I am totally off - above, i misread the question.

10 boxes each contain n balls numbered 1 to n

x balls are removed from each box (x < n), leaving n-x balls in each box, such that:

a: any combination of 5 boxes contains balls 1 to n (at least once each); and

b: any combination of 4 boxes does not contain balls 1 to n (at least one missing)

What is minimum n, and what is resulting x?

n = 5 and x = 4

seems easy unless I am missing something here.

10 boxes with 5 each = 50 total balls. n = 5.

remove 4 from each. x = 4.

we now have 10 boxes with 1 ball each. take any 5 boxes and you 1 to 5 (1 to n).

it can't be less than 5 because you need at least one ball in each box and 5 boxes to choose.

while I agree 5/9 is close I think the actual probability is slightly less.

I think it is actually .551913 instead of .555556

This is because there are actually 61 points at which a person could arrive from 12 noon to 1PM. So the potential data points would look like this

Point 0: Person A arrives at 12:00 there are 20 out of 60 minutes where they could meet.

Point 1: Person A arrives at 12:01 there are 21 out of 60 minutes where they could meet.

.

.

Point 20: Person A arrives at 12:20 there are 40 out of 60 minutes where they could meet.

.

.

point 41: Person A arrives at 12:41 there are 39 out of 60 minutes where they could meet.

.

.

point 60: Person A arrives at 1:00 there are 20 out 60 minutes where they could meet.

notice i zero based this (I can't help it, I am a programmer). so that is 61 points. hence my numbers add up to

2020/3660 or 101/183 = .551913

let me know if I have gone astray here.

thanks.

after using deductive reasoning i have determined that there is no way of winning. am i wrong?

the game is very winnable - once you figure it out - its easy.

give it a try.

Pearls before swine

but seriously I came up with 8 max and maybe less if you get lucky.

5 races to rank horses A through D

A vs B

C vs D

winners vs each other

losers vs each other

2nd round losers vs each other.

then race E vs 2nd or 3rd it's your choice. and move E up or down the list from there. This will be a max of 3 more total races (E beats 3rd place, E beats 2nd place, E races 1st place).

hmm funny wording

most characters is 1888 (i think)- which would be MDIIILXXXVIII for a total of 13 characters.

Most Roman Numerals though could be thought of as the largest value which of course would be 2009 MMIX

here goes

If E were to live through the first round he would most certainly shoo the person with the next best percentage (D if he is still alive, followed by C, then by B..) that gives E the best chance at continuing in subsequent rounds.

A should almost certainly shoot for E, A does not need to worry about someone coming after him in the first round, so he needs to take out the best shot if he can.

now assuming A misses

B should shoot for C. If he kills C, then D must go for E and then B would continue to the next round (either D kills E and we start at A again, or D misses and then E kills D).

if he misses C, well his chances remain the same as if he decides to shoot as someone else because ...

C should always shoot at B. This leaves D/E to shoot each other as described above.

now assuming A kills E.

B should shoot for A. he would then be guaranteed to move on as before (either C kills D, D kills C, or both C & D survive)

C of course tries to kill D

D of course tries to kill C.

my guess is C probably is the spot to be in if you had a choice.

Education

A piece of paper is an "ink-lined plane". An "inclined plane" is a "slope up". A "slow pup" is a lazy dog

This reminds me of an oldie I heard 30 yrs (or so) ago.

How is a piece of paper like a lazy dog?

[Replicant]

what numbers can you put at least 2 different mathematical symbol between them (1 at a time)and the result is the same?

give at least two nos.

do you want different numbers?

4 - 2 = 2

4 / 2 = 2

1 and any other number like:

3 / 1 = 3

3 * 1 = 3

3 ^ 1 = 3

Chair

(hair) above

(air) around

answer is "no"

could be the truth teller saying "no", therefore the middle person is NOT the random answerer

could be the liar saying "no", meaning he would really say "yes" if you asked the answer outright - so the middle person is NOT the random answerer

BUT, if the person you ask that of is the random answerer and the answer is "no", there is no logic by which to conclude anything.

