Just 3 questions/puzzles that seem obvious but aren't 

Get the "Trivial" shirt here: stemerch.com/collections/trivial and the mandelbrot set wall tapestry here!: stemerch.com/collections/mandelbrot/products/mandelbrot-fractal-wall-tapestry

blud stopped making shitposts so he can tell us we're dumb and I'm here for it

I forgot you also have an educational channel

My head hurts less when I watch your other channel.

So he has both a brain rot channel and a brain education channel. 😂 love it. EDIT: I realize that some people who are reading this comment have not seen Zack’s other channel. It is not brain rot comedy, but it is stuff that I would listen to with headphones on. Still though, the guy is an absolute genius and is comedic gold.

That was an illegal cliffhanger. I need to know more.

This is your second channel now xD

When you flip two coins, the options are HH, HT, TH, or TT. However, when the person holding the coins says he knows that at least one coin is tails, the order of the coins doesn’t matter, therefore the sample space becomes 1. a tail and a heads or 2. two tails. Another way of thinking about it is when the coin holder guarantees one tails, you can remove that coin from the question altogether since it has a 100% probability of being tails, leaving only a 50% chance for the other coin to be tails.

Bro made this video with the SOLE purpose of making us feel dumb 💀💀💀

The first puzzle can also simply be solved with symmetry. Both glasses contain the same amount of liquid, because we've moved one tablespoon from each glass to the other. So whatever the distribution of coke and juice in the first glass is, the distribution in the other glass must be exactly the same but reversed.

During the last one, I feel like I spotted the problem with it. By revealing the suit, you're able to know specifically which set it's in. Without revealing the suit, you still know it's going to be in a set of that size, you just don't know which one. In your 4 card example, knowing you have one Ace doesn't make it a 1 in 5 until you know the suit. There's only 3 other cards. The probability was always 1 in 3, but from two possible sets. I'm having a hard time organizing this thought. If your 4 card deck was shuffled, and you drew two cards from the top of it without looking at them, the odds of you drawing both Aces is 1 in 6. If you draw the first card and look at it, and it's an Ace, your odds of drawing the second one become 1 in 3. You inherently know the suit of the card by having looked at it, so you can narrow down which set you're working in. However, if you draw that card and show it to a friend, and they only tell you it's an Ace, and not which one it is, that doesn't change the fact that you have a 1 in 3 chance of drawing the other Ace as the next card. Holding two cards, knowing one of them is an Ace, you're holding 1 of 5 hands. You will land in a set that does not contain two of those possible hands, no matter which Ace you're holding. You're really only holding 3 possible hands, (Ace - Ace, Ace - 2, Ace - 7), and caring about which Ace it is in the back half is changing the question subtly.

Probability changes depending on what information the observer has. Doesn't matter if any new information feels insignificant, the observer must check how the new information changes the sample space. About your coin problem, Let's say you have a 100 copies of yourself who all flipped two coins and are holding one in each hand. So, 25 people with two tails, 25 with two heads, 25 with tails in left and heads in right, and 25 with tails in right and heads in left. 1) If you ask "All those who have at least one Tails, stand up", 75 of them would stand. Out of these 75, 25 would have both coins as tails, hence the probability equals 1/3. 2) If you ask "All those whose left hand's coin is Tails, stand up" only 50 of them would be standing. Out of this 50, 25 of them will have both tails. Hence probability equals 1/2. 3) If you repeat 2 with "right" instead of "left", you will get the same probability. 4) Finally, If you say something like "All those with Tails in THIS hand, stand up", but we (the observers) do not know which hand you are referring to, it does not matter. We know that no matter if you asked for the left hand or the right, there will be 50 people standing. And 25 of them will have two tails, making the probability equal to 1/2 again. Which hand is revealed does not matter, the fact that one of the hands was revealed reduces the sample space to 50. Also, it is important to note that 1) is fundamentally different from 2) or 3) because no hand is revealed, instead it is only revealed that two heads are not possible.

The two coins problem is actually asking two completely different questions depending on what information you give. When you say _"one of these two coins is tails, what's the chance the other one is too?"_ you're actually asking *"What is the probability that both of these coins are tails given that at least one of them is?"* When you say _"this coin is tails, what's the chance this other one is?"_ you're actually asking *"What is the probability that this coin is tails given that another (unrelated) coin is tails?"* Looking at it this way it makes sense that they have different answers. The answer to the latter question is obviously 50% because the two coins are unrelated, but in the first question we have no way to separate the states of the two coins, which complicates the probability. In the final example when the screen blacks out to prevent us from seeing which hand's coin you say is tails, we are not given the ability to distinguish between the two coins, so it's equivalent to the first question.

I dont know which is better, zach star uploading or zach star himself uploading

For the first one, you can reason from the end state. Since the volume end up the same, the volume of foreign liquid has to be the same in each glass.

For anyone who doesn't know, the reason why the probability 'changes' is because we're applying the wrong theory to find the probability in both of these scenarios. Let's take the coin scenario: 1. In the presented scenario, we're looking for an ORDERED pair. Let's assume the order is in the format [x, x]. 2. No matter what, one coin must be tails. 3. All possible ORDERED pairs for this scenario are [T, T]; [T, H]; and [H, T]. This means that, assuming we don't know the position of the coins, we'll have a 1/3 chance of the ORDERED pair being [T, T]. If we do know the position, let's say the second value is T, we can rule out any pair where the second value is not T. This leaves two pairs and a 1/2 chance of the ordered pair being [T, T]. Now, the reason why I say that we're applying the wrong theory is because the question asked is, "If I have two coins and one is tails, what are the chances that the other is heads?" This means that we are not, in fact, supposed to be looking for an ordered pair, and instead we should be looking for the value of the OTHER coin. In this case, we ignore the known coin. Because there are only two states a coin can be in, both tails and heads will have a 1/2 probability, and so, no matter what the position of the first coin is, the second coin will have a 50:50 chance to be tails. All we have to do to apply this same logic to the card example is twist it a bit. Sorry if I spoiled the mystery for anyone who was waiting for the second video! tl;dr: The probability doesn't actually change, we're just applying the wrong theory for the situation. Also, Zachstar is doing God's work here showing people how statistics can really be extremely skewed

This just confirms my theory that people never tell me all the info, but still expect a full/perfect answer

i'm so proud of myself for getting the first question with the soda and juice right, the rest, i don't play cards, so idk

Now I want to test the probabilities in a real world application and what the results end up being.

I thoroughly enjoyed this. You know why? I got all answers wrong. I love it when I am wrong and learn something new. Zach, you are amazing and love your videos, no intro, no outro, no music, to the point and the topics I like.

