Comments by "\/" (@joebazooks) on "TED-Ed" channel.

  1. if you pull out a blue ball first, you are now left with 1 red ball and 1 blue ball, which is 50/50. if you pull out the red ball first, you are 100/100 going to pull out a blue ball next. pulling out a blue ball first does not increase the likelihood of then pulling out a red ball on the second pull.
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  2. Lucas farach-colton lol i suggest you go back to the drawing board and actually think about what you're saying since it's nonsensical. according to your math... you don't pick red 2/3 * 1/2 = 33.33% you do pick red 1/3 * 1/2 = 16.66% none of this even adds up LOL i'm not even sure how this analogy is similar to the initial scenario. all of this demonstrates that you're a troll with completely absurd ideas.
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  3. 3rd one is the best
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  4. whatttt?>! why have i not heard of this other channel heretofore
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  5. fantastic riddle. the first decent one you've posted in a while
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  6. Thief with world's largest chocolate bunny robs town. = With world's largest chocolate bunny thief robs town. ≠ Thief robs town with world's largest chocolate bunny. The last example, which was used in the video as an ambiguous sentence really isn't ambiguous at all, and should be interpreted as 'A thief robs the town with the world's largest chocolate bunny'. Again, 'Robbers who steal chocolate bunnies rapidly attract the outrage of onlookers' isn't that ambiguous in my opinion, but maybe that's just me. 'Rapidly' modifies 'attract', otherwise the sentence might be structured as one of the following: (1) Robbers who rapidly steal chocolate bunnies attract the outrage of onlookers; (2) Robbers, who steal chocolate bunnies rapidly, attract the outrage of onlookers; and not as: (A) Robbers who steal chocolate bunnies rapidly attract the outrage of onlookers; or (B) Robbers who steal chocolate bunnies attract the outrage of onlookers rapidly.
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  7. Justin Grindley de nada
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  8. Justin Grindley no, pero tu me puedes comprar un cafe si estas una persona agradable
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  9. de donde estas aprendiendo espanol?
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  10. Justin Grindley muere una muerte dolorosa
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  11. Justin Grindley nono, tu capacidad es impresionante. duolingo.com es un buen recurso si quieres aprender y practicar mas. ya estoy muerto
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  12. Pablo Augusto Palma Terán haha thanks :) and thanks for the info
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  13. the guy/girl standing at the back announces the sequence aloud? lol
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  14. ***** that's because the actual example is ironically an illogical example. just imagine if there were only 2 ppl held hostage; let's say one woman and one man. the observer tells the two individuals that at least one of them has green eyes, if nobody leaves that night, if neither the woman or the man leaves that night, it's because each of them already knows that the other individual has green eyes, which therefore caused each of them to be uncertain as to whether or not he himself/she herself has green eyes because "AT LEAST ONE (of them)" does not mean "JUST ONE (of them)" but rather "ONE AT LEAST". does that make sense? in other words, the other individual's uncertainty, or reluctance to leave, is indicative of the fact that they know you have green eyes.
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  15. wtf... how do you get 4 possibilities from two frogs whereby one frog is male?! Male and female and Female and male are duplicates. if 1/2 frogs is definitively male, the only possibilities become Male and male or Male and female.
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  16. David Williams It's still 50/50 because this ratio was established as a condition in the beginning for ALL frogs. You CANNOT discount the 3rd frog, which is on its own, and then INCLUDE the 2nd frog; The 3rd Frog is STILL information that has to be accounted for, even though it's a variable, and even though you choose not to go in that direction in order to lick it. (1) MX | X (2) MX | Y (3) MY | X (4) MY | Y In other words... (1) MM | M (2) MM | F (3) MF | M (4) MF | F If you consider MF | M and FM | M as unique conditions, then you must also consider MM | F and MM | M as unique conditions. You cannot exclude the 3rd frog because the ratio of 50:50 is true for ALL frogs, including just the frogs you choose lick.
