 | | Re: On a scale of 0 to 10?
|
|
(...) Oh yeah! (...) Oh... yeah. (...) Hm. Double checked the math-- it appears solid [1], which would mean that there's no viable solution for: s = Ae^(q + B) + C for coordinates (-1,0), (0,5), (0.25,10). Darn. Hm. I guess that also means that any (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: On a scale of 0 to 10?
|
|
(...) *snip* (...) summed exponents like that can be re-written in this way: e^B * e^.25 + e^B * e^-1 = 2e^B (e^.25 + e^-1) * e^B = 2e^B Uh oh. Divide both sides by e^B and we've got (e^.25 + e^-1) = 2 1.652 = 2 Which ain't right. I think your math (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: On a scale of 0 to 10?
|
|
(...) Thought about that... If necessary I guess, but I'm pretty sure there's a way to do it. I think what I need is: s = Ae^(q + B) + C And at the moment, I've solved: C = -Ae^(B - 1) A = 10 / (e^(B + .25) - e^(B - 1)) And I'm scratching my head (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: On a scale of 0 to 10?
|
|
(...) How about a double-linear solution? from -1 to 0, translate linearly from 0 to 5. From 0 to .25, translate linearly from 5 to 10. You won't have a single equation, but that seems to accomplish what you want, right? Adrian (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | On a scale of 0 to 10?
|
|
I've got a function that represents a quality rating, based on 2 parameters, which are on a scale of 0 to 1: (d-k)*k => quality (q) Notice that this yields a sort of wacky result, insofar as the quality can vary between -1 and 0.25, not -1 and 1 or (...) (22 years ago, 26-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: Website move
|
|
(...) The domain name is the fully qualified host name. If you need DNS to look it up, it's a domain name - the "host" in the RFC. That's what it's a "Domain Name Service" :) Anything after the / is part of the path, as defined in the HTTP RFC (...) (22 years ago, 26-Mar-03, to lugnet.space, lugnet.publish, lugnet.off-topic.geek)
|
| |
| | Re: Backwards Compatibility (Was Calling all Meta-commands)
|
|
(...) ROFL. I'm sorry, I can't reply in kind. My brain is too fogged. Steve (22 years ago, 22-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: Backwards Compatibility (Was Calling all Meta-commands)
|
|
(...) you mean $puctuation++ > ! $puctuation ? $goodness++ : $goodness-- ; ? (22 years ago, 22-Mar-03, to lugnet.cad.dev, lugnet.off-topic.geek)
|
| |
| | Re: connect four
|
|
(...) This interested me for a while-- I wrote a recursive program once that would look X ply into the game, though because I never stored state information, it took ridiculously exponentially longer with each value of X. Something like 6 ply was (...) (22 years ago, 19-Mar-03, to lugnet.org.ca.rtltoronto, lugnet.off-topic.geek)
|
| |
| | Re: Access Help
|
|
James & All, (...) Yes, I have a total query, where it is running a sum of all the data from a date range (say 1/1/02 to 12/31/02 or whatever). The value I am trying to evaluate is a calculated sum of various percentages of our recovery numbers, in (...) (22 years ago, 7-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: Access Help
|
|
Scott, Can you post and example of the type of queries you have been trying? I know you said they where giving you errors, but at least we'd have an idea of what you are trying to do. jt (...) (22 years ago, 7-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Access Help
|
|
To All, I have been working on a project at work for over a year now, converting an Excel Production Recap program into an Access program. I have been successful converting it, outside of some small issues, which I have finally resolved in the last (...) (22 years ago, 7-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: Now the stakes are higher....
|
|
(...) Effectively, he's right but he's not giving things the probability they're due. In his solution, the (r1 r2) choice is GUARANTEED that no matter which one you reveal (r1 or r2), the one revealed will be red. However, with, say, (r1 b), there's (...) (22 years ago, 6-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: Now the stakes are higher....
|
|
In lugnet.off-topic.geek, David Eaton writes: <snip> (...) That's perfect! I owe you a lunch! Now whetehr or not my co-worker succumbs to 'outside sources' as valid enuf to change his position remains to be seen I'm trying to explain his fallacy and (...) (22 years ago, 6-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: Now the stakes are higher....
|
|
(...) HA! Congrats on winning 100 bucks :) (...) Proof number one: (pure stats) #!/usr/bin/perl for(1..1000) { @marbles = ("r","r","b","w"); $pick = ''; for(1..2) { $n = int rand(@marbles); $pick .= $marbles[$n]; splice @marbles,$n,1; } $tot++; (...) (22 years ago, 6-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Now the stakes are higher....
