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Little tip for those who don't want to pay in order to now the answer: it's number one. Why you ask? This is due to the one pair revealed. If you look at the second possibility (two different cards) you will see that this lowers your chance to get a pair on your own. But here is the twist: the first one lowers yours as well but only from (let's say 1-ass) 14 possibilitys to 13. The second however lowers your chance from 14 to 12 due to the cards getting out of the deck.
You can't touch this You can't touch this You can't touch this You can't touch this My, my, my my music hits me so hard Makes me say, "Oh my Lord" Thank you for blessin' me With a mind to rhyme and two hype feet It feels good, when you know you're down A super dope homeboy from the Oaktown And I'm known as such And this is a beat, uh, you can't touch I told you homeboy (You can't touch this) Yeah, that's how we livin' and ya know (You can't touch this) Look in my eyes, man (You can't touch this) Yo, let me bust the funky lyrics (You can't touch this) Fresh new kicks, and pants You gotta like that, now you know you wanna dance So move outta yo seat And get a fly girl and catch this beat While it's rollin', hold on! Pump a little bit and let 'em know it's goin' on Like that, like that Cold on a mission, so fall on back Let 'em know, that you're too much And this is a beat, uh, they can't touch Yo, I told you (you can't touch this) Why you standin' there, man? (You can't touch this) Yo, sound the bell, school's in, sucka (You can't touch this) Give me a song, or rhythm Makin' 'em sweat, that's what I'm givin' 'em Now, they know You talkin' 'bout the Hammer, you're talkin' 'bout a show That's hyped, and tight Singers are sweatin', so pass them a wipe Or a tape, to learn What's it gonna take in the 90's to burn The charts? Legit Either work hard or you might as well quit That's word because you know (You can't touch this) (You can't touch this) Break it down! Stop, Hammer time! Go with the flow, it is said That if you can't groove to this then you probably are dead So wave yo hands in the air Bust a few moves, run your fingers through your hair This is it, for a winner Dance to this and you're gonna get thinner Move, slide your rump Just for a minute let's all do the bump (Bump, bump, bump) Yeah (You can't touch this) Look, man (You can't touch this) You better get hype, boy, 'cause you know you can't (You can't touch this) Ring the bell, school's back in Break it down! Stop, Hammer time! You can't touch this You can't touch this You can't touch this Break it down! Stop, Hammer time! Every time you see me, the Hammer's just so hyped I'm dope on the floor and I'm magic on the mic Now why would I ever stop doin' this? With others makin' records that just don't hit I've toured around the world, from London to the Bay It's "Hammer, go Hammer, MC Hammer, yo Hammer" And the rest can go and play You can't touch this You can't touch this You can't touch this Yeah (You can't touch this) I told you (Can't touch this) Too hot (Can't touch this) Yo, we outta here (Can't touch this)
Oh once I got paralysed 10 times in a row from a shinx in pearl, so I couldn’t move, then I missed razor leaf (95% accurate) 3 times. I wanted to train grass type that I can’t remember.
The problem with scanning the same rock multiple times is that we don’t know whether the reason it has a 10% false positive rate is due to a stochastic error (extra scans could give different results) or a systematic error (extra scans give the same results). For example, if the device is sensitive to another mineral that exists in 10% of rocks, extra scans won’t fix it.
Or think it more simply: if there is element in the worthless rock trigger a positive reading from the detector, it will ALWAYS report the false positive.
That is very true, and from what another Commenter has pointed out Tricky Joe probably was aware of the fact that the rock most likely was worthless hence he was willing to trade it for a mere $200, like that Commenter said, neither a rational or irrational person would sell something they thought was more valuable then what they were selling it for for what they are selling it for.
@@dalooneylass5561 -- if Tricky Joe had some method for being aware of whether or not the rock was worthless, he wouldn't need to build a detector. By the rules of the puzzle he couldn't know -- and he doesn't have to. We can calculate the _average value_ of the rock the detector gives a positive result for, and it's (10*$1000 + 99* $0)/109, which equals about $91.74. If Tricky Joe is making this deal frequently, then he's always going to come out on top so long as he asks for _more_ than this amount. If he asks for _less_, then the trade starts making sense, as over a longer number of trades you'll eventually make money (as the losses from the times it was a false positive are vastly outweighed by the gains from those times when you make $1000 from it being a true positive).
Me: watches and doesn’t try to solve it Solution: Bla bla bla Me: DAMNIT I COULD’VE DONE THAT! I’LL NEVER WATCH LNE OF THESE AGAIN! 5 seconds later... OOH ANOTHER RIDDLE VIDEO! REPEAT UNTIL I DIE
Okay, Tricky Joe is a trickster, but we have to give him credit for his invention. It is actually very useful. Having to examine 109 rocks is much better than having to examine all 1000 rocks for unobitanium
Now that you mention it, you're right! Even if it's not a guarantee that the rock that's detected is unobtainium, the fact that it can even detect it at all with relatively decent accuracy allows you to narrow your search quite significantly. With that in mind, it might actually be better to do a slightly different deal: Increase the offer to either $300 or $400, but include the device itself along side the rock. By having the device involved with the equation, in the short term you'll likely take a bigger hit, but in the long term it would actually be more beneficial!
There's a reason his name is trickey Joe. Most likely he built a device that always gives a false positive but told you the story of a 10% chance. Otherwise why would it react with literally the first rock he tested.
Doing a second round of testing would make it even more accurate, you would need to scan the 109 rocks, get positives on 10 of them and false positives on another 10. Two rounds of testing and you have narrowed the search down from 1000 rocks with 1% accuracy to 20 rocks with 50%.
@@13g0man Of course, that assumes that the cause of the false positive is entirely random. If it's the result of some other mineral or a structure of minerals that the detector cannot tell from Unobtanium, it's still no better. If nothing else, those 109 rocks would let you figure out which is the case and refine the detector further.
For the bonus riddle, picking scenario 1 is slightly better because there’s more pair combinations with 2 and 4 cards left (in the example there’s 2 queens and all 4 5’s left, giving 7 combinations) than 3 queens and 3 5’s left (6 combinations). This is because the rate at which n cards gives x combinations also changes (2 cards have 1 possible combinations, 3 have 3, 4 have 6, and so on, basically triangular numbers)
@@jatinchandwani8427 we already know / assume that she has a pair or not, so the question isn’t “is drawing two pairs more likely than drawing a pair and a non-pair?” It’s “is drawing a pair given that a pair has been drawn more likely than not drawing a pair given a pair has been drawn?”
@@jatinchandwani8427A ) 2Qs Theres 12 packs of 4 and a pair of Qs left in the deck, so the probability of drawing a pair is: 12× (4/50 × 3/49) + (2/50 × 1/49) = 5,959% B) Q + 5: Theres 11 packs of 4 and 2 packs of 3 So 11× (4/50×03/49) + (3/50×2/49) × 2 = 5,878%
To put it another way, at the start there are 78 possible pairs in the deck, that is 6 pairs for each of the 13 card values. This is because there are 6 ways the suits can pair up: (♣♠), (♣♥), (♣♦), (♠♥), (♠♦), (♥♦). So with that in mind, we look at how many pair combinations are left for each scenario. Scenario 1 eliminates (Q♥) and (Q♦) leaving the only pair option (Q♣ Q♠). This means 5 options were eliminated Scenario 2 eliminates (Q♥) and (5♠). From the Queens this leaves only (Q♦Q♣), (Q♦Q♠), and (Q♣Q♠). From the 5's this leaves only (5♣5♦), (5♣5♥), and (5♥5♦). This eliminates 3 pairs from each Queens and 5's for a total of 6. So in Scenario 1 the deck will have 73 pairs left for you to hit while in Scenario 2 there will be only 72 pairs left for you to hit.
Because this was a math problem, not a logic problem. We understand probability and money better than what “Ozo” means, or how to create a statement both true and false without saying a paradoxical statement. 😅
then you go to the third position from disky disk A and select the Noether 9000 and buttons D and E to destroy the laser robot ants and cut the antidote into fifths and use parity to find out what place your in and what hat you are currently wearing with a 10% false positive rate then you get silver by moving counterclockwise then go to locker 64 to find out who ate which crystal then go in groups and add 5 and 7 and square root to escape with 60 bees and at least 16/30 rubies (marking the heavier ones with a +) on planet 3 and after all of that you have to question my sanity, and the answer is ozo.
But it doesn’t state that the device has a limited use. Theoretically, you can check those 109 rocks again. And again. And again. In the real world I would do that
Striker The idea was to demonstrate the risk of false positive in e.g. medical screening, and the concept of conditional probability. Don't take the riddle so literally, it was just an example
Exantius E I know but my luck would tell me if I repeat this again and again, I have a high chance of getting a false positive based on my win/lose ratio in games
It doesn't mean that the detector has 90% chance of returning a negative for the same rock. 1 out of 10 rocks will give you a false positive, and that rock will give you a false positive no matter how many times you test it. At least that's how I interpret the scenario. Perhaps the detector is measuring something like radioactivity as an indicator of whether unobtanium present. A false positive would be a rock that shows high radioactivity due to some substance that is not unobtanium.
