Can you solve the false positive riddle? - Alex Gendler 

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I won't consider doing business with someone named Tricky Joe.

I've played enough Pokémon in my days to know that even a 90% chance are terrible odds.

"We just found this really rare stone. What should we call it?" "Can'touchisium"

Step 1: Ask Tricky Joe: If I have green eyes, would you say Ozo? Step 2: Ask the unobtainium to leave the island

I said no because tricky joe’s name alone is untrustworthy 😂😂😂

"Would you make the trade?" *No because I have trust issues*

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.

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

*_Me_* : No because my *"friend"* named _tricky joe_ wouldn't sell me an authentic, and genuine ore for a fifth of its price.

this one doesn't seem nearly as hard as some of ted-ed's other riddles

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)

But shout out to Joe for increasing the chances of getting the ore from 1% to 9%. That's pretty insane.

Why you shouldn't take the deal: *THE GUY SELLING YOU THE ORE IS NAMED TRICKY JOE. THINK ABOUT IT.*

*Why is it so hard to obtain Unobtanium*

“Should you make the trade?” Me: ozo

1: confirm you have green eyes 2: ask tricky joe to leave 3: figure out how to escape the white void

Me: Alright, im gonna watch this video to see if i can solve it. Me: *fails miserably* Also me: ooh, another one *cycle repeats*

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

"At first, this may seem like a good deal" It doesn't. "But why did it seem like a sure bet?" It didn't.

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.

We just got bamboozled by Tricky Joe.

"Well this video is called false positive so I;m gonna go with no." Context clues are the best.

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)

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.

Answer: No. Reason: His name is Tricky Joe.

Honestly, the correct answer to this riddle seemed obvious and intuitive so I thought I was wrong

Through amazing deductive reasoning and actual dedication, i finally got one ted ed riddle right

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.

Solution: Shoot the mineral 10 times. See whether it goes off all 10 times. Crash the price of unobtainium and get rich!

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)]

Everyone who has played XCOM2 will know that 90% chance means certain failure.

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.

Plot twist: The device gets it wrong 100% of the time and its all a scam

Me in math class: I won’t use all this other than grocery shopping Me after watching couple of TED-Ed videos: nvm...

I wish this is the kind of stuff we actually learn in school instead of the same thing every year

One of the only riddles I have ever solved from ted ed.

Yeah, no. I would never trust someone named “Tricky Joe” Like seriously.

- reads title - probably not but I'll still learn something so might as well click on it

this is literally the only ted ed riddle i’ve gotten

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).

"Stick around for an extra riddle" -TedEd

First riddle I've ever solved on this channel under a minute. Highschool's finally doing its job

First ted ed riddle I solved, this one is defiantly way more simple than the others

"False positives in medical testing are preferable to false negatives" - well not exactly nowadays.

i won't take anything from joe.he looks damn suspicious..

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.

This is the first TED riddle I actually manage to correctly solve, yay!

Thankfully, this is the easiest kind of question to solve in overall probability questions as long as I draw charts accurately.

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.

I just said no because Tricky Joe is Tricky Joe and seems like he's pretty shady. None of the fancy math stuff lol.

for the bonus riddle, both options have a probability of (odds of not picking a queen or a 5 and getting a double) + (odds of getting a queen or a 5 and getting a double), for both the left side of that sum are identical (both have the same 44 non q/5 cards in the dec and the same total q/5 cards). This means we can ignore them and directly compare the probabilities of qqq555 and qq5555. for the first, no matter what you pick there are two remaining cards to draw from out of the 5 to get a pair so p=2/5. For the second option, p(q) = 2/6 and getting a second becomes 1/5. for 5 it's 4/6 * 3/5. so the second option, the odds of a double is 2/5 +1/15 - very slightly better

First riddle i gto exactly right, down to the calculations

Vibranium’s weird, distant cousin

Why is Unobtanium so cheap if it's so rare?

This was easier than the other riddles, bonus one too! scenario 1 is my answer

Love how the mine is just a large circular room with lone rocks sitting there

Just tell him that at least one unobtanium has green eyes

PLOT TWIST: Instead of Unobtanium. He got the largest diamond instead

using conditional probability formula really helped me solve this

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

Just scan the rock again.

The first riddle i was able to do! :)

This is the first time *ever* that I managed to solve a Ted Ed riddle without looking :p

Time to block Tricky Joe on Discord 😤😤😤

*clap* BONUS *clap* RIDDLE

p(getting a pair) =when there are 2 queens gone, there are 1+12*6 pairs availble and when a queen and a 5 is gone, it leaves 2*3 +11*6 pairs in the deck. so when two queens are gone, the deck has 73 pairs and when a queen and a 5 is gone, there are 6+66 pairs which is 72 pairs in the deck. Therefore, it shows that if amy has a pair of queens, your chance of getting a pair is slightly higher than is she has a queen and a five.

Regarding the final question, if pair is defined as both the same color, scenario 1 is the better likelihood, but if that isn't a requirement, the question becomes much more complex.

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.

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

I think it’s rather weird knowing that “unobtanium”, a very rare mineral, would be worth $1000 for something the size of a boulder

First riddle I’ve ever solved on the channel :D

So... Fallacies... Need a video on those.

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.

According to me in scenario 1 we have slightly more chance of getting a pair. I used P and C to find the number of pairs we can't make in both the scenarios and I am getting that we can make 1 more 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 whole time I’m just thinking shoot the rock 10 times then

Love these riddle videos

There are 50 cards left for me to pick, therefore 1225 combinations that I can possibly have. I do not play card games, however, and therefore I don't know if "pair" means ANY two of the same number/letter OR that they ALSO have to be of the SAME COLOUR. _____________________________________________________________ First assumption: Case 1: the only way to have double queens is to have the other two, but there are 6 possible pairs for each of the other 12, so 73. Case 2: There are 3 ways to have queens, 3 ways to have the other one, and 6 ways to have any of the other 11, so 72. There is a slightly bigger chance you'll have a pair if Amy has a pair of her own. _________________________________________________________ Second Assumption: Case 1: still one way of getting queens, but only 24 other ways, a total of 25 Case 2: 1 way of getting queens, 1 way of getting the other number that Amy has, 22 ways to get other pairs, total of 24 Still more chances for a pair if Amy has a pair of her own. _________________________________________________________

Thank you Ted Ed I am learning alot from your puzzled videos. Love you and cant thank enough for the learning I am getting.

Taking the rock was obviously a bad decision... but people fall for obvious scams every day

This one certainly hits differently in 2021.

I finally solved a riddle the right way!!!! I’m so proud of myself

I think the secret riddle is figuring out whether it’s “unobtainium” or “unobtanium” 🙊

3:38 me doing maths

First time I actually got the riddle right! :D

The answer to the problems at the end is scenario 1. There a total 1225(50C2) possibilities in both cases. In the first case, there are 73 (12×4C2+1) possibilities where you get a pair, while in the second case there are 72(11×4C2+2×3C2) possibilities you get a pair. Proving that the odds of getting a pair is greater in the first case

Ok i admit Ted is smarter than me

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

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

One of the best viedeos from you

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.

Its always great to have a riddle that i can never solve.

Tricky Joe looks at me, notices my eyes are green, then just gives me the rock for free.

If you find a rock with unobtainium, just test the rocks a few more times. If it contains the mineral, it will go off every time. If it doesn't go off even once again, then it's a normal rock.

Wow I acually got a riddle right for once in my intire life :D

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??

First time solving a TED-Ed riddle!

Weird how thinking of this more has already made me immune to this bias

I probably would say no because his name is Tricky Joe