-step shuffle dance moves.

My theory is most people will step to the dominant side. That being the right hand side as most people are right handed. So my question is are you ambidextrous???

This is a Canadian stalemate. Both wants to move for the other and not look rude, so we spend hours just going back and forth saying "you first, no you first, no I insist you first". And "sorry. No I'm sorry."

If you're going in the opposite direction, it works fine when both step to the right or to the left. It only fails if one person steps to the right and the other steps to the left.

I just grab them by the scruff of the neck and beat the living daylights out of them, then I turn around and walk back the way I came from, it's far from ideal but who's perfect right

My favourite explanation as well. Or in the case of 100 doors, the choice is keep your 1 door or choose all of the 99 other doors.

Totally agree

That’s a great insight

Alright, I can live with that. But people want to believe in their luck so that's where my problem was.

Exactly. It doesn't matter when you decide to switch, the probability locks in when you choose the door. 1/3 you're right and 2/3 chance you're wrong. Switching is basically just saying: I think it's not the one I picked first.

Yeah, it especially causes confusion if those assumptions you're stating are not even mentioned. Why do I have to go along with these assumptions then?

I agree... When Monty opens a curtain he is revealing information.

@@insignificantfool8592 It doesn't actually depend on those factors though. By choosing a door, you yourself are fixing a condition. The fact that one of the doors doesn't contain the goat that was chosen while you were holding a door means the odds are now 2:3 for the last goat. This is irrespective of the fact the host will never open your door or the goat's door.

@@Demmrir many people think so, but it's not true. Imagine the host really doesn't want the contestant to win the car (maybe he has to pay for the prizes himself and is low on cash). What exactly prevents him from opening the contestant's door if he chose a goat door and only offer the switch when he chose the car door?

The tendency to stick with one's original choice is also reinforced by the contestant assuming that the host knows he or she chose the $1 million and is intentionally trying to throw him or her off. The other assumption being that if you had chosen wrong, he wouldn't give you an opportunity to change your mind.

whelp, atleast you got your name right.

The circuit output 3% of what you expected? Well that is you :) Circuit theory is only an approximation of circuit behaviour. V=IR, I=C dv/dt, V = L di/dt, KVL, KCL, etc are all only valid given assumptions about the size of the circuit, no interference from outside, and the speed of stimulus. They don't even describe the physical processes that lead to those quasi-steady state conditions described by those equations. A similar thing is going on with mathematical models describing semiconductor behaviour. They are simplified approximations whose equations are only considered valid when corresponding assumptions are valid. A single device can have multiple equations modelling behaviour and you can make the model more or less accurate at the expense of complexity. The models that humans use to design with are broadly simplified and will get you in the right ballpark. Computer simulation can use more complicated models and get you closer still. But this is before manufacturing variability is factored in, not to mention environmental factors that you may not be aware of or choose to ignore. Long story short, designs are broadly correct if your assumptions are correct but building something in the environment it is intended to operate in is always necessary to make sure things operate as intended and something(s) hasn't been overlooked. Getting an output 3% of expected is a significant error someplace. If everyone's designs were 97% off, it would be a fluke that anything worked. Electronic design would be more like witchcraft and the models used would be essentially worthless.

Exactly. That one bit of information changes the whole game. I just feel sorry for the Noble Prize winning physicists who got the answer wrong because he just assumed it was a completely random game.

You are right, and what’s worse is that not that the crucial piece of information was left out inadvertently. Is that they didn’t understand its implications and their whole reasoning about the answer is wrong, as evidenced by the way he proposes to experiment with cards, and pick one at random to reveal from the 2 that are not the chosen one.

Exactly, the probabilities depend on the host strategy.

Yepp, please!

Yes!

You probably shouldn't watch videos on the channels Stand-Up Maths or Dr Brian Keatong after a 12 hr shift then either lol

The way I'd look at it is that much of mathematics is a toolkit for manipulating information. But not in arbitrary ways; the manipulations have to follow various properties like being self consistent and following deductive logic. It's not surprising that tools for deriving information -- essentially augmenting our intuitive reasoning -- would be useful. And then, further to this, the question of whether maths will always be useful comes down to whether our universe has properties like self consistency. Well, if it doesn't, then all bets are off.

Same here. I also thought it was wrong when I first heard it, wrote a computer simulation, and had figured out the same thing before I even hit run.

You can't put the two stage MH problem into code, since we don't know what the host will decide to do. How do you model the host's decision as to whether to open the player's door or whether to offer a switch?

​@@insignificantfool8592 the host knows which door is the winning door. If you picked the winning door, then the eliminated door is irrelevant. If you didn't then there's only one door that can be eliminated. So basically your code is - identify which 2 doors are left, if the first isn't the winning one, eliminate it, if it is, eliminate the other non-selected one.

@Jonathan Newman that would be assuming the host always has to offer the switch. That was not mentioned and thus shouldn't be assumed.

@@insignificantfool8592 the premise of the problem is that the switch is always offered.

Or, Monty doesn't want to give away a million bucks, so he shows you a sure thing after you picked the best door, hoping you'll switch and save him $999,000.

Now, this example I understand! Thanks

The problem with that intuitive way to understand the problem is it's not actually understanding the problem. If the empty door is arbitrarily revealed, there's no advantage in switching. The door not containing the prize isn't the only thing that factors into why it's advantageous to switch.

