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

Post by ttctciyf » Sat, May 05, 2018, 5:59 PM UTC

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.

Post by konschubert » Sat, May 05, 2018, 6:15 PM UTC

konschubert wrote: post #82

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.

Is the range of integers limited? Does she choose them randomly? If the range isn't limited, hoe would we even define a normalized probability distribution?

Post by tomp » Sat, May 05, 2018, 6:35 PM UTC

OscarCunningham wrote: post #100

tomp wrote: post #94

Earlier quoted context omitted.

I agree that it’s not entirely strictly specified - in particular, there’s no random (uniform) distribution over integers. I guess you could get around this by specifying in in a way you did, or “parametrized” over some parameter, or say “for any distribution” (and mandate a single turn of the game)... Anyways, this is also highly unintuitive for me, so I’m far from certain that the solution below is correct; having…

I think the "orthodox" answer is that your method wouldn't work, because if Alice picks 10 and 20 and opens 10 first then you have a 0% chance of winning. Instead (they say) you should pick your number at random, so that it has at least some probability of being any given integer. For example you could pick the number n with probability 2^|n|/3. That way no matter which two numbers Alice picks there's always some pro…

Post by OscarCunningham » Sat, May 05, 2018, 6:42 PM UTC

tomp wrote: post #105

OscarCunningham wrote: post #100

Earlier quoted context omitted.

I think the "orthodox" answer is that your method wouldn't work, because if Alice picks 10 and 20 and opens 10 first then you have a 0% chance of winning. Instead (they say) you should pick your number at random, so that it has at least some probability of being any given integer. For example you could pick the number n with probability 2^|n|/3. That way no matter which two numbers Alice picks there's always some pro…

> if Alice picks 10 and 20 and opens 10 first then you have a 0% chance of winning. No because you pick the hand (i.e. it’s essentually random), so you still have 50% chance of first revealing 20 and gettingit right.

Post by iopuy » Sat, May 05, 2018, 8:21 PM UTC

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.

    11 if it's the sequence of N such that N=2^n for natural numbers n, where N does not contain a 24