|
|
So now he says he's willing to put 100 dollars on the 1 in 5... I'm swayed back to the 1 in 3... So here's the original question-- If you have 4 marbles in a bag, 2 are red, 1 is blue, 1 is white You grab 2 marbles from the bag at the same time One (...) (22 years ago, 6-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) Didn't mean to offend. Sorry 'bout that. I read the site thru again and your synopsis of the explanation above and I concur that it should be 2/3, but I like to work things thru... :) If I were to walk in off the street right after one of the (...) (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) Ok, here's where I'll take it from. The car is behind one of the doors, so the probabilities must add up to 1. 3 doors, 1/3 of a chance for each door. You pick door 1. There is a 1 in 3 chance the car is behind that door. There is a 2 in 3 (...) (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) Saw that ;) but understood... but if we're talking one instance--one event--and that's what we're, indeed, talking about--you are on a game show, one event--you have one chance of winning one car behind one door.... so you have 3 doors I won't (...) (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) Saw that ;) but understood... but if we're talking one instance--one event--and that's what we're, indeed, talking about--you are on a game show, one event--you have one chance of winning one car behind one door.... so you have 3 doors I won't (...) (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) I think it's a great idea! Dave K (22 years ago, 5-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) This analysis assumes there is a percievable difference between the red balls, which as originally stated, there isn't. It's similar to the difference between combinations and permutations. The two permutations you've listed with a red ball in (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) If you keep going on and on and on... well I probably still will, unless someone explains why I should not. (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) Bleargh! typo. r2 w (r1 is the only red ball, it is removed) (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) Why not? There are two different red balls, so each of those red balls is equally likely as the white ball to be in my hand. (...) No, because you only tried to remove r2. We are only removing *a* red ball from the bag; because there are 2 red (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) Clear as mud, for then if the red is irrelevant, then there cannot be (...) there can only be (...) for if we remove either red ball from the bag, that very same red ball in this specific instance cannot be in the person's hand. We cannot take (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) Basically, each column decreases in liklihood as more choices are added. Start with one colum: Prob. Prize 1/3 1 1/3 2 1/3 3 Now, you get to choose a door. Since the prize door doesn't affect the probability: Prob. Prize Choice 1/9 1 1 1/9 1 2 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) Nope, you're assuming that the marble pulled out of the bag is r2. It might be r1. Better to leave off the numbers, as the red marbles are not distinct. hand bag r r w r r w w r r remove a red from the bag and it reduces to: hand bag r w r w w (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) I think I may have found the fallacy... (indulge me if you please) Here's a bag with 3 marbles in it--2 are red (goats, whatever), one is white (brand new car) Pick a random one out of the bag and hold it in your hand--don't look at it! What's (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) This one here? (URL) many trials? It is a *probability* after all, no guarantees about a short run of trials. I ran 30 "stay" and 30 "switch" and got roughly 40 and 60 as the two percentage results. Moving to the serious number cruncher, which (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) think that way, though I'll note I passed Calculus). I come up with 2/3 if you switch. A 1/3 chance that you have the right choice initially. The other two doors are an aggragate 2/3 chance. Even though one is eliminated, it is a wrong one, so (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) though I admit it has been a long time since I have taken my probablity and statistics class. (...) Gosh, everyone is seizing on this: I didn't say it was and it has nothing to do with my answers!!! :-) (...) Sigh. Yes, I know that. (...) Yes, (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) If I were to think about it like a sequence of events, logically leading up to the end choice, it would go something like this-- The event starts-- Contestant picks a door -- 1, 2, or 3 If this is all there is to the contest, we have 33 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) See this one for a clearer picture: (URL)No, because he won't use the same algorithm since he then becomes (...) That's not the problem presented, though. According to the problem, it's KNOWN that he ALWAYS reveals a zonk prize door that you (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) Thank you for repeating the message I already read and applied to the situation. I did illuminate that it is not a valid Let's Make a Deal scenario, but it's still the one I'm talking about. -->Bruce<-- (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) Doh. This is just coming back to haunt me. I should never have included the "door revealed" column. It's entirely irrelevant, and is only serving to confuse. A better list: Prize Choice Switch or Stay? 1 1 stay! 1 2 switch! 1 3 switch! 2 1 (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) That is not a cheat. -->Bruce<-- (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) <mucho snippage> (...) Bruce - I see the confusion. You aren't solving the problem as presented: "Now, here's a trickier one: Ever watched the old game show "Let's Make a Deal"? Sometimes, they'd offer people things similar to this scenario. (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) Whew. You are way off in your probablity and statistics analysis: if you want to double everything, then you would have to list 112 and 113 twice also. They are *not* the same selection. (...) and 1 1 2 (...) and 1 1 3 (...) (etc. snip) (...) (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
| |
| | Re: math question (or pattern... whatever...)
|
|
(...) That doesn't make sense at all-- If I choose door 1 and the prize is behind door 1, the host can choose to show me door 2 *or* door 3 - 2 possibilities If I choose door one and the prize is behind door 2, the host can *only* show me door 3 - (...) (22 years ago, 4-Mar-03, to lugnet.off-topic.geek)
|