I came up with a modification to the riddle. imagine you COULD zap it multiple times but each time you did, it lost 10% of it's current value how many times would you zap it
For the card riddle: It is option one (ca. 5.959% chance of success). Option B has ca. 5.878% chance of success. The reason is that in the first option you are only decreasing your chances of getting a pair of queens but still have the same chance of getting any other pair. In the second option your chances of getting a pair of queens and a pair of fives are decreased. Calculation: Option A: [12*(4/50)*(3/49)]+[(2/50)*(1/49)] Option B: [11* (4/50)*(3/49)]+[2*(3/50)*(2/49)]
There are four queens (and four fives) in total So the second option you have higher chances of getting a pair than on the first one. You are more likely to draw a five (or a queen) out of three remaining fives (or queens) out of 62 cards than you are to draw a queen out of two remaining queens out of 62 cards
This video doesn't just do a good job of explaining false positives; of all videos, this one does the best job of connecting the idea back to the formula. (3:23)
@@youtubeuniversity3638 he's assuming that the gun gives a false positive due to the gun failing on an individual shot basics, rather than say the false positive coming from the rock mimicking the wavelength of unobtainium
@@youtubeuniversity3638 right, so the assumption he makes is this: the machine is 90% accurate, shot to shot, regardless of other factors. HOWEVER, it is completely possible that the inaccuratracy is caused by the fact that other components of the rock cause it to read the same on the detector as unobtainium. So, lets say the reader detects based on wavelength, and false source produces a wave close to that off unobtainium, then shotting it 10 times isn't going to improve your chances of knowing it isn't what you want just because it read 9 out of 10 times. This is a good question though! :)
My thoughts exactly. I immediately thought "but there's 10 rocks that don't have anything that will trigger it, so the chance of finding the correct one is slim"
For the bonus riddle, my hypothesis is that the likelihood of you having a pair is greater in scenario 1. In scenario 2, the likelihood of having a pair decreases because there are two fewer options of two cards matching up, whereas in scenario 1, the likelihood of having a pair would be only impacted if you had a queen in your hand.
I have written a comment stating that there are 78 possible pair combinations in a deck of 52 cards. Scenario 1 eliminates 5 combinations while Scenario 2 eliminates 6. Therefore #1 is more likely that u have a pair with 73 combinations vs 72 in #2
@@amna7491 if you use baye's theorem will change the answer to the scenario 2, because you already draw a pair and the probability to draw a second pair in a row will deacrese substantially
If its wrong 100% of the time then you would know if its unobtainium or not 100% of the time because if it does not beep it means that is unobtainium because its wrong and it didnt beep and if it beeps you can just disregard it and ignore it
As for the bonus riddle: Let's examine the probability of obtaining a pair in either of the cases- 1) When Amy has 2 Queens - Out of the 50 remaining cards, we can either firstly choose any 2 pair cards from any of the other other 12 quadruples, i.e. 12*(4C2) ways. To get a queen pair there is only one way as there are only 2 queens available so they have to be picked anyway. Implies: total number of ways in this case is 12*(4C2) + 1 = 73 2) When Amy has a Queen and a 5 - Out of the 50 cards, we can either firstly choose any 2 cards from any of the other 11 same-numbered quadruples, i.e. 11*(4C2) ways. To get a Queen pair or a 5 pair, we have have 2 choices (Queen/5) and then we gotta choose 2 out the available 3 cards. Implies: total number of ways is 11*(4C2) + 2*(3C2) = 66 + 6 = 72 73 > 72, thus, we have a very slightly greater chance of getting a pair in the first case.
Since the first case removes two queens from the deck, this reduces your chances of getting a pair if either of your cards is a queen, but also reduces your chances of getting a queen in the first place. The second case creates two bad cards, since now both queens and fives have a reduced chance of being pairs.
@@randomentirely1006 It does. Every card Amy has is less likely to be a card you have. If Amy has a pair, that's only one type of card that's harder to match, but if she has two different cards then that's now two cards that will be harder to match.
They would have been better to just call him Joe. the shady looking appearance plus the name Tricky Joe immediately put me off. It's a fun thought experiment, but TedEd biased it here, which kind of sucks.
I said no because a rational actor would not offer to sell it for $200 if he thought it was worth more than $200. Similarly, a deceitful actor would not offer to sell it for $200 if he thought it was worth more than $200.
Solution to the extra riddle is simple: your chance of a pair is higher if Amy has a pair of queens because neither of the queen cards can be one of your cards. If a queen was one of your cards you cannot have you cannot have a pair.
For the playing card riddle i'd guess the two queens would make it more likely for you to get a pair. That makes it less likely to get a pair of Queens again, but a queen and another card make two different sets less likely.
Yes. We can simplify the question by considering a deck of only Queens and Fives. If Amy has a pair of Queens, the number possible pairs you can have is 1 + 4C2 = 7. 1 for another possible pair of Queens, and 4C2 for all the possible pairs of Fives. If Amy has a Queen and a Five, the number possible pairs you can have is 3C2 + 3C2 = 6. 3C2 being the number of possible pairs from 3 Queens/Fives. In other words, you are more likely to have a pair if Amy also has a pair.
I agree and also think that’s the correct answer but until someone does the maths I wouldn’t be surprised if it’s equal. Because sure, the pair of queens only makes queens less likely and the two options make two less likely... but less is not quantitative... the less in the queens scenario is more sizeable. In one scenario there are only 2 queens less making that particular pair very unlikely but in the other scenario two numbers are cut to 3... there’s still a decent chance for a pair with three (although all in all the pair is likely to come from a number with 4 cards in either scenario) //I realise this is written terribly
edfreak9001 Yes, the pairnof queens is better. I did the math: It is the same if your first card is anything other than a queen or a 5, so just look at those sets of cards: Scenario 1: First card is a queen 2/50 times. 2nd card is a queen then 1 out of 49 times if the first one is a queen. First card is a 5 4/50 times and 2nd card is also a 5 3/49 times. That's [(2*1)+(3*4)]/(49*50) or 14/2450 In scenario 2, you get a queen 3/50 cards and another queen 2/49 of the remaining cards, or 6/2450. It's the same for a 5, so that doubles to 12/2450. That's less tban 14/2450, so you are better off with the scenario where the other side draws a pair. Unless you are trying to play head to head. Your odds of beating a pair of queens with 2 cards are way lower than your odds of beating a queen five off suit.
I thought there might be an equity break down of this problem as well. If we measure the value of the rock to be 9% of 1000 then we could consider buying it for 90. However if you made this trade every time over a long period, you would only break even since 89% of your rock is paid for by the other 9%. So if you can haggle Joe below 90 you have a decent chance of profiting.
As for the bonus riddle, as an avid poker player, I knew the answer immediately. First, notice that the chances of holding a pair other than queens or 5s is the exact same in either scenario - so we can discount them completely, as if the deck only had 8 cards, 4 Queens and 4 5s. When Amy has 2 Queens, there is 1 possible combination of pocket queens left and 6 possible combinations of pocket 5s, making a total of 7. When Amy has 1 Queen and 1 5, there are 3 possible combinations of pocket 5s and 3 combinations of pockets Queens, making a total of 6. Therefore, it's more likely you have a pair when Amy has a pair of Queens than when she has a Queen and a 5.
Am i the only one who immediately thought the 10% chance of being wrong is obviously bigger than the 1% chance of being right and saw the scam straight away
No I was shaking my head as soon as he started framing the question and I still thought we were only dealing with 100 rocks let alone 1000. Terrible odds clearly not written for gamblers.
This was the first puzzle I've been able to solve in full iirc (of the ones not solvable by a logic table, which I've been able to do pretty easily and I think pretty much anyone could eventually).
Bonus Riddle solution: Total no of pairs=78 No of pairs left in scenario 1= 73 No of pairs left in scenario 2=72 So there is a higher probability of getting a pair in scenario 1
(Thanks to "yoshimo" for pointing out my error). Too long and not enough time to read? The first senario gives you slightly better odds of getting a pair (5.959% against 5.878%). THE EXPLANATION. A deck of playing cards contains 52 cards (this excludes the Jokers and Tarot trumps). These can be sorted in four stacks of thirteen cards (Hart, Spade, Clover, Diamond); or in thirteen sets of four (ace, 1, 2, 3, ect.). One card can make twelve different kinds of sets (4 x 3 = 12). Meaning that the thirteen sets of cards combined gives us 156 available sets (13 x 12 = 156). The chance the first card you draw can make a pair is always 100 %. SENARIO ONE. Our friend Amy has drawn two cards, the Queen of Harts and the Queen of Diamonds. The only way you can make a set with a Queen now is to draw the only other Queen (2 x 1 = 2). The other 12 sets aren't affected (12 x 12 = 144). This leaves us with 50 cards and 146 available sets. What is the chance you'll draw a NEW card (not a Queen)? 48 / 50 = 0.96 (96 %). And what is the chance you'll draw the SAME card (a Queen)? 2 / 50 = 0.04 (4 %). Either way, this leaves us with 49 cards. In the case of a new card, this means there are three cards left to match to. In the case you have drawn a Queen, there is only one card left. You drew a NEW card. What are your odds to get a pair? 3 / 49 = 0.061224489 (6.1224489 %). And your odds to get a card that doesn't match? 46 / 49 = 0.93877551 (93.877551 %). You drew a QUEEN. What are your odds to get a pair? 1 / 49 = 0.020408163 (2.0408163 %). And your odds to get a card that doesn't match? 48 / 49 = 0.979591836 (97.9591836 %). RESULTS. Your chance of getting a pair in this situation. (0.96 x 0.061224489) + (0.04 x 0.020408163) = 0.059591835 (5.959183596 %). Your chance of NOT getting a pair in this situation. (0.96 x 0.93877551) + (0.04 x 0.979591836) = 0.940408163 (94.0408163 %). SENARIO TWO. Our friend Amy has drawn two cards, the Queen of Harts and Fifth of Spades. This changes the odds of two sets - the Queen and five - and not one like in the previous senario. The Queen and fives can now only make 6 sets (3 x 2 = 6) each. The other 11 sets aren't affected (11 x 12 = 132). This leaves us with 50 cards and 144 available sets. Which is, to be honest, not that much of a difference. The chance you draw a NEW card changes slightly. Since Amy has drawn two different cards - which both leave three of their sets in the deck - we can't draw those six cards. 44 / 50 = 0.88 (88 %). And the chance we draw the same card (either a Queen or five)? 6 / 50 = 0.12 (12 %). One card drawn leaves us with 49, in case you can't count :P You have drawn a NEW card. What is the chance the next card will match? 3 / 49 = 0.061224489 (6.12244898 %). And the chance it won't match? 46 / 49 = 0.93877551 (93.87755102 %). You have drawn either a QUEEN or FIVE. Because Amy also has both a Queen and a Five, there are only two of each left in the deck. (Thanks again yoshimo). What is the chance you will get a match? 2 / 49 = 0.040816326 (4.081632653 %). And the chance the card won't match? 47 / 49 = 0.959183673 (95.91836735%). RESULTS. Meaning in this situation, your chance to gather a pair is: (0.88 x 0.061224489) + (0.12 x 0.040816326) = 0.058775509 (5.877550944%). The chance you won't get a pair however, is: (0.88 x 0.93877551) + (0.12 x 0.959183673) = 0.941224489 (94.12244896%). CONCLUSION You probably won't get a pair either way and Amy most likely cheated in the first senario.