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  17. David Williams You're mistaken. The position of the Male frog on the left side is irrelevant, as are the positions of all 3 frogs. You're allowing this impertinent information to confuse you. The only pertinent facts are: (1) there are two frogs of a given species; (2) one is male, the other is unknown; (3) the quantity of individuals of the species is unknown; and (4) of the entire species, half are male and half are female. The fact that 1 out of only 2 frogs of an entire species is male is impertinent information since we do not know how many individuals of said species there are. Let me simplify this for you. If only 4 individuals composed the entire species, and there was a randomly sampled set, whereby we knew 1/3 was M, the data can be represented as so: M + M/F | M/F The only possibilities are as follows. (1) M M | M (2) M M | F (3) M F | M (4) M F | F Therefore it's still a 50% chance. You cannot simply change the position of the frogs on the left side and pretend that this creates new possibilities. In other words, if M F is unique from F M, then Ma Mz must also be unique from Mz Ma.
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  18. David Williams He says verbatim, "...(And to make matters worse,) the male and female occur in equal numbers", which means the sample from which you take your quantity of 3 is split evenly 50/50. This also means that you cannot use an odd number of pennies. You must begin with an even number of pennies with an equal distribution of heads-up and tails-up among them. Thus, if you wanted to use, say, 11 pennies instead of 3, you would start with 12 pennies and then remove 1 from that sample. Man, at least know what you're talking about before you begin talking, otherwise you're just demonstrating your ignorance.
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  19. David Williams Well, if there an equal probability of choosing, say, a male frog at random, then the population must be split evenly, otherwise there would not be an equal probability. This is why, for instance, there cannot be an equal probability of choosing a male or female frog at random from an odd number of frogs, even if the sample from which you are selecting a frog is substantial, say 1000 frogs; if there are 501 males and 499 females, there is not an equal probability of selecting one or the other, as it is slightly more probable that you will select a male frog at random, however insubstantial this difference may be and despite variance. I'm sorry to break it to you but if you have an equal probability of selecting heads-up or tails-up coins from a massive sample, you're no more likely to pick one over the other; which means even if you are to select just 2 coins from 500 coins, probability dictates that you'll end up with 1 heads-up coin and 1 tails-up coin. I get what you're saying, and I agree with what you've said, but your conclusion is simply wrong.
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  20. David Williams that's a faulty and terrible analogy... try creating an analogy that is adequately analogous and you will quickly find yourself running into problems. no, really, i'm challenging you to create an adequate analogy and then share it with me!
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  21. David Williams it's not that it wasn't clear or that i did not understand what you were saying. it's an awful analogy since it is barely if at all analogous to the initial the conditions. such conditions are not similar because (1) the quantity from which the 3 frogs appear is unknown (not that this matters). however, what is known is that (2) there is an equal quantity of male and female frogs and therefore an equal probability of selecting one or the other. you're job is to create an analogy that demonstrates why knowing 1 out of 2 (out of 3 frogs) is male skews the probability of the 2nd frog in the group of 2 in favor of being female.
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  22. David Williams in the real world, MF and FM are not different... if you have two frogs and one is male, there are only two possibilities: MM and MF. in the real, physical world the sequence or position of the frogs changes nothing, yet you continue to pretend that it does.
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  23. David Williams even though you now know FF is not a possibility, it only grants you the understanding that your probability is not 100% so the change is insignificant.
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  24. you're confused as hell, my friend; hence you have troubles expressing yourself and your perspective.