If you had no tails in your hand, then you would say "one of them is heads", so you need to include the cases "one of them is heads". then in 50% of all choices the other one is the same.

For anyone having existential issues with the last puzzle, it always helps to remember that probabilities are not a 'concrete thing' that we are 'measuring.' They are just an approximation that can freely change as more information is added. Think of it this way: the True chance of drawing an ace is always either 0% or 100% -- the deck is already there and the universe already knows whats on top. But the probability we're calculating is just an estimate based on what info we have, and as info is changed then the estimate changes. When we could have ANY ace in our two-card hand, then there are tons of ways it could fail (A❤ 5♠️, A♦️J♦️, etc. etc.) But when we hear the suit, even if we dont "care," it still helps us greatly reduce the number of ways our hand can fail (any failure now MUST look like A♠️ and something). Any information helps our estimate, even if we don't "care" about the info. Whether the dealer is right or left handed might matter, the weather might matter. In the problem we just assume that these factors have absolutsly no effect whatsoever. But its important to remember that the universe already decided the top card of the deck, and ANY info on how it got there can help refine our probability estimation.

For the last problem: It's the classic Monty Hall problem; or at least, bears much similarity to it. The trick is that while it doesn't technically matter which one you reveal, it does matter HOW you choose to reveal it. We are looking at whether the following two scenarios have the same probability: Scenario 1: I look at both coins. If exactly one of them is tails, I reveal that that one is tails. If both are tails, I randomly choose one to reveal that it's tails. What's the probability that both are tails, given I reveal the left coin is tails, and what's the probability given I reveal the right coin is tails? Or, Scenario 2: I look at both coins. If at least one of them is tails, I say that at least one of them is tails. What's the probability that both are tails, given that I say at least one of them is tails. In both Scenarios, there's a 25% chance of double tails, 25% chance of left coin being tails, and 25% chance of right coin being tails. Zach incorrectly reasons that in Scenario 1, when I reveal the left coin, it's a 50-50 chance of whether that's due to having double tails or only left coin being tails. In actuality, two-thirds of the time when I reveal the left coin is tails, the right coin is heads - as the 25% chance of double tails is shared evenly between revealing the left and right coins, meaning though there's a 25% chance of double tails, there's only a 12.5% chance of double tails AND I reveal the left coin is tails, as there's another 12.5% chance of double tails AND I reveal the right coin is tails. Therefore, we're actually comparing the 12.5% chance of double tails AND I reveal the left coin is tails, VS only the left coin is tails. In Scenario 2, as Zach says, it's one-third because it's a 33-33-33 chance of having double tails, left tails, or right tails. Edit after reading the comments: The chance is one-third for both of the above scenarios of having double tails. However, we can also imagine Scenario 3: I look at one coin (randomly choose whether to look at the left coin or right coin). If that one coin is tails, I say that at least one coin is tails. Scenario 4: I look at one coin same as in 3. If that coin is tails, I reveal that the coin I looked at was tails. Scenario 3 and 4 would indeed have a probability of 50% that both are tails. Scenario 4 is easier to explain: If I reveal it, then there's indeed only 2 options, Tails-Heads or Tails-Tails. I did not even look at the other one so it's impossible for the other coin toss to affect the information I give you. Therefore, it's exactly the same as just flipping one coin and asking what's the probability it's tails. Scenario 3 is trickier, but it's still 50%. You might think there are "three scenarios" - Tails-Heads, Heads-Tails, Tail-Tail - but if you think of it this way, the probability that it is Tails-Tails is twice as likely. If I flip Tails-Heads or Heads-Tails, I only have a 50% chance of seeing the coin with Tails when looking at a coin at random. I assumed that Zach was referring to Scenarios 1 and 2, but I don't think he made it clear. So just to cover all scenarios, and to clarify what I mean by it mattering HOW you reveal it, I'm adding this edit

I wish I had a teacher like Zach. His charisma really made thinking about these questions/puzzles fun.

I’m glad you’re able and willing to educate us

2:00 that's how solved it initially too (I did x/(x+1) × (1/(x+1)) but then realized it's obviously the same cause they both have the same volume before and after. It's like that card trick where knowing the number of red cards in one pile tells you how many black cards in the other

0:44 I got it wrong, but I just noticed there is a simple explanation of why it has to be the same. The total amount of soda and juice is one cup total and each cup has the same amount of liquid. If there is "1- a" soda and "a" juice in one cup, there must be "1-a" juice and "a" soda in the other. How we mix the liquids is irrelevant, if both cups end with the same amount of liquid, the composition must be mirrored.

For the cards (and coins), I figured there's a quite easy way to calculate it: You divide the overall probability of the event (for example both cards being aces) by what you already know (at least one card is an ace). Both cards being an ace would be 4/52 * 3/51 (for a standard deck of 52 cards with 4 aces). The probability that at least one of the cards is any ace can be calculated with 1 - (48/52 * 47/51). The total calculation therefore would be 4/52 * 3/51 / (1 - (48/52 * 47/51)), which as mentioned results in 1/33 or roughly 3%. For the second example, where we know that one of the cards is the ace of spades, also the overall probability changes - we need the probability of one card being the aces of spades, and the other one being any ace. This probability is 1/52 * 3/51 * 2. (notice that the factor 2 is there because the position of the aces of spade, either first or second, doesn't matter). The given probability is, that one cards is the ace of spades, that's 1 - (51/52 * 50/51). Once again, if you calculate this you get 1/52 * 3/51 * 2 / (1 - (51/52 * 50/51)) which results in 1/17, or roughly 6% and almost double of the first probability. Please correct me if I'm wrong, I just came up with these calculations on the spot, but as the numbers do seem to be close to the (not precisely mentioned) actual answers from the video I'm fairly confident.

5:05 What if they're not from the same pack?

Verbally said “f*ck off” several times watching this, positively and endearingly

I feel like the last problem has strong under-currents of the monty-hall problem. By revealing which of the hands/doors has it (or doesn't have it in the monty-hall case), you narrow the probability space for the other hands/doors. Though that being said, this problem is still breaking my brain somewhat, so it hasn't solved it for me yet.

