Get the "Trivial" shirt here: https://stemerch.com/collections/trivial and the mandelbrot set wall tapestry here!: https://stemerch.com/collections/mandelbrot/products/mandelbrot-fractal-wall-tapestry
@anni1961 Says:
The warmup was so fun and easy and then it got so much more difficult😅
@kdcpo6105 Says:
I have such a love/hate relationship with these brilliantly crafted ads. (Shakes fist angrily)
@Reuraku Says:
The problem I find with the last one is showing the variations is misleading since 2 of the options are in effect the same with regards to the question. Specifying one is tails and posing the question what are the odds both are means one hand regardless of which is irrelevant, since the question really boils down to what are the odds of the other coin being tails. The statement leads to a reduction in the question and 2 of the options of outcomes are effectively the same so the odds are always 50/50.
With the cards it comes down to ace vs non-ace, no matter what one ace is out of the pool so the odds of the other one comes down to number of aces remaining/number of cards total specifying which ace was removed from the pool chances nothing as just like with the coin several options are effectively the same.
@luissilva6119 Says:
two cards from a standard deck, it's given that one of them is an ace, what are the odds that the other is also an ace? 1[other card] X (3[possible positive results] / 51[total results possible]) = about 5.9% nowhere does the suit come into this equation. the odds do not change, that is rediculous.
@m64982 Says:
had to stop watching half way through. your stochastic probability theory is completly wrong.
@Sofanthiel-OwO Says:
3:45 That's an average, not the median.
@Susanmugen Says:
Regarding the "Holding an ace" paradox.
At 6:00 you say the probability of holding an ace if you say "hey I'm holding an ace" is there's two aces in 1 of the 5 hands displayed. But you're not displaying ace of diamons first and ace of spades second, so one of the combonations you need for the calculation is missing. When you say "i'm holding an ace" there are two different aces that could be, a spade or a heart. We have to add the probably of a 2nd ace for both spades and hearts. It's 1/3 for spades as displayed and for hearts, it's 1/3 the 2nd card is ace of spades. So if the graphic showed a hearts-first spades-second combo on there, it'd be easier to see that the probably went from 2/6 (which reduces to 1/3) to 1/3 by specifying the suit.
@youtubersdigest Says:
Casinos hate Zach after this😂
I'm loving this channel i just discovered. Thank you Zach
@meea1971 Says:
I wish he was my teacher in college. I loved math but he would make learning math and calculus more fun so i would understand it better in relation to physics!
@blindspy Says:
Coke and apple juice sounds kind of good
@wingotplays Says:
Intuitively, the coins makes so much sense. It isn't you saying that the other one is tails that is relevant. If the question was "Is the coin in my right hand tails?" the answer is obviously 50/50. By specifying the value of the right hand, you're reducing the problem space to "what is the value of the left coin?" hence 50%.
It's an inverse Monty Hall problem imo.
@Abolish_The_S-N-T_NOW Says:
Ohhh, my head hurts now.....need the conclusion
@CosmicFilms-007 Says:
I feel dumb now
Sometimes I forget he has an educational channel because of how funny he is
@VendoCitroenSaxo1.6iVTS16vCup Says:
The second problem is bounded if you have the same number of populations and if it's impossible to reveive negative salary, in this case the maximum of women's increase is 5% making the option E the right one, what am I not getting? (In this case the women salary is infinite with respect to men's)
@ionex18 Says:
2:50 Isn't it supposed to be "mean" salary? Because "median" refers to the middle value when you arrange it in ascending order...
@FlameLord050 Says:
With the 2 cards, you are mistaking two different questions.
Question 1 is what is the probability of drawing 2 aces, if you are guaranteed to always draw 1?
Question 2 is what is the probability of drawing an ace after having already drawn 1 ace?
Im not actually certain what the correct phrasing of question 1 should be because I can't think of the question in which you would ignore the possibility of not having drawn an ace at all, while still accounting for the probability of drawing it.
@FlameLord050 Says:
Man, i got question 2 wrong because I was thinking mean.
@danielname3420 Says:
No, it doesn’t make a difference that we don’t know which was revealed. Either way it’s 50/50, just not sure which 50/50
@szymonanonim7214 Says:
The thing about probability is that we either assume each individual happening has it's own probability , that does not interact with other probabilities , or we say that there is a connection between probabilities , qnd each impacts the other. There are problems with both stances. One , imagine you flip a coin. It is 50/50. Now imagine other person flipping a coin , and landing tails each time. Now , theoretically , that makes our flip almost certain to land on heads. Cause what is probability of something falling on tails a hundred times of your friend and one time you. That is implying all probabilities are connected and everything can happen only even amount of times. Which is wrong. Lets say probabilities does not impact each other. Then me landing tails after landing it hundred times over is a 50/50. Me landing it continously is always 50/50 no matter how many times I do it. Which is wrong. Funny to think about.
