Can you solve the false positive riddle? – Alex Gendler


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. Pay him 50$ to rescan it, if it is a false alert i save 150$ if it goes off I get 750$ if it has two false positives in a row (very unlikely) I loose 250$ “Mmm… I like those odds…” -Homer.J.Simpson

  3. If you use the gun again for a second and third round and so on you take the original 10% error down to maybe just 2 rocks and buy those for 200 each walk away with 1600 and the pick axe ⛏

  4. I understood false positives after they compared it to innocent people being jailed(i think). So, if there's a murder and a suspect, but the details aren't easy to find(no CCTV, etc) People focus on the fact that there's a 90% motive for murder and don't focus on the fact there's a 5% percent that they could've been near the crime scene.

  5. Before seeing the answer here's what I solved:
    90% of the time the device shows the correct reading.
    So if u buy 10 rocks which the guy says they have unobtanium -> 1 out of 10 does not actually contain it.
    2000 invested
    9 * 1000 for the good ones = 9000
    Profit = 9000-2000 = 7000
    So, my answer is YES.

    Edit after seeing answer:
    I did not consider the first 1% chance of appearance.
    Stricly calculated the odds if you buy 10 rocks from the guy.

  6. Well, the real idea would be to calculate out the average return on investment, which would be about $90.90, which is less than $200, so it's a bad idea. Now, if the unobtanium was worth more to the point that average return exceeded $200, you should do the bargain (especially if you can keep getting it).

  7. Same as antivirus. It obly works 80% of the time cause my exploits are considered malware when its literally not

  8. The answer to your secondary riddle is option 1. If we assume that there's 2 card packs and that it has to be the same card, option 2 rules out both options it highlights. Meanwhile, option 1 only rules out those specific queens.

  9. In the case with 2 queens, you remove the possibility of getting any pair with the queen of hearts and the queen of diamonds. This removes the following pairs :

    -queen of hearts and queen of diamonds
    -queen of hearts and queen of spades
    -queen of hearts and queen of clubs
    -queen of diamonds and queen of spades
    -queen of diamonds and queen of clubs

    You don’t remove queen of diamonds and queen of hearts, because it’s the same thing as queen of hearts and queen of diamonds, which we have already removed. Because of this, you only remove 5 pairs.

    Now for the case with the queen and the 5. Like the last case, you remove all the pairs with the queen of hearts and the 5 of spades(it could also be any card that isn’t a queen). This removes 6 pairs, because there are 3 other queens, and 3 other 5s. So in scenario 1, you have a higher chance of getting a pair than in scenario 2.

  10. easier way: ok you have 99% chance it is correct, but say there are a million rock.. 1% of million

    10,000 fake beep rocks.. so Tricky weasel Joe got one that beeped it could be very possible that it is one out of the 1k false reading rocks which is out of that million and the metal is rare in that million rocks. .. the chance is very rare to risk 200 dollars for winning 1000 dollars. Because in that million there are 10k error. I exaggerate the number to see the difference clearly
    but if it were 10 rocks.. and he selected one that beep the chance is very high that it is right rock!
    the higher the initial number the HIGHER the odd it is fake stone due to error reading.. etc etc.

  11. “You guys says mining unobtanium is hard work?” – Said me holding my iron pickaxe looking for diamond in Minecraft

  12. 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

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

  14. Here are the actual calculations for the problem (P stands for probability):

    First, determine all the probabilities:
    P(rock contains unobtainium) = 0.01
    P(rock doesn't contain unobtainium) = 0.99

    If rock contains unobtainium:
    P(detector detects unobtainium) = 1
    P(detector doesn't detect unobtainium) = 0

    If rock doesn't contain unobtainium:
    P(detector detects unobtainum) = 0.1
    P(detector doesn't detect unobtainium) = 0.9

    Therefore,
    P(rock contains unobtainium and detector detects unobtainium) = 0.01 * 1 = 0.01
    P(rock contains unobtainium and detector doesn't detect unobtainium = 0.01 * 0 = 0
    P(rock doesn't contain unobtainium and detector detects unobtainium) = 0.99 * 0.1 = 0.099
    P(rock doesn't contain unobtainium and detector doesn't detect unobtainium) = 0.99 * 0.9 = 0.891

    Now for some conditional probability:
    P(rock contains unobtainium given that the detector detected unobtainium) = P(rock contains unobtainium and detector detected unobtainium)/P(detector detected unobtainium) = 0.01/(0.01+0.099) ≈ 0.09174
    P(rock doesn't contain unobtainium given that the detector detector detected unobtainium) = P(rock doesn't contain unobtainium and the detector detected unobtainium)/P(detector detected unobtainium) = 0.099/(0.01+0.099) ≈ 0.90826

    Now we use expected value to calculate the expected payoff if we accept the trade. Multiply the payoffs by the probabilities that you'll get those payoffs & add them to each other:
    Profit if the rock contains unobtainium * P(rock contains unobtainium given that the detector detected unobtainium) + amount of money lost * P(rock doesn't contain unobtainium given that the detector detected unobtainium) = $800 * 0.09174 + (-$200) * 0.90826 = $73.39 – $181.65 =
    -$108.62

    So, our expected payoff if we accept the trade is -$108.62, so we obviously wouldn't accept the trade. If you have any questions reply and I'll try to answer them 🙂

  15. 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.

  16. 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

  17. 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

  18. 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)

  19. 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.

  20. 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.

  21. 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

  22. 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.

  23. Question : Should you buy the Unobtanium?

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

  24. 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

  25. 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).

  26. 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.

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