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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]

#61

Post by sirclueless »

mulur wrote: post #60
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.

Hmm I'm not exactly good with maths (or puzzles for that matter) but here is a blind stab: By break even, I think you mean that the game is played multiple times, as long as necessary. If you are not following a martingale like strategy (edit: you can't anyways, the bets are fixed) and respond randomly (or with full faith that this is your lucky day, doesn't matter really), you are expected to break even with 50% cha…

You can do better than that. You can guarantee there is > 50% chance you win on the first guess.

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

#62

Post by mulur »

sirclueless wrote: post #61
mulur wrote: post #60

Earlier quoted context omitted.

Hmm I'm not exactly good with maths (or puzzles for that matter) but here is a blind stab: By break even, I think you mean that the game is played multiple times, as long as necessary. If you are not following a martingale like strategy (edit: you can't anyways, the bets are fixed) and respond randomly (or with full faith that this is your lucky day, doesn't matter really), you are expected to break even with 50% cha…

You can do better than that. You can guarantee there is > 50% chance you win on the first guess.

Oh that's interesting. Thank you for the clarification.

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

#63

Post by gjm11 »

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)

Yes, if you liked these you'll probably enjoy Winkler's books. If you're going to get exactly one of them, I suggest the first rather than the second, but both are good.

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

#64

Post by maaaats »

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)

Book of Enigmas by Fabrice Mazza has puzzles similar to some of these in the PDF.

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

#65

Post by [deleted] »

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.

[deleted]

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

#66

Post by Hoagy »

sirclueless wrote: post #61
mulur wrote: post #60

Earlier quoted context omitted.

Hmm I'm not exactly good with maths (or puzzles for that matter) but here is a blind stab: By break even, I think you mean that the game is played multiple times, as long as necessary. If you are not following a martingale like strategy (edit: you can't anyways, the bets are fixed) and respond randomly (or with full faith that this is your lucky day, doesn't matter really), you are expected to break even with 50% cha…

You can do better than that. You can guarantee there is > 50% chance you win on the first guess.

Wow, that is very counterintuitive..

I suppose you could set any threshold x, and play so that if the first number is bigger than x you stick, otherwise you switch. This would be strictly better than random (break even) if x is within the range of numbers that A is picking from and the same as random if not.

It depends on when you measure your chances: before A has picked a generator this gives you >50% chance but if A has already picked a range to draw from that doesn't contain x then you're back to 50/50. Still, I think this would count as guaranteeing >50%?

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

#67

Post by jmblpati »

I think there's an easier to visualize solution to the box problem.

Let A have dimensions (a,b,c) and let B have dimensions (x,y,z). Assume A fits inside B.

We have (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc. This is the sum of the A's hypotenuse squared plus its surface area. The same holds for B.

Note that A's hypotenuse is at most that of B-- the hypotenuse of a is its longest axis, and it needs to fit in B somehow. Further, note that the surface area of A is less than that of B. To see this, consider the nesting of A inside B and realize that both boxes' interiors are convex sets. Imagine inflating A inside of B by taking the sets A_t consisting of all points within B that are within distance t of a point in A. It is not hard to see that this inflating operation can only increase the surface area of A, and since the maximum surface area we can get is that of B we have that A has smaller surface area than that of B. Thus,

(a+b+c)^2 = (A hypotenuse)^2 + (A surface area) <= (B hypotenuse)^2 + (B surface area) = (x+y+z)^2. The claim follows.

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

#68

Post by pyk »

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 as well.

On a number line, the cases are below.

If A > C then the six originally equally possible cases are narrowed to three cases:

A-----B-----C (impossible)

A-----C-----B (impossible)

B-----A-----C (impossible)

B-----C-----A B C-----A-----B B > A

C-----B-----A B So you would guess B If A A-----B-----C B > A

A-----C-----B B > A

B-----A-----C B B-----C-----A (impossible)

C-----A-----B (impossible)

C-----B-----A (impossible)

So you would guess B > A -- the first hand's number is lower with probability 2/3.

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

#69

Post by mtreis86 »

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.

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.

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

#70

Post by mulur »

sirclueless wrote: post #61
mulur wrote: post #60

Earlier quoted context omitted.

Hmm I'm not exactly good with maths (or puzzles for that matter) but here is a blind stab: By break even, I think you mean that the game is played multiple times, as long as necessary. If you are not following a martingale like strategy (edit: you can't anyways, the bets are fixed) and respond randomly (or with full faith that this is your lucky day, doesn't matter really), you are expected to break even with 50% cha…

You can do better than that. You can guarantee there is > 50% chance you win on the first guess.

Ok, my solution is needlessly complicated because unbounded numbers between -Infinity and Infinity has a mean of 0, so if we just pick 0 and say "if hand1 pyk's solution above works 66.6% even for the first try...
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