The flaw in this solution is with your probabilities for drawing a queen or a five in the second scenario. It is true that the chance of drawing a queen or a five is 6 out of 50; there are 3 queens and 3 fives in the deck. However, it is important to realize that the drawn card cannot change its value. Once it is drawn, it is that value and cannot change. This means that the drawn card has only 2 out of 49 cards with which it can be paired (1 is owned by Amy; 1 is already in your hand). This means that the probability of the two drawn cards being a pair in the second scenario is about (0.88 * 0.0612) + (0.12 * 0.0408) ≈ 0.0588 (5.88%) So taking all of the other information (which is correct; I checked), it is actually Scenario 1 which has the better odds. Also, probably not a great idea to insult the reader before even getting to the explanation. Have some respect!
Oh snap, you are right about that! I just really wanted that senario 2 was the right one, you know? So I guess that's why I missed that (or because it has been 4 years since doing any kind of math :P ). Glad to know the rest was correct, though. Also, I didn't mean to insult but if someone doesn't want to read the math and only the answer, which I put above the whole explanation for their convenience, I think it is justified to call that person lazy (or at least, uninterested). Honestly, that TL;DR is saying basically the same thing. But since you pointed out my error I will edit it out as a thank you. Thanks for pointing out the error.
Oh jeez, I didn't think about it that way at all, lol. I'm not very good at this kinda thing yet, so my logic was probably plenty faulty, though I did conclude the first scenario was more advantageous to you. I considered just the Queens and 5s and the pairs available using just those suits. Scenario one can give you *Q♠ Q♣, 5♣ 5♠, 5♥ 5♦, 5♠ 5♦, 5♣ 5♦, 5♣ 5♥, 5♠ 5♥,* Q♠ 5♣, Q♠ 5♠, Q♠5♥, Q♠ 5♦, Q♣ 5♣, Q♣ 5♠, Q♣ 5♥, or Q♣ 5♦. Seven out of the fifteen combinations are pairs. Scenario two can give you *Q♠ Q♣, Q♠ Q♦, Q♣ Q♦,* Q♦ 5♣, Q♦ 5♥, Q♦ 5♦, *5♥ 5♦, 5♣ 5♦, 5♣ 5♥,* Q♠ 5♣, Q♠5♥, Q♠ 5♦, Q♣ 5♣, Q♣ 5♥, or Q♣ 5♦. Only six of the fifteen combinations are pairs.
For the last card thingy, I think it's the first one. In scenario 1, a pair of card was made. Which means that out of 52 playing cards, those 2 isn't counted anymore. It means that for the 1st scenario, it would be the chance to have a pair of card out of 50 cards. In scenario 2, the cards were different. While taking 2 cards still reduces the amount of card to 50 (just like the 1st one), it also increases the chance of not getting the same pair. Those 2 cards are different, meaning that their pairs would still be in the rest of 50 cards. While on 1st scenario the ratio of the card pairs is 25 : 25 (or 1 : 1), in the 2nd scenario it would be 27 : 23, which increases the chance of not getting the same cards. While it doesn't seem like much and it's still gonna be hard to get a pair of same cards anyway, it still affects the probability.
4:35 Answer: If your opponent has a pair, as the difference is the number of QQ and 55 in both scenarios. if A=QQ, then You=1xQQ + 6x55, whereas if A=Q5, you=3xQQ+3x55
@@lunarao That's true but the more times you shoot it the more certain you can be that it actually is unobtainium. but regardless this trade is bad because you don't know how much of the rock is actually unobtainium and how much consist of other minerals.
Perfection is a matter of perspective; it depends on what you seek to achieve. Sure, if you're trying for a 100% accurate device, than achieving that would be perfection. But if you want a device that's slightly inaccurate, to try and scam people out of their money, then that would be perfect for your purposes.
Bonus riddle solution It's more likely when she has 2 queens. That gets rid of only 1 pair. When she has a queen and a five, that's 2 cards that can't be paired up.
No its not, i think. The probability of getting a pair when theres 3 card that would make a pair is up. Yeah theres probabilty to take the lone card but its less probable, and remember that you have 6 card with more ptobability for a pair and only 2 lone cards that would screw your chances.
There are (48*3+2*1)/2!=73 ways to make a pair if Amy has a pair. There are (44*3+6*2)/2!=72 ways to make a pair if Amy has a non-pair. So it is more likely that you'll have a pair if Amy has a pair, but only slightly. There are C(50,2) = (50*49)/2! = 1225 ways to choose any random two cards. So the odds of getting a pair if Amy has a pair is 5.96% vs. 5.88% if she doesn't have a pair.
Steven Chabot wouldn’t even do it with out all the math it’s obviously too unreliable the video itself tries to trick us by saying “it works most of the time”
Steven Chabot see it's funny. if tricky Joe scams me (the miner) i scam you (yokel at market). no matter whether the rock is real deal or not, i still make a profit off morons like you. Jerry.
Scenario 1 right? As there is only one pair that yours cannot be while in scenario 2 there are 2 sets of pairs you can't have. This wouldn't make that much difference though??
If discrete math taught me anything, it is to use Bayes Theorem, which tells me the chances of this being a good deal is ~.1% Edit: ok watching the video, used a different form probability and am now wondering why Bayes Theorem doesn't work. Used %chance any rock is unobtanium for branch 1 and %chance detector resulted in positive or negative reading for branch 2. Tell me if you get something else.
RanDumSocksGames Messed uo my last answer. Here is the correct one. Let U be the event that a rock is unobtainium, and let D be the event that the detector goes off. P(U|D)=P(D|U)*P(U)/[P(D|U)*P(U)+P(D|~U)*P(~U)]=1*.01/(1*.01+.1*.99)=.0917 or approximately 10% chance that rock is unobtainium given that the detector went off.
You’re forgetting expected value. If the rock was worth $2,223, then it would be worth the trade for $200. Why? Let’s examine an extreme case. Let’s say the rock is worth $1B (billion), and Tricky Joe wants to sell it for $1, then it would be worth taking the 91% chance of it not being unobtanium, because there’s a 9% chance you’ll walk away with $1B minus the $1 that you paid for the rock. So what’s the formula? The expected payout (the price of the rock X minus the $200 cost of the rock) times the probability that you will get a payout (9%) has to be greater than what you are paying for the rock ($200) times the probability that you will not get a payout (91%) (X-$200)*0.09>$200*0.91 X*0.09>$200 X>$2,222 So in this video they said that the cost was $200 and the payout was $1,000. That isn’t worth it. What they failed to mention is that if the cost was $90 or less, or if the payout was greater than $2,222.22, then it would be worth trading.
Bayes Theorem is P(A|B) = P(B|A) * P(A) / P(B). A is unobtainium. B is a positive reading. B happens 100% of the time given real unobtainium so P(B|A) = 1. 1% of all rocks are unobtainium so P(A) is 0.01. B happens 100% on the 1% unobtainium and 10% on the other 99%, so for any rock, P(B) = (1*0.01) + (0.1*0.99) = 0.109. => P(A|B) = 1 * 0.01 / 0.109 = 0.0917.
I finally got it right!!! I watched this in the past and didn't remember, but recently I've watched a medical video about how the relatively rare occurrence of the corona virus in a population makes the low false negative rate in tests exorbitant and the tests unreliable
Similarly, there was a scare some time back claiming that eating more than a certain amount of red meat each week increases the risk of a particular form of cancer by 30%, but what they didn't tell you was that that that form is so rare anyway that even a 30% increase doesn't actually make that big a difference!
My immediate thought: "SHOULD YOU MAKE THE TRADE?" "Well, of course not, otherwise this wouldn't be an interesting problem and you wouldn't have asked in the first place." But yeah, I did the math anyway XD
Tricky Joe is the expert of this device. He would know that it's just likely a false positive but he didn't test it multiple times, so perhaps it was his plan all along
Bonus riddle: There are 78 possible pairs (13 card values and each has 4 card types means there are 6 possible pairs for each card value, 13*6 = 78). Let X be number of possible combinations of 2 cards. There are now 50 cards total (she already has 2 cards and we use the same deck), so X will be a pretty big number, the precise number is not needed. In scenario A: She already has a pair, which means there are only 2 cards of that value left in the deck, therefore we have 73 pairs to choose from (78-5=73). In scenario B: She has 2 cards of different values, which means there are 3 cards of each specific value left. This removes 3 possible pairs from each value (3*2=6 pairs), therefore we have only 72 pairs to choose from. Result: Divide the numbers by X and compare them: 73/X > 72/X Probability of having a pair is bigger in scenario A.