But the 2/3 probability of the second door is also not good

@@strifera yes, of course, this is assuming knowledge of how the show always works, which is fair to assume in this case. That's part of the stated problem, and it's an event that is repeated every time on the show. The host always knowingly picks an empty door to reveal

And thank you for watching :)

Agreed. This was very solid.

Oh? I've read numerous examples just today, where people point out that Monte would never open the big prize door.

I don't understand wy the words "game show" don't automatically infer this to people. In what game show in the world is the big prize given away at random?

If he reveals at random it's a different (and rather unsatisfactory) game. "Here's $1m; would you like to switch to that?"

It doesn't matter what the host chooses to reveal if it's a fair game, though. In the video's example, you still either chose the $1M or the banana, it's not like the banana was what was revealed. The remaining 2 curtains becomes that psychological poison chalice dilemma with the host at that point. It would only be unfair odds if the host can cheat by changing what prize was behind the chosen curtain when asking if you're sure.

Simon's explanation is wrong. He said: "He [the host] reveals that concealed behind curtain number 3 is a thousand dollars", but he did not establish that the host always reveals a curtain. That is actually a huge difference, because without this additional information, the problem is unsolvable. The host could always reveal a curtain, he could reveal a curtain because you guessed right and it is too early for someone to win the main prince. Many versions of the Monty Hall problem make this mistake.

The host knows which curtain hides the million, and he will never open this curtain, and he will never reveal what's behind the curtain you chose, so there are three scenarios : 1 : If you chose the Million - he can reveal what's behind either of the other curtains at random - if you change you lose 2 : If you chose the Thousand - he can *only* reveal the Banana - if you change you win 3 : if you chose the Banana - he can *only* reveal the thousand - if you change you win He has knowledge and restricted choices, so what is revealed does change the odds ..

@@davidioanhedges The problem with your idea is there will never be a 2nd curtain reveal out of 3. If the 1st one revealed shows $1000, what's the 2nd reveal going to show? If it reveals the $1M, you choose the revealed $1M. If it reveals the banana, you choose the only unrevealed curtain remaining since it will be the $1M.

@@davidioanhedges Precisely.

new channel for him: CinemaGraphics

@@alyssinwilliams4570 lol

That can't be more than the third time I've heard that from Simon. Although thousands of times it's wtf? Lol

All 3of them!

Because he's let us know that while we may not know how to calculate the velocity of a baseball, our brains are fully aware of how the baseball is moving.

Excellent explanation

You did the best explanation but I'm still confused 😆

Excellent explanation. I've always stuck with "yeah nah it's now 50/50", even after watching the video, but 10 minutes ago something happened in my brain and something made sense but I didn't know why. This explained it well.

Pretty bad explanation here

What if you happen to have picked the right door anyway? The host will be forced to pick crappier doors, and in that case switching would be bad.

Your gold, my red.

I use the Hamming Distance in anomaly detection frameworks. Uncanny how it works!

When I was a kid, most of my friends thought I was crazy when I brought up this question... I was always happy being that dude, and now my bubble was popped 20yrs after the fact; when I find out it was just my friends that were...................... Lmao 😂🤣😂

I think he must have explained it wrong, because the way he showed it still comes out to a 50/50 choice, by my calculations. Step 1. Pick between three. Step 2. Throw that choice in the garbage because it doesn't matter. Step 3. Pick between two.

@@johndaniel7161 the odds don't change because an incorrect choice is revealed. That is why grouping visually helps explain. [1] [2, 3], odds of 1 are 33.3%, odds of 2 or 3 are 66.7% because they are grouped. Revealing 3 is a goat does not change the odds, 1 remains 33% and the group is 66% even though one in the group is invalid.

​@@johndaniel7161 I should add that the host knows which is the winning choice and which are the non-winning choices. When the host reveals a door in the grouped set they intentionally and knowingly reveal a non-winning choice (which actually matters a bit).

@@bretmcdanel8595 It technically does not matter if the host revealed a winning choice, if the format was still "Do you want the best prize from among the doors you did not pick?", it would just make a bad gameshow - because that's what the host is *actually* asking, just phrased to hide it, and why the grouping helps. You're not being given a chance to switch within two doors, you're being given the chance to pick the best from two thirds of the doors, just you already know one of them is a dud.

@@iskierka8399 But that's kind of the point. The host can't do that, because it's against the rules of the game. When you're playing blackjack, the dealer cannot refuse to deal you a card, because then the game cannot commence. Similarly, the game host cannot reveal the winning door, because then the game is over. It's because the rules are set in stone that this is a mathematics problem instead of a sociology experiment.

The confusion becomes, why doesn't that same probability exist for the door selected first? Why does revealing one door increase the probability of only the door not selected?

@@dudeinoakland It doesn't. Whoever thinks that revealing one door gives you better chances if you change your mind is delusional

@@dudeinoakland The key to understanding the problem is that _the prize does not move._ Starting out, there are X doors (where X = 3 in the classic problem), each with a prize probability of 1/X. All the doors are in one group and the probability of the prize being behind a door in that group is 1. However as soon as you choose a door, you have split the doors into two groups: we will call them C (door you chose) and R (the doors you rejected). There is only one door in group C, so the probability of the prize being in C is 1/X. There are X - 1 doors in R, so the probability of the prize being in R is (X - 1)/X. No matter which doors in R the host opens the probabilities of the two groups remain as they were, 1/X for C and (X-1)/X for R, because _the prize does not move._ Eventually there will be only one door left unopened in group R so that door must have the same probability of having the prize as group R does (and always had).

