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Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]

math.dartmouth.edu

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Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]

#71

Post by cabaalis »

mtreis86 wrote: post #69
Moodles wrote: post #57

Earlier quoted context omitted.

Reminds me of this: 1, 2, 3, 4, 5, 6, 7, 8, 9,... What comes next? 10 if it's the sequence of natural numbers 13 if it's the sequence of N such that N=2^n for natural numbers n, where N does not contain a 0 153 if it's sequence of N such that the the sum of the digits of N each raised to the power of the number of digits in N equals N.

Those problems always bothered me. I think that for any sequence of numbers there is an infinite number of next-in-sequence solutions regardless of the sequence length or numbers contained. One may be more obvious but you can put any number next and find a pattern that matches. Example - what if those are a sequence of digits in pi.

I think "What comes next?" is an incomplete question, without any context. "What comes next in the sequence of (blah)?" is a complete question with full context. But that question wouldn't make anybody feel superior.

"Identify as many sequences as you can which fit this set of numbers, and tell me the next number in each sequence" is the non-trick actually being asked.

Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]

#72

Post by 3131s »

arachnids wrote: post #50
mikenew wrote: post #49

Spent quite a while on the Wimbledon problem before giving up and reading the answer. But the answer is wrong, because they don’t play tiebreakers at Wimbledon (or any of the grand slams besides the US Open). Pretty frustrating TBH. Kind of felt the same about the padlocked boxes one too, although at least that one is more just a little ambiguous vs outright wrong.

They don't play tiebreakers in the fifth set. All other sets work normally. The answer is correct because it's only talking about the first three sets.

Yeah, but it's so easy! Why is that considered a hard problem? Just because most people are unfamiliar with tiebreakers in tennis?

Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]

#73

Post by OscarCunningham »

pyk wrote: post #68
tomp wrote: post #55

Another very counterintuitive (for me) problem: how do you do better that break even in the following game: 'A' chooses two distinct integers, writes them on slips of paper and holds one out in each hand in a fist. You choose a hand and reveal a number. You must then guess whether the other number is higher or lower than the revealed one, winning $1 if you guess right and losing $1 otherwise.

After writing this one out, this reminds me of the Monty Hall problem. In this case my guess is that you use a prior -- assume the two unknown numbers are A & B, and then assume a random integer yourself C. From there, if A (the first revealed number) is less than C, then that narrows the remaining cases giving you a 2/3 chance. If A is greater than C, that also narrows the remaining cases and gives you a 2/3 chance…

This works, but you don't get a probability of 2/3, because the cases aren't equally likely. There isn't a probability distribution for C such that for any A and B the probability of A<C<B is 1/3. We would have to have P(0<C<10)=1/3 and P(10<C<20)=1/3 and P(0<C<20)=1/3, which is impossible.

Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]

#74

Post by 3131s »

zawerf wrote: post #40

Where can I find more of these? Do you guys recommend this author's books such as https://www.amazon.com/Mathematical-Puzzles-Connoisseurs-Pet... ? (It's hard to find the right search term for this that doesn't return a lot of non-mathematical brainteasers or stuff aimed at kids)

The logician Raymond Smullyan wrote many books full of logic puzzles, this one being the most well known:

https://en.wikipedia.org/wiki/To_Mock_a_Mockingbird

Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]

#75

Post by OscarCunningham »

tomp wrote: post #55

Another very counterintuitive (for me) problem: how do you do better that break even in the following game: 'A' chooses two distinct integers, writes them on slips of paper and holds one out in each hand in a fist. You choose a hand and reveal a number. You must then guess whether the other number is higher or lower than the revealed one, winning $1 if you guess right and losing $1 otherwise.

This problem frustrates me, I'm not quite convinced that it's well defined as written.

My first instinct is to say "Based on your knowledge of Alice, assign a probability distribution over the pairs of integers she might pick. Then when one is revealed you should condition on that fact. Then just pick whichever of higher or lower is most likely."