It’s just common sense though, if she has just one pair that means more pairs for you but if she has different cards then you wouldn’t be able to get pairs of those cards.
For the Amy riddle, sorry for not showing my working: Scenario 1 gives you a 5.96% chance of receiving a pair. Scenario 2 gives you a 5.88% chance of receiving a pair. The answer is Scenario 1.
the answer is obvious, and i don't even need to hit Atlantic City to know. don't fuckin hit if the house has 20. scenario 2 is best cuz house has 15, so they gotta hit, chances of bust are high at that point. that's where counting helps.
Just hijacking your comment to show the work since we got the same answer. scenario 1 gives (4/50)*(3/49)*(12)+(2/50)*(1/49) = 5.96% scenario 2 gives (4/50)*(3/49)*(11)+(3/50)*(2/49)*2 = 5.88% looking at the chance of getting a pair from cards not held by Amy first card is 4/50 second is 3/49. In scenario 1 that's 12 different ranks then calculating the odd of getting the other pair of queens in 2/50 and 1/49. In scenario 2 there are 11 different ranks for the 4/50 and 3/49 draws then for 5 and queen 3/50 and 2/49.
![Seungmin "그렇게, 천천히, 우리(As we are)" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/kAzmhLHePqU/mqdefault.jpg)
Sign up for free at brilliant.org/TedEd/, and Brilliant will email you the solution to the bonus riddle! Also, the first 833 of you who sign up for a PREMIUM subscription will get 20% off the annual fee. Riddle on, friends!
TED-Ed thanks😀😁😂
Thx
I love u
Not a fan of having to sign up to get the anset.
Keep up the good work!
I really enjoy this series :-)
Little tip for those who don't want to pay in order to now the answer: it's number one. Why you ask? This is due to the one pair revealed. If you look at the second possibility (two different cards) you will see that this lowers your chance to get a pair on your own. But here is the twist: the first one lowers yours as well but only from (let's say 1-ass) 14 possibilitys to 13. The second however lowers your chance from 14 to 12 due to the cards getting out of the deck.
"We just found this really rare stone. What should we call it?"
"Can'touchisium"
DU NUN NUN NU NU NU NU NU
CANT TOUCH THISSSSS
DU NUN NUN NU NU NU NU NU
You can't touch this
You can't touch this
You can't touch this
You can't touch this
My, my, my my music hits me so hard
Makes me say, "Oh my Lord"
Thank you for blessin' me
With a mind to rhyme and two hype feet
It feels good, when you know you're down
A super dope homeboy from the Oaktown
And I'm known as such
And this is a beat, uh, you can't touch
I told you homeboy (You can't touch this)
Yeah, that's how we livin' and ya know (You can't touch this)
Look in my eyes, man (You can't touch this)
Yo, let me bust the funky lyrics (You can't touch this)
Fresh new kicks, and pants
You gotta like that, now you know you wanna dance
So move outta yo seat
And get a fly girl and catch this beat
While it's rollin', hold on!
Pump a little bit and let 'em know it's goin' on
Like that, like that
Cold on a mission, so fall on back
Let 'em know, that you're too much
And this is a beat, uh, they can't touch
Yo, I told you (you can't touch this)
Why you standin' there, man? (You can't touch this)
Yo, sound the bell, school's in, sucka (You can't touch this)
Give me a song, or rhythm
Makin' 'em sweat, that's what I'm givin' 'em
Now, they know
You talkin' 'bout the Hammer, you're talkin' 'bout a show
That's hyped, and tight
Singers are sweatin', so pass them a wipe
Or a tape, to learn
What's it gonna take in the 90's to burn
The charts? Legit
Either work hard or you might as well quit
That's word because you know (You can't touch this)
(You can't touch this)
Break it down!
Stop, Hammer time!
Go with the flow, it is said
That if you can't groove to this then you probably are dead
So wave yo hands in the air
Bust a few moves, run your fingers through your hair
This is it, for a winner
Dance to this and you're gonna get thinner
Move, slide your rump
Just for a minute let's all do the bump
(Bump, bump, bump)
Yeah (You can't touch this)
Look, man (You can't touch this)
You better get hype, boy, 'cause you know you can't (You can't touch this)
Ring the bell, school's back in
Break it down!
Stop, Hammer time!
You can't touch this
You can't touch this
You can't touch this
Break it down!
Stop, Hammer time!
Every time you see me, the Hammer's just so hyped
I'm dope on the floor and I'm magic on the mic
Now why would I ever stop doin' this?
With others makin' records that just don't hit
I've toured around the world, from London to the Bay
It's "Hammer, go Hammer, MC Hammer, yo Hammer"
And the rest can go and play
You can't touch this
You can't touch this
You can't touch this
Yeah (You can't touch this)
I told you (Can't touch this)
Too hot (Can't touch this)
Yo, we outta here (Can't touch this)
what is this thread? I don't get it
I won't consider doing business with someone named Tricky Joe.
Rishabh Kumar "I don't trust like that"
Κοκκινης Eric Andre!
Rishabh Kumar true
lmao! True!
Or honest joe
I've played enough Pokémon in my days to know that even a 90% chance are terrible odds.
Want some 90% real water
@@danielgrace4283 nein, that 10% could be cyanide
Oh once I got paralysed 10 times in a row from a shinx in pearl, so I couldn’t move, then I missed razor leaf (95% accurate) 3 times. I wanted to train grass type that I can’t remember.
Haha
90% donut
10% poison
The problem with scanning the same rock multiple times is that we don’t know whether the reason it has a 10% false positive rate is due to a stochastic error (extra scans could give different results) or a systematic error (extra scans give the same results). For example, if the device is sensitive to another mineral that exists in 10% of rocks, extra scans won’t fix it.
Or think it more simply: if there is element in the worthless rock trigger a positive reading from the detector, it will ALWAYS report the false positive.
That is very true, and from what another Commenter has pointed out Tricky Joe probably was aware of the fact that the rock most likely was worthless hence he was willing to trade it for a mere $200, like that Commenter said, neither a rational or irrational person would sell something they thought was more valuable then what they were selling it for for what they are selling it for.
@@dalooneylass5561 what else would you expect from a guy named Tricky Joe?
@@dalooneylass5561 -- if Tricky Joe had some method for being aware of whether or not the rock was worthless, he wouldn't need to build a detector. By the rules of the puzzle he couldn't know -- and he doesn't have to. We can calculate the _average value_ of the rock the detector gives a positive result for, and it's (10*$1000 + 99* $0)/109, which equals about $91.74. If Tricky Joe is making this deal frequently, then he's always going to come out on top so long as he asks for _more_ than this amount. If he asks for _less_, then the trade starts making sense, as over a longer number of trades you'll eventually make money (as the losses from the times it was a false positive are vastly outweighed by the gains from those times when you make $1000 from it being a true positive).
@@dalooneylass5561 *than
*_Me_* : No because my *"friend"* named _tricky joe_ wouldn't sell me an authentic, and genuine ore for a fifth of its price.
Has none of you heard of CRAAAZZYYYY REDD?
_tricky joeee_
Except neither of you know if it is yet
Also Me: now if you excuse me I have to now go on a blind date with Serial killer Susie and invest in a business with Money laundering Lenny
me: no cuz i don’t have friends in the first place
Me: Alright, im gonna watch this video to see if i can solve it.
Me: *fails miserably*
Also me: ooh, another one
*cycle repeats*
The vicious cycle😂😂..samehere, my friend
Hey, at least you tried.😁
Better trying than zero.😁😄
Me: watches and doesn’t try to solve it
Solution: Bla bla bla
Me: DAMNIT I COULD’VE DONE THAT! I’LL NEVER WATCH LNE OF THESE AGAIN!
5 seconds later...
OOH ANOTHER RIDDLE VIDEO!
REPEAT UNTIL I DIE
That's how we learn xxx
Same-except I actually get half of the,
"Would you make the trade?"
*No because I have trust issues*
Underrated
Brilliant
Me 2
On mah phone
Trust tissues, trusted you, but betrayed
What happened?
Okay, Tricky Joe is a trickster, but we have to give him credit for his invention. It is actually very useful.
Having to examine 109 rocks is much better than having to examine all 1000 rocks for unobitanium
Good idea, bad pricing. At 60 dollars price, buyer can still make 30 dollars profit.
Now that you mention it, you're right! Even if it's not a guarantee that the rock that's detected is unobtainium, the fact that it can even detect it at all with relatively decent accuracy allows you to narrow your search quite significantly.
With that in mind, it might actually be better to do a slightly different deal: Increase the offer to either $300 or $400, but include the device itself along side the rock. By having the device involved with the equation, in the short term you'll likely take a bigger hit, but in the long term it would actually be more beneficial!
There's a reason his name is trickey Joe. Most likely he built a device that always gives a false positive but told you the story of a 10% chance.
Otherwise why would it react with literally the first rock he tested.
Doing a second round of testing would make it even more accurate, you would need to scan the 109 rocks, get positives on 10 of them and false positives on another 10. Two rounds of testing and you have narrowed the search down from 1000 rocks with 1% accuracy to 20 rocks with 50%.
