Riddles

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gimp
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Riddles

Post by gimp »

Riddle #1

This riddle is very math oriented.

I work supply chain and today I had an interesting problem. After I figured it out I was quite proud of myself and realized I could easily turn my problem into a simply stated math riddle.

Scenario:
You have a bag with four cards colored green, yellow, blue, and red. Without looking, you will pull a card out individually until you pull a green card. To clarify, anytime a green card is pulled, the round is over, and you need to place all the cards back in the bag. Anytime a different card is pulled, you need to leave that card out of the bag and continue to pull until you grab a green card. Even if there is one card left and you know it is green, it still needs to get counted as a "pull".


Question:
In a perfect world*, if you reached into the bag and pulled a card out 1,000,000 times, how many rounds would a green card be pulled first? Second? Third? Fourth?
*Basically means on average; I.g. in a “perfect world” if I flipped a coin 10 times it would land tails 5 times.

Answered first by Zweq and analcactus
answer: 100k, 100k, 100k, and 100k

Riddle #2

The last one was warm up, but this one has tricks abound.

Scenario:
Imagine you start at point Z, and you need to find an item that could be at 4 different destinations A, B, C, and D. Every interval takes a certain amount of time in minutes.
Z-A = 1
Z-B = 2
Z-C= 3
Z-D = 4
A-B = 5
A-C = 6
A-D = 7
B-C = 8
B-D = 9
C-D = 10

Question:
If I left point Z in search of the item on a whim, with no bias towards time or which point I travel to next, what is the expected value of time it would take me to find the item?

EDIT: There is bias in the fact that you wont return to a point in which you have already been in the course of a single trip. I.g. you wouldn't go Z-A-B-A. After B youd have to go to C or D if you still haven't found the item in that trip.

Hint: A-B = B-A, C-D = D-C, etc.

Answered by sunl
Answer: 13.75 minutes

Riddle #3

Scenario:

The Chemical Shipment

I have four workers (Madness, Ville_J, Bjenn, and Orcc) who carry dangerous chemicals in a special chemical vat from A to B. At point A they load the chemicals, and at point B they unload the chemicals. Only one or two people may travel at any given time and when there are two they must travel together. There is only one vat, and the chemicals must always be in the vat when traveling (once the vat is unloaded at point B, it must travel back to point A to pick up another shipment).

1. Madness pilots an airplane that can carry the vat, but Bjenn, Orcc, and Ville_J are afraid of flying with him and refuse to passenger. It takes the plane 2 hours to fly from A to B.
2. Bjenn has a license for a two seater truck that may carry the vat, but will only let Madness ride with him because he thinks he is too good for Ville_J and Orcc, he is also secretly infatuated with Madness. It takes the truck 4 hours to go from A to B.
3. Orcc likes to use his spacious two seater forklift to make the trek while carrying the vat. However Ville_J insists his way is faster and will not ride on Orcc’s forklift. It takes the forklift 10 hours to go from A to B.
4. Ville_J pushes the vat on a dolly with the help of another. This takes 15 hours to go from A to B, you are also able to catch thousands of pokemon going this way.

Point A has zero planes, one truck, one forklift, and one dolly. Point B has two planes, zero trucks, one forklift, and one dolly.

Question:
What is the shortest amount of time I can make three shipments of chemicals and successfully end up with all four workers at point B, assuming they all start at point A? Also, who is the best kuski out of the four? You must explain your answer either in PM or here to get credit for the answer.

Hint: A-B = B-A. Modes of transportation may be used more than once or not at all. Don’t Zig-Zag.

Answered by sunl
Answer: 29. Bjenn
Last edited by gimp on 27 Apr 2018, 17:19, edited 14 times in total.
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Lousku
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Re: Riddles

Post by Lousku »

250000, 250000, 250000 and 250000? What did I miss?
then again i don't know anything
maybe easier not to think abouut alöl things thought than not things thought ... or something..=?
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Re: Riddles

Post by gimp »

Clearly you missed something. In order for a green card to even be pulled 250000 times on the fourth try youd also have to at a minimum pull a blue, red, and yellow 250000 times each before it. You would already be at 1,000,000 pulls.
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Re: Riddles

Post by Lousku »

Oh right, I thought it was about finding the green one 1000000 times... nvm xd
then again i don't know anything
maybe easier not to think abouut alöl things thought than not things thought ... or something..=?
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Re: Riddles

Post by Zweq »

i sucked at probability & statistics courses for a reason but what i think is that we can say beforehand it's equal chance for green card to be pulled out on nth pull, because it's like if you shuffle 4 cards, you have 1/4 chanc for it to be on spot x

so therfor you need to calculate the number of rounds needed in average to reach millionth pull and this i cannot do

when you know the number of rounds needed in average then the answer is number of rounds in average / 4 for each pull

so for exzamppel if rounds needed hapend to be 400k the answer would be 100k/100k/100k/100k

just aplying lousku logick here dont sue me plz..
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Re: Riddles

