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

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I forgot you also have an educational channel

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

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.

This is your second channel now xD

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.

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.

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.

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

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

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.

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

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

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

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.

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.

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

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