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While Mike the Wiz created a "humor" topic, it seems we don't have anything for puzzles and such.   Now we do.

I saw this and thought it is interesting .....

Last week, you were seated in an audience, when T-shirts were being launched via cannon in your direction. The rows of seats in the audience were all on the same level, they were numbered 1, 2, 3, etc., and the T-shirts were being launched from directly in front of Row 1.

Additionally, there was no air resistance, and the particular model of T-shirt cannon being used was able to launch T-shirts to the very back of Row 100 in the audience, but no farther.

If the T-shirt was launched at a random angle between zero degrees (straight at the unfortunate person seated in Row 1) and 90 degrees (straight up), which row should you have been sitting in to maximize your chances of nabbing the T-shirt?

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10 hours ago, piasan said:

If the T-shirt was launched at a random angle between zero degrees (straight at the unfortunate person seated in Row 1) and 90 degrees (straight up), which row should you have been sitting in to maximize your chances of nabbing the T-shirt?

I don't know if my answer is silly given the speed it might fire, but the first row, so you can catch it when it fires. ;) Or else if yo mean, "nabbing" literally, again row 1, in order to "nab" it. 

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Assuming all shots at all angles are fired with equal force then would Row 100 be the place to be ? Unless the t-shirt was fired at zero degrees straight at Row 1, in the absence of air resistance all would follow a curve to the back row ?

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Of course if it's Fourcade doing the shooting I'll say it right now; we have little chance of survival. ;) (and those tiny targets with a diameter of 4.5 inches are set at 50 metres and I think 1.8 inches for shooting in the prone position after skiing flat out and your heart going like the clappers.)

 

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4 hours ago, wibble said:

Assuming all shots at all angles are fired with equal force then would Row 100 be the place to be ? Unless the t-shirt was fired at zero degrees straight at Row 1, in the absence of air resistance all would follow a curve to the back row ?

Based on my "back of the envelope" spreadsheet calculations, the last row (row #100) is by far the best place to be.   I get that the last row has about 9% chance of getting the T-shirt; the 99th row has about 3.7% chance, and it goes down from there.

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9 hours ago, wibble said:

Assuming all shots at all angles are fired with equal force then would Row 100 be the place to be ? Unless the t-shirt was fired at zero degrees straight at Row 1, in the absence of air resistance all would follow a curve to the back row ?

No.  The absence of air resistance means the landing row depends only on the angle of launch and gravity.

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10 hours ago, mike the wiz said:

I don't know if my answer is silly given the speed it might fire, but the first row, so you can catch it when it fires. ;) Or else if yo mean, "nabbing" literally, again row 1, in order to "nab" it. 

I'm not too sure.   If we place the rows 1m apart, the cannon would need to launch at 31.4 m/s to reach the back row.  That's 113 km/hr (71 mph).  You'd have about 0.032 seconds to react and catch the shirt before you (literally) eat it.

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1 hour ago, piasan said:

I'm not too sure.   If we place the rows 1m apart, the cannon would need to launch at 31.4 m/s to reach the back row.  That's 113 km/hr (71 mph).  You'd have about 0.032 seconds to react and catch the shirt before you (literally) eat it.

Yeah I thought, "nabbing" stood a better chance, I thought because that means, "stealing", it may be one of those red-herring type riddles where they make you think it's more complicated an answer than it really is. Every time I expect a red-herring riddle I don't get one and every time I don't expect one I do. :gaah:

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35 minutes ago, mike the wiz said:

Yeah I thought, "nabbing" stood a better chance, I thought because that means, "stealing", it may be one of those red-herring type riddles where they make you think it's more complicated an answer than it really is. Every time I expect a red-herring riddle I don't get one and every time I don't expect one I do. :gaah:

Hey, it works if you have really, really fast reflexes.

You overthought the puzzle.   You are sentenced to 3 70mph t-shirts from a distance of one meter. :rotfl3:

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11 hours ago, wibble said:

Assuming all shots at all angles are fired with equal force then would Row 100 be the place to be ? Unless the t-shirt was fired at zero degrees straight at Row 1, in the absence of air resistance all would follow a curve to the back row ?