Post by analcactus »

imo 100k per:

eg 100k green first
100k green second (+100k fail cards first)
100k green third (+200k fail cards)
100k green fourth (+300k fail cards)
sums up to 1m
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Re: Riddles

Post by kuchitsu »

gimp wrote:Clearly you missed something. In order for a green card to even be pulled 250000 times on the fourth try youd also have to at a minimum pull a blue, red, and yellow 250000 times each before it. You would already be at 1,000,000 pulls.
Not really? If the fourth card is not green, that means I won't pull it at all since the green card will come before.
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Re: Riddles

Post by kuchitsu »

The "problem" is basically:

there are 4 slots
1 apple and 3 bananas are put in them randomly
what's the chance of an apple appearing in the slot 1/2/3/4?

1/4 obviously

---

Another way of looking at it:
1st pull - I think we all agree that it's 1/4
2nd pull - now it is 1/3 but not just 1/3, instead it is 1/3 of the "remaining chance", so 1/3*3/4=1/4
3rd pull - now it is 1/2 of the "remaining chance", so 1/2*1/2=1/4
4th pull - now it is 1 of the "remaining chance", so 1*1/4=1/4
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Re: Riddles

Post by Lukazz »

kuchitsu wrote:The "problem" is basically:

there are 4 slots
1 apple and 3 bananas are put in them randomly
what's the chance of an apple appearing in the slot 1/2/3/4?

1/4 obviously
But that's not the question. ;) imo analcactus is right.
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Re: Riddles

Post by kuchitsu »

Obviously he isn't. No matter what, you WILL pull the green card sooner or later because it is in the bag. So the chances of pulling it at 1st, 2nd, 3rd and 4th pulls should sum up to 100%. But in his answer they only sum up to 40%. And there can't be a "fail card" at the 4th pull. If you got to the 4th pull, that means the remaining card can be only green.
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Re: Riddles

Post by gimp »

i posted but deleted. figure ill wait a bit longer
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Re: Riddles

Post by Lukazz »

kuchitsu wrote:So the chances of pulling it at 1st, 2nd, 3rd and 4th pulls should sum up to 100%.
100.000 rounds, where you pull the green card first = 100.000 pulled cards in total
100.000 rounds, where you pull the green card second = 200.000 pulled cards in total
100.000 rounds, where you pull the green card third = 300.000 pulled cards in total
100.000 rounds, where you pull the green card fourth = 400.000 pulled cards in total

100.000 + 200.000 + 300.000 + 400.000 = 1.000.000
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Re: Riddles

Post by analcactus »

i summon anpad, he's mathboys

kuchi, we agree that 1/4 is probability for green card to appear in each slot.
the gimmick is that you can't see all slots, you open them in order (1->2->3->4)
that's why when the card is eg in 3rd slot, you'll have to take two fail cards beforehand on this try.
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Re: Riddles

Post by kuchitsu »

Oops, I thought that there are 1,000,000 rounds, not card pulls. I'll have to rethink everything.
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Re: Riddles

Post by kuchitsu »

Oke, I get it now, I think anpalcactus might be right.
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Re: Riddles

Post by gimp »

Yes mans are right gj, I think zweq was first to answer but anal had good explanation too. new riddle posted.
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Re: Riddles

Post by Lukazz »

Can you return to a location that you've been to before? So go from Z to A, to B, and then back to A and then Z?
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Re: Riddles

Post by gimp »

Very good questions. No, you cannot return back to the same location, in this way it is much like the card riddle, once you have gone to point A, it is out of the bag. and you never return to point Z.
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Re: Riddles

Post by gimp »

Okay I added one more riddle. This one not as hard.
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Re: Riddles

Post by sunl »

Riddle 2:

Assume that the item I is located at a random point {A,B,C,D} with equal probability 0.25

Let X be your location. X0=where you start (Z), X1=your first destination, X2=your second destination, X3=your third destination, X4=your last destination
For example if you do a trip ZA AB BC BD (ZABCD), then X0=Z, X1=A, X2=B, X3=C, X4

1) Let you set out on a trip to find the item. There is a 1.00 probability that you go from X0 to X1. You have a 0.25 chance of finding the item, so 0.75 of the time you will go from X1 to X2, and 0.50 you will go X2 to X3, and 0.25 you will go from X3 to X4.