Regarding the two coins/two cards issue. The probability changes based on how much information the revealer has, how much they reveal, and what logic they use to reveal it. For example lets take 2 coin flips, the revealer knows the result of both flips, and will reveal the first flip that is tails. If the revealer reveals that the second flip is tails, the first flip MUST be heads, so the coins are HT If the revealer reveals that the first flip is tails, the second flip is a 50% chance, so TH or TT are equally likely. If the revealer chooses at random when there are 2 tails, we have these possibilities: TH reveal 1 = ⅓ HT reveal 2 = ⅓ TT reveal 1 = ⅓×½=⅙ TT reveal 2 = ⅓×½=⅙ So, regardless of whether they reveal 1 or 2, the odds are ⅓ that the other coin is T. If the revealer randomly chooses one coin to look at and declare and will declare whatever side is showing, the odds are 50% the other coin matches. That is why the Monty Hall problem is hard. The host knows everything and choosing to reveal a goat affects the probabilities post reveal. If the host randomly revealed a door without knowledge (sometimes revealing the prize), then naive probability works.

I was not convinced with question 1 so I did the math myself if you're interested : Lets assume both solutions of soda an juice have an initial volume of 1000ml. We'll respectively call these solutions A and B. Step 1 : You take a tablespoon that contains X ml (where X is a small number, say 10ml or 5ml) from solution A and pour it in solution B. After step 1, - solution A contains 1000-X ml of soda and 0 ml of juice - solution B contains X ml of soda and 1000 ml of juice Step 2 : You mix perfectly solution B, that now contains 1000+X ml. Now, every sample from solution B has to be in the same proportions of soda and juice as the whole solution B. After step 2, solution B contains : - a proportion p_soda of soda, where p_soda = X/(1000+X), with 0

Blud left us with a cliffhanger

4:06 If anyone is have problem here just as me, the median is the value that is in the middle when they are sorted from lower to grater for example “2,3,6,9,10” in here 6 is the median and if the number of values is even like “1,3,5,8” you need to get the average of the 2 middle values in this case (3+5)/2=4

Zack can make me laugh so hard I stop breathing, AND teach me useful problem solving skills. Respect.

The Monty hall problem only works as we assume it does if the host has inner knowledge of the game and we know he undergoes certain rules (He always reveals a goat etc). If those rules aren’t in place (ie the host doesn’t know what’s behind the doors and just picks randomly) then the Monty hall problem as we understand it isn’t the same as in some scenarios the host might just open the one with the car. In this example I feel like it must be to do with the set up. How do we conduct this game? Do we flip both coins over and over until at least one of them is tails? And what rule does the flipper follow in revealing the tails? Let’s say that he flips both and if there’s at least one tails he rolls a die. If there is only one tail he tells the viewer about it regardless of the die roll. If there are two tails he’ll tell the viewer about the left tail when then die is odd and the right tail when the die is even (the viewer can’t see the die he just knows the assignment is random). Under these strict rules we can eliminate the one in four events when they’re both heads as nothing will come of those. Since our guy reveals tails every time in this scenario nothing has changed. The probability of the other being tails is still 1 in 3. I think the confusion comes from vagueness in the rules of choosing. Oh and in this new version of the game him saying it in the dark is just the same as him saying it to us directly. Since we know the rules and everything’s consistent it’s 1/3rd regardless. Revealing the tail literally does nothing whether we see it or not. To get the scenario like the second one with the 1/2 chance we’d have to have a situation where our flipper only announces a tails when he has it in his left hand. The reason this is now 1/2 rather than 1/3rd is now in the heads left tails right case which before would have lead to an announcement nothing happens. So we only get the tails left heads right or the tails left tails right cases. So the flipper’s strategy is what matters and that determines probability.

Thank you for uploading on your second channel. This is much easier to understand than the most recent video on your first channel!

If the coins are randomly flipped repeatedly and only presented to us when the left coin is tails, then the probability of guessing correctly that both are tails is 50%. If instead the coin flips are presented when and only when at least one coin is tails, then the probability of guessing correctly that both are tails is 33%.

Probability is where information about the world is taken in and a number describing a state of that world is given out. If we were to simulate the world a bunch of times then we can count of the number of times things happened and use them to get probabilities based on the worlds we could be in, and the worlds where an event happened. We restrict the space of worlds to those where someone had two coins in their hands and secretly "revealed" which hand held a tails. These worlds can be split in two: those where that person knows a tail is in their left hand, and those where that person knows a tail is in their right hand, but we have no information to determine which one of those worlds we are in. We just know that we are in a world where there is at least one tails, and it is 1/3. The person holding the coins has different information available, the subset of worlds they could be in is different, and so they would get 1/2. I found that probability starts making a lot more sense when you think about what state space the situation is occurring in, and what subspace the event of interest occurs in. These spaces are determined entirely by the information, and as different observers can have different information, their state spaces can also be different leading to different probabilities.

The median was not properly represented in the second example. Median is a powerful tool that more people should be aware of. The value difference between median and average can tell a lot about a data set; loan figures are a perfect example for this indeed.

The cards are an inverse Monty Hall. Put yourself in the position of the dealer at the six minute mark. There are five combinations of cards, each with a twenty percent chance of being dealt. You choose a suit for the ace that you will tell the player. What is the chance that each combination will be in the possible remaining combinations? The answer is 50% for each, except the pair of aces, for which it is 100%. There are only three combinations left, but that doesn’t mean that each had an equal 1/3 chance of being left. The chance of the pair being dealt was 20% from the outset, and the dealer’s statement about the suit of one of the cards does not change this.

Card problem was wrong. In both cases of the 4 cards problem shown, it's 1/3rd. You have artificially thrown away one of the two possibilities for the two ace pairing by saying 'the order doesn't matter', but it actually does IF AND ONLY IF you have an ace already but haven't disclosed which ace it is - it can either be Ace of Hearts and then the Ace of Spades, or, if the ace you are holding is the Ace of Spades, it can be the Ace of Spades then the Ace of Hearts. In the Ace of Spades example, the possibility of the Ace of Hearts then Ace of Spades option doesn't exist so you don't get the probability wrong for that one. The coin problem is based on the same logical error. If you say one coin is tails, yes, the possibilities are limited since you removed HH, but you should have removed either the TH or the HT as well, since one of the coins cannot be heads anymore. Thus you are left with TH and TT, for 50%. Specifying which coin makes no difference, except that it decides whether TH or HT was correct to remove.

this tripped me up so hard, i had to stop the video and just kinda think for a minute

With the first question, the mixing wouldn't matter right? The amounts would be equivalent regardless of the proportion of juice to soda in the tablespoon

In the last question, it depends on how the situation came about, *not on what information you reveal* about the situation. Did you choose heads for one and then flip the other one? (50%) Did you flip them until you got at least 1 heads? (33%) Did you flip them and then (out of the 1 or 2 sides that you saw), chose between calling heads and tails (or the left or right hand's result)? (50%) ...And if there's both heads and tails, will you reliably prefer saying heads? (33% if you said heads and 100% if you said tails.) Did you choose both? (Then for you the probability is either 0% or 100%, but for us it depends on how well we know how you'd choose.) Same with the cards: Did you draw 2 cards (and discarded both) until you got at least one ace? Then it's 3%. Did you draw 2 cards (and discarded both) until one is the ace of spades? Then it's 6%. Did you draw 2 cards, and then discard *one* and draw a replacement until you got at least one ace? Then the chance that you have a second ace is basically zero. (Since it can only happen in the initial draw.) Did you choose the ace of spades (or any of the others) and _then_ draw a second card; or did you draw a first card and _then_ choose one of the aces as the second card? The probability of having an ace goes up by 1/3rd. Did you draw one card each from 2 _separate_ decks of 52 cards (until one is an ace/ the ace of spades)? Did you draw two cards from a _combined_ pile of two 52-card decks (until one is an ace/ the ace of spades)? Did you choose both cards?

