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Post by mtntigger on Aug 18, 2011 16:06:38 GMT -5
Virgil - For you... Without using a search engine, a dictionary, or any other resource, can you provide the literal meanings of the following words (or at least your best guess)? 11. Chortle. I learned something new today; now I can go home. ![:)](//storage.proboards.com/forum/images/smiley/smiley.png)
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Virgil Showlion
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Post by Virgil Showlion on Aug 18, 2011 18:59:58 GMT -5
I've always thought of "chortle" as a stuttering laugh.
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ՏՇԾԵԵʅՏɧ_LԹՏՏʅҼ
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Post by ՏՇԾԵԵʅՏɧ_LԹՏՏʅҼ on Aug 18, 2011 19:09:07 GMT -5
Isn't a chortle more like a supressed snicker? (and no, I don't mean the candy bar)
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Post by mtntigger on Aug 18, 2011 19:10:09 GMT -5
No karma for you. ![:(](//storage.proboards.com/forum/images/smiley/sad.png) It is literally a chuckle + a snort. Lewis Carroll made up that word in his Alice in Wonderland books. ;D
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Virgil Showlion
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Post by Virgil Showlion on Oct 17, 2011 15:49:36 GMT -5
Congratulations to Beerwench and Jarhead... or is that Beerhead and Jarwench...? ![;)](//storage.proboards.com/forum/images/smiley/wink.png) ... for collectively annihilating problem #61 shortly after it's introduction. 2 karma for both. ![](http://syonidv.hodginsmedia.com/vsmileys/thumbsup.png) But get your thinkin' hats out for... Problem 62: Odd Sex(Ha! You thought 'odd' meant something other than 'probability', and 'sex' meant something other than 'gender', but you were wrong! ![](http://syonidv.hodginsmedia.com/vsmileys/muhaha.png) You must now solve this brainteaser as penance.) Suppose you're a guest lurking on EE late at night. Because you're a guest, you can't see who's logged in. But you can see how many members are logged in. On the board listing, you see that Rick (a male) just posted something in EE a few seconds ago. The InfoCenter reports that three members are logged onto the board, and hence Rick must be one of those three members. You also know that Rick is never ever around unless there's at least one EE lady to flirt with. Hence, you know that at least one of the other two members online is female. "Gee," you think, "since there are three members online, Rick is online and at least one woman is online, that means there's a 50% chance there's two men and one woman online, and a 50% chance there's two women and one man online." But... is you reasoning correct? Are the odds of having 2 men, 1 woman vs 2 women, 1 man as you say they are? And if not, why? ( Note: Suppose that the probability of a "random" EEer being male is exactly 50%.)
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Colleenz
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Post by Colleenz on Oct 17, 2011 15:54:54 GMT -5
Isn't this the same problem as "The prize is behind one of three doors"?
After you pick your door, the game show host opens one of the remaining two doors. Of course the one he opens has no prize (because it was not random). Then he asks if you want to switch doors.
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Colleenz
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Post by Colleenz on Oct 17, 2011 15:55:50 GMT -5
Note to Rick - I am not saying you are not a prize. You know I ![](http://syonidv.hodginsmedia.com/vsmileys/heart.png) you.
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Apple
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Post by Apple on Oct 17, 2011 15:58:01 GMT -5
I sense a bit of Schodinger's Cat theory. Did someone watch The Big Bang Theory recently?
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Befferz
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Post by Befferz on Oct 17, 2011 16:00:34 GMT -5
Or does the system count the guest, even though they're not logged in? ![???](//storage.proboards.com/forum/images/smiley/huh.png)
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Colleenz
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Post by Colleenz on Oct 17, 2011 16:01:34 GMT -5
Virgil said member.
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Befferz
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Post by Befferz on Oct 17, 2011 16:02:26 GMT -5
Heh heh heh #muhaha#
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Apple
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Post by Apple on Oct 17, 2011 16:03:11 GMT -5
lmao!
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Virgil Showlion
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Post by Virgil Showlion on Oct 17, 2011 16:15:55 GMT -5
Augh! That darn thing burned my eyes. Thank youuuuu avatar blocker. Anyway, to answer your questions: "you" are not a factor in this problem. You're not counted as a member. And this problem does involve conditional probability, but it's up to you to determine whether it's similar to the Monty Hall problem or not. Think about it for a while and venture a guess. It turns out to be a very tricky problem and so there are no "dumb" guesses. Except for maybe, "There's a 120% chance of three men," or the like. ![;)](//storage.proboards.com/forum/images/smiley/wink.png)
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Post by ՏՇԾԵԵʅՏɧ_LԹՏՏʅҼ on Oct 17, 2011 16:28:12 GMT -5
Since one person is female, and the other is Rick, (who is male) wouldn't that count as three members present?
[Yes my mind immediately went there].
[ETA: I expect extra Karma for my correct answer] ;D
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Colleenz
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Post by Colleenz on Oct 17, 2011 16:30:43 GMT -5
Ahh- So we know Rick is a prize, but not all men are prizes. ![;)](//storage.proboards.com/forum/images/smiley/wink.png)
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Virgil Showlion
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Post by Virgil Showlion on Oct 17, 2011 16:32:58 GMT -5
Rick counts as one member. At least one female member is present. A total of three members are present. "You" are a guest and do not count as a member. You are not included in the three and your gender is unimportant.
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Post by ՏՇԾԵԵʅՏɧ_LԹՏՏʅҼ on Oct 17, 2011 16:34:53 GMT -5
Read my answer again - Rick is Male - hence his presence alone counts as 2 "members" being in the room - the female makes the count Three.