When you do X0->X1, you choose X1 randomly; therefore ZA,ZB,ZC and ZD are all equally likely. Therefore, the expected amount of time to go to X1 is average(ZA,ZB,ZC,ZD)*1.00

When you do X1->X2, you are at a random point A,B,C,D going to a random point A,B,C,D, excluding (AA,BB,CC,DD), so all directions are equally likely. As you have only 0.75 chance of making the trip, the average expected time from X1-X2 is
average(AB,AC,AD,BA,BC,BD,CA,CB,CD,DA,DB,DC)*0.75
=average(AB,AB,AC,AC,AD,AD,BC,BC,BD,BD,CD,CD)*0.75
=average(AB,AC,AD,BC,BD,CD)*0.75

When you do X2->X3, you are at a random point A,B,C,D going to a random point A,B,C,D, excluding (AA,BB,CC,DD) and excluding X1->X2. However. X2 is equally likely to be any combination of {AB,AC,AD,BC,BD,CD}, and so there is no bias in terms of which route you cannot take. Therefore, X2->X3=average(AB,AC,AD,BC,BD,CD)*0.50

X3->X4=average(AB,AC,AD,BC,BD,CD)*0.25


average(ZA,ZB,ZC,ZD)=(1+2+3+4)/4=2.5
average(AB,AC,AD,BC,BD,CD)=(5+6+7+8+9+10)/6=7.5
Expected trip=2.5*1.0+7.5*0.75+7.5*0.50+7.5*0.25=13.75

Therefore, the expected trip length is 13.75 minutes
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Re: Riddles

Post by sunl »

Riddle 3:

->To skip the complicated proof and read easyman's strategy, go to end.

-----------

You must make 3 trips. In order for 4 people to be at the other side
Trip 1: 2 go, 1 return; Trip 2: 2 go, 1 return; Trip 3: 2 go. =2*3-1*2=4

Therefore there must always be 2 people heading to point B and 1 person heading back to point A. Unless you want to make more than 3 trips.

We have
Madness: alone 2 hours airplane
Bjenn: w Madness 4 hours truck
Orcc: w Madness/Bjenn 10 hours forklift
VilleJ: 14 hours w anyone dolly

Let's rename to
Madness=A, Airplane=α
Bjenn=B, Truck=β
Orcc=C, Forklift=γ
VilleJ=D, Dolly δ
Point A=(X)
Point B=(Y)

So now we have
C uses γ, takes him 10 hours, and accepts A & B as passengers
=C: γ 10 AB

Summarizing the problem:
A: α 2
B: β 4 A
C: γ 10 AB
D: δ 14 ABC
The vehicles available are:
(X)={β,γ,δ}
(Y)={α,α,γ,δ}

----

Solution:
The difficult challenge is getting everyone to (Y).

(logic 1): Nobody likes D and so D cannot ride on α,β,γ. The only way for D to go to (Y) is by taking δ. Therefore we will make a trip using δ. A,B or C can be passengers. This will be called trip T(D+?), where ? is the passenger.

(logic 2): A is on the opposite side of α. Therefore, A must be carried to (Y) to be able to use his α. However, once he is carried to the other side, he basically can use α to either go forward or back because there will be 1 α on each side. Therefore, we "unlock" A's abilities by bringing him to the other side one time.

(logic 3): B is on the same side as β. Therefore, B travel between (X) and (Y). However, as soon as someone brings B across (i.e. C or D bring B across = T(C+B) or T(D+B)), B is on the opposite side of β and so B cannot travel anymore. This is bad if we need someone fast to travel, unless A can travel back as A is faster. However, it's impossible to bring A to the other side and make him stay there, since he'll always be unlocked once he reaches the other side (logic 2), and it's always fastest to bring him back. Therefore, we only want to do T(C+B) or T(C+D) on the last trip, if at all.



(1)
Let's say D takes A as a passenger T(D+A). Then obviously, A will be coming back by α. The second trip can be T(B+A) or T(C+B) or T(C+A). T(C+B) is not a good idea because (logic 3). Therefore, the two possibilities are T(B+A) or T(C+B).

So the two reasonable routes are
T(D+A)+T(A)+T(B+A)+T(A)+T(C+A) and
T(D+A)+T(A)+T(C+A)+T(A)+T(B+A); which are the same thing in a different order. Therefore, the best route starting with T(D+A) takes 14+2+10+2+4=32

(2)
Let's say D takes B as passenger T(D+B). By (logic 3), this has to be the last trip. Since we want D to take B across, we cannot have T(B+A) happen because then B will have already crossed and D will be all sad alone on the original side (X). Therefore, we start with
T(C+A)+T(A)+...
The next trip has to be T(B+A) or T(D+B). We already decided that T(D+B) is a bad idea since it is not the last trip. However, we also decided that T(B+A) cannot happen because then, T(D+B) is impossible. Therefore, there are no good options with T(D+B)
->No reasonable routes

(3)
Let's say D takes C as passenger T(D+C). Then, C has to come back. But then, the only person that wants to ride with C is D and so C will have to come back. That already makes T(D+C)+T(C)=24, meaning we have 8 hours left to do 1.5 full trips, but since C is back on (X), C has to cross again by way of γ, which costs 10 hours. T(D+C)+T(C)+T(C+?)=more than 32. Even if you do T(D+C) as the last trip, C has to do at least 3 trips so it's not worth it.