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  25. David Williams You're wrong because it does not matter to any extent which frog is male. There are 2 frogs and 2 possible outcomes: MX and MM or MF (50/50 odds). If it did matter which frog was male and which frog was female, then necessarily you must account for that and consider the possibility of FM in addition to MF, otherwise you're complicating the situation unnecessarily. Again: a. Get 4 pennies. b. Place 2 heads-up and the other 2 tails-up. c. Remove 1 heads-up penny from the group (this will be your Male). d. Put on a blindfold. e. Mix up the remaining 3 pennies. d. Take 1 penny from the active pool of 3 and remove it from the game while still blindfolded. e. Distribute 1 of the 2 remaining pennies to the left. f. Distribute the last penny to the right. g. Remove your blindfold. f. Repeat steps A to G 100,000 times If you actually did this, you would see that over time your odds will settle around 50/50 due to the fact that you're more likely (66% likely) to remove 1 tails-up (Female) altogether from the game in order to form the set (2 on the left, 1 on the right) since 2/3 are tails-up (Female), which leaves you with 2 unknown pennies, 1 of which is likely to be heads-up, 1 of which is likely to be tails-up. This works for any sample size. Seriously think about it!
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  26. Btw, that's a false distinction: the distinction between possibilities and outcomes. We're calculating the probability of outcomes, and we do this by considering the possibilities and calculating their likelihood in relation to one another.
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  27. Lugh Summerson or you could just do the experiment i outlined above :]
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  28. +marina Ⓥ. the back guy could announce 'black black black' or 'white white white white white white' for however many there are with black or white hats
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  29. Ameer A saying 'Black Black Black' is virtually the exact same thing. ppl still have to deduce whether or not theyre wearing a black or white hat and this clue isn't prohibited by the conditions
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  30. ***** awww, did mommy not give you a hug before she sent you off to school yesterday morning?
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  31. Ameer A the tallest guy wouldn't say out all of the hat colours. he would say either "black black black" or "white white white white white white". if he said "black black black" and somebody didn't see 3 people with black hats, then that person would have to deduce which colour his/her hat is. and yet nobody was guessing the colour of their hat... lol
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  32. Михайло Ткачук exactamente
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  33. 王庭茂 what a funny guy he is, eh?
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  34. rockybingutu which is why the person in the back says either "black black black" or "white white white white white white". it's the same concept as the suggested solution and still abides by the conditions
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  35. the GBO vlogger i'm not interested in how you've interpreted the video or the conditions therein. i'm asking you which explicit condition prohibits one from repeating either the word "black" or "white"?
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  36. rik.yan it shows a sequence of black and white hats, which of course is illegal since one must "not even try saying words other than black or white". again, there isn't any explicit instructions or conditions that prohibit saying black or white multiple times. there is no reference whatsoever to quantity, but, as WarpZone10  said, it's really not in the spirit of the riddle.
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  37. 王庭茂 i'm shocked he's still composing comments. it's as if he thinks people are still reading them lol
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  38. ***** exactly :D "you can say `black` or `white`", once, one hundred times, or however many times you wish to
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  39. Cheshire Cad it shows a multi-coloured sequence, which has absolutely nothing to do with saying either black or white multiple times... god, ppl r so dense
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  40. Leo 1618033 i really didn't come up with a legitimate answer, though i knew it had something to do with the visibility of the person standing at the back, so i said he should just announce the entire sequence to everybody else (knowing full well that this "solution" isn't acceptable) and then skipped to the answer :)
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  41. Cheshire Cad i'm obfuscating things? yeah right lol that's exactly what you're doing
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  42. Corsican cheese nomnomnomnom
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  43. Ero Shyn@@bi it's definitely allowed since it wasn't explicitly prohibited
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  44. Ero Shyn@@bi "Black black black black" isn't explicitly prohibited. in fact, the method that was endorsed isn't much different.
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  45. Ero Shyn@@bi i have no idea what you're trying to say
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  46. Ero Shyn@@bi no, your english or capacity for expression is sad ;)
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  47. John Madison haha yep. it's pretty disappointing that nearly all of the riddles presented on here lack decent explanations.
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  48. Andrew Ness BOO! lol
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  49. Douglas Phillips well said
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  50. a/the 'mind' is just an abstraction. what does that, 'mind', even mean?
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