@@dudeinoaklandyou literally can switch from your initial one door to both the other doors and Monty also shows you, which of the other two doors is a loss.

Most people think it's an even chance. They don't use logic

That's a useful explanation, thanks. Maybe you can answer a follow-up question or two please? First question: Let's use Simon's 100 curtain example, but instead of the host opening curtains that they choose, a number of curtains are revealed but randomly so. The host is just there for show but doesn't actually choose anything. Does this affect the outcome? Second question: Let's say that after I pick a curtain, the host reveals 98 curtains (either randomly or selectively, I don't know if/what difference it makes), then I switch to the other curtain, as offered by the host, and then a friend of mine walks onto the stage. However, my friend hasn't watched the show and isn't told which curtain I originally picked and isn't told if I switched or not. He just sees 98 opened banana curtains and 2 closed curtains. Then he's told to pick one. Intuitively I'd say my friend has a 50/50 chance of picking correctly, but how can it be that he and I seemingly have the same choice but might have different probabilities of outcomes? Thanks in advance.

@@charlesq7866 1. It only changes the outcome if the prize is accidentally revealed. If the Random Curtain Opening Device opens only one curtain, and it is a banana, you should still switch to one of the other 98 closed curtains. Your probability of winning will increase from 0.01 to 0.01010204... If the RCOD opens two curtains with bananas, switching to one of the other 97 closed curtains increases your probability of winning to 0.01020619..., and so on. 2. Your intuition is correct. If your friend walks in, has no information of what has happened ("what's with all the bananas?"), and is asked to choose a closed curtain, they have a 0.5 probability of winning. The two of you have the same choice, but you have _more information._ Your friend only knows that there are two curtains and one prize. You know that the prize was somewhere behind one of 100 curtains. When you chose your curtain you split the curtains into two groups: your one curtain, with a 1/100 probability of winning, and the other 99, each with a 1/100 probability but _collectively_ with a 99/100 probability of concealing the prize. As curtains are opened that 99/100 becomes concentrated onto fewer and fewer curtains until only one remains in the group you did not choose. With no information, your friend's choice is an _independent trial_ and I think this is where a lot of people get tripped up by this problem. They think that your second choice, to stick or switch, is an independent trial when it is not. For it to be an independent trial for you, you would have to lose the information you had already been given. The only way that could be done is if after the reveal of all the bananas the prize was somehow randomly redistributed between the two remaining closed curtains. Here is another example of the problem, a simple card game with a standard deck of cards. The goal is to get the ace of spades, if you have it when all the cards have been turned over, you win some money. The deck is shuffled and spread out face down. You are asked to pick a card, any card. The card you pick is placed just aside, still face down, you do not get to look at it. Then the dealer flips over the remaining cards, one at a time for dramatic effect. If it isn't the ace of spades the card is tossed in a garbage can under the dealer's table. The dealer gets through all the cards until just two remain: your card, and the dealer's last card. Let's now consider three scenarios. In the first scenario, you are given the option to stick with your card or switch to the dealer's last card. You know now that the correct choice is to switch: your card has a 1/52 probability of being the ace of spades, the dealer's card has a 51/52 probability. In the second scenario your friend walks up to join the game. They have no idea what has happened so far - not even which card is "yours" - they just know the goal is to find the ace of spades and that one of the two cards on the table is it. From their perspective, with the information they have, the probabilities are equal. In the third scenario, instead of your friend coming by, the dealer shuffles the last two cards (in such a way that you cannot tell which is which), places them back on the table, then asks you if you want to stick with "your" card, or switch. Your card isn't really "your" card anymore, is it? The prize has been randomly redistributed. You have lost all the information you had. Your choice has become an independent trial. Now you are just like your friend: the probabilities of either card being the ace of spades are equal.

@@charlesq7866 YES. Exactly. Most explanations of the MHP assert that somehow when an option is eliminated from a set (a door opened in this case) that its initial "share" of probability is passed only to some sub groups and not the others. This is arbitrary and problematic. If this is the case, I want to include MY door in the group with the opened door, and have MY door receive special treatment.

@@JonMartinYXD Thankk you very much. I think I understand the concept better now :). Took me an hour or so in bed pondering it and varying the thought experiments, but I think I got it!

@@khemkaslehrling3840 The reason that your door doesn't get special treatment is that when you make your initial choice you've split the groups of doors into two groups: the one you chose and the ones you rejected. If something happens to one of the doors in the rejected group (like the door being opened) then that affects the rejected group, not the one you chose, which is why the "success probability" (SP) of your chosen door doesn't change from its initial position. However, in a way, your door is included: if (as discussed in my reply to Jon Martin) you have a friend who walks in after the door in the rejected group is opened, and they don't know which door you chose, and they see the opened door and understand the game concept, then they'll know that the door opening has just now changed the SP on all the doors (for them but not for you), meaning your door is receiving the treatment of a change of SP, which you wanted. It might help to know that your advantage in information over your friend still puts you in a better position. Let's say there are 10 doors. You picked one. The SP of the right door being in your chosen group (a group of one) is 1/10=0.10, and 9/10=0.90 for the rejected group. A door in the rejected group is then opened, meaning the SP of the 8 remaining closed doors in the rejected group is now 0.90/8=0.1125 Your friend then walks in and just sees 9 closed doors, 1 open door, and doesn't know which door you chose. They ignore the open door, so their SP is 1/9=0.1111111111 and this is less than your SP.

