Timeline for I know that you know...
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Aug 29, 2018 at 18:26 | review | Suggested edits | |||
| Aug 29, 2018 at 19:14 | |||||
| Dec 17, 2012 at 2:23 | comment | added | fedja | OK, I had hardly any time for anything last 3 weeks (I still run some debts even on this site), but I'll think a bit of it now and, if I have any bright idea, I'll post. What seems clear to me is that the problem is treated in a slightly wrong way: it is clear that if you allow people to play the continuous version for long, the strategies will converge to 0. However, there may be a limiting density shape that is just scaled by the drift and I wonder whether some reasonable assumptions would allow us to determine it. | |
| Nov 27, 2012 at 17:18 | comment | added | Tony Huynh | I was assuming that the players have to pick integer values, in which case I (think) it is a pure forecasting game if there are many non-colluding players, since there is only one correct answer. I also think the continuous game is quite interesting (possibly it is more interesting to start from 0 in this case). I am happy to discuss further either here or AoPS. | |
| Nov 27, 2012 at 13:34 | comment | added | fedja | Agree. Actually, this game looks interesting enough to try to discuss it separately in more detail. Note that it is not a pure forecasting game no matter how many the players are: you do not win just when you are sufficiently close to the target. What gives you the prize is being closer than everyone else. The distinction is quite subtle but it may be crucial for understanding what's really going on. Would you like to do it here or, say, on AoPS? | |
| Nov 26, 2012 at 11:43 | comment | added | Tony Huynh | Good point. However, if we assume that the number of players is large in comparison to the max value, then no single player (or a small group) can affect the average. It then effectively becomes a forecasting problem (unless we assume that a large chunk of the group is colluding against you, but then you are indeed pretty screwed if that is the case). | |
| Nov 26, 2012 at 0:38 | comment | added | fedja | ---Now, it is clearly foolish to write down any number greater than 666667, since 2/3 of the average cannot be more than 666667.--- Not if you playing with a partner. Of course, you won't take the prize yourself, but you can raise your partner's chances quite a bit by writing a large number if you agree upon a decent strategy. Now, how do I know that there are no coalitions out there? And if I don't, the second step of the induction fails... | |
| Nov 24, 2012 at 2:33 | comment | added | Qiaochu Yuan | In practice when actually playing this game it is important to model everyone else you're playing with (especially if they don't know how to compute Nash equilibria). I played a version of this game once and I intentionally chose the highest number to throw off everyone else's strategies (and I wasn't alone in doing this). | |
| Nov 23, 2012 at 21:47 | comment | added | Paul Reynolds | Thanks very much Richard! I will take the time to read this as it's something I've wanted to know for a while. | |
| Nov 23, 2012 at 21:09 | comment | added | R Hahn | Paul & Emil: such "beauty contest" experiments reveal baffling play: for r = 0, not everyone plays 0. Some do not play increasing functions of r. Many people seems to play roughly linearly in r relative to their play at r=1. Here is my analysis of some data I recently collected from web surveys: faculty.chicagobooth.edu/richard.hahn/SPCH_Oct12.pdf. Plots of observed strategies: faculty.chicagobooth.edu/richard.hahn/individualSummaries.pdf | |
| Nov 23, 2012 at 20:02 | comment | added | Paul Reynolds | Interesting! I wonder how this varies with the fraction $r = 2/3$... | |
| Nov 23, 2012 at 19:58 | comment | added | Emil Jeřábek | IIRC, I’ve read that in human experiments the 2/3 of the average tends to be around 20% of the maximum. | |
| Nov 23, 2012 at 19:43 | comment | added | Paul Reynolds | A agree that choosing $1$ is the only rational choice, but still I would like to actually do this experiment with a bunch of non-mathematicians and see what $2/3$ of the average turns out to be. I suspect it changes after the first go, or if you give them much time to think, but such empirical information could have been useful in a pub quiz I used to go to. | |
| Nov 23, 2012 at 17:59 | history | edited | Tony Huynh | CC BY-SA 3.0 |
added 18 characters in body
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| Nov 23, 2012 at 17:24 | history | answered | Tony Huynh | CC BY-SA 3.0 |