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Posts posted by plainglazed


20 hours ago, BMAD said:
This contradicts the solution below. As your a > b > c > d > e > f > g solution does not follow the condition that
a + b must equal the sum of the rest
yes, but...
Spoiler...was referring to the specific case of only five weights to prove a solution cannot exist

Spoiler
Flip coins to make the bazillion coin the center of the longest string of the same value (heads or tails). Would expect about nine consecutive heads or tails would be enough to be both easily spotted and not take very many flips. At some point it could make more sense to break up other existing long strings vs adding to the indicating string. The downfall of this strategy is if the valuable coin is near either end, but you could strategize to flip the first or last coin to heads if the money coin is in the say first eight coins and the next three coins then tell at what position taking at most five flips, otherwise flip (if necessary) the first and last coins to tails.

Spoiler
Flipping a coin eleven times results in _{11}C_{8}+_{11}C_{9}+_{11}C_{10}+_{11}C_{11 }or 232 trials that have 8 or more heads. There are _{11}C_{7} or 330 trials that would have exactly seven heads, half of which would have eight heads with one more flip:
(_{11}C_{7}/2+_{11}C_{8}+_{11}C_{9}+_{11}C_{10}+_{11}C_{11})/2^{11}=.193848

for a set of five...
Spoiler...thinking it is not possible. Label the five integers in increasing order a,b,c,d,e:
e+d has to equal a+b+c and e+c must equal a+b+d but e+d>e+c and a+b+c<a+b+d 
another seven
Spoiler1,2,3,4,5,6,7
and I think an example of six?
Spoiler8,7,6,5,4,2
still working on five...

3 hours ago, CaptainEd said:
Or is it counterbalanced by the last card?
yeah, think I agree with that statement, dealer.

better than 50%?
SpoilerHalf of the time there will be a point where there are more black cards than red cards flipped, in which case the odds the next flip is red is greater than 50%. If that does not occur, the long term odds that the last card is red is still fifty/fifty? May be faulty logic  conditional probability is a bear.
For example: Odds of black, black are .5(25/51) = ~.2451 then the odds the next card flipped is red is (26/50) or .52. the other 1.2451 of the time, the odds the last card is red is .5? So with this scheme the combined odds of picking a red are (.2451)(.52)+(1.2451)(.5)=.5049.
But if this logic stands, how to optimize?

I think I see the pattern here and am encouraged by the beauty of binary symmetry so here goes:
SpoilerFor integers i and the number of prisoners n equal to 2^i1, the probability of success p is (2^nm)/2^n or n/2^i where m is the number of strings needed to memorize and is 2^(ni).
i n=2^i1 m=2^(ni) p=(2^nm)/2^n or n/2^i
1 1 1 1/2
2 3 2 3/4
3 7 16 7/8
4 15 2,048 15/16
5 31 67,108,864 31/32
.
.
. 

hey bn  Thanks for getting this one started. You are correct in all you say above including your initial summary of the problem. Have edited the OP to include that explanation. I assure you there is a scheme in which one can guarantee finding seven heavy coins. Perhaps start off by finding one guaranteed heavy coin in two uses of a balance scale. Or not. I have no doubt you or others here will discover the method for finding seven heavy dimes.

You are shown a pile of dimes all of which have one of two distinct weights differing by a small amount not detectable by feel. Forty eight dimes are separated from this pile and you are told of these forty eight, light ones are a dime a dozen (literally  i.e. 44 heavy dimes and 4 light dimes). Using a balance scale twice, find seven heavy dimes.
EDIT: for clarification
 1

okay, not too proud to start this off with some probably way simplistic maths:
Spoilersix or seven clusters. if you can have a fraction of a cluster would guess 6 5/9.
Think the time and speed intervals are red herrings. Once the slowest car enters the highway all other vehicles will line up behind/cluster with it, eventually. The slowest cars expected position would be 50. Then the same scenario repeats so the second slowest expected at 25; then 12 1/2, 6 1/4, 3 1/8, 1 5/9. But can you have a fraction of a car thus six or seven or on average 6 5/9?