Try modeling all the scenarios. Imagine a game with 3 doors, and 1 car and 2 goats. Assume the contestant chooses door 1. There are now 4 scenarios and two options each, resulting in eight outcomes. And the split of win/lose for stay/switch is 50/50. 1. Car / Goat / Goat - host shows you door 2 - switch lose / stay win 2. Car / Goat / Goat - host shows you door 3 - switch lose / stay win 3. Goat / Car / Goat - host shows you door 3 - switch win / stay lose 4. Goat / Goat / Car - host shows you door 2 - switch win / stay lose

@@khemkaslehrling3840 I know perfectly well the scenarios, and worked all this out long ago and even tested my mathematics with a simulation, just to make sure. However, the point you completely miss is that the Monty Hall problem is often not fully specified. The extra conditions are a) the host must know what is behind each door b) the host must only open another door with the lesser prize c) most importantly of all, the host must do this every single time Without (a) & (b) we are in a different scenario as either could lead to the main prize being revealed. If (c) is not met, then the host might only decide to open another door when the competitor has already picked the main prize. Of they might only do so if the competitor chose a lesser prize, or maybe some mixture of the two. The host might act to help, or to hinder. Like many probabilistic problems, it's essential to be very precise or it is easy to be lead astray.

@@TheEulerID I am forever mystified that people hear the words "game show" and do not immediately understand that the host will never reveal the prize. Thus, the game show host must know what is behind the doors. And with this it is inferred thhat the host always never reveals the prize. Perhaps some people really don't understand how game shows work?

The believers of the "switching is better" solution will tell you that Monty will always offer the switch. In that case there is no room for deception. But as you say, this is counter everything that makes up human Interaktion, so obviously people tend to interpret it in a way that it was Montys decision and not a forced play. In that case, staying with your door is indeed not a bad idea.

Andrew Well, it helps some people.

@@SigFigNewton Or if they know it, they don't see that they selective revelation creates a sampling bias. I am not proud, I admit that it took me a long time to understand it.

I don’t think you understand the Monty problem

By which you mean, I think, that if Monty Hall always opens 98 doors, none of which have $ 1,000,000, then it is more intuitively obvious that you should switch.

You could also just list all the possible situations you can end up in after you have chosen your door and the host has revealed a door: Door in [] is chosen, door in || is revealed by the host, L looser door and W winner door. Assuming the host always opens a looser door, we get these cases: 1. [L] |L| W 2. [L] W |L| 3a. [W] |L| L 3b. [W] L |L| Now, we can say that 3a and 3b are the same and count them as one single case, because the order of the remaining doors (the L's) doesn't matter in this case. This leaves us with 3 cases. 2 of which switching would lead to you winning and one which wouldn't. This is how you can reach the 2/3 probability of winning by switching. Edit: Another way too look at it, is that you have 2/3 probability to chose one of the L's at first. And whenever you have chosen one of those L's, you're in case 1 or 2. Thus, switching wins in 2/3 of all cases.

I've heard it before where someone says they proved it on a computer. I'd suggest that you already biased your initial conditions . It's a 50/50 chance. The problem posed is the probability of success of picking a concealed object when given two choices. Forget about the razzmatazz and music and audience reactions.

@@philip5940 I also proved it mathematically. And it's not, that's the point of the video

@@AaronWhiffin the probability of success in picking a concealed object when given two choices is 50/50 . It's always been so . There's no such thing as magic . The events prior to and during the decision making do not change anything concerning the abstract of the decision making .

@@philip5940 Tell me you don't understand the video without saying it hahaha

@philip5940 You have a blind spot in this, Philip. Yes, a choice between two is a 50/50 choice. BUT: When you pick the first curtain, it has 1/3 chance of being correct. Meanwhile... *The other two curtains together* have 2/3 chance of holding the jackpot. The host removing one non-jackpot curtain of these two is just a detail, a clarification, of what is in this second 'set'. That 'side of the equation' is still 2/3 .

You can take three cards, two marked 'Goat/switching wins', one marked 'Car/staying wins', and simply write down what you first picked.

Nope - try again

The new player chooses with a 50/50 chance of what he'll choose. However, if he selects the first curtain (preselected by the previous player), he will still get only 1/3 chance of the million; and if he picks the other, 2/3! Heh heh!

@@twotubefamily9323 Actually, secheltfishmarket6419 is correct. In isolation, a 2nd player who doesn't know about which door had been picked really DOES have 50-50 odds. Seriously. Because there are TWO problems operating simultaneously in the true Monty Hall problem. The 3-door game (where a zonk has been identified) retains the 1/3 and 2/3 odds. The 2-door game, however, does not have the same odds. There, it really is 50-50 - but you get fooled because you only see the second choice and (incorrectly) put aside the fact of a first pick.

Yeah, it’s not very well explained here. The Monty Hall Problem relies on the host always revealing from the other two curtains and never revealing the grand prize.

The confusion comes in when forget that all of math comes down to counting. It's just that what is being counted has become more and more complex. The symbol "4" isn't real, but is used to mark that "there are 4 things". "1/4" means "1 thing out of 4 things", "4/2" is "4 things separated into 2 groups", and so on. The numbers themselves are not real, they're just labels for convenience that represent what IS real.

@@Vaeldarg Symbols are actually real, they represent real things. When we say they aren't real we mean they aren't concrete, but right now I'm writing letters that convey to your mind real things.