@@13g0man Of course, that assumes that the cause of the false positive is entirely random. If it's the result of some other mineral or a structure of minerals that the detector cannot tell from Unobtanium, it's still no better. If nothing else, those 109 rocks would let you figure out which is the case and refine the detector further.
Why you shouldn't take the deal:
*THE GUY SELLING YOU THE ORE IS NAMED TRICKY JOE. THINK ABOUT IT.*
Grandom your profile pic is outdated
I thought Huberdale grew up
Hubert cumberdale`
Hi Hubert
Yeah, the name made me very suspicious.
Also he made it and than was like hey wannna buy this 1000 rock for 200$
Step 1: Ask Tricky Joe: If I have green eyes, would you say Ozo?
Step 2: Ask the unobtainium to leave the island
why are there so many comments that pop up using ''the green eye riddle'' and when i comment them it gets no likes
Only true fans will get that reference
But are there indestructible robot ants in there?
@@had0j wait they will come
Avatar reference??
For the bonus riddle, picking scenario 1 is slightly better because there’s more pair combinations with 2 and 4 cards left (in the example there’s 2 queens and all 4 5’s left, giving 7 combinations) than 3 queens and 3 5’s left (6 combinations). This is because the rate at which n cards gives x combinations also changes (2 cards have 1 possible combinations, 3 have 3, 4 have 6, and so on, basically triangular numbers)
awesome bro
But If you look at it another way, Isn't It less probable to draw two pairs in exactly two draws. So shouldn't Scenario two be the better bet?
@@jatinchandwani8427 we already know / assume that she has a pair or not, so the question isn’t “is drawing two pairs more likely than drawing a pair and a non-pair?” It’s “is drawing a pair given that a pair has been drawn more likely than not drawing a pair given a pair has been drawn?”
@@jatinchandwani8427A ) 2Qs
Theres 12 packs of 4 and a pair of Qs left in the deck, so the probability of drawing a pair is:
12× (4/50 × 3/49) + (2/50 × 1/49) = 5,959%
B) Q + 5:
Theres 11 packs of 4 and 2 packs of 3
So 11× (4/50×03/49) + (3/50×2/49) × 2 = 5,878%
To put it another way, at the start there are 78 possible pairs in the deck, that is 6 pairs for each of the 13 card values. This is because there are 6 ways the suits can pair up: (♣♠), (♣♥), (♣♦), (♠♥), (♠♦), (♥♦).
So with that in mind, we look at how many pair combinations are left for each scenario.
Scenario 1 eliminates (Q♥) and (Q♦) leaving the only pair option (Q♣ Q♠). This means 5 options were eliminated
Scenario 2 eliminates (Q♥) and (5♠). From the Queens this leaves only (Q♦Q♣), (Q♦Q♠), and (Q♣Q♠).
From the 5's this leaves only (5♣5♦), (5♣5♥), and (5♥5♦). This eliminates 3 pairs from each Queens and 5's for a total of 6.
So in Scenario 1 the deck will have 73 pairs left for you to hit while in Scenario 2 there will be only 72 pairs left for you to hit.
this one doesn't seem nearly as hard as some of ted-ed's other riddles
John Michael Yeah, that’s what I felt!
Because of the 2 narrowed choice you have to take. Most if the hard riddles are full of options and even some of em have 2 or more answers.
This is also a probably based on mathematic probability, something you’re very likely to learn about in school.
Its just conditional probabilities which people should’ve learned in highschool:/
Because this was a math problem, not a logic problem. We understand probability and money better than what “Ozo” means, or how to create a statement both true and false without saying a paradoxical statement. 😅
*Why is it so hard to obtain Unobtanium*
because is unabtaunible
Because it cannot be obtained
Because ted had said so
@_Bob McCoy, how badly did it hurt that people didnt seem to understand that joke?
So, is there a punchline or is that it
We just got bamboozled by Tricky Joe.
Thrill Cosby Do you watch Jacksfilms or do you just happen to use that expression? I'm curious
I just happen to use that expression.
Thrill Cosby just shoot the right rocks 10 thousand times and u see if it is real or fake
This seems to be the most obvious way out of the situation.
"We" I think you mean "I"
1: confirm you have green eyes
2: ask tricky joe to leave
3: figure out how to escape the white void
Instead conform tricky Joe has green eyes, then ask him to leave
1.-Confirm tricky joe doesn’t have green eyes
2.-Throw him into the volcano
DONT YOU DARE WANT TO LEAVE THE WHITE VOID
then you go to the third position from disky disk A and select the Noether 9000 and buttons D and E to destroy the laser robot ants and cut the antidote into fifths and use parity to find out what place your in and what hat you are currently wearing with a 10% false positive rate then you get silver by moving counterclockwise then go to locker 64 to find out who ate which crystal then go in groups and add 5 and 7 and square root to escape with 60 bees and at least 16/30 rubies (marking the heavier ones with a +) on planet 3 and after all of that you have to question my sanity, and the answer is ozo.
@@mathguy37 umm, you also have to figure out that the German stole the prized fish, and that you need 6 chrono nodules to escape the white void
Honestly, the correct answer to this riddle seemed obvious and intuitive so I thought I was wrong
Happens to me sometimes
It was obvious to me too, but only because I've seen similar problems before.
It didn't seem like a good deal to me either. Glad I'm not the only one.
That happens too often to me, but this time i actually got it right-
Same
But it doesn’t state that the device has a limited use. Theoretically, you can check those 109 rocks again. And again. And again.
In the real world I would do that
Striker The idea was to demonstrate the risk of false positive in e.g. medical screening, and the concept of conditional probability. Don't take the riddle so literally, it was just an example
Exantius E I know but my luck would tell me if I repeat this again and again, I have a high chance of getting a false positive based on my win/lose ratio in games
It doesn't mean that the detector has 90% chance of returning a negative for the same rock. 1 out of 10 rocks will give you a false positive, and that rock will give you a false positive no matter how many times you test it. At least that's how I interpret the scenario. Perhaps the detector is measuring something like radioactivity as an indicator of whether unobtanium present. A false positive would be a rock that shows high radioactivity due to some substance that is not unobtanium.
Pretty sure the $200 dollar offer stands as is, Striker. It's Joe's detector, after all.
I came up with a modification to the riddle. imagine you COULD zap it multiple times but each time you did, it lost 10% of it's current value
how many times would you zap it
"Well this video is called false positive so I;m gonna go with no."
Context clues are the best.
"Tricky Joe" didn't give it away? 😋
“Should you make the trade?”
Me: ozo
Or ulu?
If i asked "ozo means yes" would you say ulu?
@@bakingcreative2414 ozo
Ulu
@@bakingcreative2414 “helicopter”
For the card riddle: It is option one (ca. 5.959% chance of success). Option B has ca. 5.878% chance of success. The reason is that in the first option you are only decreasing your chances of getting a pair of queens but still have the same chance of getting any other pair. In the second option your chances of getting a pair of queens and a pair of fives are decreased. Calculation:
Option A: [12*(4/50)*(3/49)]+[(2/50)*(1/49)]
Option B: [11* (4/50)*(3/49)]+[2*(3/50)*(2/49)]
Or, option 1 there are 73 pairs and options 2 there are only 72 pairs
George Snyder Harvard wants to know your location
I appreciate the comment was looking for the answer
You could've just said that there are less pairs you can get in option b because Amy has 2 cards that you can make a pair with...
There are four queens (and four fives) in total
So the second option you have higher chances of getting a pair than on the first one.
You are more likely to draw a five (or a queen) out of three remaining fives (or queens) out of 62 cards than you are to draw a queen out of two remaining queens out of 62 cards
Me in math class: I won’t use all this other than grocery shopping
Me after watching couple of TED-Ed videos: nvm...
- reads title - probably not but I'll still learn something so might as well click on it
Every time..
Actually this was the first one I was able to solve
yeah it was pretty easy tbh
Easy
This video doesn't just do a good job of explaining false positives; of all videos, this one does the best job of connecting the idea back to the formula. (3:23)
First riddle I've ever solved on this channel under a minute. Highschool's finally doing its job
idk this one was very easy though haha
ikr
+TheWildCard Good Pun😂😂👍
idk haha you look so proud for solving a junior school riddle
Aditya Waghmare what pun?
Solution: Shoot the mineral 10 times. See whether it goes off all 10 times. Crash the price of unobtainium and get rich!
Why would repeat shots of the same rock give differed results?
@@youtubeuniversity3638 he's assuming that the gun gives a false positive due to the gun failing on an individual shot basics, rather than say the false positive coming from the rock mimicking the wavelength of unobtainium
@@bobbobson3098 And I'm trying to ask why he would make the assumption he is, as opposed to the opposite.
@@youtubeuniversity3638 right, so the assumption he makes is this: the machine is 90% accurate, shot to shot, regardless of other factors.
HOWEVER, it is completely possible that the inaccuratracy is caused by the fact that other components of the rock cause it to read the same on the detector as unobtainium.
So, lets say the reader detects based on wavelength, and false source produces a wave close to that off unobtainium, then shotting it 10 times isn't going to improve your chances of knowing it isn't what you want just because it read 9 out of 10 times.
This is a good question though! :)
@@bobbobson3098 So, assuning that we all understand *what* he assumed, *why* dod he assime it, as opposed to assuming something else?
Answer: No.
Reason: His name is Tricky Joe.
Nate well said
😂😂😂😂
i was just thinking that
Nate hahaha
I already was looking for this comment
"At first, this may seem like a good deal" It doesn't.
"But why did it seem like a sure bet?" It didn't.
