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Join date
31-Aug-2004
Last activity
4-Dec-2019
Posts
421

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Post
#90387
Topic
Riddles
Link
https://originaltrilogy.com/post/id/90387/action/topic#90387
Time
What are the parameters on the special rule? Can I just have one filosopher (generally we spell it philosopher, after the Greek spelling) grab right then left? Or is that a hard constraint on the problem?

Otherwise, can we say he only grabs left if he can see a fork on his right?
Post
#90380
Topic
The DanielB Prisoners and Lightbulbs Thread
Link
https://originaltrilogy.com/post/id/90380/action/topic#90380
Time
So my question on the table is:

At some point, you're going to have two guys with points, the counter and some other guy. In order to transfer them, the one will have to go directly before the other. On any two days, each has a 1 in 100 chance of being selected. The consecutive chance is 1 in 10,000, even if you are trying every single day (which you can't, because the day holds a variable). So, even to do just that last exchange, I can't see how it is mathematically possible to average 3,000 days?

Daniel, have you found a solution to this, or am I just too pessimistic about the probability? Does the math work out like the "two people in the room with the same birthday" problem? No...it can't. Hmmm...
Post
#90364
Topic
Riddles
Link
https://originaltrilogy.com/post/id/90364/action/topic#90364
Time
(I'm triple posting because I'm changing subjects. Sorry)

How is it possible for a common person to throw a ball, have it stop in midair, and return to her (I've been reading law books that use "her" as the neuter singular to be both grammatically and politically correct...), with nothing attached to it and no special equipment?
Post
#90362
Topic
Riddles
Link
https://originaltrilogy.com/post/id/90362/action/topic#90362
Time
What confuses me, DanielB, is that at somepoint, you're going to have two prisoners with tokens, or points, or whatever. To dump them off, the one, the counter, will have to directly follow the other one. Each has a 1 in 100 chance in any day, so the chance of a consecutive sequence is 100 x 100 = 10,000. That means that once you get to that point, even if you are able to check every day, the minimum is 10,000 days (I understand it will often happen significantly sooner than that). To execute in 3,000 days, on average, you have to get around this last step somehow, because 3,000 average doesn't come from a 10,000 probability, which doesn't even start on day 1.

?

Should Ric, Daniel, and I discuss this in a separate thread since it killed the riddles-proper?
Post
#90360
Topic
Riddles
Link
https://originaltrilogy.com/post/id/90360/action/topic#90360
Time
A solution without going to an arm’s length:
Light starts off. All prisoners are assigned the number 1.
Days go in a 20 day cycle. Days are numbered 1-20. It starts over day 21.

Days 1-19:
If light is off –
If your number is 0, add the number of the previous day. Leave the light on.
If your number is less than 5, subtract one from your number. Leave light off.
If your number is greater than 5, add the previous day number to your number. Leave light on.

If light is on –
If your number is 0, leave it on.
If your number is less than the day in the cycle, keep your number and leave it on.
If your number is greater than the day in the cycle, subtract the day from your number, turn the light off.

On Day 20:
If light is off: add 20 to your number. Leave light off.
If light is on: keep your number, leave light off.

If your number is greater than 50, always leave the light on and never subtract.
If your number is 100, make your assertion.

In English:
Basically, if you leave the light on you are communicating “all the previous days in the cycle are accounted for.” Likewise, if you leave the light off, you are saying “count me and all the days before me.”

At all time there will be 100 “points” totaled. These points are stored either in the day number or by each prisoner. Initially, 1 point is stored by each prisoner. On day 1, he transfers his point from himself to the day (now day = 1 and prisoner = 0). On day 2, supposedly the same thing happens (day=2, another prisoner = 0). Suppose the first prisoner re-enters the room – he assigns himself the previous day (2) and turns the light on (if light is on, the ‘day’ value is 0, if light is off, the ‘day’ value is the day number.) Hooray! We’ve just transferred a point to another prisoner. The goal is to get one guy all 100 points.