Ok, so i feel like the problem here is in the fact that we use two different probability distributions in the two examples. I.e. if you choose the two coins randomly, then look at them, confirm that one of them is a tails, and then reveal it, it's still 1/3, since you revealed the one that you saw that was a tails, not picked a tails in the first place. I feel like it is pretty much the same problem that in Monty Haul paradox but in reverse

Simpler explanation for problem 1 without calculating volumes: After the switch we end up with 2 solutions of exactly the same volume as before since we moved the same volume back and forth (1tbs in this case). Since the total volume of liquid in the soda bottle didn't change, whatever the amount of soda missing (aka the amount of soda now in the second bottle) must have been replaced by exactly the same amount of juice and vice versa. Thus the amounts are equal.

Yessir the boy/girl paradox drama is back raaaaaa🗣️🗣️

The answer to your last question is still 1/3 because when the screen blacks out, we receive no new information about which hand has the tails hence the probability doesn’t change for us. This example reminds me of quantum mechanics where two or more particles can be in an entangled state and have a joint probability distribution (1/3 probability case) but when the state function of one collapses, the probability distribution of the other changes (revealing which hand has tails hence probability changes to 1/2). This is very similar to the famous double-slit experiment where the simple act of observation changes the result.

You entered the wrong answer in the brilliant ad and it turned green.

I once gave the first riddle to a friend of mine who was a chemist. And the first thought she made was that the volume decreases as soon as you pour two different liquids together. Therefore the second tablespoon contains more liquids (when you messure the mass) than the first one. And therefore there is more juice in the soda. Since then I always use mass instead of volume when asking this riddle.

With the two aces problem, I assumed a standard 52-card deck. If we know the first card is an ace, there are 3 aces remaining out of 51 other cards that can be drawn, which equals a 1 in 17 chance.

How did you get 3%? When I calculated it, I got 5.8% the first time and I have no idea what numbers I'm supposed to plug in the calculator to get 3%. Here's my thought process: If I have an ace, that's 1 card out of 52. For the other card to be an ace, it has to be any of the other 3 from the total of 51 cards left. So the probability is 3/51 which is about 5.88%.

People tend to forget that probability and randomness is strictly linked to how much you know about the process of "randomization". If you knew how a coin is thrown exactly, you could predict its outcome, if you knew parts of it, you could come up with more accurate probabilities of the outcome, ones that make you right more often.. Same thing in the last exemple of the video, depending on what you know, the random experience changes, and the probabilities associated with each issue evolve too, strictly based on your own knowledge of the situation.

Thinking about it, I think the card/coin thing is actually way more intuitive if you think about what the inverse, the scenarios that you must reject* for the premise to be true. By saying "one of these is a tails", you're essentailly saying "I flipped both coins until either one (or both) of them was tails". By saying "_This_ one is tails", you're essentially saying "I flipped two coins until the left one was tails". In the first scenario, the only time you'd re-flip both coins is if they were both heads. In the second, you'd re-flip both coins if the left one was heads, no matter what the right coin was. So you rejected more outcomes. Similarly, when you're drawing two cards, when you say "One of these is an ace", you're saying you rejected all the outcomes where you didn't draw an ace. There's 1326 different two-card hands in a standard deck of 52 cards, but only 112 of them contain one or more aces, so you're rejecting 1214 aceless hands. Of the remaining 112, only 12 of them have 2 aces, or just under 11%. When you say "One of these is the ace of spades", you're saying you rejected all the outcomes where you didn't draw _specifically_ the ace of spades. Out of those 1326, there's only 30 that have the ace of spades, so you're rejecting 1296 of them, and of those remaining 30, 6 have two aces, or exactly 20%. Since there are more non-aces than aces in the deck, rejecting more outcomes means rejecting more one-ace outcomes than two-ace outcomes. *When I say "reject", I don't necessarily literally mean reject, so much as "it did not happen, because if it had, the puzzle would not be structured this way"

For the first one, you can take the "table spoon" to the extreme max and use a full cup instead. So basically you would pour 1 cup of cola into the juice, then pour 1 cup of the solution back into the cup that originally contained the cola. Now it's very obvious that it's equal. The reason this works is because there's nothing special about the "table spoon", it's just a arbitrary amount, any amount would work

I watched this earlier today, left the house and did a bunch of errands, gleefully awaiting my return to see the follow up video and ........ I hope you upload it soon! Great content Zach, I follow you here and on the comedy which is my all time favourite comedy.

Was anyone else constantly waiting for a joke or a punchline? Zach just sounds so snarky all the time even when he’s not even trying to be, it’s really hard to take him seriously.

For the third quiz, the problem is that the statement you make is affected by the coin flips. If you pick a random tails whenever there is one flipped up to say this one is tails. When you do pick a coin to say it is tails, the two possibilities T H and T T are not equally likely since in T T case you had a 50% chance to say the other coin was tails making that half as likely and the probability the second coin is tails returns to 1 in 3. Hopefully Zach explains it better than I did.

Please stop saying "odds" when you mean "probability." Although they are inextricably linked and often yield the same information, they do have important differences. Prob. = likelihood an event occurs; Odds = P(event happens) / P(event doesn't happen). This error is like mixing up standard deviation with variance. Yes, they typically reveal the same info, but an error is an error.