@@konroh2 Where were you going with this?

@@Vaeldarg Concepts are real things, so language and symbols are real. An idea of something is real. Real things aren't just material, what is invisible is also real. When someone says maths isn't real because it's just a symbol it's denying the spiritual reality of ideas, which are also real.

@@konroh2 "the spiritual reality of ideas" is easy to deny, because no it isn't real.

He HAS framed the problem totally incorrectly. His explanation is a reasonable approach to explain the actual Monty Hall problem, but it’s a pity that he did not properly describe the Monty Hall problem in the first place, and the video really should be corrected - I’m just surprised that there aren’t more comments pointing this out!

I'm guessing math is tricky for you

​@@twotubefamily9323 No, I have a science degree and went all through calculus. The point is that the "chances" are calculated by telling us that after selecting a door and one of the other doors is exposed, should we change our selection to the other door? BUT there is nothing in this set of "rules" that says the second door MIGHT expose the million dollars, and nothing in the rules that we cannot select the opened door showing the million dollars. THAT change the odds dramatically. But we ASSUME the rules are that the million dollars cannot be exposed at the second door opening and that we can't select the open door.

@@twotubefamily9323 blaster is correct. This is yet another sloppy/incorrect formulated Monty Hall problem. The host does not "cheekily decides" to open another door. In classic Monty Hall, the host opens another door knowing that it is not the big price. And he would have done this regardless which door you initially choose. THEN you will have a 2/3 chance to win if you switch. For example, if the host "cheekily decides" to open one of the other doors randomly, discovers it to not be the price and therefore offer you to switch. THEN you will still have 50/50 chance between the two remaining doors.

@sverkeren Respectfully, no! When you pick the first curtain, it has 1/3 chance of being correct. Meanwhile... *The other two curtains together* have 2/3 chance of holding the jackpot. The host removing the non-jackpot curtain of these two is just a detail - that 'side of the equation' is still 2/3 .

@@Michael_Arnold Respectfully, yes! In classic Monty Hall, if you first choose wrong (with 2/3 chance), then the host will practically show you where the price is by AVOIDING that door. If the host does not actively avoid the winning door, then you don't get that probability boost. Instead that probability get evenly distributed on the remaining doors. What could have happened matters.

Astonishingly, for every day life most of us probably learn all the math we're ever going to need by about grade 3 or 4. (Or, college in Florida) Throw in a bit of percentages and knowing how to average that comes in handy sometimes, and in some backwards countries fractions which I think are taught a little later, and that's about it. Which is not to say learning beyond add subtract multiply and divide is a waste of time. Depending on what you end up doing, you may need algebra or geometry or calculus, but the vast majority of us don't. Where we bump into it in day to day life, it's done for us.

What makes it that your initial choice ISN'T the million dollars? Nothing. The odds don't change.

@@lgmediapcsalon9440 Which is why you still have 1/3 chance you were right ... The new information means that the 1/3 and 1/3 for the other two options merge, but your option does not change

There is NO new information provided by opening one of the unchosen doors - at least no useful information. (Using the car/goats/doors format):- You knew that there was at least one goat behind the other two doors, and you knew that Monty would show you a door with a goat. The reveal tells you nothing useful, only that the revealed door was a door that had a goat behind it. Knowing that does not help you decide. If there was no reveal, and you were offered both the other two doors over your first choice, you would swap - and you would know that you are getting at least one goat.

@@user-gy4yq8om3e The new information is that the host will not open a door to the prize so his options are limited and so you gain new information

I still dont think that is true. When you are given the chance to either keep your current choice or choose another, you are actually just rechoosing based on what is still unknown. Just because they say "keep your current choice" does not mean that you are keeping your initial probability. Please respond, because there might be a high probability that I'm missing something :P

@Wuddigot Simplest way to think about it. You know how the game works. You know that you will be shown a losing door after you choose a door. Beginning of the game you have a 1 in 3 chance of picking the winning door. A 2 in 3 chance of picking the wrong door. Would you agree 2 in 3 odds are better than 1 in 3? So you are most likely to choose a wrong door. So knowing how the game works, you pick your first door knowing you're more likely to pick one of the 2 out of 3 that are losers. Because of this, when a door is removed, you pick the other. Definitely not a fool proof way to win Just gives you better odds. You gotta look at the odds of picking the wrong door to determine the best odds of picking the correct door.

@@Wuddigot I just reread you comment. Keeping the first door keeps your original 1 in 3 odds. So keeping your first choice does keep your original probability. Gotta bring the odds of choosing one of the wrong doors into the equation. 2 in 3, which is more likely. Start of game you're thinking "I have the best chance of picking a wrong door" (2 in 3) So after you pick a door (odds are its a loser) you know one of the other two doors are more likely to be the winning door. Right? Since one of them doors are removed from the game, you pick the other. Giving you 2 in 3 odds of it being the winning door.

@@al1383 That just isnt true though. The odds of you choosing the wrong door is now less because there are less wrong doors that you would choose from. You will not choose a door you know is wrong, and therefor the conditions (and probability) have changed. Am I hearing the initial problem incorrectly or something? guy chooses door, before result is revealed he is told one of the wrong doors and given the choice to choose again... Is that it? he now still does not know which of the doors is right, but he does know that one of the two remaining doors is wrong. If I am not missing a detail, this is an iterative problem, as in each completion of the action loop starting at the beginning with a different set of conditions.