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Colleenz
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Post by Colleenz on Oct 17, 2011 16:39:50 GMT -5
I am still thinking 33% - just a twisted version of the Monty Hall problem. Before I knew it was Rick and his chick, they were all 33%. Defining Rick and his chick does not alter the gender probability of member #3. I miss SBS ![:(](//storage.proboards.com/forum/images/smiley/sad.png)
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Colleenz
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Post by Colleenz on Oct 17, 2011 16:41:10 GMT -5
Wait - I don't even make sense to myself anymore. I'm going to go pet my snow leopard!
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Colleenz
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Post by Colleenz on Oct 17, 2011 16:42:06 GMT -5
33% chance it's a dude.
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Post by hoops902 on Oct 17, 2011 16:43:04 GMT -5
There's only a 33% chance there are 2 females online.
Given random probability of 2 people (since we know there is Rick) there is an equal chance of MF, FM, MM, & FF. We know the MM is not possible (since Rick wouldn't be on if it were) so there is now only an equal chance of FM, MF, and FF.
So only 33% chance that both guests are female.
ETA: To make it clear to the problem. 66% chance that there are now 2 males and 1 female on. and a 33% chance there are 2 females and 1 male on.
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Post by Peace Of Mind on Oct 17, 2011 16:43:53 GMT -5
Virgil, you're so cute. SL meant Little Rick as the other member to Rick. ;D
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Colleenz
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Post by Colleenz on Oct 17, 2011 16:44:05 GMT -5
87% chance its debMD
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Post by ՏՇԾԵԵʅՏɧ_LԹՏՏʅҼ on Oct 17, 2011 16:51:09 GMT -5
... but then again, he DID say "odd sex" - those two things kind of go hand-in-hand ;D
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Virgil Showlion
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Post by Virgil Showlion on Oct 17, 2011 16:52:42 GMT -5
*sigh* I guess I'm not up to par on my sexual innuendo humour. ![:P](//storage.proboards.com/forum/images/smiley/tongue.png) Hoops provides a short proof for an interesting alternative: a 33 1/3% chance of two women and a man, and a 66 2/3% chance of two men a woman. So is the OP (50%/50%) correct, hoops (33%/67%) correct, or are they both incorrect? ![;)](//storage.proboards.com/forum/images/smiley/wink.png)
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hoops902
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Post by hoops902 on Oct 17, 2011 18:06:30 GMT -5
*sigh* I guess I'm not up to par on my sexual innuendo humour. ![:P](//storage.proboards.com/forum/images/smiley/tongue.png) Hoops provides a short proof for an interesting alternative: a 33 1/3% chance of two women and a man, and a 66 2/3% chance of two men a woman. So is the OP (50%/50%) correct, hoops (33%/67%) correct, or are they both incorrect? ![;)](//storage.proboards.com/forum/images/smiley/wink.png) We're both wrong. I failed to account for the timing of everyone's arrival to the board. I believe there's actually only a 60% chance that there are 2 M and 1 F. Possible combinations based on order of arrival: F-R-M F-R-F F-M-R M-F-R F-F-R So 3 of 5 combinations include a M plus R (so 2 males).
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Virgil Showlion
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Post by Virgil Showlion on Oct 17, 2011 19:19:32 GMT -5
You forgot F-R-F, which would make it 3 of 6 combinations with two men and one woman. If you stand by this logic, you would therefore be agreeing with the problem statement. But what about combinations starting with R? Such as R-F-M?
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hoops902
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Post by hoops902 on Oct 17, 2011 22:35:13 GMT -5
You forgot F-R-F, which would make it 3 of 6 combinations with two men and one woman. If you stand by this logic, you would therefore be agreeing with the problem statement. But what about combinations starting with R? Such as R-F-M? F-R-F was my 2nd combination above. I'm assuming we are starting from a zero-member point and therefore Rick could not be first. If we're not you have to add in the 3 Rick-first combinations of R-F-M, R-M-F, & R-F-F...which gets us to 5 out of 8 combinations of 2 men and 1 woman. If we're not starting from a zero point the logic gets dicey since Rick can never ever "be around" without a female (so if a female came, then rick, then another female came before the first left we're assuming Rick signs off at the precise time the first female does if no other female arrives prior to her leaving). Since the only way that logic can work is a start-from-zero and once you sign on you're not signing off, I assumed that structure rather than a simultaneous sign-off. Otherwise it's no longer really random, it's heavily dependent on how long each member stays signed in prior to Rick's arrival. So R-F-M would need to be a higher weighting than R-M-F since it's more likely that 2nd member signs on while there is still a female signed on while R-M-F is less likely (at least to some degree in a random sign-on pattern as F signs-on later in the pattern so more chance for previous female members to sign off and for Rick to leave due to no females). Yes I realize this is getting incredibly complicated for a brain-teaser...hence why I assumed a simplified structure of zero start and then sign-on with no sign-offs.
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Post by ՏՇԾԵԵʅՏɧ_LԹՏՏʅҼ on Oct 17, 2011 22:54:04 GMT -5
My head hurts after reading that. ![](http://syonidv.hodginsmedia.com/vsmileys/confused.png)
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Virgil Showlion
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Post by Virgil Showlion on Oct 17, 2011 23:22:37 GMT -5
Hoops, I specified that "Rick is only around if at least one lady is around" to make the problem seem less contrived. It's a way for "you" (the guest) to know that at least one woman is online when in reality there is no way for a guest to determine the gender(s) of online members. I understand your grievance about the complication of Rick being the first, but I'll stipulate that you can safely ignore it. The fundamentals of the problem are: - three board members are online and can have logged on in any order. You have no idea about the probability of any login order. (Ultimately the answer to the problem is the same regardless of the order, but if it helps your analysis you can assume that the probability of all login orders is the same.)
- one of the members is Rick
- at least one of the remaining members is female
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