The Unofficial NSX Puzzler Thread (in honor of Tom Maglozzi)

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One of my favorite radio shows is Car Talk. If you are a fan, you know that one of the Click ‘n’ Clack Tappet Brothers, Tom Magliozzi, passed away in 2014 due to complications of Alzheimer’s disease.

In honor of Tom, I start this Unofficial NSX Puzzler thread. Feel free to post puzzlers. While any puzzler is fair game, I would encourage you to come up with your own puzzlers. Let’s see how inventive we can be.

I’ll start:

On a beautiful spring Sunday, at just about noon straight up, a young couple was traveling home in their SUV, returning home from Grandma’s house. In the backseat were their two young children, and the family dog all the way in the back. It was a lovely day, Mom was driving, Dad was in the front passenger seat. They were on an Interstate highway, traffic was really light, and Mom was enjoying the relaxed drive, traveling just a few MPH below the posted speed limit. But. . . .

Since Grandma had pumped the kids full of M&M’s, they were still on a sugar high and were rambunctious. Dad was playing the car radio loudly, and the dog was in the back barking indiscriminately just to join in on the noise.

Mom wanted a peaceful ride, so she turned off the radio, and said to the kids “Let’s play the quiet game. If you guys can stay quiet for exactly ONE MINUTE, we’ll stop at the park playground on the way home.”

So, Mom was driving along, and counted down. “OK, 3 . . . 2 . . . 1 . . . START!” The kids shut up, the dog laid down, and all that was heard was the quiet thrum of the engine and the tires rolling down the road.

EXACTLY one minute later, Mom said “OK, YOU DID IT, so we’ll stop at the park on the way home!”

Now, Dad says “How did you know that was exactly one minute? I know your wristwatch doesn’t have a second hand on it, you didn’t look at your smart phone, and the car’s digital clock has been broken for six months.”

Mom says “Oh, I knew.” Dad says “Oh, yeah?!? How?”

So, how did Mom know when exactly one minute had passed?
 
She was driving 60mph, which means one mile per minute. Looking at the odometer, when one mile was covered then one minute had passed.
 
That's what I was thinking too, except she was "traveling a just a few MPH below the posted limit".

Yup, that's it, 60 MPH, one mile per minute. Actually, I had passing mileposts in mind rather than using the car odometer (or even better yet, I've seen "Begin Measured Mile" and "End Measured Mile" signs but it's been years since I've seen those), but odometer works, too.

OK, anyone have another puzzler, maybe a bit harder than this one?

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OK, here's another one.

What do these words have in common?

- Education
- Reputation
- Evaluation
- Encouraging

- Renumeration
- Questionable

The first four words comprise a special case of these words in common. What is it?

And finally, while this word shares commonality with all the other words listed here, what makes this word even more special?
- Simultaneously
 
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All the words have all the vowels in them "A,E, I, O, U". The last word has "and sometimes Y"

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This was actually a puzzler a while back and I had to call a math PhD friend of mine to explain the correct answer to me:

There are two playing cards. One card is red on one side and black on the other. The other playing card is red on both sides. You are handed a playing card that is showing the red side. You can not see the other side of the card nor can you see the other playing card. What is the probability that the other side of the card you are holding is red??
 
All the words have all the vowels in them "A,E, I, O, U". The last word has "and sometimes Y"

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This was actually a puzzler a while back and I had to call a math PhD friend of mine to explain the correct answer to me:

There are two playing cards. One card is red on one side and black on the other. The other playing card is red on both sides. You are handed a playing card that is showing the red side. You can not see the other side of the card nor can you see the other playing card. What is the probability that the other side of the card you are holding is red??

25%
 
Lessee, the card in my hand, the side facing me is red. So that means that the other side can be either black or red, since the red side facing me is inconclusive as to whether the card I have is red/black or red/red. So, I'd say it's 1 in 2, or 50/50.

And yes, all those words I posted had all the vowels, and the "special" word even had the ubiquitous "and sometimes y". :biggrin: BTW, the first four words use each vowel only once.
 
How about 2/3

Bingo. It was explained to me that probability riddles are very counter intuitive. The math PhD that explained it to me told me that at a party he and a fellow math PhD argued over a probability riddle for over an hour with neither side admitting defeat (that must have been some party!!!). So the way it has to be looked at is not like there are just two cards but four sides of the cards. Since there are three red sides and you are seeing one red side that leave two red sides to one black side left. 66% of seeing a red side.
 
I have to disagree.