@@Wuddigot you going to get me racking my brain on this again! 😃 To tired today. I'll respond later

On a similar game show, there are four doors. Each hides a pair of siblings; one has two boys, and one has two girls. The other two have one boy and one girl; one where the boy is older, and one where the girl is older. You win if you pick a door with only a single gender. You pick a door, and the host opens another that has two girls. So you know that your door has at least one boy, but what is the probability that it has two? By your logic here, "your original choice" was correct with either a 1/4 probability or a 1/2 probability. It's ambiguous, because "your original choice" could mean a winning door, or a door with two boys. But neither matches the popular answer to the equivalent "boy or girl" paradox problem, 1/3. Discuss.

@@jeffjo8732 In your game, the host reveal one of the winner doors. Let's say the host reveal all the doors, what's your winning chance? It's 50%, right? What is the difference if the host reveal one door? Also, the host can clearly see the result even before he open any door.

@@Araqius "In your game, the host reveal one of the winner doors." But he can't reveal all of the winning doors. The way I stated it is how we make it identical to the Boy or Girl problem since in it you are told "there is at least one..." I could have made winning be choosing a mixed door, but while the principles are the same the situations change. And my point is that, while your answer is right with a necessary assumption, the reason you give is wrong. In the Monty Hall Problem, here are the possible ways the game can progress after you pick door #1: A) Probability 1/3: The car is behind door #3, and the host must open door #2. B) Probability 1/3: The car is behind door #2, and the host must open door #3. C) Probability 1/3: The car is behind door #1, and the host can open either #2 or #3. The problem is that WE DON'T KNOW HOW HE CHOOSES. But if we assume he opens door #3 with probability Q, case C separates into: C1) Probability Q/3: The car is behind door #1, and the host chooses to open #3. C2) Probability (1-Q)/3: The car is behind door #1, and the host chooses to open #2 So, if you see him open #3, the probability that the car is behind #1 is (Q/3)/[(1/3)+(Q/3)], or Q/(1+Q). And the probability that it is behind #2 is (1/3)/[(1/3)+(Q/3)], or 1/(1+Q). Now, if we don't know how he chooses, we can only assume that Q=1/2, and so these become 1/3 and 2/3, as you said. But the point is that this isn't because #1 started out at 1/3. It has to be re-calculated, and can be anything between 0 and 1/2

@@jeffjo8732 The host can clearly see the result even before he open any door. If he see you pick BB, he know you will win. If he see you pick GG, he know you will win. Unless he do something like revealing the door with GB and BG.

So the host opens a single gender door (a WINNING one), which is different from the Monty Hall where the host opens a LOOSING one. Anyway. In this case, I would definitely choose the winning door because it is winning. Remember the strategy is to win (obviously) and I am shown a door which leads to win so I take it (1/1) and in this case it is possible. So the host always shows a wrong one (otherwise the game ends with the succes of the participant) Bsed on this premises let's go further. So half of the time if I pick well first (two boys or two girls (2/4) -> (1/2)), I am expecting the host to show me a loosing door ( not a winning, otherwise I win) So let me rephrase it. A game with 4 doors : (M/M) (M/F) (F/M) (F/F) Participant chooses one of the four. The host show the participant one doors : option 1) Host show a winning one (1/4) : we already ruled it out, otherwise the game is stupid and always won by the participant. option 2) Host show a loosing one (2/4) (more interesting) and in this case, the participant is shown a wrong door and he originally choosed a door which is (1/2) chance to win. So two cases for option 2 where a wrong door is showned : a) original choice is right (1/2): changing -> one the two remaining door one is right and the other not (1/2) not changing -> Participant stays with (1/2) chances b) original choice is wrong (1/2): changing -> definitely good because Participant original choice was bad and he is shown a bad one, so if he changes he wins not changing -> definitely bad because Participant original choice was bad and he stays with his choice and because changing does not affects a) but affects b) in a good way, CHANGING strategy seems to be the better what are the odds now a) 1/2 whatever you do one half of the time b) 1/1 if you change the other half so : 1/2 * (1 / 2) + 1/2 * (1) => 3/4 You may wander why the solution to Monty Hall which is (2/3 best strategy) is more little than /4 best stategy for your problem) It is because there are more good solutions proportionaly (2 on 4 in this case) than in the Monty Hall (1 on 3) This raised two questions which I found interesting. How to explain... I took a lot of words to try to explain it. How can I decrease the number of words to explain it (and in a better way). The actual challenge is here to explain. As I review the above lines, I am not pleased. Please help me to improve my explanation. First on the Monty Hall problem Second on this 2 kids and four doors problem. And I would say, this hundred curtain problem presented in the video which leads to more perplexity... It's says that 98 (or 99 if you originally picked the right one) curtains are shown with bananas behind than it is obvious for you to change. Ok, nice way to pull you in the right direction but definitely drag you in the wrong direction. What if, (with the same problem , "hundred curtains") I only showed you, as a host, a curtain ( on the hundred which are present) which contains bananas, would you think it is worth to change or to stay. And Thank you to say 'discuss' in your post. It is rarer and rarer to read this.

I don’t think you understand the Monty problem

Ouch.

Doesn't sound like your calculus teacher was in any danger of getting a Fields medal.

The great Feynman, who was gifted in math, admitted that when it came to music he couldn't carry a tune. My father, born about the same year as Feynman, was a violinist in a major symphony. He could carry a tune. So, not all of us have the same abilities.

