petals around the rose

[Raryn]

You are found guilty of murder and given the death penalty. Your state has an interesting way of determining how you will die. The judge says "You can tell me one statement right now. If it is true, you will be hung. If it is false, you will be drowned."

What do you tell him to avoid being killed?

---

Members don't see this ad.

 

[GoldShadow]

You are found guilty of murder and given the death penalty. Your state has an interesting way of determining how you will die. The judge says "You can tell me one statement right now. If it is true, you will be hung. If it is false, you will be drowned."

What do you tell him to avoid being killed?

"I will be drowned."

 

[Raryn]

"I will be drowned."

 

[vadd0]

You are found guilty of murder and given the death penalty. Your state has an interesting way of determining how you will die. The judge says "You can tell me one statement right now. If it is true, you will be hung. If it is false, you will be drowned."

What do you tell him to avoid being killed?

You're sexy.

 

[JackInTheBox]

thats not a good analogy for this system. we are not dealing with a coin in isolation, we are dealing with the coin as part of a system. its not "a coin", its "the other coin"

Exactly, we're not dealing with theoretical coins in isolation. The odds of the second coin being heads is not 1/3. It's part of a system that at the time the student observes it, has already collapsed into one of the 3 possibilities.

If you repeat the experiment thousands of times and only look at all the times at least one coin is heads, the number of times that second coin is also heads will approach 1/3, but after any single flip, that second coin is either heads or it isn't. It doesn't exist in some nebulous quantum state until the student observes it.


Let's try another analogy. I have a jar full of red and blue balls. I randomly pick one out and ask you to guess which color I have. The odds of you randomly guessing correctly is 1/2, but the odds that I have a red ball is either 0 or 1 because I've already picked it. It's not going to spontaneously change color 1/2 of the time. The same goes for the second coin. It's already been flipped.


It really is all semantics, but I'm just really anal about the wording of probability riddles because I love statistics

Edit: That was post 666 for me. How fitting that it should be one where I'm playing Devil's Advocate

 

[deleted74029]

I've got one, everyones probably already heard it though.

3 guys check into a hotel thats charges 30 bucks a night. Each guy pays 10 dollars and heads to the room. The desk clerk realizes that the room is actually 25 bucks and not 30 so he hands the bellboy 5 bucks to give back to the 3 guys. Realizing that he cannot split 5 dollars 3 ways he gives each guy 1 dollar back and keeps the other 2 for himself. So instead of paying 10 for the room each guy actually paid 9 for the room. 9 x 3 = 27 + the 2 bucks the bellboy kept = 29 so where's the other dollar?

 

[ChubbyChaser]

actually 66% not 50%

I still dont understand why its 66% and not 50%.???

 

[Raryn]

I still dont understand why its 66% and not 50%.???

Because the host always opens a door that was false.

Think of it in an expanded way:

There are a billion doors, each one of which has a 1/1 billion chance of having a car behind it. You pick one, and then the host opens all of the other doors except one, knowing each open door will not have the car.

He offers to switch his one remaining door for yours. Should you do it?

Your door has a 1/1000000000 chance of having the car. The other door has a 999999999/1000000000 chance of having it. Why? Because you had that chance of NOT picking the right door.

 

[tdittyx2x3]

This took me about 50 rolls and an hour.. then I go downstairs and give my mom the challenge and she gets it in 3 rolls.

 

[michigator04]

Because the host always opens a door that was false.

Think of it in an expanded way:

There are a billion doors, each one of which has a 1/1 billion chance of having a car behind it. You pick one, and then the host opens all of the other doors except one, knowing each open door will not have the car.

He offers to switch his one remaining door for yours. Should you do it?

Your door has a 1/1000000000 chance of having the car. The other door has a 999999999/1000000000 chance of having it. Why? Because you had that chance of NOT picking the right door.

You and I seem to approach these finite number problems differently. I look at that situation as there is 1 door with a car behind it, and it's either yours or his last one. The original odds on any door, including both yours and the other remaining door, were 1/10^9 but with the elimination of the other 999999998 (incorrect) doors, there is now a 1/2 chance with either door.