"no" could mean he is lying and would have said "yes" (which would be a lie)

"no" could mean he is telling the truth and would have said "no"(which would be a truth)

answer is "yes"

could be the truth teller saying "yes", therefore the middle person IS the random answerer

could be the truth teller saying "yes", meaning he would really say "no" if you asked the question outright

BUT, if the person you ask that of is the random answerer and the answer is "yes":

"yes" could mean he is lying and would have said "no" (the truth)

"yes" could mean he is telling the truth and would have said "yes" (a lie)

as you stated in your logic... the random answerer cannot be counted upon for a definitive true or false answer. and so if you happen to ask the random answerer first (your logic assumes you will only ever ask either the liar or the truth teller first), you are likely presented with a lie of some form or antother (as presented above) but cannot logically deduce which form the answer takes: a lie about the truth, a truth about the truth, or a truth about the lie.

plasmid's logic works

you are just trying to eliminate the the random answerer in with the first question.

It doesn't matter if you ask the random person first, since the second question is always posed to one of the other two. So if the random person tells you a lie, or the truth it does not matter because the second person will always be either the truth teller or the liar.

Also his first question is worded in the double negative way, so that both liars and truth tellers effectively answer the question truthfully.

Okay I have an idea

Ask A: "Will B tell me the left path leads to heaven?"

If A is the truth teller he will answer like this

B=Liar: Yes or No (i don't care which)

B=Random: No answer because he does not know and therefore unable to tell the truth.

If A is the liar he will answer like this.

B=Truth: Yes or No (i don't care which)

B=Random: No answer because he does not know and therefore unable to tell a lie.

If A is the random answerer he will answer like this.

B=Truth: Yes or No

B=Liar: Yes or No

So If i get an answer I can safely assume B is either a liar or a truth teller, if I do not get an answer then C is either a liar or a truth teller.

based on that I will ask either B or C (the one I know not to be the random answerer).

"Would you tell me the Left path leads to Heaven if I asked you?" If he answers Yes - it is the path, if he answers No, the right path is the correct path.

Ive been thinking about this for a little while

I think we need to eliminate the random answerer with the first question so that the second question is posed to either the liar or the truth teller (you don't care which). That way we reduce riddle to the old Liar/Honest person/1 question/2 doors riddle.

So along that line we need a question that we could pose to either a truth teller, liar, or someone who answers randomly and get all of them to answer in a way that we know that a specific person is either the liar or the truth teller.

does any of this make sense, and am i barking up the right tree?

The Good, The Bad and the Ugly - They formed an alliance of hate to steal a fortune in dead man's gold

yep

Okay - two more hints (although i think these will give it away) - they have nothing to do with the tagline, just the movie

There are two kinds of people in the world

genre - western

yours

wait i was totally wrong here. but this one should work

another

yours

I've been thinking of number 4 since yesterday, but can't think of anything. From the sounds of the clue, it sounds like two enemies need to team up to do something, possibly travel through time and perform a robbery. Either that or steal some ancient heirloom or artifact. I think we probably could use a hint.

Well reading your line of thinking maybe my original clue is a little misleading. There was no time travel involved in this movie. Let me start with a slight revision of the riddled tagline - changes in bold.

4. Adversaries combine to relieve the passed of its value. (note possessions would still work here).

Are the blocks perfect cubes (H/W/D all the same length)?

No takers on #4? Do you all need a hint?

does it take Dan 8 minutes to go through the entire tunnel or 3/4 of it? I assume the 30 meters on Dan's time was a mistake and it takes 8 minutes for him to go through the entire tunnel.

If so my answer is

16 minutes of swimming.

Adam & Dan - 8 min

Adam back - 1 min

Adam & Carl - 4 min

Adam back 1 min

Adam & Britney 2 min

totals 16 minutes.

if you meant 8 minutes for 30 meters for Dan - his tunnel time becomes 10 2/3 minutes so that total time would be 18 2/3 minutes.

Wow you guys are good.

Here is what we have so far

1. Reversed directions, inverted colors, I'm confused. - Miller's Crossing (palmerc7)

2. Captivate you by fright, optimism escapes. - The Shawshank Redemption (ruledbysecrecy, Docta92)

3. Labor is unpleasant. - Office Space (Pickett, Docta92)

4. Adversaries combine to relieve the past of its possessions.

5. I can never recall not dreaming of being a nefarious criminal. - Goodfellas (TheWwwyzzerdd)

6. Any wall will be ascended. - The Great Escape (Pickett - nice one I thought this would be last gotten).

Still looking for #4