The fact that there are three sides left does not make the number of possibilities remaining three. The number of remaining possibilities is only two. It's either red or it's black. That there are three sides left is irrelevant. If there are really three possibilities left, please list them. :biggrin:
 
I have to disagree.

The fact that there are three sides left does not make the number of possibilities remaining three. The number of remaining possibilities is only two. It's either red or it's black. That there are three sides left is irrelevant. If there are really three possibilities left, please list them. :biggrin:

+1

I think 2/3 would only work if the faces of the cards were completely independent. However, they are not. Basically, you have 2 cards: red/red and red/black. You have a 1 in 2 chance of getting either card. Now, prior to getting a card, you have a 3 in 4 chance of having a red face displayed (assuming the cards could be randomly tossed so that any one side has an equal chance of being shown) but you still have a 1 in 2 chance of receiving either card. Having been dealt a card with a red face, it still remains a 1 in 2 chance that the other side is red.

This is completely different than, say, having 4 coins: 1 black and 3 red. Since they are completely independent of each other, then you have a 3 in 4 chance of drawing a red coin. If you do draw a red coin, then you have a 2 in 3 chance of drawing another red, and so on.
 
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I am not a math wiz by any stretch of the imagination but, like I said, these probability riddles don't make sense. The answer is 2/3 like it or not. If anyone has some spare time and would like to conduct an experiment with a statistically significant number of samples (say 1000) you will see that the number of red sides you get is in the area of 663 not 500. The artificial part is where someone has to hand you a card with a red side.

Check for yourself.

http://www.braingle.com/brainteasers/teaser.php?op=2&id=279&comm=0

Try this one on for size. It's even worse:

Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1 , and the host, who knows what’s behind the doors, opens another door, say #3 , which has a goat. He says to you, "Do you want to pick door #2 ?" Is it to your advantage to switch your choice of doors?
 
I am not a math wiz by any stretch of the imagination but, like I said, these probability riddles don't make sense. The answer is 2/3 like it or not. If anyone has some spare time and would like to conduct an experiment with a statistically significant number of samples (say 1000) you will see that the number of red sides you get is in the area of 663 not 500. The artificial part is where someone has to hand you a card with a red side.

Check for yourself.

http://www.braingle.com/brainteasers/teaser.php?op=2&id=279&comm=0

I checked that link. Here is the explanation:

"If you think of the 3 red sides as labeled r1, r2, and r3, then 2 of the 3 times you see a red card, it will be red on the other side."

If the side facing me is R1, then the other side could be R2, R3, or black.
If the side facing me is R2, then the other side could be R1, R3, or black.
If the side facing me is R3, then the other side could be R1, R2, or black.

So there are the three possibilities I was looking for. BUT, let's say R1 and R2 are on the same card. Since both sides are RED, R1 and R2 could be facing you AND IT DOESN'T MATTER, since R1 is the same as R2, so even though it's possible for either R1 or R2 to be facing you, they need to be combined into once instance of a possible outcome. I agree, if I had them detailed and said you have R1 facing you, what is the chance you have either R2 or R3 on the other side, but since it's generic which Red is facing me, it's not a unique case.

I dunno, I'm starting to think maybe this thread wasn't such a good idea. My head is starting to hurt. :biggrin: Now I am 50/50 on which answer is correct.
 
It's 2/3.

It's a straightforward instance of conditional probability:
"what's the probability of A given that B has occurred?"
In this case, A is "the other side is red", B is "the side you can see is red".

As described on the page linked above (which goes into much more detail than you need),
the answer is calculated as a quotient: P(A∩B) / P(B).
The numerator (bear with me if your browser doesn't render the intersection sign) is the
probability that both A and B occur; the denominator is the probability that B occurs.

P(A∩B), the probability of both A and B occurring, is 1/2.
It happens if and only if you've drawn the red-red card,
which you have a 50-50 chance of doing.

P(B), the probability that the side you see is red, is 3/4.
There are four sides that could have been facing up,
all four equally likely to be drawn, and three of them are red.

To calculate P(A given B), divide 1/2 by 3/4 to get 2/3.


As often happens with probability questions, there are multiple ways to reach the
right answer. The explanation on the braingle.com page cited earlier is sound;
each of the three cases ("R1", "R2", "R3" showing) happens 1/3 of the time.

I checked that link. Here is the explanation:

"If you think of the 3 red sides as labeled r1, r2, and r3, then 2 of the 3 times you see a red card, it will be red on the other side."