That is a ridiculous and poor way to teach or incentivize anyone. If your calculus teacher could not rebuild an car engine or write a symphony... is he a simpleton?? Different types of smarts.. All of which are important.

George Bernard Shaw said ‘Those who can, do; those who can’t, teach.’

You might enjoy looking into the Theory Of Forms by Plato.

@@bramvanduijn8086 Thanks for the tip. I might give it a try, but I don't think I'm qualified enough to understand things like this. What I just described were just the thoughts of some kid about things he learned in school.

The Earth is much-much smoother than, say, a billiard ball - blown up to the same scale, the pits and bumps on a billiard ball would be much larger than the mountains and sea depths of the earth...

Her answer is to be expected from the mind of a person conditioned to answering IQ questions. The correct answer is indeed 50/50 and not ⅓ to ⅔ . Interestingly Tibees does a video where she answers an IQ question as might a person with an IQ of 300 by coming up with a polynomial equation to generate the pattern presented for the trend given for which the next step in the series needs to be given . Her polynomial equation yields the original starting point instead of an obvious intuitive next number for the series.

​@@philip5940 It is not 50/50. Why would it be 50/50?

Have you *read* what vos Savant wrote? She had a web page about the whole thing, though it seems you can now only find it in archives. In her own words, "So let’s look at it again, remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose". The answer defines the conditions for answering the question? No. The question defines the conditions for answering the question. If the answer is defining the conditions for reaching the answer, it's circular reasoning.

@@hughobyrne2588 Wrong. In order to always open a losing door, whoever opens that door has to know which is a loosing door. But again, the video above shows the answer is correct, and a lot of grade schools participated in an experiment where it worked. There are even web pages that demonstrate this. I shared one link, but apparently that post vanished. I did 200 tests where I always changed, and hit real close to a 66% win rate.

@@philip5940 Not sure what happened to my response, possibly removed because it had a link. But the link was to where you could test it yourself. There are a few web pages set up to test this, and I ran 200 samples, always choosing to switch, and won real close to 66%. Also watch the video I am responding to. They say it works.

By what magic does the share of probability of the opened curtain transfer to only one of the other curtains, and not be split across both? Why not then just declare that the curtain you chose is grouped with the curtain the host opens, giving it "the entire probability of the opened curtain" and making your door 2/3rds? This is the fallacy of this Monty Hall thing.

​@@khemkaslehrling3840 Think of it this way; 3 items behind 3 curtains, 1 prize & 2 duds (worthless), no matter which door you choose the host can reveal a dud. Lets go through all 3 choices, Scenario #1 lets say you pick the prize curtain then the host reveals a dud curtain, if you switch you get another dud (1/3). Scenario #2 lets say you choose a dud curtain, the host can only show you the remaining dud so if you switch you pick the prize (1/3). Scenario #3 you choose the other dud, yet again the host can only choose a dud, so if you switch you pick the prize curtain (1/3). So in closing in only 1 out of 3 scenarios do you lose by switching, but if you chose to switch you'd be right 66.7% (2/3) of the time.

​@@justblaze6802 that explanation just gave me the light bulb moment, thank you

Another help to those who aren't yet convinced: when Monty opens one door, think of it as him opening *ALL* the other doors except the one you chose and one other one.

​@@khemkaslehrling3840 You choose one door out of 3. If I offer you to let you switch for the two remaining doors, would you take the deal? Of course you would, because your odds would be 2/3. That's what's happening here, except that the host is opening one of the two doors first. But you are still effectively getting to choose 2 doors by switching.

Yes, I tried that too. When you pick the first curtain, it has 1/3 chance of being correct. Meanwhile... *The other two curtains together* have 2/3 chance of holding the jackpot. The host removing one non-jackpot curtain of these two is just a cosmetic detail - that 'side of the equation' is still 2/3 .

It doesn't matter what you pick initially because it's ignored and a new scenario is presented. The host's reveal is never random. It must and will always be a goat or the game is ruined. The player always and only makes a real choice between one of two options, a car and a goat.

@@lrvogt1257 That's not true. Try the scenario with 100 doors. Your first pick is 1% chance of success. Then Monty opens 98 doors with goats. Now there are two doors left, Your door (1%) and Monty's door (99%). It is NOT 50-50 ;)

Hmm I don’t think you understand the Monty problem

You still get the same odds. All doors are 1/3. So the host reveals one... you still have 50% chance with the remaining doors. You gain nothing in terms of odds, you only get to see what was behind one door that you did not choose anyway. There's no 66% chance in this scenario, ever.

@@lgmediapcsalon9440 No. The key to understanding the problem is that _the prize does not move._ Starting out, there are X doors (where X = 3 in the classic problem), each with a prize probability of 1/X. All the doors are in one group and the probability of the prize being behind a door in that group is 1. However as soon as you choose a door, you have split the doors into two groups: we will call them C (door you chose) and R (the doors you rejected). There is only one door in group C, so the probability of the prize being in C is 1/X. There are X - 1 doors in R, so the probability of the prize being in R is (X - 1)/X. No matter which doors in R the host opens the probabilities of the two groups remain as they were, 1/X for C and (X-1)/X for R, because _the prize does not move._ Eventually there will be only one door left unopened in group R. What is the probability that that door hides the prize?

@@JonMartinYXD Where did I write that the prizes move? If there's 3 doors and one is open, the chance is 1/2.

