24 Sep 04
1 edit
Originally posted by royalchickenCheck for sandals.
My parents and I are about to pay large sums to have a guy in a cardigan explain that to me 😉.
Cardigans aren't sufficient you fool!!!
Jeez, cardigan-only status hints at impostershipness. Sheesh.
To summarise:
Cardigan only = careful...he may not be fully committed or may even have transferred from a non-mathematical department
Cardigan + sandals = likely to be real
Cardigan + sandals + socks = definite genuine article
Cardigan + sandals + socks - minus wedding ring = an absolute object specimen, entrust this man with your mathematical career and don't look back
24 Sep 04
Originally posted by T1000LOL 😀. However:
Check for sandals.
Cardigans aren't sufficient you fool!!!
Jeez, cardigan-only status hints at impostershipness. Sheesh.
To summarise:
Cardigan only = careful...he may not be fully committed or may even have transferred from a non-mathematical department
Cardigan + sandals = likely to be real
Cardigan + sandals + socks = definite genuine article
...[text shortened]... ute object specimen, entrust this man with your mathematical career and don't look back
Someone who might be one of my teachers may provide a counterexample:
http://www.ma.ic.ac.uk/~buzzard/stuff/pic.jpg
24 Sep 04
Originally posted by royalchickenHmm, orange clothes? He's got potential though looks a little on the young side to be a lecturer.
LOL 😀. However:
Someone who might be one of my teachers may provide a counterexample:
http://www.ma.ic.ac.uk/~buzzard/stuff/pic.jpg
PS: I fear for the Moby look-alike...that seagull look like it's going to go straight into the side of his head.
24 Sep 04
Originally posted by T1000What's the word? Tchar? The seagull IS the lecturer, obviously.
Hmm, orange clothes? He's got potential though looks a little on the young side to be a lecturer.
PS: I fear for the Moby look-alike...that seagull look like it's going to go straight into the side of his head.
29 Sep 04
1 edit
Originally posted by telerionIs there any stakes? Is this two diffrent balls, or two bets on the same ball?
Answer the following two questions.
Please limit your answers to the options provided. No explainations from you needed at this point. Just a one-letter response is sufficient. Please do not post questions. Instead message them to me.
...[text shortened]... ally and fill every interested party in on this in a day or two.
* same ball, two stakeless bets: AC (or BD) will yeild 1000 any which way..
*two diffrent balls, first AB second CD: A,D
29 Sep 04
1 edit
Originally posted by chasparosNo, there are no stakes (if by stakes you mean some sort of fee to play or penalty for not winning). You've got nothing to lose really. It's also not a combination gamble. That is you can't combine the options for the first and second question.
Is there any stakes? Is this two diffrent balls, or two bets on the same ball?
* same ball, two stakeless bets: AC (or BD) will yeild 1000 any which way..
*two diffrent balls, first AB second CD: A,D
Really it's about your perception of expected utility. Given the setup (urn, information about color distribution of balls), would you prefer the gamble A or the gamble B? Now removing gamble A and Gamble as options would you prefer gamble C or gamble D?
No balls are actually drawn. But if one were would would you want A instead B? How about C instead of D?
So basically your second answer.
Then we go back and look at how you answered and what that says about you expected utility function (If such a thing exists 🙂 )
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29 Sep 04
Originally posted by telerionB and D (I like taking chances)
Answer the following two questions.
Please limit your answers to the options provided. No explainations from you needed at this point. Just a one-letter response is sufficient. Please do not post questions. Instead message them to me.
Ok here goes.
The following description applies to both questions.
There is an urn that contains 300 balls. 100 ...[text shortened]... rticipation.
I'll keep a tally and fill every interested party in on this in a day or two.
29 Sep 04
Ok, so far I have.
oishi = BC
seraphimvultur = BC
chasparos= AD
iamtiger=BD
We need more responses. Let me know if I missed one. Even if you think it is arbitrary or are indifferent between the gambles in one or both of the questions just slap one of the gambles down. I mean if you were really in the position to take these gambles surely you wouldn't balk at a free chance to win $1000?
06 Oct 04
Originally posted by telerionAssume: Blue & Green balls are selected randomly. I.e., as far as I know, there is an equal probability of each of the 200 blue/green balls being blue or green - or stated another way, I have no reason to believe that the person selecting the balls to go into the urn is either inclined to give away money (he is more likely to put in blue balls), or is inclined to keep his money (he is more likely to put in green balls).
Ok, so far I have.
oishi = BC
seraphimvultur = BC
chasparos= AD
iamtiger=BD
We need more responses. Let me know if I missed one. Even if you think it is arbitrary or are indifferent between the gambles in one or both of the questions just slap one of the gambles down. I mean if you were really in the position to take these gambles surely you wouldn't balk at a free chance to win $1000?
Gamble A/B - equal probability, no preference
Gamble C/D - equal probability, no preference
However, if we throw out my assumption....
My personal preference in A/B is A since I know that I at least have a 1 in three chance of picking up a grand, and I haven't risked anything, versus anywhere from a 0% to 2 in 3 chance with B. In C/D it's C, since I know I have a 2 in 3 chance, where as with D my odds range from 1 in 3 to 100% based on the kindness of random chance and/or the ball selector.
Short answer: AC (but only for emotional reasons) "Do you feel lucky, punk?"
14 Oct 04
I think there is a non-zero probability that the house would try to cheat by manipulating the blue/green proportion after I chose.
Therefore:
P(a) = 1/3
P(b) = 1/3 - P(cheating)
P(c) = 2/3
P(d) = 2/3 - P(cheating)
P(a)=P(b)+P(cheating) => P(a) > P (b)
P(c)=P(d)+P(cheating) => P(c) > P (d)
My choice: AC!