If the side facing me is R1, then the other side could be R2, R3, or black.
If the side facing me is R2, then the other side could be R1, R3, or black.
If the side facing me is R3, then the other side could be R1, R2, or black.

You're allowing for all kinds of variations in labeling, which isn't necessary.
They're just labels; it's OK to settle on one labeling and consider that.
Let's say one card has R1 and R2, the other card has R3 and black.
Then a version of the cases you list would read like this:

If the side facing me is R1, then the other side must be R2.
If the side facing me is R2, then the other side must be R1.
If the side facing me is R3, then the other side must be black.

So there are the three possibilities I was looking for. BUT, let's say R1 and R2 are on the same card. Since both sides are RED, R1 and R2 could be facing you AND IT DOESN'T MATTER, since R1 is the same as R2, so even though it's possible for either R1 or R2 to be facing you, they need to be combined into once instance of a possible outcome. I agree, if I had them detailed and said you have R1 facing you, what is the chance you have either R2 or R3 on the other side, but since it's generic which Red is facing me, it's not a unique case.

It's just not that complicated. If you see a red face,
there's a 1/3 chance it's R1, a 1/3 chance it's R2, and a 1/3 chance it's R3.
 
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I'm sorry I am not trying to be obstinate. I cannot wrap my head around this.

If I have any of the red sides facing me, then there are only two possibilities left. 1) I am holding the red / red card. 2) i m holding the red / black card. The other side can be only red or black. Sorry I cannot see it any other way. If I am wrong then it is my problem.

P.S. I have a math degree. At this point wondering if it were undeserved. :biggrin:
 
I'm sorry I am not trying to be obstinate. I cannot wrap my head around this.

If I have any of the red sides facing me, then there are only two possibilities left. 1) I am holding the red / red card. 2) i m holding the red / black card. The other side can be only red or black. Sorry I cannot see it any other way. If I am wrong then it is my problem.

Yes but those two possibilities aren't equally likely.

If you draw a card, look at only one side, and see that it's red,
it's twice as likely that you drew the red-red card.

If you're not convinced, consider a more lopsided example:
There are two card decks in front of you: a deck with 52 red cards,
and a deck with one red card and 51 black. You pick one of the
decks, draw a single card from it, and see that it's red. At that point,
you know there's an excellent chance you picked the all-red deck.
 
This reminds me of the "Monty Hall" type of puzzlers.

Suppose you're on a game show, and you're given the
choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?


Even after reading several explanations on this problem, including the one on Wikipedia (link below), I still can't wrap my brain around it. But I've tested it experimentally and it absolutely works as advertised.

http://en.wikipedia.org/wiki/Monty_Hall_problem
 
OK, here’s a classic, very “oldie but goodie” and it’s been around. A lot. Hopefully some of the young pups here can appreciate it.

Three guys in a band travel to a motel and each rent a room. They each pay $10 and head off to their rooms. (Yeah, it’s a really old puzzler).

The owner of the motel thinks to himself “Man, those three guys were really nice and friendly.” So he takes a $5 bill out of the cash register and asks the bellhop to “go give this to those three guys that just checked in.”

On the way to the rooms, the bellhop thinks “Heck, I can’t divide $5 into three.” So he digs in his pocket and pulls out five singles to break the $5 bill, then goes to each room and gives each guy $1. He keeps $2 as his own tip.

So, each guy now paid $9. Three times $9 = $27 paid. Plus $2 that the bellhop kept for himself, adds up to $29.

They originally paid $30 for the three rooms. So, where’s the other $1?
 
Hotel manager has $25, bell hop has $2 = $27. Each guest has $1 =$30.

Bunch of smart people here on this forum. Yes, the amount paid is $25, not $30, so from $27 you have to SUBTRACT to get to $25, not ADD to get to $30.

Not much gets by this group. I am honored to be among you. :biggrin:
 
A Car Talk classic:

Kid shows up at the car parts store in a panic. He goes up to the counter and tells the clerk behind the counter, "I borrowed my dad's car without his permission while he is out of town. I was checking under the hood to make sure everything looked ok before he gets back tonight and I'm missing a part. I need a new 710. I don't know what happened to the old one. I guess it just fell off."

Puzzled, the clerk asks "What's a 710?" The kid answers, "It's this round black thing that says '710' on it."

The clerk still can't figure it out so he asks the kid to draw a picture of it. "Oh, I can get you one of those," says the clerk upon seeing the drawing. He goes to the back and comes back with a part. "That's it," the kid says.

The question is, what was it?
 
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