Can you solve the false positive riddle? – Alex Gendler
100 Comments


Mining unobtainium is hard work. The rare mineral appears
in only 1% of rocks in the mine. But your friend Tricky Joe
has something up his sleeve. The unobtainium detector he’s been
perfecting for months is finally ready. The device never fails
to detect unobtainium if any is present. Otherwise, it’s still highly reliable, returning accurate
readings 90% of the time. On his first day trying
it out in the field, the device goes off, and
Joe happily places the rock in his cart. As the two of you head back to camp
where the ore can be examined, Joe makes you an offer: he’ll sell you the ore for just $200. You know that a piece of unobtanium
that size would easily be worth $1000, but any other minerals
would be effectively worthless. Should you make the trade? Pause here if you want
to figure it out for yourself. Answer in: 3 Answer in: 2 Answer in: 1 Intuitively, it seems like a good deal. Since the detector
is correct most of the time, shouldn’t you be able
to trust its reading? Unfortunately, no. Here’s why. Imagine the mine
has exactly 1,000 pieces of ore. An unobtainium rarity of 1% means that there are only 10 rocks
with the precious mineral inside. All 10 would set off the detector. But what about the other 990
rocks without unobtainium? Well, 90% of them,
891 rocks, to be exact, won’t set off anything. But 10%, or 99 rocks,
will set off the detector despite not having unobtanium, a result known as a false positive. Why does that matter? Because it means that all in all, 109 rocks will have
triggered the detector. And Joe’s rock could be any one of them, from the 10 that contain the mineral to the 99 that don’t, which means the chances of it containing
unobtainium are 10 out of 109 – about 9%. And paying $200 for a 9%
chance of getting $1000 isn’t great odds. So why is this result so unexpected, and why did Joe’s rock seem
like such a sure bet? The key is something called
the base rate fallacy. While we’re focused on the relatively
high accuracy of the detector, our intuition makes us forget to account for how rare the unobtanium
was in the first place. But because the device’s error rate of 10% is still higher than
the mineral’s overall occurrence, any time it goes off is still more likely
to be a false positive than a real finding. This problem is an example
of conditional probability. The answer lies neither in the overall
chance of finding unobtainium, nor the overall chance
of receiving a false positive reading. This kind of background information
that we’re given before anything happens is known as unconditional,
or prior probability. What we’re looking for, though,
is the chance of finding unobtainium once we know that the device did
return a positive reading. This is known as the conditional,
or posterior probability, determined once the possibilities have
been narrowed down through observation. Many people are confused
by the false positive paradox because we have a bias
for focusing on specific information over the more general, especially when immediate decisions
come into play. And while in many cases
it’s better to be safe than sorry, false positives can have
real negative consequences. False positives in medical testing
are preferable to false negatives, but they can still lead to stress or
unnecessary treatment. And false positives in mass surveillance can cause innocent people to be
wrongfully arrested, jailed, or worse. As for this case, the one thing
you can be positive about is that Tricky Joe is trying
to take you for a ride.

100 thoughts on “Can you solve the false positive riddle? – Alex Gendler

  1. Sign up for free at https://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!

  2. 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.

  3. It is better for amie to draw 2 queens because if she draws 2 different cards there are 2 cards in the deck with no match

  4. 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

  5. Here’s another example of this type of riddle.
    You are an Aircraft mechanic in WW2, and the Air Force needs you to reenforce the plane, however, you can only chose one place to reenforce the plane, otherwise the weight will be off and it won’t fly as well. You look and most of them have bullet holes in the tail / rudder, a few have bullet holes in the wings, some have holes at the body, and almost none have holes in the engine. Where do you reenforce the plane? (Answer will be in my reply)

  6. I don't know much about math, but intuitively I knew these were not good odds. My logic is if the machine can misfire 10% of the time and there's a 1% chance to find unobtanium; it will misfire more than be accurate. If it was 10% and 10% the odds would be better.

  7. But what if he sold it for 100… Then I personally would buy it, because I have about 1 in 10 chance to get the rock that is worht 1000 (100 x 10). You don't find odds that high anywhere.