#3 is just updating our priors, isn't it? It's like Monty Hall kinda

Revealing the suit of the ace or the hand the coin is in doesn’t change the probabilities. You know that there is at least one ace. Because we don’t know the suit, the guaranteed ace has a 1/4 chance of being spades, a 1/4 chance of being hearts, a 1/4 chance of being clubs, and a 1/4 chance of being diamonds. If the ace is spades, the chance the other card is an ace is 3/51 If it’s any of the other suits, the chance is still 3/51 So the probability calculation is 1/4*3/51+1/4*3/51+1/4*3/51+1/4*3/51 = 3/51 You have 4 individual sample spaces depending on what the suit of the guaranteed ace is. Each one has equal probability of the 2nd card being another ace, so the suit of the first doesn’t affect the probability. The coin example is similar, but a little more tricky to wrap your head around. You know that at least one of the hands has a guaranteed tails. There’s a 1/2 chance the guaranteed tails is in the right hand and a 1/2 chance it’s in the left. If it is in the right hand, there is a 1/2 chance the left hand coin is tails. If it’s in the left hand, it’s a 1/2 chance the right hand coin is tails. So the probability calculation is 1/2*1/2+1/2*1/2 = 1/2

There’s a logic error on the Ace puzzle. In a deck of 4 cards that contains 2 Aces, if you say one card is an Ace, then the odds the other card is an ace is NOT 1/5 as you claim, it is actually 1/3. Why, you ask? Start with 4 cards, and remove one which you know is an Ace leaving you with 3 cards. The odds that other card is an ace is 1/3. It is unnecessary to specify the suit of the card, since it actually has no bearing on the result.

Me: My answer is either that there's more soda in the juice or more juice in soda. Also Me: Still ends up being wrong

For the soda coke exchange: The easier explanation is, that the amount of liquid is the same after ferrying one spoon back and forth, so all the coke that ended up on the other side has been replaced by apple juice. It doesn't even matter if the coke was properly mixed into the apple juice. Also the amount would have been different if the spoons would have been exchanged simultaneously (but still as much coke in the apple juice as vice versa). The coins example: It sounds like after asking "Which one is tails" and getting an answer the probability goes up, but that doesn't take into account that the person holding the coins then makes a choice based on the coins held, so the probability actually doesn't change. It's kind of the reverse goat problem A different question would be "Is the right hand tails?", depending on the answer the probability for the other hand changes to either 1 for "no" or 1/2 for "yes" from 2/3 before. So it depends how the coins are prepared, either flipping coins until at least one shows tails, then telling where a coin with tails is vs. determining a hand, flipping coins until that hand holds tails, then announcing that.

Problem 3 reminds me of the principle of restricted choice in bridge. When trying to locate the KQ of a suit in the opponents' hands and you see one, (they should play K or Q at random if they hold both), the options are that that opponent started with KQ, Qx or Kx, all equally likely cases. The chance they have the other as well is closer to 1/3.

1) These are independent events. The odds of "the other hand having tails" will always be 50%. It doesn't matter if the other hand was tails or not. 2) The question doesn't care about the ordering but you're including all permutations instead of all combinations. The possible COMBINATIONS are {H,H}, {T,T}, {H,T}. If one of them is T, then you get what you'd expect: {T,T} and {H,T} are the possible results. This is basically about leading questions. We wouldn't have assumed ordering mattered unless you put permutations on the screen instead of combinations.

In the game of bridge, each player is dealt 13 cards. If we deal out hands until the first player gets a hand with at least one Ace, the probability they have a second Ace is slightly less than 1/2. If we deal out hands until the first player gets a hand with the Ace of Spades, the probability they have a second Ace is slightly greater than 1/2.

Card/coin one: when you reveal a particular card, you’re no longer sampling uniformly.

1) Mixed solutions are gonna represent the same fractional proportion 2) Median salary will only depend on the middle person’s salary. You could have any of these scenarios and that still hold; for example: Median of 1, 4, 2 is 2. Median of 1, 2, 10000 is still 2 Median of 1, 3, 10000 is 3 Problem 3: Appears to be an analogue of the sister/brother problem wherein we are told that at least one is a girl or something like that. If you used Bayes formula the specification actually increases the odds. Will check my work now. Edit: For statistics majors these problems are bread and butter in any undergraduate intro probability course. For the very last problem my initial thought without writing out the work is that this is an analogue of the monty hall class of problems but Im not sure why. My intuition is still that the specification is useful as compared to the base case

This is pretty fascinating. It reminds me of the old question: "If a tree falls in the forest and there's no one to hear it, does it make a sound?" I wonder if the answer changes if a deaf person watches it happen...?

I've figured out the coin trick. Start with some assumptions about the implied rules: first, if there is a single tails you must reveal it and second, if there are 2 tails you must reveal one of them at random with equal probability. We start with what appears to be 4 possible outcomes: HH, HT, TH, TT. When you show tails in the left hand, you eliminate the possibility of HH and HT, which leaves only TH and TT. Presumably with 2 options remaining, it's a 1/2 chance right? No, cos here's the trick: there are actually 5 possibilities, not 4: HH, TH, HT, TT-L (in which you randomly revealed left coin) and TT-R (in which you randomly revealed right coin). The first 3 outcomes are 1/4 each, but the last 2 are 1/8. When you show tails in the left hand, in addition to eliminating HH and HT, you also eliminated TT-R. This leaves TH and TT-L, the latter of which is half as likely, therefore it's still 1/3. A similar thing applies to the cards.

So I went back and watched your video on the boy/girl problem (from 5 years ago). Currently scrawling on a paper plate at 4am trying to make sense of it. I found that if you add another slight tweak to the question, the probability continues to increase. Say the parent said "I have two kids. One of which is a girl, who is named either Julie OR Natalie." Seeing that it'd be perfectly logical to have two girls, one named Julie and the other Natalie, you add more possible outcomes, and thus more outcomes in which both children are girls. I drew up a punnett square and Watched the probability slowly climb, as if it were approaching 100%. I'd love to hear/see you talk about this or at least tell me where I slipped up to get the ever increasing percentage. Sorry this was so long!!

The first question is easy to solve if one considers what happens if the two cups start with 1tbsp of fluid each. Then generalize that same reasoning to any starting amount.

I have a different approach for the last problem. Obviously the way you showed it makes sense, but there is another way that can help understand the final question. Basically we have two independent experiments (throwing two coins and seeing if they are heads or tails). Once you tell me the actual result of one of them, what we are basically doing, is reducing the amount of experiments to one (one coin toss with either heads or tails). And everyone knows that a coin toss is 50:50. However, if you just tell me that at least one landed on tails, you didn't actually give me the result of one specific experiment. You just eliminated a certain set of possible results (in this case two heads). With this idea in mind, we can understand the final question. Assuming the question is again: "What is the probability that both of them are tails?", then yes, it is going to be 50:50, because you are basically just asking me for the probability of one coin landing on tails. Even if I don't know exactly which coin you are talking about.