Not Rats
SpoilerUrgh. I mean Erg. Ha ha. That's a good one bn.

had thought perhaps
Spoilerkilometer  since also measure of distance but not fitting the clues
then perhaps
Spoilerdyne  assume from some unknown force but now one too many letters. or the bug's obvious lack of energy could be the result of a foot pound.
rats, still thinking

Spoiler
a dead bug
Spoilerdistant relative
Spoilerdiameter
Spoilerwentameter  or beentameter
SpoilerSchrodinger's Cat

ooh, nice one Cyget. Spoiler for some additional thoughts:
SpoilerPerhaps the oddly adulating could be thumbs up and dispensing slight could be thumbs down (or the bird if not referring to the same finger). Woven invoking serendipity = fingers crossed? But am also having trouble with the last couplet  maybe the ok sign? Still think fingers could very well be it.

1.
SpoilerReplace the three fifty/fifty counters with one black and one white then add another white counter to keep the odds of drawing the original sole white counter at 25%. So now starting with 5 white and 3 black counters I think gives the same odds as the original scenario. And after picking two white, there are three of each left so 50%?
EDIT: yeah, as I continue to wake up here, am less convinced of my logic above.

Not a generalized formula but better than strategy for three?
SpoilerLabel prisoners AG. Label Yellow hats 0 and Green hats 1.
Each prisoner must memorize these eight strings and their inverses as well as their relative position within the group:
A B C D E F G
S1  0 0 0 0 0 0 0
S2  0 0 0 1 1 1 1
S3  0 0 1 0 1 1 0
S4  0 0 1 1 0 0 1
S5  0 1 0 0 1 0 1
S6  0 1 0 1 0 1 0
S7  0 1 1 0 0 1 1
S8  0 1 1 1 1 0 0These sixteen strings and all strings created by changing just one bit of each of the sixteen comprise all 128 possible two
color hat combinations with seven prisoners.
So now the strategy: Each prisoner sees six bits (hats) and can create two seven bit strings by inserting a 0 and a 1 at
their respective position. If neither of those two strings match any of the sixteen memorized strings, abstain. If one of
those two strings matches one of the memorized strings, choose the other bit (hat color) that did not create a match.
This effectively packs seven wrong answers in each of the sixteen memorized strings and gives one correct response in each
of the other 112 cases. So with 7 prisoners (or more) freedom is possible 7/8ths of the time.

dang, like that answer MM  nice
Spoilerhad been toying with the idea of a gold wedding band but couldn't make the jump to your answer and make it all fit.

Spoiler
Given the constraint of the OP that all pairwise distances are distinct, one such distance must be the greatest. And since there are an odd number of soldiers, that soldier (n) with the greatest distinct pairwise distance is watching another soldier (m) but that soldier(m) is watching another leaving no one watching soldier (n).

Spoiler
Sixty years. Perhaps splitting hairs but I'm a word guy. The OP states that the town's population has already "doubled" (past tense).

Silently be in doubt and subtle...
Spoilerthe number in the OP should be read with a silent B
Note: Although the word "number" as it pertains to pain (killer) exists as an adjective as in "I can still feel that Doc, can you make it number?" it does not formally exist as a noun. Although in the spirit of a riddle I feel its use here is perfectly acceptable and quite clever. Certainly I think all would understand what is meant if one says, "I need a number Doc, the pain is killing me." Nice riddle Biotop!

legalmente?
Spoilerdieciocho? as written in English  Dieciocho means eighteen in Spanish.

perhaps
Spoiler111 could be read as oneeleven
or
Spoiler10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 might be onegoogol
Who finished 2nd?
in New Logic/Math Puzzles
Posted · Edited by bonanova
Gold Star
answer
Charlie. Thought I was close this morning in the shower but it wasn't until my commute into work that I realized that possibility was indeed unique. Will leave the details to MM .Very fun puzzle my friend.