My thoughts exactly. I immediately thought "but there's 10 rocks that don't have anything that will trigger it, so the chance of finding the correct one is slim"
For the bonus riddle, my hypothesis is that the likelihood of you having a pair is greater in scenario 1. In scenario 2, the likelihood of having a pair decreases because there are two fewer options of two cards matching up, whereas in scenario 1, the likelihood of having a pair would be only impacted if you had a queen in your hand.
Yeah I came to the same conclusion
I have written a comment stating that there are 78 possible pair combinations in a deck of 52 cards. Scenario 1 eliminates 5 combinations while Scenario 2 eliminates 6. Therefore #1 is more likely that u have a pair with 73 combinations vs 72 in #2
Its actually really close:
The probability for scenario 1 is (73/1225)
The probability for scenario 2 is (72/1225)
But i am only 90% sure ;)
@@amna7491 it's the right answer, got the same calculation :)
@@amna7491 if you use baye's theorem will change the answer to the scenario 2, because you already draw a pair and the probability to draw a second pair in a row will deacrese substantially
Plot twist: The device gets it wrong 100% of the time and its all a scam
If its wrong 100% of the time then you would know if its unobtainium or not 100% of the time because if it does not beep it means that is unobtainium because its wrong and it didnt beep and if it beeps you can just disregard it and ignore it
@@lukastefanovic5378 Thank you smart person.
That is what it would seem... he is called "Tricky Joe" after all!
Yea
Then why would tricky Joe buy it for 200$?
As for the bonus riddle:
Let's examine the probability of obtaining a pair in either of the cases-
1) When Amy has 2 Queens -
Out of the 50 remaining cards, we can either firstly choose any 2 pair cards from any of the other other 12 quadruples, i.e. 12*(4C2) ways. To get a queen pair there is only one way as there are only 2 queens available so they have to be picked anyway.
Implies: total number of ways in this case is 12*(4C2) + 1 = 73
2) When Amy has a Queen and a 5 -
Out of the 50 cards, we can either firstly choose any 2 cards from any of the other 11 same-numbered quadruples, i.e. 11*(4C2) ways. To get a Queen pair or a 5 pair, we have have 2 choices (Queen/5) and then we gotta choose 2 out the available 3 cards.
Implies: total number of ways is 11*(4C2) + 2*(3C2) = 66 + 6 = 72
73 > 72, thus, we have a very slightly greater chance of getting a pair in the first case.
Since the first case removes two queens from the deck, this reduces your chances of getting a pair if either of your cards is a queen, but also reduces your chances of getting a queen in the first place. The second case creates two bad cards, since now both queens and fives have a reduced chance of being pairs.
@@TheGregamonster yours doesn't answer which one is worse
@@randomentirely1006 It does. Every card Amy has is less likely to be a card you have. If Amy has a pair, that's only one type of card that's harder to match, but if she has two different cards then that's now two cards that will be harder to match.
@@TheGregamonster that doesn't mean it outweighs the other's probability. Just cause you have two things doesn't make it better than 1
@@randomentirely1006 Right. Two things would be worse, because that's more chances to draw a card with a lower chance of forming a pair.
I just said no because Tricky Joe is Tricky Joe and seems like he's pretty shady. None of the fancy math stuff lol.
I just went with "Better safe then sorry."
They would have been better to just call him Joe. the shady looking appearance plus the name Tricky Joe immediately put me off. It's a fun thought experiment, but TedEd biased it here, which kind of sucks.
Lela Hamilfan same
would u trust jack from minecraft story mode tho?
Yes
i won't take anything from joe.he looks damn suspicious..
lmao😂😂
Well he is *Tricky* Joe
True
Love how the mine is just a large circular room with lone rocks sitting there
Through amazing deductive reasoning and actual dedication, i finally got one ted ed riddle right
I said no because tricky joe’s name alone is untrustworthy 😂😂😂
Joe mama
or he derived the nickname from a history of electronic skills, geological knowledge, and general wit
@@flamingart55 no
COPIED COMMENT!
I said no because a rational actor would not offer to sell it for $200 if he thought it was worth more than $200. Similarly, a deceitful actor would not offer to sell it for $200 if he thought it was worth more than $200.
Yeah, no.
I would never trust someone named “Tricky Joe”
Like seriously.
Yeaaa i would. Never
@Max Belcher 😂
@@iamabeautifulpensi _same_
"Stick around for an extra riddle" -TedEd
This is how Bill Cosby got me
Solution to the extra riddle is simple: your chance of a pair is higher if Amy has a pair of queens because neither of the queen cards can be one of your cards. If a queen was one of your cards you cannot have you cannot have a pair.
I was just going through the comments hoping someone came to the same solution as I did. Thank you.
Konseq there are 4 queens in a deck....
A- M- the chance of a pair is still higher based on my explaination.
For the playing card riddle i'd guess the two queens would make it more likely for you to get a pair. That makes it less likely to get a pair of Queens again, but a queen and another card make two different sets less likely.
edfreak9001 yeah that's what I was thinking :)
Yes. We can simplify the question by considering a deck of only Queens and Fives.
If Amy has a pair of Queens, the number possible pairs you can have is 1 + 4C2 = 7. 1 for another possible pair of Queens, and 4C2 for all the possible pairs of Fives.
If Amy has a Queen and a Five, the number possible pairs you can have is 3C2 + 3C2 = 6. 3C2 being the number of possible pairs from 3 Queens/Fives.
In other words, you are more likely to have a pair if Amy also has a pair.
edfreak9001 it's exactly the opposite
I agree and also think that’s the correct answer but until someone does the maths I wouldn’t be surprised if it’s equal. Because sure, the pair of queens only makes queens less likely and the two options make two less likely... but less is not quantitative... the less in the queens scenario is more sizeable. In one scenario there are only 2 queens less making that particular pair very unlikely but in the other scenario two numbers are cut to 3... there’s still a decent chance for a pair with three (although all in all the pair is likely to come from a number with 4 cards in either scenario)
//I realise this is written terribly
edfreak9001 Yes, the pairnof queens is better. I did the math:
It is the same if your first card is anything other than a queen or a 5, so just look at those sets of cards:
Scenario 1:
First card is a queen 2/50 times. 2nd card is a queen then 1 out of 49 times if the first one is a queen.
First card is a 5 4/50 times and 2nd card is also a 5 3/49 times. That's [(2*1)+(3*4)]/(49*50) or 14/2450
In scenario 2, you get a queen 3/50 cards and another queen 2/49 of the remaining cards, or 6/2450. It's the same for a 5, so that doubles to 12/2450. That's less tban 14/2450, so you are better off with the scenario where the other side draws a pair.
Unless you are trying to play head to head. Your odds of beating a pair of queens with 2 cards are way lower than your odds of beating a queen five off suit.
I thought there might be an equity break down of this problem as well.
If we measure the value of the rock to be 9% of 1000 then we could consider buying it for 90.
However if you made this trade every time over a long period, you would only break even since 89% of your rock is paid for by the other 9%.
So if you can haggle Joe below 90 you have a decent chance of profiting.
Vibranium’s weird, distant cousin
Natalia c yes because I could watch BlackPather... BUT NO ONE COULD WATCH AVATAR
Isn't Vibranium made with unobtainium?
As for the bonus riddle, as an avid poker player, I knew the answer immediately. First, notice that the chances of holding a pair other than queens or 5s is the exact same in either scenario - so we can discount them completely, as if the deck only had 8 cards, 4 Queens and 4 5s. When Amy has 2 Queens, there is 1 possible combination of pocket queens left and 6 possible combinations of pocket 5s, making a total of 7. When Amy has 1 Queen and 1 5, there are 3 possible combinations of pocket 5s and 3 combinations of pockets Queens, making a total of 6. Therefore, it's more likely you have a pair when Amy has a pair of Queens than when she has a Queen and a 5.
Just tell him that at least one unobtanium has green eyes
Why do I see this green eyes reference in these riddle videos?
@@sugar2000galaxy it's a reference to a different riddle
This is actually one that I was able to get the answer intuitively, so I was surprised when they said intuition would lead to the opposite conclusion.
*clap* BONUS *clap* RIDDLE
Isaac Rox thanks Brad
Bonus meme is dead😭😢
Wibe99 it’s back bro 😭
Thanks
Pewdipie
First riddle I’ve ever solved on the channel :D
you just gotta say no
Same!
Javan Chear give them a rest
sames
Am i the only one who immediately thought the 10% chance of being wrong is obviously bigger than the 1% chance of being right and saw the scam straight away
Sam Sihite no your not
No I was shaking my head as soon as he started framing the question and I still thought we were only dealing with 100 rocks let alone 1000. Terrible odds clearly not written for gamblers.
The second one was easy too. It's obviously scenario 1
Ikr! How is this a riddle?!
@@orderandkhos6269 I'm not even a gambler. I'm an eighth grader, and immediately this was easy. Same with the card one at the end.
I wish this is the kind of stuff we actually learn in school instead of the same thing every year
Why is Unobtanium so cheap if it's so rare?
a 1% chance of finding it doesn't seem that rare tbh
Because people have dedectors for it
Tyler Mustard no industrial uses for it. Gold is used in computers but platinum is needed for chemical processes. They are almost as equally rare.
Tyler Mustard maybe its the 1890
1% is about the rarity of copper in copper deposits. Copper. Gold is an order of magnitude more rare, IIRC.
Thank you Ted Ed I am learning alot from your puzzled videos. Love you and cant thank enough for the learning I am getting.
Just scan the rock again.
Not generally how a false positive works. i.e. the scanner may unintentionally detect other materials that occur in said rocks.
Ry B Finally! My initial reaction was "meh just zap the thing ten times and you are set". I was certain the comments would be flooding with this idea.