I think of it like in the NCAA BBall tournament. There are 64 teams to begin with. If you assume that all teams have an equal chance of beating any other team, then a bet on Team A (at the start of the tourney) would have a 1/64 chance of succeeding. If that team makes the championship game against Team B, the odds Team A winning do not remain at 1/64 (or else you would have a very poor bookie), because 62 other teams were eliminated. Likewise Team B would not have a 63/64 chance of beating you just because you didn't pick them initially. There are two possible outcomes: team A wins or team B wins.

 

[Lukkie]

in your basketball example, it would require that Team A could NEVER lose until the final game. so your analogy doesn't work, because indeed Team A will play in each round and can lose.

 

[variablistic]

So, instead of jumping to a Billion, let's just leave it at 3.

Let's say you always pick door #1.

Let's run through the possibilities.

1st possibility, car is behind door 1, host opens door 2 or 3, you keep your choice, you win.

2nd
car is behind door 2, host opens door 3, you keep your choice, you lose.

3rd
car is behind door 3, host opens door 2, you keep your choice you lose.

_____________________________________________

You can also think of it this way:

When you pick first, you have a 1/3 chance of being right. This means there was a 2/3 chance of being right among the other choices. When the host shows you which door it's not, that 2/3 chance of it being the other two doors still stands, and it would behoove you to change your choice.

 

[Raryn]

You and I seem to approach these finite number problems differently. I look at that situation as there is 1 door with a car behind it, and it's either yours or his last one. The original odds on any door, including both yours and the other remaining door, were 1/10^9 but with the elimination of the other 999999998 (incorrect) doors, there is now a 1/2 chance with either door.

I think of it like in the NCAA BBall tournament. There are 64 teams to begin with. If you assume that all teams have an equal chance of beating any other team, then a bet on Team A (at the start of the tourney) would have a 1/64 chance of succeeding. If that team makes the championship game against Team B, the odds Team A winning do not remain at 1/64 (or else you would have a very poor bookie), because 62 other teams were eliminated. Likewise Team B would not have a 63/64 chance of beating you just because you didn't pick them initially. There are two possible outcomes: team A wins or team B wins.

Its more like:

I give you a list of 64 teams, and say "one of these teams has already won." You pick a team and I say, "Well, these 62 teams didn't win," leaving us with two teams. The important part here is, I will always eliminate 62 losing teams, no matter if you pick a winner or a loser.

So, if you pick one of the 63 losers, and I eliminate 62 teams, then the final remaining team is a winner and you should switch.

If you pick the one winner, and I eliminate 62 teams, then the final remaining team is a loser, and you shouldn't switch.

Since you have a 63/64 chance of originally picking a loser in this case, you should definitely switch.

 

[michigator04]

So, instead of jumping to a Billion, let's just leave it at 3.

Let's say you always pick door #1.

Let's run through the possibilities.

1st possibility, car is behind door 1, host opens door 2 or 3, you keep your choice, you win.

2nd
car is behind door 2, host opens door 3, you keep your choice, you lose.

3rd
car is behind door 3, host opens door 2, you keep your choice you lose.

_____________________________________________

You can also think of it this way:

When you pick first, you have a 1/3 chance of being right. This means there was a 2/3 chance of being right among the other choices. When the host shows you which door it's not, that 2/3 chance of it being the other two doors still stands, and it would behoove you to change your choice.

You guys may be right, but it goes against my basic process of elimination test taking algorithm. If I'm straight up guessing on a MC exam, and the prof eliminates one answer (other than the one I was leaning toward), I would not change mine to a different one just based on this.

 

[michigator04]

Fun thread, though. I've done more critical thinking while looking at all these than I have in the last year or so of perusing SDN.

 

[Lukkie]

Its more like:

I give you a list of 64 teams, and say "one of these teams has already won." You pick a team and I say, "Well, these 62 teams didn't win," leaving us with two teams. The important part here is, I will always eliminate 62 losing teams, no matter if you pick a winner or a loser.

So, if you pick one of the 63 losers, and I eliminate 62 teams, then the final remaining team is a winner and you should switch.

If you pick the one winner, and I eliminate 62 teams, then the final remaining team is a loser, and you shouldn't switch.