@@lgmediapcsalon9440 The only way the odds can "reset" is if the prize can move after Monty opens a door. Let's put it another way. What if before Monty asked you to choose a door, he asked you if you would like to choose one door or two doors? You'd be crazy to restrict yourself to just one door, right? Choosing one door gives you a 1/3 chance of winning, choosing two gives you a 2/3 chance of winning. Well that is exactly what Monty is offering you after he opens a door. Three doors: A, B, and C. From your perspective, A has a 1/3 chance of hiding the prize, B has a 1/3 chance of hiding the prize, and C has a 1/3 chance of hiding the prize. You choose door A. You have a 1/3 chance of winning. Monty opens door B or C to reveal a goat (the goat is not the prize). Monty now offers you the chance to stick or switch. This is equivalent to you previously being given the option of choosing door A or choosing _both B and C._ Since you are the one determining if the choice is between A or B+C, B or A+C, and C or A+B, and since the prize is already in place, Monty cannot change the probabilities. If you choose only one door you have a 1/3 chance of winning. If you choose two doors, by switching, you have a 2/3 chance of winning.

@@JonMartinYXD You're moving the goal post with your "pick 2 doors". Monty opens one door, two remain. You have 50% chance of winning. End of story.

ooh, I can explain that. Super easy. One of them is wrong. ;)

Math is the subject , Maths are the stuff you do in the subject. i.e. '' There are many maths to do today , lets get mathing''.

The science is called Mathmatics. One side shortens it to Maths the other to Math. This also changed over the years which one was the more accepted one where.

I think it's fair to say the originator of the problem is the person who asked the problem, Craig F. Whitaker. He wrote a letter to vos Savant while she was writing a regular column in Parade magazine. In Whitaker's letter, as in the question as phrased at the beginning of the video, there is no indication that a compulsion is placed on the game show host to always reveal a door with the booby prize, a critical component in the chain of logic which leads to the conclusion "it's better to switch".

Eh, no! When you pick the first curtain, it has 1/3 chance of being correct. Meanwhile... *The other two curtains together* have 2/3 chance of holding the jackpot. The host removing one non-jackpot curtain of these two is just a detail - that 'side of the equation' is still 2/3 .

@@Michael_Arnold I assume that - in a game driven only by chance - if Monty happened to reveal the jackpot , which would then be out of the mix, the game would be pretty much ruined: The chance of winning would go to zero no matter what the contestant did. On the other hand, if Monty randomly and unknowingly selected a curtain that contains a booby-prize, then the remaining curtains each would retain their 1/3 chance of being the jackpot......even odds to win with either one. But it is NEVER the case that the jackpot is revealed prematurely. Human intervention.

@bobpourri9647 Well, if the host reveals the $1M, there's no game, really. But that's all irrelevant to the mathematical probabilities involved, though. It doesn't matter whether your goal is to win the $1M or the bananas, your *chances* of doing so double if you change, *provided the host reveals an unwanted prize, either before or after you change.* Here are the three possibilities of what lies behind the doors: 0 0 $ or 0 $ 0 or $ 0 0 (where $ is a coveted prize, and 0 is unwanted). If you select the first choice each time, and eliminate a 0 from what remains, sticking to your choice gives you a win 1/3 times. Changing your choice gives you a win 2/3 times. *As long as the host reveals a dud prize,* the odds tilt to 2/3 for a change of choice - mathematically and forever! ; )

What's worse, she didn't solve it. This was a well established reality before she ever said anything. Marilyn vos Savant just answered the question for someone in an article in 1990. This probability problem was solved over a century earlier. Then a bunch of twits responded without actually looking into it. So, the same phenomenon as social media comments. Give a large enough sample of humanity a forum, and you'll get a lot of very stupid responses. Those who already understand likely won't comment, those who learn only might comment, but those who hold close their ignorance will shout from the rooftops.

Yeah, the bad logic comes in that the extra 1/3 chance was added to the 2nd curtain in the video's example, but that's only because in the math it could only be added to 1 of the 2 remaining curtains. In reality, you still wouldn't know which curtain became more likely (the 2/3 chance) to have the $1M behind it.

That's an excellent way to describe the problem!

I remember reading about the Monty Hall problem when I was young. The article said that the chance of the other option being right remained two thirds even as one of the curtains was eliminated but never bothered explaing why. It was so frustrating.

Yes I agree, it’s a common misconception of probability, I have actually applied this method to various “betting” situations and guess what? I lost 6 times out 7. I actually studied probability when doing my BSc in Electronic Engineering and touched on it more during my MSc. It’s actually a very common misconception and understanding of quantum physics.

I like it.

montecarlo simulation. nice

Simply put, yes!

That is not correct. You would have a 2/3 chance of picking a wrong door regardless as to the host's knowledge. In that case if he opened one of his two doors that reveals a non-prize there would be no advantage in switching.

That is an explanation that makes straightforward sense.

But by removing one door and making that door 100% wrong then it is a 50/50 it may be one of the remaining doors. 6ou can come u0 with some pseudo intellectual nonsense explaining how the myths shows that no to be true but in real world application its 50/50 not 66/33

@@williamdoherty7824 Create an experiment to test it. You can do it with playing cards and a friend who knows the cards in the hole. But you need to repeat the experiment a number of times to get an approaching probability.

@@klaus7443how can you start a sentence correctly sayin your initial chance is 2/3 to be wrong and then ending it with the conclusion that going away from that choice gives you no advantage? Weird.