The first one can be made more obvious by noticing two facts: 1) The amount of each individual liquid across the whole experiment is constant. 2) The total volume of liquid in each container is constant between the start and end of the experiment. By (1) any cola that is missing from the cola glass must be in the juice glass, and any juice that is missing from the juice glass must be in the cola glass, whatever those amounts may be By (2) if there is X amount of cola missing from the cola glass, it must instead be replaced with X amount of juice. Since that juice can only have come from the juice glass, there must by X amount of juice missing from the juice glass. But again, by (2), if there is X amount of juice missing from the Juice glass, it must be replaced with equal amount of cola to make up the same volume. As such, no matter how you mix up the juice and color by moving any amount of it back and forth, as long as both glasses have equal volume to their starting volume, the same amount of juice will be in the cola as there is cola in the juice. This remains true even if you do not completely homogenously mix the liquids when transferring samples - that aspect is a red herring. Second one: Yeah. Medians are weird, and they don't behave nicely when you segment data. Arithmetic means play a *little* nicer, but they're still prone to counter-intuitive results, especially in multivariant data. Third puzzle highlights an interesting counterintuitive result involving Bayesian Inference, in which 'probabilities' are interpreted not as the chance of some random event happening, but instead interpreted as your *confidence* in a certain outcome. In a very real sense, once you have the cards, there is absolutely no probability that they are going to spontaneously change into other cards, just because you didn't give us full information about them; the only thing being measured is our confidence in some result being true based on the information we are given. in *that* context, it makes sense that the Confidence in the other being an Ace would be not only different, but higher, when we are given more complete information. Knowing only that there is 'an Ace' present is less information than knowing that 'there is an Ace of Spades' present. Simply having more information allows us to have more confidence in certain guesses, and less confidence in other guesses. It is an interesting question as to whether we are correct to try to use this method of assigning probability-like traits to confidence, but it does seem to give useful answers which are generally borne out by experiment.

Aren't both 1/2 and 1/3 chance possible depending on how he decides to call out possibilities? (or any other probability >= 1/3, actually) Options are TT, TH, HT, HH. Obviously, when it's TT he will say "at least one of the coins is tails". By symmetry if it's HH, he will say "at least one of the coins is heads". But what about TH/HT? Case 1: If he says "one of the coins is tails" if there is a tail, then both options result in him saying that at least one is tails. Then there are three scenarios which result in him saying "one of the coins is tails", giving 1/3 chance that the other is a tail. Case 2: If instead for TH/HT he says "one of the coins is tails" 1/2 of the time, and the other 1/2 he says "one of the coins is heads", then we have a different scenario. 1/4 chance that it's TT and (1/4+1/4)*1/2=1/4 that it's TH/HT AND that he says "one of the coins is tails", giving 1/2 chance that the other coin is a tail. Case 3: (symmetrical to Case 1) If he always says "one of the coins is heads" if there is at least one head, then the only possibility of him saying "one of the coins is tails" is if it's exactly TT. Giving 1/1 chance that the other coin is a tail. I think case 2 is most consistent with the setup, looking only at a single coin and saying if it's heads or tails, giving 1/2 chance for the other coin to be tails (when he says the first one is tails). Saying "at least one coin is tails" knowing both coins leaves all 3 cases possible.

One interesting thing I noted is that the probability of the other card being an Ace given that one of them is the Ace of Spades surprisingly is the same as just the probability of the SECOND card being an Ace given that the FIRST is the Ace of Spades (3/51 is approximately 6 percent as well). I don't really get why that would be the case.

Your ace of spades question is dependant on what assumptions we make about what you'd have said (and I'd say the most natural assumption means it isn't correct) It's only if we assume you'll ask about the ace of spaces in advance, or if some other independent process meant we knew you'd ask about the ace of spaces (eg. say I ask you "Is one of the cards the ace of spades?" and you answered "yes"). That way we can assume your card is sampled from the population of 2-card hands that contain the ace of spades. But if the fact that you asked about the ace of spades was DEPENDENT on the cards you have (which it must be - you wouldn't tell us you had the ace of spades if neither card was it), the answer changes: it's no longer a sample of 2-card hands containing the ace of spades, but the probability is not the dependent one of P(chose to talk about ace of spades | hand).' Ie. for all I know, if you'd held the ace of hearts and the ace of spades, you'd have been equally likely to phrase the question as "One of them is the ace of hearts, what's the chance the other is an ace?". So of the double-ace hands, I should only expect the question to be about the ace of spades 50% of the time. That's why the "this one" part (arguably) makes a difference with the 2-coin case. It makes it clear you're basing the question you asked on a coin you picked (even if you don't show it), so have twice the odds to pick heads in a 2-heads situation than in a heads/tails one. In fact, I'd say even without the "this one" part, it runs into the same issue of assuming you'd talk about "one is heads" rather than "one is tails" in advance: it'd be more natural to assume what you ask about depends on what's in your hand (since again, you wouldn't say "One is heads" if they were both tails, so you must be doing **some** kind of examination of the coins to base your question on.)

I don't understand how the probabilities worked for the cards. assuming he's referring to a 52-card deck, shouldn't both of them be 1/17 (roughly 6%)? in the first example, he asks for the probability that the other card is an ace, given that the first is also an ace. since there are 4 aces in a standard deck, knowing the first card is an ace pulls one from your deck, leaving you with 3 aces in 51 cards, or 1/17. in the second example, he adds the info that one of the cards is an ace of spades. this shouldn't change anything however, since knowing what the suites of the cards are doesn't change their outcomes. ((As, Ad), (As, Ah), (As, Ac)) has the same size as ((Ad, As), (Ah, As), (Ac, As)), so the outcome should be the same regardless of if the ace of spades is the known card or not. edit: if he's asking what's the total probability of pulling both cards as aces, it's 4/52 x 3/51 or 1/221, roughly 0.45%. still lost how he got 3% anywhere?

I'm glad you were able to take time out of your skit acting career to learn a little engineering stuff!

I feel smart that I instantly realised that the third one was wrong and why

The reason that the last puzzle is different is because the event sizes of 'at least 1 coin is tails' and 'this coin (unknown) is tails' (will be interpreted as coin 'i' is tails with 'i' being 1 or 2 uniformly) are different. at least 1 coint is tail = {TT,HT,TH} coin i is tails = {TT, TH} or {TT, HT} with 50% each Pr[case 1]=Pr[TT|at least 1 point is tail] = Pr[TT and {TT,HT,TH}] / Pr[{TT,HT,TH}] = 1/3 Pr[case 2]=Pr[TT|coin i is tails] = Pr[TT and coin i is tails] / Pr[coin i is tails] = Pr[TT] / (1/2) = 1/2 * case 2 is actually a sort of 'expected value' of probabilities between the cases of different 'i's, a different way to calculate this is defining a different probability with i included with the condition Pr[i=1]=0.5 (i saw a lot of people giving more weight to TT, this is pretty similar). in total, 'at least 1 coin is tails' =/= 'an unknown coin is tails' = 'one of these coins is tails' (notice that there is no use of the word EXACTLY one) why? because 'at least 1 coin is tails' tells us something about the correlation between the value of the two coins, while saying that 1 of these coins is tails does not.