No.  You can hit every row by choosing the right elevation angles. For example you can hit the 1st row with a near vertical launch angle and a near horizontal launch angle.

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So, this is a well known probability problem called the Monte Hall Problem and can be found on the internet.  But let's see how well people can think this one through:

Monte presents you three closed doors where behind one door is a brand new car...and behind each of the other two doors is a dish of doggie doo doo (most people want to win the car).

  • You pick one of the three doors
  • Monte lifts up one of the two remaining doors that you did not choose that contains a dish of doggie doo doo (he can always do this given that there are two doors with the doggie doo doo).
  • Question:  Should you switch your choice to the remaining door (the one you didn't choose originally and was not lifted up by Monte) or does it not matter??

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I'm very familiar with it and actually used this one in math classes.  

First point .... normally the other two doors weren't doggie doo doo.   One was doggie doo doo and the other was used cat litter.   :P

The Myth Busters did an episode on this one too.

It's a good one.   I'll send the answer to SN privately .....

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1 hour ago, StormanNorman said:

So, this is a well known probability problem called the Monte Hall Problem and can be found on the internet.  But let's see how well people can think this one through:

Monte presents you three closed doors where behind one door is a brand new car...and behind each of the other two doors is a dish of doggie doo doo (most people want to win the car).

  • You pick one of the three doors
  • Monte lifts up one of the two remaining doors that you did not choose that contains a dish of doggie doo doo (he can always do this given that there are two doors with the doggie doo doo).
  • Question:  Should you switch your choice to the remaining door (the one you didn't choose originally and was not lifted up by Monte) or does it not matter??

It's got to have something to do with the fact that he chooses a door he knows has doggie doo doo. (what the heck is doggie doo doo? Poop?)

If I choose the one with the car, then he can pick any and that tells me nothing. But if I chose one with doggie doo and he picks the other with doggie doo, then the one I didn't choose would be the car. But how to know what he knew?

So it seems to me if you switch and he knew his choice was the only remaining doggie then you get the car, but if he knew both remaining were doggie you get the doggie and have lost the car.

If you don't switch and he knew his choice was the only remaining doggie you get the doggie, but if he knew both remaining were doggie you get the car. 

(and that's really hard to keep in my brain without getting confused because of a learning disability with memory that I have, slight dyslexia, I tend to flip things around a lot accidentally)

So he either knows there's two remaining or knows there's one remaining, there doesn't seem to be a way to determine which one he knew but it seems to me you can't escape that it's a 1 in 2 chance so it doesn't matter if you switch.

But I just wonder if probability changes here, I am confused as to whether switching would count as two bites of the cherry within the notation of probability given we now know what is behind one door? Would switching then count as 2 attempts because the first attempt brought knowledge to us about which one wasn't the car? 

But if you don't switch doesn't that make it one attempt? In which case you should switch.

I think I need to take Piasan's probability class. It would help if I wasn't so rusty on probability. :blush:

(my numbers are probably wrong; with two attempts 1 in 3 and 1 in 2, so 2 in 5 is it?)

If you switch I make that 1 in 2.5 and if you don't 1 in 3. In which case you should switch but I may have done that wrong with the division, I've forgotten how you figure that out.)

(Don't laugh, I am self-taught and know little to nothing about probability and am aware I've probably confused myself.) Lol 

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1 minute ago, mike the wiz said:

It's got to have something to do with the fact that he chooses a door he knows has doggie doo doo. (what the heck is doggie doo doo? Poop?)

If I choose the one with the car, then he can pick any and that tells me nothing. But if I chose one with doggie doo and he picks the other with doggie doo, then the one I didn't choose would be the car. But how to know what he knew?

So it seems to me if you switch and he knew his choice was the only remaining doggie then you get the car, but if he knew both remaining were doggie you get the doggie and have lost the car.

If you don't switch and he knew his choice was the only remaining doggie you get the doggie, but if he knew both remaining were doggie you get the car. 