Now let’s jump ahead and suppose that 10 prisoners have 10 points each. On day one of a cycle, some poor sap with 0 will walk in and turn the light on, keeping 0 (day 1 – 1) as his number. On day five, a prisoner with 10 walks in. He turns the light off and subtracts 5 from his number. The next day, another prisoner with 10 walks in and sees the light off, and adds 5 to his number. Hooray! We’re still adding up points!

The PROBLEM is, that this algorithm has a distributing effect on the points. That is, the prisoners without points will outnumber the prisoners with points, and this algorithm will favor distributing the points evenly rather than consolidating them. Since the prisoners don’t know who has how many points, they have to give freely; I have not worked out a solution to this problem. I did add in two “valves” of sorts, at 5 and 50, to favor the consolidation rather than distribution of points. That is, once you reach a critical level of points, there are inhibitions to distributing them.

The other problem is, this will work *eventually*. It will take a really long time, and the prisoners would be better off waiting 5 years and taking their chances. That is statistically more advantageous than waiting for my algorithm to get them out; no one would live long enough. In the end, the guy with 1 point would have to go in on day 1 of the cycle and then Mr. 99 would have to follow (which is a 1 in 10,000 chance every 20 days…200,000 days…) and that’s just after we’ve gotten to that point.

I read some more and will look into the solution of changing the arbitrary count. However, two caveats: 1 - it's just less elegant. In a situation like this, it's obviously practical, but I like to find a single recursive solution. 2 - it get's into practical math. If you are going to change the default count number, you have to be assured that everyone is on board. You have to decide ahead of time when to change the count, so you get into probabilities. You can be pretty sure, but you've left the realm of logical solutions. There's always the possibility that the universe will screw you, so you haven't really solved the riddle. Maybe you've found a way around this... my solution doesn't guarantee a solution, but you ARE guaranteed to know when you have or haven't.

I’m enjoying this; I’ll be thinking. It’s comforting to know that other teams of people haven’t found a good solution either, so I don’t have to feel bad giving up eventually.

Does someone want to put out a normal riddle to keep this going?
Post
#89957
Topic
Empire of Dreams - what was your imagined Star Wars story...
Link
https://originaltrilogy.com/post/id/89957/action/topic#89957
Time
Yeah, I had a bunch of pre-conceived notions. I thought that the Emporer would be the one to f*** anakin up, and Anakin would worship him for his power but nurse a long time grudge/jealousy that would boil over in ROTJ, and tie things together that way.

I thought Anakin would have a father.

After Episode I, I thought that Anakin would return and find that his mother had died, which would begin his turn to the darkside. The way I wanted it, though, would be that Watto had somehow killed her, either through negligence or general slave owning abuse, and Anakin would then kill Watto. Killing him would shame him and devastate him, but it would also open the floodgates of his strength. I got that one pretty close.

At this point, I picture Padme being pregnant and Anakin somehow thinking that she has been killed (or her actually being killed) and assuming the twins are dead as well. Then he rushes off in a rage to confront whoever he's blaming, and spends the next however many years of his life thinking they're all dead, only to discover in the Death Star trench...

I did a lot more thinking, but those are still on the top of my head...

Oh, and a side note about Tatooine...it's supposed to be this non-republic, inconsequential ball of dirt far from the center of the galaxy. How come Naboo just happens to be next to it, and the start of the Empire? How come the rebellion is flying right by some ball of dirt far from anything else important when it is overtaken in episode 4? It just seems a little too involved for me; it comes up in every storyline and every game...
Post
#83269
Topic
Myths
Link
https://originaltrilogy.com/post/id/83269/action/topic#83269
Time
I was under the impression that we had no idea where Mount Ararat was. And I have difficulty believing that a wooden structure could be preserved over more than 4000 years. Where did you hear about this? I'm inclined to disbelief...

