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.

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!

Tricky Joe,Tricky Joe.

With his ragged clothes,

And his finger in his nose!

No

1 in 8 people who die the electric chair are inassinat

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.

I would take my pickaxe and knock trick joe out cold and bootgang that rock and detector

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).

”Unobtainium”Get it? You can’t obtain it XD

Can`t you just shoot the rock again and see if it goes of?

Even gold is worthless? Ok

The answer is: Kill Tricky Joe and take his gun. Easy.

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

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.

I wish I was a false positiveIn 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.

Just get him to shoot it twice.

no

Q pair Amy have a higher chance

12*6+1 for Q pair

11*6 + 6 for Q and 5

UN obtain um.. you will not obtain it nor the meaning of this comment.

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.

Just check the stone like 5 times more then you will be sure

That 3000!

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

Never trust a mfer named Tricky Joe, bet he tricky

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 WE BACK IN THE MINE-

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

2019 anyone

Me: Alright, im gonna watch this video to see if i can solve it.

Me:

fails miserablyAlso me: ooh, another one

cycle repeatsCould be 100 rocks in said mine, sure would take my chances then

Finally one I got right!

Yeah, no.

I would never trust someone named “Tricky Joe”

Like seriously.

Why is that dude a miner, he could be a fuckin CEO with that level of intelligence

I would just blast each rock 10 times

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 🙂

So they back in the mine

tricky joe is fiddleford from gravity falls

Just mine the rock before you buy it.

simplewow i actually got it right

I just always say the opposite of what I think

If it’s unobtainium it can’t be obtained,duh.

When she has pairs right?

Still trying to wrap my head around detecting false positives.

I would just test it more time and if every time it says its unobtainium ill buy it

What if he rigged the detector?

Huh?

In all of these riddles this is the one I got it correct

bonus: first one, consider the deck only contain QQ and 55

When there r 2 queens there is higher probability for getting a pair

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 got it!

and 41 others wrong

Just double check it.. shoot it again

One of the only problems I actually got on here lol

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

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

I thought Joe was selling the detector was 200$

Can't you check the rock like, twice?

or joe could scan t multiple times and if it goes off all of those times it's really unobtanium

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)

I wouldn’t do any trade with a guy named ‘Tricky Joe’

You never mentioned a 10% error rate in the beginning!

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.

Senerio 1

finally the first ted ed problem i was able to solve! i used simple conditional probability

In scenario 1 you have one possible pair more than scenario 2

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.

The whole time I’m just thinking shoot the rock 10 times then

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

Tricky Joe?

I met him in the

gangFirst riddle I solve

Is brillant Australian ?

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.

Question : Should you buy the Unobtanium?

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

This was posted on my birthday 🙂

no bc yes makes sence.

EDIT: new it! 😀

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

Yay, finally a riddle i solved xD

"Nah thanks, I'm a grinder."

Say no, because that name alone is a sign that you can’t trust him

Lol this is the first riddle I solved on my own:)

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).

0:06

That's because it's called UNOBTAINium

Q 5

Me:notice that the ore is actually plutonium

Me:accept trade

Me:become a trillionare

Woah, i actually solved a riddle right

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.

Ted ed: Yes, but actually no

no! his name is "Tricky Joe"! duh of course he's deceiving you

What if it is 100% gold

Yeah I did the correct guess

Nah, lets just mine normally, like a human

Sir this is a McDonalds

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

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

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

Tricky joe: —>

trys it for the first time<—90% correctThat dude: oh Im not gonna buy that, it’s fake

My thinking was the only 90% accuracy, coupled with the low price would be a scam.

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.

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.

Want some 90% real water

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.

Jo mama