@paeul4982 Says:
If math was taught like this in school I would probably have been 100% starter.
@Teophrast Says:
I want to wright down my "research" about boy and girl paradox. If host pulling out pair of coins 120 times, 90 times there will be at least one tail. From thos 90 only 30 will be two tails. If host reveals only one side, either left or right selectively, there will be 60 cases of right hand containing tail and 60 cases of left hand containing tail, because that 30 times when two hands contain tail overlaps with each other. If host reveal all 60 cases when left hand contain tail, and we bet all the time that both hands contains tails, it will be exacctly 50% - 30 times from 120. But it works only when host never reveals from second hand. If he reveal from both hands - 90 times, we will still win 30 times, probability 33.333...%.
@meowmura349 Says:
With the ace thing, I thought of it as, if I have one card in hand that's and ace, and I draw one card from the deck, what's the chance the other catd is also an ace. And the chance should be 3/51 since there are three aces left and 51 cards left overall. In this case the chnace would be about 6%. I also can't see how in this case knowing the suit would change the math.
@Icewall956 Says:
Wait, I thought median was the middle value in an ordered dataset- but for the second problem it looks like you took the mean/average? I feel like I’m not getting something
@TheDoc73 Says:
Look, sometimes math is bs. You know it, I know, and the American people...well, they're wrong, but you and I still know it.
The probabilities don't change when you assert which coin is which or what suit the ace is. The probability changed when you announced the single ace or the single coin result, because the "change" you are referring to is the removal of juxtaposed results, which is to say identical (but reversible) outcomes which are considered as part of probability calculations because they are the result of two separate events which could each equally be one of the possible outcomes. Once you announce the outcome of one of the events, you have made it so that the probability is defined by only a single event which happens independent of the other. If we know one coin is tails, your question isn't actually "...chances the other coin tails?" but rather "what is the chance of a coin flip being tails?" The same goes for the cards, the question becomes "out of three cards, what is the chances of an ace?"
If you said what is the probably of a specific ace being in a specific position or your asked WHICH hand the tails is in, then you put juxtaposition back in play. For example, if you said: "At least one coin is tails, which hand is it in?" The chances are back to 1 in 3, because the answers are left, right, or both.
@Ri5olu Says:
the first one is literally just algebra XD it took me 2 mins to solve (no difference by end)
@Sayan_9001 Says:
I still have trouble with the heads/tails because, if you don't know which one is Tails but that one is, as you've said we have 3 scenarios: TT, TH and HT. My problem here is that TH and HT are functionally the same and I don't see why they should be considered separate, meaning there are only 2 possibilities and it's a 50% chance. You could rephrase it as "One of them is tails, and the other one is either heads or tails, 50% chance. Does that make sense or am i missing something?
@PhatesDemise Says:
That Ace problem reminds me of the monty hall problem. I just don't like it.
@charles7623 Says:
you didn't say that both cards came from the same deck so the probability stays at 1/13 (discounting jokers)
@MonkeySimius Says:
I feel like when you are telling me the suit of the card or which hand holds a tails it isn't actually giving me any more information than just telling me you have an ace of some undisclosed suit or that you have a tails in one of your hands... Because you didn't tell me before hand you would be revealing to me if you have an Ace of spades or if you have a tails in the left hand. So... That probability diagram you showed isn't accurate. Long story short you'd have to add another option to that table to account for the fact, had you had a heart instead of a spade, you would have told me you had an ace of hearts. In the end, you would have to add enough options to account for your options of what to reveal where the probability isn't effected even with the information about which hand it is in or which suit it is.
Ya know?
@JimmyVermeer Says:
The question about the probability of the other card being an ace depending on whether or not you say which ace it is can be explained by asking how you found out that at least one of the cards was an ace. Did a person randomly select a card to see what it was, and it just happened to be an ace, or did someone ask "Is at least one of those cards an ace?" to which Zach would have to answer yes or no (assuming Zach knows what both of the cards are). In the first case, the odds are 1 in 3, but in the second case, the odds are 1 in 5. And now the suit doesn't matter, what matters is how you found out there was an ace
@Jack22themax Says:
do... you hate when our brain is like not confused or smt...
@HomicidalTh0r Says:
This is why Jack the Ripper killed all those people.