(and that's really hard to keep in my brain without getting confused because of a learning disability with memory that I have, slight dyslexia, I tend to flip things around a lot accidentally)

So he either knows there's two remaining or knows there's one remaining, there doesn't seem to be a way to determine which one he knew but it seems to me you can't escape that it's a 1 in 2 chance so it doesn't matter if you switch.

But I just wonder if probability changes here, I am confused as to whether switching would count as two bites of the cherry within the notation of probability given we now know what is behind one door? Would switching then count as 2 attempts because the first attempt brought knowledge to us about which one wasn't the car? 

But if you don't switch doesn't that make it one attempt? In which case you should switch.

I think I need to take Piasan's probability class. It would help if I wasn't so rusty on probability. :blush:

(my numbers are probably wrong; with two attempts 1 in 3 and 1 in 2, so do you divide by the 2 attempts? 3 divided by 2 is 1.5)

If you switch I make that 1 in 1.5 and if you don't 1 in 3.

(Don't laugh, I am self-taught and know little to nothing about probability and am aware I've probably confused myself.) Lol 

Spot on, Mike....good job.  Most people believe that it doesn't matter if you switch or not.  But, like you said, Monte Hall has a priori knowledge of what is behind the doors.....and that is important.  Originally, you have a 1/3 chance of being right and 2/3 chance of being wrong.  That doesn't change after Monte reveals a door with doggie doo doo.  So, that 2/3 probability falls on the remaining door.

And, yes, doggie doo doo is an American version of doggie poop.

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7 minutes ago, StormanNorman said:

Spot on, Mike....good job

Thanks for the puzzle Norman, I found it very interesting.

Of course I had to stare at the screen thinking about it until I was red in the face. :rotfl3:

But it's nice to get one for a change, I can't believe I go it right for once it's so easy to get confused by these tricky puzzles. 

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Problem .... how to turn a bus around ..... solution below .....

Edited by piasan
photoshop maybe ????
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On 5/18/2020 at 9:11 PM, piasan said:

roblem .... how to turn a bus around ..... solution below .....

If that video is authentic, then that is the craziest dude on the planet....Evel Knievel move over....

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10 hours ago, StormanNorman said:

If that video is authentic, then that is the craziest dude on the planet....Evel Knievel move over....

Yeah.... I'm suspicious that's why I went back and made the edit about photoshop you'll see under the video clip.  

One thing is you have a really narrow road on top of what looks like a knife-edge ridge and a fairly steep hill approaching it.   The topography just doesn't look right.

The other thing..... where is the camera located to get such a good perspective?

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3 hours ago, piasan said:

Yeah.... I'm suspicious that's why I went back and made the edit about photoshop you'll see under the video clip.  

One thing is you have a really narrow road on top of what looks like a knife-edge ridge and a fairly steep hill approaching it.   The topography just doesn't look right.

The other thing..... where is the camera located to get such a good perspective?

:rotfl2::rotfl3:WOW!!! I can NOT be a bit suspicious, OH NO i'm on the "IT'S FAKE" camp 1000% LMBO!!! My gut was in knots just watching this 'Fiasco'!!!!

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On 5/20/2020 at 3:33 PM, StormanNorman said:

If that video is authentic, then that is the craziest dude on the planet....Evel Knievel move over....

omg... I hope the bus was empty... that is really scary :(

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On 5/18/2020 at 9:11 PM, piasan said:

Problem .... how to turn a bus around ..... solution below .....

 

and i thought i was crazy.

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indeed the solution is there but... too risky right? Why not go a little bit more in front or just go back and turn in a safer place...? :(

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2 hours ago, LizaMiller said:

indeed the solution is there but... too risky right? Why not go a little bit more in front or just go back and turn in a safer place...? :(

Like maybe go forward to where the camera is mounted?  Surely that would be some kind of wide spot in the road.  :think:

I think photographic evidence ain't what it used to be.

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On 5/25/2020 at 12:16 PM, piasan said:

Like maybe go forward to where the camera is mounted?  Surely that would be some kind of wide spot in the road.  :think:

I think photographic evidence ain't what it used to be.

haha indeed :)

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