This was the first puzzle I've been able to solve in full iirc (of the ones not solvable by a logic table, which I've been able to do pretty easily and I think pretty much anyone could eventually).
I never get the ones solvable by a logic table... because I'm too lazy to write it out 😂
Bonus Riddle solution:
Total no of pairs=78
No of pairs left in scenario 1= 73
No of pairs left in scenario 2=72
So there is a higher probability of getting a pair in scenario 1
So... Fallacies...
Need a video on those.
Arif Good idea. But there are so many! How to choose?
Brenda Rua
Just a general video with few of the most prominent ones would be good, like "no true Scotsman" or "the fallacy fallacy".
Dont forget our friend "The Strawman"
Read David Kahneman's ''Thinking, Fast and Slow''
(Thanks to "yoshimo" for pointing out my error).
Too long and not enough time to read?
The first senario gives you slightly better odds of getting a pair (5.959% against 5.878%).
THE EXPLANATION.
A deck of playing cards contains 52 cards (this excludes the Jokers and Tarot trumps).
These can be sorted in four stacks of thirteen cards (Hart, Spade, Clover, Diamond); or in thirteen sets of four (ace, 1, 2, 3, ect.).
One card can make twelve different kinds of sets (4 x 3 = 12).
Meaning that the thirteen sets of cards combined gives us 156 available sets (13 x 12 = 156).
The chance the first card you draw can make a pair is always 100 %.
SENARIO ONE.
Our friend Amy has drawn two cards, the Queen of Harts and the Queen of Diamonds.
The only way you can make a set with a Queen now is to draw the only other Queen (2 x 1 = 2).
The other 12 sets aren't affected (12 x 12 = 144).
This leaves us with 50 cards and 146 available sets.
What is the chance you'll draw a NEW card (not a Queen)?
48 / 50 = 0.96 (96 %).
And what is the chance you'll draw the SAME card (a Queen)?
2 / 50 = 0.04 (4 %).
Either way, this leaves us with 49 cards.
In the case of a new card, this means there are three cards left to match to.
In the case you have drawn a Queen, there is only one card left.
You drew a NEW card.
What are your odds to get a pair?
3 / 49 = 0.061224489 (6.1224489 %).
And your odds to get a card that doesn't match?
46 / 49 = 0.93877551 (93.877551 %).
You drew a QUEEN.
What are your odds to get a pair?
1 / 49 = 0.020408163 (2.0408163 %).
And your odds to get a card that doesn't match?
48 / 49 = 0.979591836 (97.9591836 %).
RESULTS.
Your chance of getting a pair in this situation.
(0.96 x 0.061224489) + (0.04 x 0.020408163) = 0.059591835 (5.959183596 %).
Your chance of NOT getting a pair in this situation.
(0.96 x 0.93877551) + (0.04 x 0.979591836) = 0.940408163 (94.0408163 %).
SENARIO TWO.
Our friend Amy has drawn two cards, the Queen of Harts and Fifth of Spades.
This changes the odds of two sets - the Queen and five - and not one like in the previous senario.
The Queen and fives can now only make 6 sets (3 x 2 = 6) each.
The other 11 sets aren't affected (11 x 12 = 132).
This leaves us with 50 cards and 144 available sets.
Which is, to be honest, not that much of a difference.
The chance you draw a NEW card changes slightly.
Since Amy has drawn two different cards - which both leave three of their sets in the deck - we can't draw those six cards.
44 / 50 = 0.88 (88 %).
And the chance we draw the same card (either a Queen or five)?
6 / 50 = 0.12 (12 %).
One card drawn leaves us with 49, in case you can't count :P
You have drawn a NEW card.
What is the chance the next card will match?
3 / 49 = 0.061224489 (6.12244898 %).
And the chance it won't match?
46 / 49 = 0.93877551 (93.87755102 %).
You have drawn either a QUEEN or FIVE.
Because Amy also has both a Queen and a Five, there are only two of each left in the deck. (Thanks again yoshimo).
What is the chance you will get a match?
2 / 49 = 0.040816326 (4.081632653 %).
And the chance the card won't match?
47 / 49 = 0.959183673 (95.91836735%).
RESULTS.
Meaning in this situation, your chance to gather a pair is:
(0.88 x 0.061224489) + (0.12 x 0.040816326) = 0.058775509 (5.877550944%).
The chance you won't get a pair however, is:
(0.88 x 0.93877551) + (0.12 x 0.959183673) = 0.941224489 (94.12244896%).
CONCLUSION
You probably won't get a pair either way and Amy most likely cheated in the first senario.
The flaw in this solution is with your probabilities for drawing a queen or a five in the second scenario.
It is true that the chance of drawing a queen or a five is 6 out of 50; there are 3 queens and 3 fives in the deck. However, it is important to realize that the drawn card cannot change its value. Once it is drawn, it is that value and cannot change. This means that the drawn card has only 2 out of 49 cards with which it can be paired (1 is owned by Amy; 1 is already in your hand).
This means that the probability of the two drawn cards being a pair in the second scenario is about (0.88 * 0.0612) + (0.12 * 0.0408) ≈ 0.0588 (5.88%)
So taking all of the other information (which is correct; I checked), it is actually Scenario 1 which has the better odds.
Also, probably not a great idea to insult the reader before even getting to the explanation. Have some respect!
Oh snap, you are right about that! I just really wanted that senario 2 was the right one, you know? So I guess that's why I missed that (or because it has been 4 years since doing any kind of math :P ). Glad to know the rest was correct, though.
Also, I didn't mean to insult but if someone doesn't want to read the math and only the answer, which I put above the whole explanation for their convenience, I think it is justified to call that person lazy (or at least, uninterested). Honestly, that TL;DR is saying basically the same thing.
But since you pointed out my error I will edit it out as a thank you.
Thanks for pointing out the error.
Oh jeez, I didn't think about it that way at all, lol. I'm not very good at this kinda thing yet, so my logic was probably plenty faulty, though I did conclude the first scenario was more advantageous to you. I considered just the Queens and 5s and the pairs available using just those suits.
Scenario one can give you *Q♠ Q♣, 5♣ 5♠, 5♥ 5♦, 5♠ 5♦, 5♣ 5♦, 5♣ 5♥, 5♠ 5♥,* Q♠ 5♣, Q♠ 5♠, Q♠5♥, Q♠ 5♦, Q♣ 5♣, Q♣ 5♠, Q♣ 5♥, or Q♣ 5♦. Seven out of the fifteen combinations are pairs.
Scenario two can give you *Q♠ Q♣, Q♠ Q♦, Q♣ Q♦,* Q♦ 5♣, Q♦ 5♥, Q♦ 5♦, *5♥ 5♦, 5♣ 5♦, 5♣ 5♥,* Q♠ 5♣, Q♠5♥, Q♠ 5♦, Q♣ 5♣, Q♣ 5♥, or Q♣ 5♦. Only six of the fifteen combinations are pairs.
Actually it is Scenario
Literally no one:
My last brain cell: *explodes*
One of the only riddles I have ever solved from ted ed.
This one was way too easy... but I still learned something new so... thnx! Ted-ed
Ohh Really ?!
What's 2+2 kid
It's such a good feeling to finally find someone in the comments who agrees.
I agree also.
Tricky Joe is a man after my own heart. I love that guy!!!
TedEd: detailed scientific explanation of why it's a scam
Me, an intellectual: His name is Tricky Joe, I'm not buying what he's selling!
this is literally the only ted ed riddle i’ve gotten
Its always great to have a riddle that i can never solve.
The Channel You can solve anything if you just believe in yourself
John Its a great motivation anyways !!
Ted ed is my fav learing site🤑🤑🤑
Clearly not learing enough, though
For the last card thingy, I think it's the first one.
In scenario 1, a pair of card was made. Which means that out of 52 playing cards, those 2 isn't counted anymore. It means that for the 1st scenario, it would be the chance to have a pair of card out of 50 cards.
In scenario 2, the cards were different. While taking 2 cards still reduces the amount of card to 50 (just like the 1st one), it also increases the chance of not getting the same pair. Those 2 cards are different, meaning that their pairs would still be in the rest of 50 cards.
While on 1st scenario the ratio of the card pairs is 25 : 25 (or 1 : 1), in the 2nd scenario it would be 27 : 23, which increases the chance of not getting the same cards.
While it doesn't seem like much and it's still gonna be hard to get a pair of same cards anyway, it still affects the probability.
4:35 Answer: If your opponent has a pair, as the difference is the number of QQ and 55 in both scenarios. if A=QQ, then You=1xQQ + 6x55, whereas if A=Q5, you=3xQQ+3x55
Wow I acually got a riddle right for once in my intire life :D
TED-ED YOU ROCK!!
Edit:- Thankyou so much TED-ED for giving heart to my comment 💜💜
*b a d u m t s s*
T'Challa thanks for your worst comment
+Zubair Ahmed And the Shittiest Reply Award goes to Zubair Ahmed
Pun had better be intended
Nice.
OMG GUYS, Y'ALL ARE THE BEST!! PLEASE CONTINUE TO UPDATE!! ONE DAY WHEN I GET A JOB, I HOPE TO BE A CONTRIBUTING MEMBER OF THIS FAM #TEDEDLOVE
This is a great video explaining Bayes Theorem! Well-made example by Ted-Ed
The whole time I’m just thinking shoot the rock 10 times then
I've already ranted, so I'll run this quickly. Yes you can do that and have impractical odds, but it's not impossible.
The probability of you getting a false positive result is still 10% no matter how many time you shoot the rock.