@artyruch Says:
In last question theproblem arises due to order? Well I am sure it is the case for coins but cards seem to be of a different level to just say the problem comes with order.
@AnnabethKosa-h3m Says:
You lost me halfway through 😅
@mightymcphee Says:
only just saw this vid but to answer the last question no. and im glad you eluded to it because the entire time your making your example with "i just have to SAY its this one" i thought "why word it so obtusely?" why not be clear and say "i just have to KNOW its this one" probability is just a mathematical equation, a measurement maybe? , it does not act alone. it is a tool. after the screen comes back from being black i still dont know which hand had the tails and so it could still be one of three options, had i known which hand held the tails i could use that knowledge to eliminate a possibility leaving only two options available. i like these kinds of rabbit holes, like how almost everyone misunderstands Schrodinger's cat
hint: what doesn't reflect light?
@theunholybanana4745 Says:
With the last one it depends on how it happened
If you look at one coin and reveal a fact about it, you are reducing your sample space by 50% (which is equivalent to already making it through the first 50/50 flip), and the second will also be 50/50, this is intuitive probability based on how we usually receive information.
But if you look at BOTH coins then reveal a fact about both of them (eg "one of them flipped tails") then this is kind of deceitful, you could've seen TH or HT and said this fact, artificially boosting the probability of jumping through the first hoop by just having to jump through one of the 2 hoops if that makes sense.
With the ace example, people usually reveal information as they gain access to it, "one of these cards is an ace" is interpreted to mean "I looked at one of these cards and it was an ace" but in your example it meant "I looked at both of these cards, and one fact about this set of 2 cards is that there is at least 1 ace".
We don't have an intuition for the second case because it never shows up in day to day probabilities, it would be like if you were listening to a football game on the radio but every single goal from team A was revealed before any of the goals from team B, you've effectively selectively reordered the data to appear more favourably.
That's just a layman's interpretation, it's not 100% accurate but it does explain why it's unintuitive.
@richardguo737 Says:
6:44 there should be two aces therefore 2/6 = 1/3 so probability doesnt change
@richardguo737 Says:
4:39 what the fucl you didn't change the overall median so you lie
@georgbiber7714 Says:
I dont think It’s subtle. For the cards the actual possibilities are ace and other ace, ace and card 1 and ace and card 2. So it’s one in three cards you can have on top of the (any) ace. It doesn’t matter whether you know which specific card it is or not. And the two coin flips are just two unrelated random 50/50s. Telling you the outcome of one does not affect the other.
@WilcoVerhoef Says:
when you say "this one is tails"
you can either mean "I've looked at both of them, concluded that there's at least one tails and choosing to hold up that hand"
or you mean "I've looked at one and it happend to be tails"
and when you say "there's a tails"
you can mean "I've looked at both and concluded there's a tails"
or "I've looked at one and it happend to be a tails"
Language is fuzzy. The words are the same but they hold a different amount of information.
@KeahanTurnbull-qy1zg Says:
The second answer is wrong. 2024 media should be 225k. Making the median increase by 50% yoy
@miholitar Says:
For the ace problem I want to know how you got a 3% probability when because I keep getting 5.88 or 7.69 depending on the scenarial and the only case where I could find the 3% probability is if we have a deck of 26 cards with 2 aces witch is not something someone would assume espeshaly with so little detail. I know this isnt the point of the question but I am curiose about it. Sorry for my bad punctioation, grammer, spelling and bad english as whole
@saadkhondoker1921 Says:
let's make the 1st one obvious. Let us think the spoon size is equal to the glass size. (then the whole coke is poured, then a 50-50 mixture is poured back. both contains 50-50.)
@TheWP120 Says:
I agree with the first problem. But I want to explain it mathematically through material balances (like we do in chemical engineering):
I get that the coke in juice bottle is
cJ=(1/100*1+0*1)/(1/100+1)
=1/101
and that the juice in the coke glass is
jC=(1/100*(1-cJ) +0*(1-1/100))/(1/100+1-1/100)
=(1/100*(1-1/101))/1
=1/101
PS. I've liked your videos for a long time.
@РеймонХатат Says:
I really don't understand why medium salary doesn't change
@coyot7959 Says:
YOU ARE EVIL 🗣️🗣️🗣️🗣️
@Trytouchinggrass Says:
I never expected Zach could be like this kinda fun...Maybe that's why you should watch the nerdy science video's first then the skits.
@wendellenthorn2205 Says:
YAWN......, were you talking to me?
@TimelordPrime Says:
It amazes me that someone so smart is also so God Damn funny.
LATEST COMMENTS