1st question: I did it out with math first. Then I realized the solution with a simpler explanation than the one you gave: We don't need to assume a perfect nixture. If we pretend it is not a perfect mixture, the answer is easier to see. Let's say it is 2 cups of 100 legos each. You take 10 legos our of the red lego cup and put them in the blue. Then you take 10 legos out of the blue lego cup and put them back into the red. Or better yet, you dump everything out onto the table and put 100 legos back into each cup of any color. The blue cup has x blues and 100-x reds. That leaves x reds left in the red cup and 100-x blues. It doesn't matter how well or poorly it is mixed or how much is moved. If the two start out the same quantity and end up the same quantity, it will always be the same. For the math, I found it easy to do with 110 red legos and 110 blue legos. The math then is simple to do in your head. Move 11 reds into the blue. Mix it perfectly. Now 11/121 are red. Move 11 back. 1 is red the other 10 are blue. Add the 1 red to the remaining 99 red and you get 100 red in the red cup and 100 blue in the blue cup with 10 of the opposite color in each. When I noticed that, I immediately jumped to the solution I explained above.

I don't think the answer to the third question was right. Let's take the simple deck you showed at 5:50 as an example. You draw two cards and say 'I'm holding at least one ace. What's the probability I'm holding two aces'? The answer is 20% as you explained later on. But then I ask 'Can you tell me the suit of an ace you're holding?' and you say 'I'm holding the ace of spades'. What's the probability you're holding two aces now? You argue it should be about 33%, but I believe it's still 20%. The possibilities of 'ace of spades + two of diamonds' and 'ace of spades + 7 of clubs' are still in the sample space. But the 'ace of spades + ace of hearts' is more complicated. If you were holding that, you could have also said 'I'm holding the ace of hearts'. So that possibility only counts as half. This makes the probability you're holding two aces as 0.5 / (1 + 1 + 0.5) = 20%. The difference comes from the fact that if you're holding two aces, it's unclear which of the two you should call out.

I did the math out for the first one and was surprised (though I shouldn't have been) that the proportions were just reversed. Right before doing the math my vague intuition said "the volume of substance A being added to container B is more unlike substance B than the volume of the mixture being added to container A is unlike substance A" ... and "so I expect container B should end up with more substance A than container A ends up with substance B" - but the latter can't be the case because the volumes end up the same with nothing lost or gained. I did notice that you specified the volume removed was "a tablespoon" but the total volume of each was not specified, but I just abstracted both of those anyway in my solve. If the volume removed is 1/n of the original volume, then the proportion of foreign substance in each container ends up as 1/(n+1). I also noticed you just said "solution of" without being more explicit about the concentration. (i.e. is it a 10% solution of coke in water (also a crime)?) I solved assuming you meant each was totally pure and also figured that it would not be an interesting dimension of variation on the problem.

I was half expecting this to be a sequel to the invention of addition.

Regarding the cards and the counselling, I feel you may be planning to touch on this more in your next video. But the answer from my perspective is that there is not enough information to answer the question. In the simplified example with the cards, then were the question 'Of the pairs which contain an ace, what proportion contain two Aces?' or 'Of the pairs which contain the Ace of spades, what proportion contain 2 Aces?', it would be unambiguous. However, when you are given the information by being told, then you can't calculate the odds without knowing what information they would choose to give you with each holding. For both the questions the answer could be anything from 0% to 100% depending on this. Supposing that they would always tell you of the Ace of hearts in preference to the Ace of spades, then the fact they told you of the Ace of spades would mean the other Ace was not present. Suppose they would always prefer to tell you of a non Ace were it present, the fact they told you of either Ace would mean both must be Aces. Maybe their preference isn't only to tell you of one card at all, so that when they do that reduces the subset.

The first one is simple. My approach would be "the amount of soda missing in the soda is the amount of juice missing in the juice since one replaced the other"

I absolutely LOVE when you do this counter intuitive stuff! Can't wait for the next video! The boy-girl vids are one of my all-time-fav YT videos.

At 5:10 you say that the probability of one card being an ace given that the other one is, is about 3% and it doesn't make sense to me. Given that we remove one ace from the deck, we've got 3 aces left in a deck of 51 cards, so 3/51 is 5.8%, so almosy 6 and definitely not 3%. How did he come up with 3?

The third one depends on the means by which the information "one of these is tails" was acquired. Since english is vague, "one of these is tails" could mean either that both were checked, and it wasn't the case that both of them were heads, or that one of them was checked and it was a tail (the one being checked could be random, or one predetermined, as long as the one being checked is independent from the outcome of the flip itself). In one case you can imagine a game where a flipper sits across from a player. The flipper flips two coins behind a screen, looks at them, and if they are both heads she flips them both again until thats not the case. She then says "at least one was a tails. I will give you $18 if they are both Tails, but you must give me $12 if they are not. You can play this game for me as many times as you would like". If the 1/3 case is right, then the expected value is negative (-$2 per game) and the player will lose money in the long run. If the 1/2 case is right, then they will gradually make money (+$3 per game). You can do this yourself or similate it to verify, but what you can imagine is a big table of results of all the flips, including the HHs cases, and of the ones where the player is given the chance to take the bet, 2/3 of those would be losing. The 1/3 case is correct here. Now you can imagine a case where the flipper flips the two coins, but only decides to look at one of them (eg. By flipping a third coin labelled "left/right"). She then offers you the bet if the one she checked was a tails, otherwise she flips them both again. In this case, there are 8 possibilities LHH, LHT, LTH, LTT, RHH, RHT, RTH, RTT. The bet is only offered in the LTH, LTT, RHT, RTT cases, of which half are TT and the expected value will be a win: the 1/2 case is correct. It should be clear that this is also the case if the flipper always checks the left coin, or chooses the coin by any procedure as long as long as the outcome doesnt influence the checked coin. For the example at the end, it's 1/2. My assumptions are that you shuffled the coins and randomised their parity, revealed only one of them, and you repeated the take of the video with that procedure until you happened to get tails. That makes it like the second flipper game.