Since you have a 63/64 chance of originally picking a loser in this case, you should definitely switch.

you sir, definitely have a gift

 

[sirius08]

3 guys check into a hotel thats charges 30 bucks a night. Each guy pays 10 dollars and heads to the room. The desk clerk realizes that the room is actually 25 bucks and not 30 so he hands the bellboy 5 bucks to give back to the 3 guys. Realizing that he cannot split 5 dollars 3 ways he gives each guy 1 dollar back and keeps the other 2 for himself. So instead of paying 10 for the room each guy actually paid 9 for the room. 9 x 3 = 27 + the 2 bucks the bellboy kept = 29 so where's the other dollar?

I'd never heard this before, but I think I got it... the number 30 doesn't apply because the room doesn't cost 30, it costs 25 + 2 (for the bellboy) for a total of 9 each. Each man pays 8.33 for the room and .67 for the bellboy.

 

[ar2388]

ugh the answer is disappointingly simple.. i just dont think this way.
whatever

 

[deleted74029]

I'd never heard this before, but I think I got it... the number 30 doesn't apply because the room doesn't cost 30, it costs 25 + 2 (for the bellboy) for a total of 9 each. Each man pays 8.33 for the room and .67 for the bellboy.

Yeah theres actually a couple of answers to this one...its not one of those aha! type answers though. So theres not much fun in answering it.

 

[sirius08]

Yeah theres actually a couple of answers to this one...its not one of those aha! type answers though. So theres not much fun in answering it.

It was fun, I had to think about it.

 

[Vihsadas]

ugh the answer is disappointingly simple.. i just dont think this way.
whatever

I kno, right? It took me about 20 mins to get it. Frustrating!

 

[MILK07]

Took me a 30 mins, pen and paper to get Einstein's riddle. Took me about 4 minutes and 7 rolls for the Petals Around the Rose.

Good stuff!

I really liked the "I will die by drowning" one. Never would have gotten that!

 

[michigator04]

Einstein's wasn't that hard for me, but it does take pen/paper, and the wording is a little vague (is the first house L or R?).

The petals around the rose one threw me off because I kept trying to come up with some significance of the different colored dice on that website.

 

[chris1264]

it took me around 2 minutes

 

[aznb0y129]

Einstein's wasn't that hard for me, but it does take pen/paper, and the wording is a little vague (is the first house L or R?).

The petals around the rose one threw me off because I kept trying to come up with some significance of the different colored dice on that website.

Same here, I was trying to come up with all these different mathematical computations. It was taking too long so I cheated and looked it up. I probably wouldn't have gotten it

 

[LizzyM]

I had fun with Einstein's riddle. It reminded me of the old "analytical" section of the GRE (now eliminated to make way for "Analytical Writing"). As I was dropping off to sleep, I thought that it would have had an little joke inside if item 13 were changed to "The Pole smokes Prince" and if the final question were changed to "Which one raises fish?"

 

[ngkats]

1 roll
174 - IQ as tested at age 16 by a psychologist
41 - MCAT

Does correlation imply causality?
No.
The name of the game gives it away.

</ego stroking>

Thanks David Hume.

 

[deleted74029]

I think I would have figured it out faster if the "it will be even or zero" wasn't in there. That made me think that something was either multiplied or divided by an even number so I was writing out all these different calculations, I was just thinking way too complex.

 

[Raryn]

I had fun with Einstein's riddle. It reminded me of the old "analytical" section of the GRE (now eliminated to make way for "Analytical Writing"). As I was dropping off to sleep, I thought that it would have had an little joke inside if item 13 were changed to "The Pole smokes Prince" and if the final question were changed to "Which one raises fish?"

Yeah, it was fun.

Theres a "logic games" section currently on the LSAT thats very similar.

 

[starfishprime]

Theres a "logic games" section currently on the LSAT thats very similar.

George requests to sit next to Karyn, and Paulo must not be seated next to Raul. Esme must sit in the front row, and Tomasz must sit in the back. Geoffrey must sit on the aisle. How tall is Felix?