OK, I checked out that site and googled it. I'll take some convincing before I go with this one. It sounds like Roswell type sightings to me, so few and far between. File me with the skeptics here.
Post
#83116
Topic
Myths
Link
https://originaltrilogy.com/post/id/83116/action/topic#83116
Time
Fascinating... the prevalence of ADD and such never sat well with me but I've never looked into it. I think of it as a deficiency in the development of children based on limited adult interaction and instant gratification provided by modern entertainment and technology, which I thought should be addressed by altering the environment, not the child.

But as I said, it's just a theory and I've never looked into it. This is the first of hearing what you're saying DanielB, so I can't jump onboard immediately, but I find what you said fascinating.
Post
#83115
Topic
Riddles
Link
https://originaltrilogy.com/post/id/83115/action/topic#83115
Time
Well, glad to see everyone was enjoying the holidays, not just me. Hasn't been too much action around here.

Anyway, the answer to my earlier riddle was "empty". You can remove the letters one by one:

m-pty.
mp-t
m-t

And it still sounds like "empty."

---

I told my family the roman numeral riddle, and it stumped them and they loved it. it's a good one.

---

And I'm out of riddles, for the moment. It's open to the next asker.
Post
#83114
Topic
Classic Games
Link
https://originaltrilogy.com/post/id/83114/action/topic#83114
Time
Sorry, I don't have time to catch up on any of this thread so I may be beating the proverbially dead horse, but I followed some strong recommendations to try "shining force" on the genesis/megadrive. It's a role playing game and turns out is a precursor for FF Tactics/Front Mission and such. It's awesome! Give it a try!

Search for "emurussia" in google if you don't have a genesis. It's a great site. That's all I'll say about it.
Post
#82403
Topic
Riddles
Link
https://originaltrilogy.com/post/id/82403/action/topic#82403
Time
Fala, Ric. I know what you mean, I prefer the logic riddles because they're universal, but this is all I could think of now. It is a common word, which is some help.

Both the first two letter and the last two letter pairs are in DanielB's list (like that helps a ton

(Fala means "what's up?" in Brasil, right?)
Post
#82365
Topic
Riddles
Link
https://originaltrilogy.com/post/id/82365/action/topic#82365
Time
Alright folks, I'm going home to San Diego for Christmas. I don' t imagine I'll be spending much time here while I'm home. Actually, I may be vastly reducing my OT time in general; coming to terms with my laziness and trying to be disciplined towards the responsibilities in my life. Because for me, OT is often an outlet for laziness at work or home.

Anyway, I'm hoping that's the right answer. If it is, I'm going to post a parting riddle now. If not, you can disregard this one and stick with the last one. Bossk, I'll pm you the solution since you're a regular here, you can confirm when you or someone gets it.

And to all, a Merry Christmas, a happy future withthe memory of Star Wars past, and all that stuff.

Alright:

There is a 5 letter word in english.
If you remove the first letter, you still pronounce it the same way.
If you then remove the last letter, you can still pronounce it the same way.
If you then remove the middle letter, even then you still pronounce it the same way.
What word is it?

Bonus question:
What's the longest one-syllable word in English? (it's 8 letters plus an 's' to make it plural)
Post
#82358
Topic
Riddles
Link
https://originaltrilogy.com/post/id/82358/action/topic#82358
Time
Similar to the black dots on the forehead puzzle.

so the first guy, gets first guess. He either sees two red, two blue, or one of each. If he sees two reds, he would know he has blue, because all the reds are already taken. But he doesn't know so the other two guys can conclude he's looking at two blues or one of each color.

Enter contestant two. He either sees blue or red. He knows there can't be two reds, since contestant one didn't know. SO, if he sees a red, he would know he has a blue. But, he doesn't know the answer, so he must be looking at a blue (thereby allowing both blue-blue and blue-red scenarios)

Enter contestant three. He concludes he's wearing a blue hat, based on the reasoning above.