I missed this type of content from your channel. I LIVE IT . Before i could understand you at all . But now that i'm older i can . It's fascinating

The change in probability in the coins game doesn't come from the reveal, but from the selection: I flipped the coins until at least one was Heads and in this case it is the "shows side" one: 33% I flipped the coins until the side i chose beforehand turned up Heads and its the "shows side" one: 50% In the first case we flipped and had to reflip at 1 of the 4 cases, leaving us with 3 end results (1 of them is HH) In the second case we flipped and had to reflip at 2 of the 4 cases, leaving us with 2 end results (1 of them is HH) Same with the cards, if I chose two cards until atleast one of them is "an Ace" it doesn't matter if I tell you the suit. But if I chose two cards until one of them is the "Ace of Spades", then thats a different probability. Selection process matters, telling you the suit doesnt. Its all about the difference in cases you need to reject and reflip. For even more magical effect, you can repeat both experiments, but not tell the other and just write it down or even just think it. Based on the selection process of your flips the probabilities will reflect which process you used even without you telling the suit/side.

The first two were fairly obvious with a little thought, although the 3rd has some bad math in it. - specifically in how you look at the odds, since you're not considering how interchangeable certain probabilities are IE, when you say "I have an ace in my hand" - we know it's one of two cards, and that is very important information, and it changes our dataset so that we're not looking at the original 6 possibilities. First "I'm holding an ace in a 4 card set, what are the chances I'm holding a second ace?" We can simplify the math here because we have redundant information. We have either Ace of Hearts, or Ace of Spades, it doesn't matter which ace we have, so when we have an ace of spades and a 2 of diamonds, or a ace of hearts and the two of diamonds, it means the exact same thing to us statistically, so we can combine them as one possibility - in Schrodinger's cat fashion, we can call this A of H/S - and likewise, the other ace we don't know about is A of S/H By simplifying the probabilities in this way, we can look at the possible probabilities/possibilities like this: A of H/S + 2 of Diamonds A of H/S + 7 of clubs, or A of H/S + A of S/H (This of course could be expanded to: A of S + 2 of Diamonds A of S + 7 of clubs, or A of S + A of H A of H + 2 of Diamonds A of H + 7 of clubs, or A of H + A of S) ^Note how this compares to the set of '5' in the video. A of H + A of S is listed twice - A certain 'break in logic' in the video, there are TWO aces we could draw from, the ace of hearts, or spades, yet only one combination of the two aces is shown in the video. (To further clarify, the possibilities was essentially changed from "What are the ways we can combine 4 different cards" to "What are the possible ways we can combine 3 cards with 2 different Aces" - it's just misleading because one of the aces, is also one of the 3 cards being combined. To explain this in an intuitive way, assume you covered up the 'suit' of the aces with stickers or something. After you pick up the first ace, you know you have an ace, but not which one. For your second draw there are 3 cards left, what are the odds you'll draw another ace? 1 in 3, correct! Aren't you smart. See, math is always intuitive when you look at it the correct way. In other words, the chance of having two spades DOES NOT change by knowing which ace it is. This is misleading math. The coin example on the other hand doesn't have such a blatant error - when he talks about 'does it matter which one is tails', it's a topic on how independent the flips are. If you flip just one coin, it's 50:50, if you flip another coin, that coin is also 50:50, however if you flip two coins together, you have a 1/2 chance to get a head and a tails, and a 25% to either get two heads or two tails: HT TH HH TT So when you say "I have a tails" you're merely eliminating the two heads possibility. HT TH TT However, the moment you identify which hand has the tails, you make the other flip independent from other, IE "This coin is a tails; so what is the probability that this other coin was a tails?" 50:50 naturally. (Think of it like a % slider. We know that we have AT LEAST 1 tails, so if my left coin is a tails, that one is 100% tails, and the other is 50:50. However, if the right coin is a tails, then its 50:50 and 100% in the other direction. If we move the slider to the middle, it's 2/3s chance either side is tails: BUT there's also clarity that there is a 0% chance for both to be heads. In other words, among the two coins, there is 4/3rds chance of a tails, and 2/3rds chance of a heads.)

I believe that the issue with the probability in the last problem is the same as that one TED-Ed video about the croaking frogs (Can you solve the frog riddle? - Derek Abbott), as pointed out by MindYourDecisions in the video TED-Ed's Frog Riddle Is Wrong. When you tell me "(at least) one of the coins is tails", the sample space is not 3, but 4. The possible scenarios are: - you have H T and told me about the coin in your right hand; - you have T H and told me about the coin in your left hand; - you have T T and told me about the coin in your right hand; - you have T T and told me about the coin in your left hand. Telling me "the left coin is tails" collapses the sample space down to 2 scenarios: - you have T H and told me about the coin in your left hand; - you have T T and told me about the coin in your left hand. The probability of the other coin being tails is always 1/2, which is also the answer you get if you are only concerned with the probability of one coin (the other one) being tails. No more information from the coin that was declared tails can alter the probability of the other one being tails. This paradox is, as far as I'm concerned, falsidical.

The probability changes when a counterfactual possibility is eliminated. If you have a choice about what to reveal, how that choice is made matters. In the Aces case, if you were always going to reveal the suit of your Ace, or choose randomly in case you had both, then we eliminate ½ of the cases where you have both Aces (because you could have chosen to reveal the other suit instead). The probability you have both Aces becomes ½ out of 2½ which is still 1 in 5. This is different than the case where you reveal the suit of one of your Aces, but spades has priority over hearts. I.e. you'll always say Ace of Spades when you have both. Now I can't eliminate any probability from the pair of Aces scenario, and the probability you have both Aces is 1/3. If instead the Ace of Hearts had priority, then when you say "Ace of Spades" I know for certain you don’t have both Aces. Another situation: I ask you "Do you have the Ace of Spades"? Now your choice is eliminated, and you might say yes and you might say no. If you say yes, the probability goes up to 1/3, and if you say no, the probability goes down to 0. In the coin paradox, it again matters how the choice is made. If you were always planning to reveal a tails, picking randomly in case you had both, the probability ends up not changing. If you were always going to reveal your right hand, and it happened to be tails, the probability does change, because you might have said it's heads. The problem is underspecified. But, since we have no information on how you choose which suit to reveal, or which hand to reveal, we have to assume you chose randomly in the cases where you had a choice. If we change the original Monty Hall problem by saying after you pick a door, the host lets you choose another door to reveal. If you open the other door and it reveals a goat, now it really doesn't matter whether you switch or not. Conditional probability can be weird and unintuitive. Another way of thinking about this is the principle that information must be discounted when the choice of what information us revealed is under the control of an adversary.

8:00 THAT is what finally helped me to understand the Monty Hall problem. Thank you.

A cliffhanger in a math video is too cruel XD