The problem setter will object that we have no way of assigning a probability distribution to Alice's actions. But if that's true, what does it mean to "do better than break even"? Apparently it doesn't mean "win with probability >50%", because there is no probability of winning.

Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]

#76

Post by louisswiss »

ryanx435 wrote: post #17
deathanatos wrote: post #5

Love in Kleptopia needs to be explained better. The problem can only be solved if you can afix two padlocks onto a box, and I was presuming the lock box had a single, normally shaped padlock eye, which would make such a thing impossible. I find this happens a lot with "thought" problems: I can't solve it (and can often prove that) because the rules of the problem are inadequately explained.

Or just attach a lock that needs a combo (ex: 3,7,15) instead of a key to open it, then tell her the combination after she gets it and the risk of theft is zero. The problem specifically says they are communicating using the internet, so why not? "Hey, I sent you a box. It's locked with a combo lock. Call me when you get it: I want to be talking to you when you see the surprise! I'll tell you the combo on the phone!"

Can't believe I didn't think of that before I got to the double padlock solution.

Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]

#77

Post by emsy »

Honestly I already lost interest after reading the solution to the first problem, because there's no indication that the prisoners are allowed to alter the boxes in advance. This is such a dumb "gotcha".

Edit: I re-read the solution and it doesn't require actual labeling (It's hard to imagine, because I'm not a native English speaker).

Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]

#78

Post by XCSme »

emsy wrote: post #77

Honestly I already lost interest after reading the solution to the first problem, because there's no indication that the prisoners are allowed to alter the boxes in advance. This is such a dumb "gotcha". Edit: I re-read the solution and it doesn't require actual labeling (It's hard to imagine, because I'm not a native English speaker).

What are you talking about? They don't alter the boxes in any way, they just say "hey, I'm number 1, you're number 2, etc...".

Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]

#79

Post by pwagland »

Moodles wrote: post #57
deathanatos wrote: post #5

Love in Kleptopia needs to be explained better. The problem can only be solved if you can afix two padlocks onto a box, and I was presuming the lock box had a single, normally shaped padlock eye, which would make such a thing impossible. I find this happens a lot with "thought" problems: I can't solve it (and can often prove that) because the rules of the problem are inadequately explained.

Reminds me of this: 1, 2, 3, 4, 5, 6, 7, 8, 9,... What comes next? 10 if it's the sequence of natural numbers 13 if it's the sequence of N such that N=2^n for natural numbers n, where N does not contain a 0 153 if it's sequence of N such that the the sum of the digits of N each raised to the power of the number of digits in N equals N.

While I fully agree with the premise that any sequence of numbers can have an essentially infinite choice of 'next number', depending on how you define your sequence, and the simplest way to prove that is simply to given a sequence of _n_ numbers solve the _n+1_ polynomial equation, given that it is incompletely specified then you have an infinite number of solutions. Or just drop in a heavy side step function (https://en.wikipedia.org/wiki/Heaviside_step_function).

However, i'm not sure that I understand your second example.

If N=2^n (assuming 2 to the power of n, not 2 xor n) then one would expect the sequence to be 1, 2, 4, 8, 16, etc.

If N=2^n using the C xor notation, then we would expect a sequence 3, 0, 1, 6, 7, 4, etc

Your third example is pure evil, and I'm glad that I never had you as a maths teacher ;-)

Re: Seven Puzzles You Think You Must Not Have Heard Correctly (2006) [pdf]

#80

Post by emsy »

XCSme wrote: post #78
emsy wrote: post #77

Honestly I already lost interest after reading the solution to the first problem, because there's no indication that the prisoners are allowed to alter the boxes in advance. This is such a dumb "gotcha". Edit: I re-read the solution and it doesn't require actual labeling (It's hard to imagine, because I'm not a native English speaker).

What are you talking about? They don't alter the boxes in any way, they just say "hey, I'm number 1, you're number 2, etc...".

you're right, thanks for pointing it out.
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