BIG BRAIN
@@lunarao That's true but the more times you shoot it the more certain you can be that it actually is unobtainium. but regardless this trade is bad because you don't know how much of the rock is actually unobtainium and how much consist of other minerals.
@@uremomisepic But even so, it might be that there's some impurity in the rock that it ALSO detects and will consistently cause a false positive.
0:06
That's because it's called UNOBTAINium
3:38 me doing maths
Oooooof so true.
Ikr
"So why is this result so unexpected?"
Well because most people don't actually think
It really is not unexpected at all
YEAH I AM POSITIVE
I CAN SOLVE IT SURELY
EDIT :THNX FOR THE HEART
I AM A BIG FAN ☺️
No you can't
Aditya Waghmare oh
I ll try for sure
0:18 The unobtaniun detector he has been *_P E R F E C T I N G_* For months is finally done.
Perfection is a matter of perspective; it depends on what you seek to achieve. Sure, if you're trying for a 100% accurate device, than achieving that would be perfection.
But if you want a device that's slightly inaccurate, to try and scam people out of their money, then that would be perfect for your purposes.
Time to block Tricky Joe on Discord 😤😤😤
First ted ed riddle I solved, this one is defiantly way more simple than the others
Bonus riddle solution
It's more likely when she has 2 queens.
That gets rid of only 1 pair.
When she has a queen and a five, that's 2 cards that can't be paired up.
So a pair is 2 cards with the same color and number?
@@kele8559
Don't need to be the same color.
Just as long as the rank/number is the same.
No its not, i think. The probability of getting a pair when theres 3 card that would make a pair is up. Yeah theres probabilty to take the lone card but its less probable, and remember that you have 6 card with more ptobability for a pair and only 2 lone cards that would screw your chances.
There are (48*3+2*1)/2!=73 ways to make a pair if Amy has a pair. There are (44*3+6*2)/2!=72 ways to make a pair if Amy has a non-pair. So it is more likely that you'll have a pair if Amy has a pair, but only slightly.
There are C(50,2) = (50*49)/2! = 1225 ways to choose any random two cards. So the odds of getting a pair if Amy has a pair is 5.96% vs. 5.88% if she doesn't have a pair.
@@KSJR1000
Wasn't expecting the math behind the answer, but thanks.
Ok i admit Ted is smarter than me
We respectfully disagree. You can do it! :)
TED-Ed now i have goal
Everyone who has played XCOM2 will know that 90% chance means certain failure.
This is the first TED riddle I actually manage to correctly solve, yay!
Taking the rock was obviously a bad decision... but people fall for obvious scams every day
Steven Chabot TRICKY Joe
Steven Chabot wouldn’t even do it with out all the math it’s obviously too unreliable the video itself tries to trick us by saying “it works most of the time”
The video never tricks you.
No i just scan it 3000 times and if not real i get another one
Steven Chabot see it's funny. if tricky Joe scams me (the miner) i scam you (yokel at market). no matter whether the rock is real deal or not, i still make a profit off morons like you. Jerry.
I CAN’T BELIEVE I ACTUALLY GOT THE ANSWER RIGHT ON ONE OF THESE. so i guess learning probability wasn’t a waste after all. 😂
Wow you got a question with only two possible answers right. Well done you know basic probability
@@Nathan-oe8ut this guy didnt get it right.
I think the secret riddle is figuring out whether it’s “unobtainium” or “unobtanium” 🙊
Science with Katie it's the latter, it is taken from a reletavely old movie called 'core'... It's good
goovind narula she's pointing out a spelling error in the video.
Katie do you use like bots?
Hmm it’s the former in the English captions
You are aware that is says in the vid that it is unabtanium
I finally solved a riddle the right way!!!! I’m so proud of myself
Scenario 1 right? As there is only one pair that yours cannot be while in scenario 2 there are 2 sets of pairs you can't have. This wouldn't make that much difference though??
Well, in scenario one there are 5 pairs that you can't get, in scenario two there are 6 pairs that you now can't get.
"False positives in medical testing are preferable to false negatives" - well not exactly nowadays.
If discrete math taught me anything, it is to use Bayes Theorem, which tells me the chances of this being a good deal is ~.1%
Edit: ok watching the video, used a different form probability and am now wondering why Bayes Theorem doesn't work. Used %chance any rock is unobtanium for branch 1 and %chance detector resulted in positive or negative reading for branch 2. Tell me if you get something else.
RanDumSocksGames Messed uo my last answer. Here is the correct one.
Let U be the event that a rock is unobtainium, and let D be the event that the detector goes off.
P(U|D)=P(D|U)*P(U)/[P(D|U)*P(U)+P(D|~U)*P(~U)]=1*.01/(1*.01+.1*.99)=.0917 or approximately 10% chance that rock is unobtainium given that the detector went off.
You’re forgetting expected value. If the rock was worth $2,223, then it would be worth the trade for $200. Why? Let’s examine an extreme case. Let’s say the rock is worth $1B (billion), and Tricky Joe wants to sell it for $1, then it would be worth taking the 91% chance of it not being unobtanium, because there’s a 9% chance you’ll walk away with $1B minus the $1 that you paid for the rock.
So what’s the formula?
The expected payout (the price of the rock X minus the $200 cost of the rock) times the probability that you will get a payout (9%) has to be greater than what you are paying for the rock ($200) times the probability that you will not get a payout (91%)
(X-$200)*0.09>$200*0.91
X*0.09>$200
X>$2,222
So in this video they said that the cost was $200 and the payout was $1,000. That isn’t worth it.
What they failed to mention is that if the cost was $90 or less, or if the payout was greater than $2,222.22, then it would be worth trading.
Bayes Theorem does work, you just need to expand P(Detector is positive) properly.
Bayes Theorem is P(A|B) = P(B|A) * P(A) / P(B).
A is unobtainium. B is a positive reading.
B happens 100% of the time given real unobtainium so P(B|A) = 1.
1% of all rocks are unobtainium so P(A) is 0.01.
B happens 100% on the 1% unobtainium and 10% on the other 99%, so for any rock, P(B) = (1*0.01) + (0.1*0.99) = 0.109.
=> P(A|B) = 1 * 0.01 / 0.109 = 0.0917.
smokey04200420 They didn't fail to mention it, it just wasn't the question
I finally got it right!!!
I watched this in the past and didn't remember, but recently I've watched a medical video about how the relatively rare occurrence of the corona virus in a population makes the low false negative rate in tests exorbitant and the tests unreliable
Similarly, there was a scare some time back claiming that eating more than a certain amount of red meat each week increases the risk of a particular form of cancer by 30%, but what they didn't tell you was that that that form is so rare anyway that even a 30% increase doesn't actually make that big a difference!
My immediate thought:
"SHOULD YOU MAKE THE TRADE?"
"Well, of course not, otherwise this wouldn't be an interesting problem and you wouldn't have asked in the first place."
But yeah, I did the math anyway XD
Same
you know who taught joe how to be tricky?
*_joe mama_*
you got me.
Well played.
Hey, early! Insteresting as always!
Z Doesn’t matter if you’re early, just WATCH THE VIDEO
Marcel de Vries am watching -_-
Jealous boiiiiii
Thankfully, this is the easiest kind of question to solve in overall probability questions as long as I draw charts accurately.
Tricky Joe is the expert of this device. He would know that it's just likely a false positive but he didn't test it multiple times, so perhaps it was his plan all along
Bonus riddle:
There are 78 possible pairs (13 card values and each has 4 card types means there are 6 possible pairs for each card value, 13*6 = 78).
Let X be number of possible combinations of 2 cards. There are now 50 cards total (she already has 2 cards and we use the same deck), so X will be a pretty big number, the precise number is not needed.
In scenario A:
She already has a pair, which means there are only 2 cards of that value left in the deck, therefore we have 73 pairs to choose from (78-5=73).
In scenario B:
She has 2 cards of different values, which means there are 3 cards of each specific value left. This removes 3 possible pairs from each value (3*2=6 pairs), therefore we have only 72 pairs to choose from.
Result:
Divide the numbers by X and compare them: 73/X > 72/X
Probability of having a pair is bigger in scenario A.
It’s just common sense though, if she has just one pair that means more pairs for you but if she has different cards then you wouldn’t be able to get pairs of those cards.
For the Amy riddle, sorry for not showing my working:
Scenario 1 gives you a 5.96% chance of receiving a pair.
Scenario 2 gives you a 5.88% chance of receiving a pair.
The answer is Scenario 1.
the answer is obvious, and i don't even need to hit Atlantic City to know. don't fuckin hit if the house has 20.
scenario 2 is best cuz house has 15, so they gotta hit, chances of bust are high at that point. that's where counting helps.
Just hijacking your comment to show the work since we got the same answer.
scenario 1 gives (4/50)*(3/49)*(12)+(2/50)*(1/49) = 5.96%
scenario 2 gives (4/50)*(3/49)*(11)+(3/50)*(2/49)*2 = 5.88%
looking at the chance of getting a pair from cards not held by Amy first card is 4/50 second is 3/49. In scenario 1 that's 12 different ranks then calculating the odd of getting the other pair of queens in 2/50 and 1/49. In scenario 2 there are 11 different ranks for the 4/50 and 3/49 draws then for 5 and queen 3/50 and 2/49.
Tricky Joe looks at me, notices my eyes are green, then just gives me the rock for free.
Yay, the 1st one I answered correctly and knew why
PLOT TWIST: Instead of Unobtanium. He got the largest diamond instead
I probably would say no because his name is Tricky Joe
What if he's a good boy though?
This is the first time *ever* that I managed to solve a Ted Ed riddle without looking :p