Therefore, the fastest trip is T(D+A)+T(A)+T(B+A)+T(A)+T(C+A) or T(D+A)+T(A)+T(C+A)+T(A)+T(B+A).

----------

Easyman's strategy:

We need to bring the vat across 3 times and bring the 4 mans across.

Nobody wants to let VilleJ ride on their vehicle. So VilleJ is sad and the only way for him to reach the other side is to take his dolly to reach point B. VilleJ takes Madness, because Madness can fly back really fast and then go with Orcc by forklift to point B, and then fly back a second time and go with Bjenn to point B by truck. If we don't take Madness across with VilleJ, then we have to go back by truck or forklift which are both really slow compared to airplane.

Therefore, the fastest trip is
VilleJ+Madness by dolly
Madness flies back
Orcc+Madness by forklift
Madness flies back
Bjenn+Madness by truck

Orcc and Bjenn can go in whatever order.

Obviously VilleJ is best for contribution to elmans, despite Pokemon Go and dolly. But Madness, Bjenn and Orcc are cool too.
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Re: Riddles

Post by gimp »

Sunl:

excellent job on the second riddle. I used a different route and got 13.61 as the expected, your 13.75 gives me reassurance though that our answers are sound, after checking using different numbers, our results differ not out of coincidence but because of rounding error from multiple calculations. If you are interested in how I did it I can send you my excel sheet. Again very impressed at your speed and accuracy with this, I thought this riddle would take much longer, but I underestimate elma mans intelligence around here!

As for the third riddle, you are wrong. Ville_Js time is 15 hours, not 14. Meaning your answer should be 33, in which case I ask you to read the end of the hint. Good luck.
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Re: Riddles

Post by Madness »

Riddle #1:

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Re: Riddles

Post by Madness »

Riddle #3:

You have four workers. Madness is the fastest because he has an airplane. Ville_j with his dolly is the slowest, but Orcc is lazy and wakes up 2 hours late and happens to be even slower. Bjenn has a truck, which is pretty fast, but he takes a break and never returns back.

Answer:
1. Madness
6. Ville_j
7. Orcc
29. Bjenn
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Re: Riddles

Post by Bjenn »

Hahhaha wtf =D
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Re: Riddles

Post by gimp »

Madness great additional input on riddle 1! I'm sorry but your riddle 3 explanation and answer is incorrect. Riddle 3 still waiting to be figured out.
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Re: Riddles

Post by sunl »

Ah okay I misunderstood as
(once the vat is unloaded at point B, it must travel back to point A to pick up another shipment)
meaning that you always had to travel with vat, therefore zigzag was mandatory

Then:

VilleJ/Orcc -> +15 (without tub)
Bjenn/Madness -> +4
Madness <- 2
Madness -> 2
Madness <- 2
Madness -> 2
=27 hours
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Re: Riddles

Post by gimp »

sunl wrote:Ah okay I misunderstood as
(once the vat is unloaded at point B, it must travel back to point A to pick up another shipment)
meaning that you always had to travel with vat, therefore zigzag was mandatory

Then:

VilleJ/Orcc -> +15 (without tub)
Bjenn/Madness -> +4
Madness <- 2
Madness -> 2
Madness <- 2
Madness -> 2
=27 hours
Sunl you do have to travel back with the vat to point A, the rules you followed in your first answer are fine, its just not the fastest order. the reason why I said don't Zig-Zag is because I anticipated that somebody would use the same logic as you and come up with the answer 33 minutes(Zig-Zag internal = 33), hence I said "don't Zig-Zag". I'm confident you will find the answer though, and don't forget to state the best kuski afterwards!
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Re: Riddles

Post by sunl »

Fine.

Bjenn+Madness
Madness
VilleJ+Orc
Bjenn
Bjenn+Madness

4+2+15+4+4=29

I messed up my logic 3
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Re: Riddles

Post by gimp »

29 is the correct answer but you didn't state the best kuski so it is still only partially correct
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Re: Riddles

Post by sunl »

Well then
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Re: Riddles

Post by gimp »

Sunl i am going to give you credit for riddle 3. The answer was 29, but i was hoping you would say 'Bjenn' after 29, to state who the best kuski of the four were. To make the complete answer be "29. Bjenn". You didn't do that, but you did all the hard work for this answer and maybe have not been privy to this inside joke and that isn't fair, so i give you kudos to your amazing intellelct so far with these riddles :)
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