 

[Vihsadas]

I had fun with Einstein's riddle. It reminded me of the old "analytical" section of the GRE (now eliminated to make way for "Analytical Writing"). As I was dropping off to sleep, I thought that it would have had an little joke inside if item 13 were changed to "The Pole smokes Prince" and if the final question were changed to "Which one raises fish?"

I agree. It was a lot of fun!
I'm not sure that I agree with Einstein in thinking that this requires some heavy logic skills and is a way to determine who is smart and who is not. I think that the most important traits to have for success in Einstein's riddle are mental organization skills and awareness.
As long as you have a way to visibly organize your data and you are aware of "If then" as well as all of the "If nots", then it's not so bad. (Still took me a while though. )

 

[savant]

I tried figuring out the pattern by looking at the dice and then seeing what the answer was for like 10 minutes.

Then I realized that probably wasn't very productive, so from then on, I wrote down the dice values and the solutions. I was lucky to get two 0's within the first 5 rolls. They seemed to be predominantly of a certain type of number. I then rolled many times to see if I could replicate that, which I could. The solution, after figuring that out, was more simple.

 

[dhbighit]

George requests to sit next to Karyn, and Paulo must not be seated next to Raul. Esme must sit in the front row, and Tomasz must sit in the back. Geoffrey must sit on the aisle. How tall is Felix?

Obviously I am missing something here. I feel like the answer must be very simple and have little or nothing to do with the seating arrangements....????

 

[officedepot]

wikipedia has all the right answers...

 

[redlight]

Einstein's wasn't that hard for me, but it does take pen/paper, and the wording is a little vague (is the first house L or R?).

The petals around the rose one threw me off because I kept trying to come up with some significance of the different colored dice on that website.

lol ditto with the different color die. i kept thinking that one of the die was the rose and the other die were the petals and you had to add or subtract their #s in some way to get the answer.. i guess i cant think straight at this hour.. odd thing is that one of the methods i came up with randomly worked for 4 straight tries so that also threw me off. ::

 

[Raryn]

Obviously I am missing something here. I feel like the answer must be very simple and have little or nothing to do with the seating arrangements....????

That question is a joke. No logical answer to it.

Its making fun of how convoluted some of the LSAT logic games passages get.

 

[HWO]

I've got one, everyones probably already heard it though.

3 guys check into a hotel thats charges 30 bucks a night. Each guy pays 10 dollars and heads to the room. The desk clerk realizes that the room is actually 25 bucks and not 30 so he hands the bellboy 5 bucks to give back to the 3 guys. Realizing that he cannot split 5 dollars 3 ways he gives each guy 1 dollar back and keeps the other 2 for himself. So instead of paying 10 for the room each guy actually paid 9 for the room. 9 x 3 = 27 + the 2 bucks the bellboy kept = 29 so where's the other dollar?

There is no missing dollar. The three men give 10 dollars each to the clerk which means the clerk is +30 and the men are -(10)(3)=-30. The clerk gives the bellboy 5 dollars which makes it, clerk: +25, bellboy: +5 and men: -30. The bellboy then gives 1 dollar to each of the men and keeps two for himself making it, clerk: +25, bellboy: +2, and men: -27.

25=-(2-27)
=25=25

The addition in the original problem is done incorrectly to misdirect you.

I've got another one for you all that I heard recently. It goes something like: Luke wants to bring his sister, mother-in-law, and wife to a restaurant. His mother-in-law and wife don't get along and can't be left alone at any time. His mother-in-law and sisters also don't get along and can't be alone together. Luke has a car that can only seat him and one other person but he must remain in the car. How can he bring all of his family to the restaurant without having to make a detour at the hospital because of a feud?

 

[amikhchi]

isn't that just like the wolf/chicken/seed river problem?

funny how the mother-in-law would be the wife's actual mom (right?)

take mother-in-law and drop her off, come back and get the wife, drop off the wife and take the mother back to the original place, leave mother-in-law at original place and pick up sister, drop off sister to restaurant, and come back for mom... everyone there safe n sound

 

[Decicco]

I cheated and now I feel bad because I could have gotten it I was trying the most ridiculous things (looking at the color of the dice, etc)

 

[circulus vitios]

I gave up after not figuring it out after 15 minutes. The solution is the most ******* thing I've ever seen. I hate riddles.