  8. The real prize is the toxicity you got rid of during the expedition. You don’t need a person like tricky joe you’re better than that

  9. I would tell him Id buy it for twice the price if he'd scan the rock 10 more times and the result is the same every 10 times.

  10. Question : Should you buy the Unobtanium?

    More like : Should your start thingking like a Crime Investigator and start solving random math and percentages

  11. for the bonus, it's quite a riddle, but one thing is certain, there are two queens, a five, and another card if there is the same amount of each card in both scenarios. in scenario 1, they have two queens, and you have a five, the fourth card is what is the problem, it's either a queen, a five, or another overall, making chances about 1/3. in scenario 2, they have the queen and the five, meaning you have one queen, and the extra card is a queen, a five, or another entirely, again making it about 1/3! it doesn't matter which scenario you have, just don't bet stuff because your chances are slim either way

  12. For the bonus riddle, it's Scenario 1 where it's more likely for you to get a pair. Since there are 26 pairs that you can get from the deck (getting each card only once), removing 1 pair only will still leave you with 25 pairs to get (which is Scenario 1), while for Scenario 2, getting 2 different cards will leave only 24 pairs (2 cards won't have a pair).

  13. I would just buy his machine and copy the materials to make my own machine and shoot the rocks myself x10 times so I can verify. Solved.

  14. Answer for the bonus riddle:
    a pack cards has 13 different cards of 4 shapes each = 52 cards. So u can make 6 pairs with each number card. For ex: with 4 queens u can make 6 pairs (spade&heart, spade&clubs, spade&diamond, heart&clubs, heart&diamond, clubs&diamond). Thus in total there is a probability of 1/78 (6pairs * 13 cards= 78). ur opponent already has 2 queens which means there are only 2 queens left in the remaining deck eliminating 5 pairs of queens above said. that will increase the probability to 1/73 (78-5). In the next scenario ur opponent has 1Queen, 1five. So the number of pairs u can make with remaining cards is 72 pairs (78-3-3). So the probability of getting a pair is better in second scenario which is 1/72 (As we all know that 1/72 is better than 1/73). Like if anyone understand this

  15. People have a habit of looking over specific things, oh you mean the fact that he’s tricky joe or we couldn’t of scanned this twenty time(or maybe it mistakes cause… it gives off a certain pulse or something), or how I’m a poor miner and I don’t have 200% or…. ummmm…. math

  16. LOL as I work in the Department of Epidemiology, I immediately thought of this as the positive predictive value and gave the instant result :))

  17. OH MY GOD I ACTUALLY SOLVED SOMETHING!! The first scenario would be the better chance I’m assuming because that only takes away one possible pair out of the deck, the Queens. But with the second scenario, it takes away two possible pairs, queen and a 5 since she has one of each.

  18. Or you could shoot the device 2 or 3 more times at the rock and there is a SUPER small chance then that you're getting a bad deal, because it ALWAYS detects actual Unobtanium.

  19. wouldn't you still be checking 90% less rocks? so instead of unobtanium being 1 percent of the rocks you carry its about 9%, isn't it a reusable tool? it would still greatly increase the chances of you finding unobtanium.

  20. If Ik the ore is in 10% or rocks and the size of the rock if it was pure would be $1,000 at best odds that’s still only $100 of ore plus the cost of labor on me! That alone is enough to sway me away, plus given the ore could have been picked up from another sample I would have him retest it a few times outside to make the probability that it is in there higher. if positive best I would give him is $50

  21. Soooo, ya sayin’ diamonds, obsidian, emerald, sapphire, aquamarine, opal and all the other precious rocks are worthless???

  22. Him: so why did joes rock seem such a great bet
    Him: about to explain stuff
    Me: it because they eat too many dum dum’s

  23. This is honestly easier than the passcode riddle. Its so intuitive. I didnt even need to think to solve this, and i got it right.

  24. It's weird, because my immediate mathematics tells me "it's only worth it if it's a $20 price tag", and I can't remember what those mathematics are after seeing the video.

  25. This make no sense you said the laser gun could find unobtanium without failure but then you said it would have a nine percent chance of it finding it

  26. Only put it in cart if it returned a positive reading four consecutive times — that increases the accuracy from 90% to 99.99%

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